Prasanna

Students can use Class 8 Math Solution Odia Medium and Class 8 Maths Chapter 3 Question Answer Odia Medium ସଂଖ୍ୟାର କାହାଣୀ to check their answers after solving exercises.

8th Class Maths Chapter 3 Question Answer Odia Medium

Class 8 Maths Chapter 3 Odia Medium

3.1 ରୀମାର କୌତୁହଳ

Page No. 53

Question 1.
ବଡ଼ ସଂଖ୍ୟାକୁ ସୂଚିତ କରିବା ପାଇଁ ଏହି ପଦ୍ଧତିକୁ ଆଗକୁ ବଢ଼ାଯାଇ ପାରିବ କି ?
Solution:
ହଁ । ଏହି ପଦ୍ଧତିରେ ବଡ଼ ସଂଖ୍ୟାକୁ ସୂଚିତ କରାଯାଇଥାଏ ।
ଗଣନା ଏବଂ ଏକ ସମୂହର ଆକାର ନିର୍ଣ୍ଣୟ କରିବା ପାଇଁ ଲିଖୁତ ସଂକେତଗୁଡ଼ିକର ଏକ ମାନକ ଅନୁକ୍ରମ ଥାଏ । ଅନୁକ୍ରମକୁ ଏକ ସଂଖ୍ୟା ପ୍ରଣାଳୀ କୁହାଯାଏ । ବସ୍ତୁଗୁଡ଼ିକର ସମାହାରକୁ ମାନକ ଅନୁକ୍ରମ ଅନୁସାରେ ଏକ-ଏକ-ମେଳକ କରି ଓ କ୍ରମ ଅନୁସରଣ କରି ଗଣନ କରାଯାଇପାରିବ । ସଂଖ୍ୟାମାନଙ୍କର ପରିସମାପ୍ତି ନଥୁବାରୁ ଅସୀମ ମାନକ ଅନୁକ୍ରମ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପ୍ରସ୍ତୁତ କରିବା ଏହାକୁ ବ୍ୟବହାର କରି ସହଜରେ ଗଣନା କରିହେବ । କାଠି ବ୍ୟବହାର କରି ଏକ ଅସୀମ ମାନକ ଅନୁକ୍ରମ ପାଇହେବ, କିନ୍ତୁ ବଡ଼ ପରିମାଣର ଜିନିଷମାନଙ୍କର ପରିମାଣ ନିର୍ଣ୍ଣୟ କରିବା ପାଇଁ ଏହା ସୁବିଧାଜନକ ହୋଇନଥାଏ, କାରଣ ଅଧିକ ସଂଖ୍ୟକ ବସ୍ତୁର ପରିମାଣ ଜାଣିବା ପାଇଁ ବହୁତ କାଠି ଆବଶ୍ୟକ ହୋଇଥାଏ ।
ଦ୍ଵିତୀୟ ପ୍ରଣାଳୀରେ ନିର୍ଦ୍ଦିଷ୍ଟ ଭାଷାର ଅକ୍ଷରକୁ ବ୍ୟବହାର କରି ଗଣନ କରିବା ସୁବିଧାଜନକ, କିନ୍ତୁ ଏହା ଏକ ଅସୀମ ମାନକକ୍ରମ / ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପାଇଁ ବ୍ୟବହାର ହୋଇପାରିବ ନାହିଁ ।

ତୃତୀୟ ପଦ୍ଧତିରେ ଦିଆଯାଇଥିବା ମାନକକ୍ରମ | ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପ୍ରକୃତରେ ୟୁରୋପରେ ବ୍ୟବହୃତ ହେଉଥିଲା । ଅବଶ୍ୟ ପରବର୍ତ୍ତୀ ସମୟରେ ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ତାର ସ୍ଥାନ ନେଇଥିଲା । ଏହାକୁ ରୋମାନ୍ ସଂଖ୍ୟା ପ୍ରଣାଳୀ କୁହାଯାଏ ।
ଶତାବ୍ଦୀ ଶତାବ୍ଦୀ ଧରି ୟୁରୋପରେ ବହୁଳ ଭାବରେ ପ୍ରଚଳିତ ଥିଲା ଏବଂ ଅନେକ ଦିଗରୁ ଏହାର ବ୍ୟବହାର ସୁବିଧାଜନକ ଥିଲେ ମଧ୍ୟ ଅତି ବଡ଼ ସଂଖ୍ୟାକୁ ପ୍ରକାଶ କରିବା ପାଇଁ ଏହି ପ୍ରଣାଳୀ ଉପଯୁକ୍ତ ନଥୁଲା, କାରଣ ଅଧିକରୁ ଅଧିକ ବଡ଼ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ସୀମିତ ଅକ୍ଷର ମାଧ୍ୟମରେ ପରିପ୍ରକାଶ କରିବା ସହଜ ନଥୁଲା ଏବଂ ସଙ୍କେତ ବ୍ୟବହାର ନକରି ବଡ଼ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଲେଖୁ ସମ୍ଭବ ହେଉନଥିଲା । ସଂଖ୍ୟା ପ୍ରଣାଳୀର ଇତିହାସରୁ ଆମେ ଜାଣୁଛେ, ଏହାର କାର୍ଯ୍ୟକାରିତା ସାଧାରଣତଃ ଏହି ତିନୋଟି ଉପାୟ, ଯଥା- କାଠି, ଗୋଡ଼ି ବା ଶରୀର ଅଙ୍ଗକୁ ନେଇ କରାଯାଉଥିଲା । କିଛି ଗୋଷ୍ଠୀର ଲୋକ ଏଥିପାଇଁ ଉଭୟ ବସ୍ତ ଓ ନାମର ବ୍ୟବହାର କରୁଥିବାବେଳେ ଚୀନ୍‌ର ଲୋକମାନେ ସଂଖ୍ୟାଗଣନା ପାଇଁ ତିନୋଟିଯାକ ଉପାୟର ଉପଯୋଗ କରୁଥିଲେ ।
ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଥିବା ପ୍ରତୀକଗୁଡ଼ିକୁ ସଂଖ୍ୟା ସୂଚକ (Numerals) କୁହାଯାଏ ।
ଉଦାହରଣ ସ୍ଵରୂପ- 0, 1, 5, 36, 193 ଆଦି ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ବ୍ୟବହୃତ ହେଉଥିଲା ସଂଖ୍ୟାସୂଚକ ।

ନିଜେ କରି ଦେଖ : (Page No. 54)

Question 1.
ମନେକର, ତୁମେ ପ୍ରଥମ ଉପାୟ / ପ୍ରଣାଳୀ ଅନୁଯାୟୀ କାଠି ସାହାଯ୍ୟରେ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଉପସ୍ଥାପନ କରିବା ପାଇଁ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ବ୍ୟବହାର କରୁଛ । ହିନ୍ଦୁ ସଂଖ୍ୟା ପଦ୍ଧତିର ସଂଖ୍ୟା ନାମ ବା ସଂଖ୍ୟା ସୂଚକକୁ ବ୍ୟବହାର ନକରି କେବଳ ଦୁଇଟି କାଠି ନେଇ ବା ଦୁଇଟି ସଂଖ୍ୟା ନେଇ ଯୋଗ, ବିୟୋଗ, ଗୁଣନ ଓ ହରଣ କରିବାର ଏକ ଉପାୟ ସ୍ଥିର କର ।
Solution:
ମନେକର ଦୁଇଟି ଦଳ ମେଣ୍ଢା ଅଛନ୍ତି । ପ୍ରଥମ ଦଳରେ ଚୈ ମେଣ୍ଢା ଓ ଦ୍ଵିତୀୟ ଦଳରେ 4ଟି ମେଣ୍ଢା ଅଛନ୍ତି । ଆମେ ପଥର ଗୋଡ଼ି ବ୍ୟବହାର କରି ଗଣନା କରିଛୁ । ବାହାର କରିବେ ।
ଯୋଡ଼ିବାପାଇଁ ସମସ୍ତ ପଥରକୁ ସମାନ ପାଉଚରେ ରଖାଯାଏ ଓ ବିୟୋଗ କରିବା ପାଇଁ ଅଧୁକ ପଥର ଥୁବା ପାଉଚରୁ କମ୍ ଥୁବା ପାଉଚରେ ଯେତେ ଅଧ‌ିକ ପଥର ଆମେ ଜାଣିବା ଯେ , ପ୍ରଥମ ଦଳରେ କେତୋଟି ମେଣ୍ଢା ସଂଖ୍ୟାର ଦୁଇଗୁଣ ହେବ ।
ଏଥ‌ିପାଇଁ ଆମେ ପଥର ବ୍ୟବହାର କରି ଦୁଇଥର ଗଣନା କରିବ ଏବଂ ସମସ୍ତ ପଥରକୁ ସମାନ ପାଉଚରେ ରଖୁବା । ମନେକର ଆମ ପାଖରେ 12ଟି ପଥର ଅଛି ଏବଂ ସେଗୁଡ଼ିକୁ ତିନୋଟି ସମାନ ଗ୍ରୁପ୍‌ରେ ବିଭକ୍ତ କରିବାକୁ ଚାହିଁବା । ଆମେ ସେଗୁଡ଼ିକୁ ତିନୋଟି ପାତ୍ରରେ ଗୋକିକ ପରେ ଗୋଟିଏ ରଖୁ, ଯେପର୍ଯ୍ୟନ୍ତ ସବୁ ନ ସରିଛି । ପ୍ରତ୍ୟେକ ପାତ୍ରରେ ପଥର ସଂଖ୍ୟା ଭାଗଫଳ ସୂଚାଇବ ।

Question 2.
ଦ୍ଵିତୀୟ ପ୍ରଣାଳୀରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟା ପଦ୍ଧତିକୁ ଆଗକୁ ବଢ଼ାଇବାର ଗୋଟିଏ ଉପାୟ ହେଉଛି, ଏଥିରେ ଏକରୁ ଅଧିକ ଅକ୍ଷର ସମାହାର ବ୍ୟବହାର କରିବା । ଯଥା – 1 ପାଇଁ a ଓ 27 ପାଇଁ ‘aa’ ନେଇପାରିବା । ସମସ୍ତ ସଂଖ୍ୟାକୁ ଲେଖିବା ପାଇଁ ତୁମେ ଏହି ପ୍ରଣାଳୀକୁ କିପରି ଆଗକୁ ବଢ଼ାଇପାରିବ ? ଏହା କରିବା ପାଇଁ ଅନେକଗୁଡ଼ିଏ ଉପାୟ ଅଛି ।
Solution:
ଇଂରାଜୀରେ a, b……….z – ଏହିପରି 26 ଟି ଅକ୍ଷର ଥାଏ ।
କିନ୍ତୁ ଆମେ ସମସ୍ତ ସଂଖ୍ୟା ମାନଙ୍କୁ ଲେଖୁ ଆଗକୁ ବଢ଼ାଇବା ପାଇଁ ଦୁଇଟି ଅକ୍ଷର ଯୋଡ଼ିକ ବ୍ୟବହାର କରି ପାରିବା ।
ଯଥା- aa, ab, ac………az ଇତ୍ୟାଦି ।
ଏହା 26 ଟି ଅକ୍ଷରଠାରୁ ଅଧିକ ହୋଇପାରିବ ।

Question 3.
ତୁମେ ନିଜେ ଏକ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପ୍ରସ୍ତୁତ କରିବା ପାଇଁ ଚେଷ୍ଟାକର ।
Solution:
ମନେକର ସଂଖ୍ୟାଟିର ଆଧାର B
∴ 3⁰ = 1 = A, 3¹ = 3 = B, 3² = 9 = C ………

3.2 କେତେକ ପ୍ରାରମ୍ଭିକ ସଂଖ୍ୟା ପ୍ରଣାଳୀ (Some Early Number Systems)

Page No. 56

Question 1.
ସେମାନଙ୍କର ସଂଖ୍ୟା ନାମଗୁଡ଼ିକ କିପରି ଗଠିତ ହୋଇଛି, ଲକ୍ଷ୍ୟ କର ।
Solution:
3 ର ସଂଖ୍ୟାନାମ, 2 ଏବଂ 1 ସଂଖ୍ୟାନାମକୁ ନେଇ ଗଠିତ ହୋଇଅଛି । 4 ସଂଖ୍ୟାପାଇଁ ସଂଖ୍ୟାନାମ 2ଟି 2ର ସଂଖ୍ୟାନାମକୁ ନେଇ ଗଠିତ ହୋଇଅଛି ।

Question 2.
ଦକ୍ଷିଣ ଆମେରିକାର ଏକ ପ୍ରାଚୀନ ଗୋଷ୍ଠୀ ଓ ଦକ୍ଷିଣ ଆଫ୍ରିକାର ବୁସମେନ୍ ଆଦିବାସୀ ଗୋଷ୍ଠୀର ଲୋକମାନଙ୍କର ସଂଖ୍ୟାପ୍ରଣାଳୀ ନିମ୍ନରେ ଦିଆଯାଇଛି । ଭୌଗୋଳିକ ଦୃଷ୍ଟିକୋଣରୁ ଏହି ତିନି ସଂପ୍ରଦାୟ ପରସ୍ପରଠାରୁ ବହୁ ଦୂରରେ ଥିଲେ ଏବଂ ସେମାନଙ୍କ ମଧ୍ୟରେ କୌଣସି ସମ୍ପର୍କ ନଥୁଲା । ଏହାସତ୍ତ୍ବେ ସେମାନେ ସମାନତା ଥୁବା ସଂଖ୍ୟା ପ୍ରଣାଳୀ ବିକଶିତ କରିପାରିଥିଲେ।

Solution:
ଯଦିଓ ଗୁମୁଲଗାଲର ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ କେବଳ 6 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାପାଇଁ ସଂଖ୍ୟାନାମ ଥିଲା, କିନ୍ତୁ ସେହି ସଂଖ୍ୟା ପଦ୍ଧତିର ଉତ୍ପରି ଅନ୍ୟ ଦୁଇ ସମ୍ପ୍ରଦାୟ ବ୍ୟବହାର କରୁଥିବା ସଂଖ୍ଯାପ୍ରଣାଳୀ ପରି ହୋଇଥିଲା । ସଂଖ୍ୟାକୁ ଉପସ୍ଥାପନ କରିବା ପାଇଁ ଟାଲି ପ୍ରଣାଳୀ ଅପେକ୍ଷା ଦୁଇ ଦୁଇ କରି ଗଣନା କରିବା (ଯୋଡ଼ା ଗଣନା) ଅଧିକ ଫଳପ୍ରଦ ଅଟେ । ବିଭିନ୍ନ ସଂଖ୍ଯାପ୍ରଣାଳୀରୁ ନିଆଯାଇଥିବା ଧାରଣାକୁ ନିମ୍ନମତେ ସାଧାରଣୀକରଣ କରାଯାଇପାରେ । ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସଂଖ୍ୟାକୁ ସମୂହ | ଦଳକରି ଗଣନା କରିବା (ଯେପରି ଗୁମୁଲଗାଲ୍ ପ୍ରଣାଳୀରେ 2 କୁ ନିଆଯାଇଛି) ଏବଂ ବଡ଼ ସଂଖ୍ୟାକୁ ପରିପ୍ରକାଶ କରିବା ପାଇଁ ଏହି ସମାହାର ସହିତ ଜଡ଼ିତ ଶବ୍ଦ କିମ୍ବା ସଙ୍କେତ ବ୍ୟବହାର କରିବା । ବିଭିନ୍ନ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ସାଧାରଣତଃ ବ୍ୟବହୃତ କିଛି ସମାହାର ହେଉଛି 2, 5, 10 ଓ 20 । ତୁମେ ରୋମାନ୍ ପ୍ରଣାଳୀରେ 5ରେ (ପାଞ୍ଚ ପାଞ୍ଚ ନେଇ) ଗଣନା କରିବାର ଧାରଣା ପାଇପାରିବ।

Page No. 58

Question 1.
ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସଂଖ୍ୟା ଆକାରର ସମୂହକୁ ନେଇ ସଂଖ୍ୟା ଗଣିବା ପ୍ରଣାଳୀର ବ୍ୟବହାର ଯୋଗୁଁ କେଉଁ ସବୁ ଅସୁବିଧା ସୃଷ୍ଟି ହୋଇଥବ ? କେବଳ 5ରେ ସମୂହରେ ଗଣନା କରି ତିଆରି ହୋଇଥୁବା ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ତୁମେ 1345କୁ କିପରି ପରିପ୍ରକାଶ କରିବ ?
Solution:
ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଆକାରର ସମୂହକୁ ନେଇ ଗଣନା କରିବା ଏବଂ ସଂଖ୍ୟାକୁ ପରିପ୍ରକାଶ ପାଇଁ ଏହାକୁ ‘ବ୍ୟବହାର କରିବା ପାଇଁ ଟାଲି ପ୍ରଣାଳୀ ଅଧିକ ଉପଯୁକ୍ତ | ଫଳପ୍ରଦ ହେଲେ ମଧ୍ୟ ଏହି ପ୍ରଣାଳୀରେ ବଡ଼ ବଡ଼ ସଂଖ୍ୟାକୁ ପରିପ୍ରକାଶ କରିବା କଷ୍ଟଦାୟକ ହୋଇପାରେ ।
1345କୁ ଲେଖୁବାକୁ 5ଟି ଦଳ | ଗ୍ରୁପ୍‌ରେ ଆମେ ଦ୍ଵାରା ଭାଗ କରିବା ।
1345 ÷ 5 = 269, 1345 = 5 + 5 + 5 + 5 +………. + 5 (269 ଥର)

ନିଜେ କରି ଦେଖ : (Page No. 59)

Question 1.
ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ରୋମାନ୍ ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କର ।
(i) 1222
(ii) 2999
(iii) 302
(iv) 715
Solution:
(i) 1222 = MCCXXII
(ii) 2999 = MMCMXCIX
(iii) 302 = CCCII
(iv) 715 = DCCXV
ରୋମାନ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଗୁଣନ ଓ ଭାଗକ୍ରିୟାର ଗାଣିତିକ ପ୍ରକ୍ରିୟା ସମ୍ପାଦନ କରିବା ସହଜ ହୋଇ ନଥୁଲା ।

Question 2.
ଉଦାହରଣ : ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ହିନ୍ଦୁସଂଖ୍ୟାକୁ ପରିବର୍ତ୍ତନ ନକରି ଯୋଗ କରିବାକୁ ଚେଷ୍ଟାକର ।
(a) CCXXXII + CCCCXIII
Solution:
ଆମେ ମୋଟ I, X ଏବଂ C ସଂଖ୍ୟାଗୁଡ଼ିକ ନିର୍ଣ୍ଣୟ କରିବା ଏବଂ ସେଗୁଡ଼ିକ ବୃହତ୍ତମ ସଂଖ୍ୟାରୁ ଆରମ୍ଭ କରି ଦଳଭୁକ୍ତ କରିବା । C ସବୁଠାରୁ ବଡ଼ ସଂଖ୍ଯାପରି ଦେଖାଯାଉଛି, କିନ୍ତୁ ଲକ୍ଷ୍ୟ କର 5ଟି C(100)ରେ ଗୋଟିଏ D(500) ହୋଇଥାଏ । ତେଣୁ ଯୋଗଫଳଟି ହେଉଛି-

(b) ସମାଧାନ କର : LXXXVII + LXXVIII
Solution:

LXXXVII + LXXVIII = CLXV

Page No. 60

Question 1.
ରୋମାନ୍ ପ୍ରଣାଳୀରେ ଦିଆଯାଇଥ‌ିବା ଦୁଇଟି ସଂଖ୍ୟାକୁ ହିନ୍ଦୁ ସଂଖ୍ୟାରେ ପରିବର୍ତ୍ତନ ନକରି ତୁମେ କିପରି ଗୁଣନ କରିବ? ନିମ୍ନଲିଖତ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ଯୋଡ଼ିଗୁଡ଼ିକର ଗୁଣଫଳ ନିଶ୍ଚୟ କରିବାକୁ ଚେଷ୍ଟାକର।
V × L, L × D, V × D, VII × IX

Solution:
V × L = 5 × 50 = 250 = CCL
L × D = 50 × 500 = 25000, ରୋମାନ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ପରିବର୍ତ୍ତନ କରିବା ଅସମ୍ଭବ।
V × D = 5 × 500 = 2500 = MMD
VII × IX = 7 × 9 = 63 = LXIII
CCXXXI × MDCCCLII = 231 × 1852 = 427812, ରୋମାନ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ପରିବର୍ତ୍ତନ କରିବା ଅସମ୍ଭବ।

ନିଜେ କରି ଦେଖ : (Page No. 60-61)

Question 1.
ପ୍ରଶାନ୍ତ ମହାସାଗରୀୟ ଦ୍ଵୀପର ଏକ ଆଦିବାସୀ ସଂପ୍ରଦାୟ ବିଭିନ୍ନ ବସ୍ତୁକୁ ଜାଣିବା ପାଇଁ ଭିନ୍ନଭିନ୍ନ ସଂଖ୍ୟାନାମର କ୍ରମକୁ ବ୍ୟବହାର କରୁଥିଲେ। ତେବେ ଚିନ୍ତା କରି କହ, ସେମାନେ କାହିଁକି ଏପରି କରୁଥିଲେ?
Solution:
ସେମାନେ ବିଭିନ୍ନ ବସ୍ତଗୁଡ଼ିକୁ ଗଣନା କରିବା ପାଇଁ ସଂଖ୍ୟାନାମର ବିଭିନ୍ନ କ୍ରମ ବ୍ୟବହାର କରୁଥିଲେ । କାରଣ ସେମାନଙ୍କର ପଦ୍ଧତିଗୁଡ଼ିକ ନିର୍ଦ୍ଦିଷ୍ଟ ସାଂସ୍କୃତିକ ବ୍ୟବହାରିକ କିମ୍ବା ଭାଷାଗତ -ଆବଶ୍ୟକତା ଅନୁଯାୟୀ ଡିଜାଇନ୍ କରାଯାଇଛି । ଏହି ପଦ୍ଧତିଗୁଡ଼ିକରେ ସେହି ବସ୍ତୁଗୁଡ଼ିକୁ ଦଳଭୁକ୍ତ କରିବା, ମୂଲ୍ୟନିର୍ଣ୍ଣୟ, ବ୍ୟବହାର କରିବା ଓ ଉପଯୁକ୍ତ ସରଳ ଭାବେ ଗଣନା କରିବାରେ ସାହାଯ୍ୟ କରିପାରେ।

Question 2.
2ର ସମୂହକୁ ନେଇ ଗଣନା କରିବା ଉପାୟକୁ ବ୍ୟବହାର କରି ରୁ ବଡ଼ ସଂଖ୍ୟାକୁ ଗୁମୁଲଗାଲ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଆଗକୁ ବଢ଼ାଇବାକୁ ଚେଷ୍ଟାକର। ଏହି ପ୍ରଣାଳୀରେ ଥ‌ିବା ସଂଖ୍ୟାଗୁଡ଼ିକରେ ବିଭିନ୍ନ ଗାଣିତିକ ପ୍ରକ୍ରିୟା (+, -, ×, ÷) ସଂପାଦନ କରିବା ପାଇଁ ଉପାୟଗୁଡ଼ିକ ସ୍ଥିର କର (ହିନ୍ଦୁ ସଂଖ୍ୟାସୂଚକ ବ୍ୟବହାର ନକରି)।
ନିମ୍ନଲିଖତଗୁଡ଼ିକୁ ମୂଲ୍ୟାୟନ କରିବାରେ ଏହାକୁ ବ୍ୟବହାର କର।
(i) ଉକାସର – ଉଦାସର – ଉକାସର – ଉକାସର – ଉରାପୋନ) + (ଉକାସର – ଉଦାସର – ଉଲ୍ଲାସର – ଉରାପୋନ)
(ii) (ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉରାପୋନ) – (ଉକାସର – ଉକାସର – ଉକାସର)
(iii) (ଉକାସର – ଉକାସର – ଉକାସର – ଉଦାସର – ଉରାପୋନ) × (ଉକାସର – ଉକାସର)
(iv) (ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଭକାସର – ଉକାସର) ÷ (ଉକାସର – ଉକାସର)
Solution:
ଆମେ ଜାଣୁ ଯେ, ଉରାପୋନ 1 ସହିତ ଏବଂ ଉକାସର 2 ସହିତ ମେଳଖାଏ ଏବଂ ଗୁମୁଲଗାଲ୍ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ସଂଖ୍ୟାଗୁଡ଼ିକ 2Sରେ ଗଣନା କରାଯାଏ । ବିଭିନ୍ନ ଗାଣିତିକ କାର୍ଯ୍ୟପାଇଁ ନିମ୍ନଲିଖୁତ ଉପାୟମାନ ଦିଆଯାଇଛି ।
ସଂଖ୍ୟାନାମଗୁଡ଼ିକ ଏକାଠି ଯୋଡ଼ି ଯୋଗକରାଯାଇପାରିବ ।
ଉଦାହରଣ : (ଉକାସର) + (ଉକାସର) = ଉକାସର – ଉକାସର
(ଉକାସର – ଉକାସର) + (ଉରାପୋନ) = ଉକାସର – ଉକାସର – ଉରାପୋନ
ଏକ ଲମ୍ବା କ୍ରମରୁ କିଛି ସଂଖ୍ୟାନାମ ବାହାର କରି ବିୟୋଗ କରାଯାଇପାରିବ ।
ଉଦାହରଣ : (ଉକାସର – ଉକାସର – ଉରାପୋନ) – ଉରାପୋନ = ଉକାସର – ଉକାସର
(ଉକାସର – ଉକାସର – ଉକାସର) – (ଉକାସର – ଉକାସର) = ଉକାସର ।
ସମାନ ଶବ୍ଦକୁ ପୁନଃରାବୃତ୍ତି କରି ଗୁଣନ କରାଯାଇପାରିବ ।
ଉଦାହରଣ : ଉକାସର × ଉକାସର = ଉକାସର – ଉକାସର
(ଉକାସର – ଉକାସର) × ଉକାସର = ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର ।
ଶବ୍ଦକୁ ସମାନ ଦଳରେ ବିଭକ୍ତ କରି ବିଭାଜନ କରାଯାଇପାରିବ ।
ଉଦାହରଣ : (ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର ÷ (ଉକାସର – ଉକାସର) = ଉକାସର ।
(i) ଉକାସର – ଉକାସର – ଉଲ୍ଲାସର – ଉକାସର – ଉରାପୋନ) + (ଉକାସର – ଉକାସର – ଉକାସର – ଉତ୍ତାପୋନ) = ଉକାସର – ଉକାସର – ଉକାସର – ଉଲ୍ଲାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର (∴ ଉରାପୋନ + ଉରାପୋନ = ଉକାସର)
(ii) (ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉରାପୋନ) – (ଉକାସର – ଉକାସର – ଉକାସର) = ଉକାସର – ଉରାପୋନ
(iii) (ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉରାପୋନ) × (ଉକାସର – ଉକାସର) = ଉସାକର – ଉକାସର – ଉକାସର – ଉକାସର – ଉସାକର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର
(iv) (ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର – ଉକାସର) ÷ (ଉକାସର – ଉକାସର) = ଉକାସର – ଉକାସର ।

Question 3.
ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀର ଯେଉଁ ବୈଶିଷ୍ଟ୍ୟଗୁଡ଼ିକ ରୋମାନ୍ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ତୁଳନାରେ ଅଧିକ ଫଳପ୍ରଦ ବୋଲି ଭାବୁଛ, ସେଗୁଡ଼ିକୁ ଚିହ୍ନଟ କର ।
Solution:
(i) ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ହେଉଛି ଏକ ସ୍ଥାନୀୟମାନ ପ୍ରଣାଳୀ, ଯେଉଁଠାରେ ରୋମାନ୍ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ନାହିଁ ।
(ii) ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଶୂନ୍ୟ ପାଇଁ ପ୍ରତୀକ ଅଛି ଯାହା ‘0’ କିନ୍ତୁ ରୋମାନ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ନାହିଁ ।
(iii) ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ବଡ଼ ସଂଖ୍ୟାଗୁଡ଼ିକ ପଢ଼ିବା ସହଜ ଓ ଶୀଘ୍ର ହୁଏ ଏବଂ ସଂକ୍ଷିପ୍ତ ଭାବେ ଲେଖାଯାଇପାରେ । ଯେତେବେଳେ ରୋମାନ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଏହା ଅତ୍ୟଧ‌ିକ ଜଟିଳ ଏବଂ ଦୀର୍ଘ ହୋଇଥାଏ ।
(iv) ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଯୋଗ, ବିୟୋଗ, ଗୁଣନ, ଭାଗକ୍ରିୟା ଭଳି ଗାଣିତିକ ପ୍ରକ୍ରିୟା ସମ୍ପାଦନ କରିବା ସହଜରେ ହୋଇଥାଏ। ରୋମାନ ପ୍ରଣାଳୀରେ କେବଳ ମୌଳିକ ଯୋଗ ଅନୁମତି ଦିଏ ମାତ୍ର ଗୁଣନ ଓ ଭାଗକ୍ରିୟା ଭଳି ଗାଣିତିକ ପ୍ରକ୍ରିୟା ସମ୍ପାଦନ କରିବା ସହଜ ହୋଇନଥିଲା।

Question 4.
ଏହି ବିଭାଗରେ ଆଲୋଚନା କରାଯାଇଥିବା ଧାରଣାଗୁଡ଼ିକୁ ବ୍ୟବହାର କରି ତୁମେ ପୂର୍ବରୁ ସୃଷ୍ଟି କରିପାରିଥିବା ସଂଖ୍ୟା ପ୍ରଣାଳୀକୁ ଅଧିକ ସୁନ୍ଦର ବା ପରିମାର୍ଜିତ କରିବାକୁ ଚେଷ୍ଟାକର।
Solution:
ନିଜେ ଅଭ୍ୟାସ କର।

3.3 ଆଧାରର ଧାରଣା (The Idea of a Base)

ନିଜେ କରି ଦେଖ : (Page No. 62)

Question 1.
ନିମ୍ନଲିଖତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ମିଶରୀୟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କର।
1023, 2660, 784, 1111, 70507
Solution:

Question 2.
ନିମ୍ନ ସଂଖ୍ୟାସୂଚକ କେଉଁ ସଂଖ୍ୟାକୁ ବୁଝାଏ?

Solution:
(i) 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 + 6 + 10 = 200 + 70 + 6 = 276
(ii) 1000 + 1000 + 1000 + 1000 + 100 + 100 + 100 + 1 + 1 + 10 + 10 = 4000 + 300 + 20 + 2 = 4322

Page No. 62

Question 1.
ପୂର୍ବ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ଯାର ସମାନ 10ଟି ସମୂହକୁ ଏକାଠି କରିବା ପରିବର୍ତ୍ତେ (ମିଶରୀୟ ପ୍ରଣାଳୀରେ କରାଯାଇଥିବା ଭଳି), ପୂର୍ବ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ଯାର ସମାନ 5ଟି ସମୂହକୁ ଏକାଠି କରି ଆମେ ଗୋଟିଏ ସଂଖ୍ୟାପ୍ରଣାଳୀ ପାଇପାରିବା କି ? ଏହି 5 ସମୂହକୁ ଯେକୌଣସି ଧନାତ୍ମକ ପୂର୍ବସଂଖ୍ୟା ବଦଳରେ ବ୍ୟବହାର କରାଯାଇପାରିବ କି ?
Solution:
ମନେକରାଯାଉ । ହେଉଛି ପ୍ରଥମ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ।
ପୂର୍ବ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା (1)ର 5ଟି ସମୂହକୁ ଏକାଠି କର ।
ଏବେ ମିଳିଥୁବା ଆକାରର ଦ୍ଵିତୀୟ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା 5 ହେଉ ।
ପୂର୍ବ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା (5)ର 5ଟି ସମୂହକୁ ଏକାଠି କର ।
ଏହାର ଆକାରର ତୃତୀୟ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା 5 × 5 = 25 ହେଉ ।
ଏବେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା (25)ର 5ଟି ସମୂହକୁ ଏକାଠି କର ।
ଆମେ ଚତୁର୍ଥ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା 5 × 25 = 125 ପାଇବା ।

Question 2.
ଏହି ନୂତନ ପ୍ରଣାଳୀରେ 143କୁ ପରିପ୍ରକାଶ କର ।
Solution:
143 ଠାରୁ ସାନ ହୋଇଥିବା ସବୁଠାରୁ ବଡ଼ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା 53 = 125
125 ଠାରୁ ଆରମ୍ଭ କରି ଆମେ ସଂଖ୍ୟାମାନଙ୍କୁ ସମୂହରେ ପରିଣତ କରିବା ।
ଆମେ ପାଇବା 143 = 125 + 5 + 5 + 5 + 1 + 1 + 1
ତେଣୁ ନୂତନ ସଂଖ୍ୟାପ୍ରଣାଳୀରେ 143 ସଂଖ୍ୟାଟି ହେଉଛି

ନିଜେ କରି ଦେଖ : (Page No. 63)

Question 1.
ଉପରୋକ୍ତ ସାରଣୀ – 2ରେ ଥିବା ସଂକେତଗୁଡ଼ିକୁ ବ୍ୟବହାର କରି ନିମ୍ନ ସଂଖ୍ୟାଗୁଡ଼ିକୁ 5 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଲେଖ । 15, 50, 137, 293, 651
Solution:

Question 2.
ସେହି ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟରେ ଏପରି କୌଣସି ସଂଖ୍ୟା ଅଛି କି, ଯାହାକୁ 5 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କରାଯାଇପାରିବ ନାହିଁ ? ଯଦି ହଁ କାହିଁକି, ଯଦି ନା କାହିଁକି ନୁହେଁ?
Solution:
ହଁ । ଶୂନ୍ୟ (0) ସେହି ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟରେ ଯାହାକୁ 5 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କରାଯାଇପାରିବ ନାହିଁ, କାରଣ ଏହାର କୌଣସି ପ୍ରତୀକ ନାହିଁ ।

Question 3.
ଗୋଟିଏ 7- ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକ ସ୍ଥିର କର ।
Solution:
7⁰ = 1, 7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401
ତେଣୁ 1, 7, 49, 343, 2401 ସଂଖ୍ୟାଗୁଡ଼ିକ 7-ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ମାର୍ଗଦର୍ଶୀ ।

Question 4.
ସାଧାରଣତଃ n-ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପଦ୍ଧତିର ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକ କେତେ ହେବ ସ୍ଥିର କର ।
Solution:
n-ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକ ‘n’ର ଘାତ ସଂଖ୍ୟା n⁰ = 1, n, n², n³, …. ।

Question 5.
ଉଦାହରଣ : ନିମ୍ନଲିଖୂତ ମିଶରୀୟ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଯୋଗକର :
Solution:

ଆମେ । ଏବଂ ମ୍ ର ମୋଟ ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କରିବା ଏବଂ ସବୁଠାରୁ ବଡ଼ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ଯାରୁ ଆରମ୍ଭ କରି ସେଗୁଡ଼ିକୁ ସମୂହଭୁକ୍ତ କରିବା । ଏଠାରେ ମୋଟ ହେଉଛି 15ଟି ଲ ଏବଂ 15ଟି | ଅଛନ୍ତି । ଯେହେତୁ 10)ର ପରବର୍ତ୍ତୀ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ହେଉଛି ୭, ତେଣୁ ପ୍ରଦର୍ଶନ ସଂଖ୍ୟା ଯୋଗ କରିବାକୁ ନିମ୍ନ ପରି ଆଉଥରେ ଦଳଭୁକ୍ତ କରାଯାଇପାରିବ ।

ଯେହେତୁ 10ଟି 1କୁ ∩ ଲେଖାଯାଇପାରିବ, ତେଣୁ ଆମେ ପାଇବା

ନିଜେ କରି ଦେଖ : (Page No. 65)

Question 1.
ନିମ୍ନ ମିଶରୀୟ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଯୋଗକର ।

Solution:

Question 2.
5 ଆଧାରବିଶିଷ୍ଟ ପ୍ରଣାଳୀରେ ଥିବା ନିମ୍ନଲିଖିତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଯୋଗକରୁ ।

ମନେରଖ, ଏହି ପ୍ରଣାଳୀରେ ଏକ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାରେ 5 ଗୁଣିଲେ ପରବର୍ତୀ ସଂଖ୍ୟା ମିଳିଥାଏ ।
Solution:

Page No. 66-67

Question 1.
ଯେକୌଣସି ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାକୁ ∩ (10) ଦ୍ଵାରା ଗୁଣନ କଲେ କ’ଣ ହୁଏ ? ନିମ୍ନଲିଖୂତ ଗୁଣନ କାର୍ଯ୍ୟ ସମ୍ପାଦନ କର ।

ପ୍ରତ୍ୟେକ ମାର୍ଗଦଶୀ ସଂଖ୍ୟା 10ର ଘାତ ଅଟେ ଏବଂ ଏହାକୁ 10ରେ ଗୁଣନ କଲେ ଗୋଟିଏ ଘାତର ବୃଦ୍ଧିହୁଏ, ଯାହା ପରବର୍ତ୍ତୀ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ହୋଇଥାଏ ।
Solution:

Question 2.
ଯେକୌଣସି ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାକୁ (10²) ଦ୍ଵାରା ଗୁଣନ କଲେ କ’ଣ ପାଇବା ? ନିମ୍ନଲିଖତ ଗୁଣନ କାର୍ଯ୍ୟ ସଂପାଦନ କର।

Solution:

Question 3.
ନିମ୍ନଲିଖତ ଗୁଣନଗୁଡ଼ିକ ସଂପାଦନ କର–

ଅର୍ଥାତ୍‌, ଯେକୌଣସି ଦୁଇଟି ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ଯାର ଗୁଣଫଳ ଆଉ ଏକ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ଅଟେ ।
Solution:

ମିଶରୀୟ ସଂଖ୍ୟା ପଦ୍ଧତିରେ କୌଣସି ସଙ୍କେତ ନାହିଁ ।

Question 4.
ଏହି ଧର୍ମ ଆମେ ପୂର୍ବରୁ ପାଇଥବା 5 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀ ପାଇଁ ପ୍ରଯୁଜ୍ୟ ହେବ କି ? ଯେକୌଣସି ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପାଇଁ ଏହା ପ୍ରଯୁଜ୍ୟ କି?
Solution:
ହଁ, 5 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀ (ଅର୍ଥାତ୍, ଯେକୌଣସି ଦୁଇଟି ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାର ଗୁଣଫଳ ଆଉ ଏକ୍ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା) ପାଇଁ ଏହା ପ୍ରଯୁଜ୍ୟ ହେବ । ଅଧିକନ୍ତୁ ଯେକୌଣସି ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପାଇଁ ଏହା ପ୍ରଯୁଜ୍ୟ ମଧ୍ୟ।

Page No. 68

Question 1.
ଏବେ ନିମ୍ନଲିଖୂତ ଗୁଣଫଳଗୁଡ଼ିକ ନିଶ୍ଚୟ କର।

Solution:

Question 2.
ଗୋଟିଏ ସଂଖ୍ୟାକୁ ( ସହିତ ଗୁଣନ କରିବାର ସରଳ ନିୟମ କ’ଣ ହେବ?
Solution:
ଦୁଇଟି ସଂଖ୍ୟାକୁ ଗୁଣନ କରିବାର ପ୍ରକ୍ରିୟାରେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକର ଗୁଣନ ସଂପୃକ୍ତ । ଯେତେବେଳେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକ ଏକ ସଂଖ୍ୟାର ଘାତ ଅଟନ୍ତି, ସେତେବେଳେ ସେମାନଙ୍କର ଗୁଣଫଳ ଆଉ ଏକ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ହୋଇଥାଏ । ଏହି ତଥ୍ୟ ଗୁଣନ ପ୍ରକ୍ରିୟାକୁ ସରଳ କରିଥାଏ । ମାତ୍ର ରୋମାନ୍ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଏହା ସତ୍ୟ ନୁହେଁ । ସେଥ‌ିପାଇଁ ସେଗୁଡ଼ିକ ମଧ୍ୟରେ ଗୁଣନ କରିବା କଷ୍ଟକର ହୋଇଥାଏ । ଅର୍ଥାତ୍, ଏକ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ଯେଉଁଥରେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକ କୌଣସି ସଂଖ୍ୟାର ଘାତ ହୋଇଥାନ୍ତି ଅର୍ଥାତ୍ ନିର୍ଦ୍ଦିଷ୍ଟ ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟାପ୍ରଣାଳୀ କେବଳ ସଂଖ୍ୟା ପରିପ୍ରକାଶରେ ସୁବିଧାଜନକ ନୁହେଁ, ବରଂ ଗାଣିତିକ ପ୍ରକ୍ରିୟା ସଂପାଦନ କରିବାର ବହୁତ ଉପଯୋଗୀ ହୋଇଥାନ୍ତି । ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀର ଧାରଣା ସଂଖ୍ୟା ପ୍ରଣାଳୀର କ୍ରମବିକାଶ ଇତିହାସରେ ଏକ ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣ ମାଇଲଖୁଣ୍ଟ ଆମ୍ଭମାନଙ୍କର ଆଧୁନିକ ହିନ୍ଦୁ ସଂଖ୍ୟାପ୍ରଣାଳୀ ଏହି ଧାରଣା ଉପରେ ପର୍ଯ୍ୟବସିତ ।

Page No. 69

Question 1.
ଉଦାହରଣ ସ୍ଵରୂପ, ସଂଖ୍ୟା 3426କୁ ନେବା।
Solution:
ଏହାକୁ ନିମ୍ନମତେ ଦଳଭୁକ୍ତ (ସମୂହୀକରଣ) କରାଯାଇପାରିବ ।
Solution:
3426 = 1000 + 1000 + 1000 + 100 + 100 + 100 + 100 + 10 + 10 + 1 + 1 + 1 + 1 + 1 + 1
ଚିତ୍ରରେ ଏହି ସଂଖ୍ୟାଟିକୁ ଦର୍ଶାଯାଇଛି । 16ଟି ଏକକ କିପରି ସୂଚିତ କରାଯାଇଛି ଦେଖ ।

Question 2.
ପ୍ରତ୍ୟେକ ଧାଡ଼ି ଓ ଏହାର ଉପରଭାଗରେ ଥିବା ଗୋଟି (ଗୋଲି)ଗୁଡ଼ିକୁ ଏକାଠି କରାଯିବ । ଯଦି କୌଣସି ଧାଡ଼ିରେ ମୋଟ ସଂଖ୍ୟା 10ରୁ ଅଧିକ ହୁଏ, ତେବେ କ’ଣ କରାଯିବ?
Solution:
ଏହି ସମସ୍ୟାଟିରେ 7 ଏକକ ଏବଂ 3 ଏକକ ମିଶି 10 ଏକକ ହେଉଛି ଯାହାକୁ 10କୁ ସୂଚାଉଥିବା ଧାଡ଼ିରେ ଗୋଟିଏ (1) ଗୋଟିଏରେ ସୂଚାଯାଇଛି ଓ ଏହା ଧାଡ଼ିର ଟିକିଏ ଉପରକୁ ରଖାଯାଇଛି ।

ନିଜେ କରି ଦେଖ : (Page No. 69-70)

Question 1.
ଏପରି କୌଣସି ସଂଖ୍ୟା ଅଛି କି ଯାହାକୁ ମିଶରୀୟ ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କଲେ ସଂକେତ 10 ଥରେ କିମ୍ବା ତା’ଠାରୁ ଅଧିକ ଥର ଆସିଥାଏ ? କାହିଁକି ନୁହେଁ ?
Solution:
ନା, ଏପରି କୌଣସି ସଂଖ୍ୟା ହୋଇପାରିବ ନାହିଁ, ଯାହାର ମିଶରୀୟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କଲେ ସଂକେତ 10ଥର କିମ୍ବା ତା’ଠାରୁ ଅଧିକଥର ଘଟେ । କାରଣ ମିଶରୀୟ ପ୍ରଣାଳୀରେ 10ର ଘାତ ଭାବରେ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ଅଛି ।
ଉଦାହରଣ : 10¹ = | | | | | | | | | | = ∩

Question 2.
4 ଆଧାର ବିଶିଷ୍ଟ ନିଜର ଏକ ସଂଖ୍ଯାପ୍ରଣାଳୀ ସୃଷ୍ଟିକର ଏବଂ 1 ରୁ 16 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାକୁ ଲେଖ।
Solution:

Question 3.
ଆମେ ତିଆରି କରିଥିବା 5 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟାକୁ 5ରେ ଗୁଣିବା ପାଇଁ ଏକ ସରଳ ନିୟମ ଲେଖ ।
Solution:
5 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟାକୁ 5 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟାରେ ଗୁଣିଲେ ଆମେ 5 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପାଇବା।

3.4 ସ୍ଥାନୀୟମାନ ପରିପ୍ରକାଶ (Place Value Representation)

Page No. 71-72

Question 1.
ଉଦାହରଣ : ଏହି ପ୍ରଣାଳୀରେ 640 ସଂଖ୍ୟାଟିକୁ ଲେଖୁବା ।
Solution:
ଏହାକୁ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାରେ ଦଳଭୁକ୍ତ କଲେ
640 = 10 × 60 + 40
ଯଦି ମିଶରୀୟ ଧାରଣାକୁ ବ୍ୟବହାର କରୁ, ତେବେ ସଂଖ୍ୟାଟିକୁ 10 S ବ୍ୟବହାର କରି ଏବଂ 40 କୁ 4 < s ବ୍ୟବହାର କରି ଲେଖାଯାଇପାରିବ ।

Question 2.
ଆମେ ଏହାକୁ ଆହୁରି ସଂକ୍ଷେପରେ ପ୍ରକାଶ କରିପାରିବା କି?
Solution:
ହଁ, ଆମେ ଏହି ସଂଖ୍ୟାକୁ ସରଳ ଭାବରେ ନିମ୍ନରୂପରେ ପ୍ରକାଶ କରିପାରିବା ।
ଯାହାକୁ ଦଶଟି 60) ଏବଂ ଗୋଟିଏ 40 ଭାବରେ ପଢ଼ାଯାଇପାରିବ
ଯେପରି ପୂର୍ବରୁ ସମୀକରଣରେ ଲେଖାଯାଇଛି ।

Question 3.
ଉଦାହରଣ : 7530
Solution:
7530 = (2) × 3600 + (5) × 60 + 30
ତେଣୁ ଏହାର ପରିପ୍ରକାଶଟି ହେବ =

Question 4.
ଯଦି ଆମେ 60ର ବିଭିନ୍ନ ଘାତ ପାଇଁ ସଙ୍କେତଗୁଡ଼ିକୁ ସମ୍ପୂର୍ଣ୍ଣରୂପେ ବାଦଦେଇ ପରିପ୍ରକାଶକୁ ଆହୁରି ସଂକ୍ଷିପ୍ତ କରିବା, ତେବେ କ’ଣ ହେବ?
Solution:

ମେସୋପଟାମିଆର ଲୋକମାନଙ୍କର ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ସବୁଠାରୁ ଡାହାଣୁ ପଟରେ ଥିବା ସଙ୍କେତ ସବୁ 1 ର ସଂଖ୍ୟା ଦର୍ଶାଉଥିଲା । ତାର ବାମ ପାର୍ଶ୍ଵରେ ଥିବା ସଙ୍କେତସବୁ )ର ସଂଖ୍ୟା ଦର୍ଶାଉଥିଲା । ପରବର୍ତ୍ତୀଟି 3600ର ସଂଖ୍ୟା ଦର୍ଶାଉଥୁଲା ଇତ୍ୟାଦି । ଯେତେବେଳେ 60ର କୌଣସି ଘାତ ଦେଖାଯାଉ ନଥୁଲା, ସେତେବେଳେ ସେହି ସ୍ଥାନକୁ ଖାଲି ଛଡ଼ାଯାଉଥିଲା ।

ନିଜେ କରି ଦେଖ (Page No. 73)

Question 1.
ନିମ୍ନଲିଖତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ମେସୋପଟାମୀୟ ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କର ।
(i) 63
(ii) 132
(iii) 200
(iv) 60
(v) 3605
Solution:

Page No. 76

Question 1.
ଉଦାହରଣ 1 : ମାୟା ପ୍ରଣାଳୀ ବ୍ୟବହାର କରି ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାଗୁଡ଼କୁ ପ୍ରକାଶ କର ।
(i) 77
(ii) 100
(iii) 361
(iv) 721
Solution:

(iii) (i) ଭଳି ଅଭ୍ୟାସ କର ।
(iv) (i) ଭଳି ଅଭ୍ୟାସ କର ।

ନିଜେ କରି ଦେଖ : (Page No. 80)

Question 1.
ତୁମେ କାହିଁକି ଭାବୁଛି ଯେ ଚୀନ୍ ଦେଶର ଲୋକମାନେ ଜୋଙ୍ଗ (Zong) ଏବଂ ହେଙ୍ଗ (Heng) ସଂକେତ ମଧ୍ୟରେ ଅଦଳବଦଳ କରୁଥିଲେ ? ଯଦି କେବଳ ଜୋଙ୍ଗ (Zong) ସଂକେତ ବ୍ୟବହାର କରାଯାଏ, ତେବେ 41 କୁ କିପରି ଲେଖାଯାଇପାରିବ ? ଯଦି କ୍ରମରେ ଥ‌ିବା ଦୁଇଟି ସ୍ଥାନ ମଧ୍ୟରେ କୌଣସି ଆଦୃଶିଆ ଖାଲିଜାଗା ନଥାଏ, ତେବେ ସେହି ସଂଖ୍ୟାକୁ ଅନ୍ୟ କୌଣସି ଉପାୟରେ ବର୍ଣ୍ଣନା କରାଯାଇପାରିବ କି ?
Solution:
ସବୁଠାରୁ ଉପଯୁକ୍ତ କାରଣ ହେଉଛି ସେମାନେ ପଢ଼ିବା ସମୟରେ ଭୁଲ୍ ସଂଖ୍ୟା ଏଡ଼ାଇବାକୁ ଚାହୁଁଥିଲେ । ସେମାନେ ସମସ୍ତ ସଂଖ୍ୟା ରଡ଼୍ ସଂଖ୍ୟା ଦ୍ଵାରା ଉପସ୍ଥାପନା କରୁଥିଲେ । ଯାହା ଲିଖିତ ପ୍ରଣାଳୀ ଅପେକ୍ଷା ସଂଖ୍ୟା ଲେଖୁବା ଓ ହିସାବ କରିବାରେ ଅଧିକ ଫଳପ୍ରଦ ଥିଲା । କେବଳ ଜୋଙ୍ଗ (Zong) ସଂକେତ ବ୍ୟବହାର କରି 41କୁ ଲେଖାଯାଇ ପାରିବ : ||||
ଯଦି କ୍ରମିକ ସ୍ଥାନ ମଧ୍ୟରେ କୌଣସି ଖାଲିସ୍ଥାନ ନରହେ, ଏହାକୁ ସହଜରେ 5 ଭାବେ ସହଜରେ ବ୍ୟାଖ୍ୟା କରାଯାଇପାରିବ ।

Question 2.
‘ଉକାସାର’ ଏବଂ ‘ଉରାପୋନ’କୁ ଅଙ୍କ ଭାବରେ ବ୍ୟବହାର କରି ଏକ ସ୍ଥାନୀୟମାନ ପ୍ରଣାଳୀ ଗଠନ କର । ଗୁମୁଲଗାଲ ପ୍ରଣାଳୀ ସହିତ ଏହାକୁ ତୁଳନା କର ।
Solution:
ଧରାଯାଉ 2⁰ = 1 = A, 2¹ = 2 = B, 2² = 4 = C, 2⁴ = 16 = D
ଉଭୟର ସମାନ 2 ଆଧାର ରହିଛି କିନ୍ତୁ 2 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀରେ ଅନେକ ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ଅଛି, ଯେତେବେଳେ ଗୁମୁଲଗାଲ ପ୍ରଣାଳୀରେ କେବଳ ଦୁଇଟି ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା ଅଛି ।

Question 3.
ତୁମର ଦୈନନ୍ଦିନ ଜୀବନରେ କେଉଁଠାରେ ଏବଂ କେଉଁ ବୃତ୍ତିରେ ହିନ୍ଦୁସଂଖ୍ୟା ସୂଚକ ଏବଂ 0 ଏକ ଗୁରୁତ୍ଵପୂର୍ଣ ଭୂମିକା ଗ୍ରହଣ କରନ୍ତି । ଯଦି ଆମର ସଂଖ୍ୟା ପ୍ରଣାଳୀ ଏବଂ 0 ଆବିଷ୍କୃତ ହୋଇନଥାନ୍ତା କିମ୍ବା କଳ୍ପନା କରାଯାଇ ନଥାନ୍ତା, ତେବେ ଆମର ଜୀବନ କିପରି ଭିନ୍ନ ହୋଇଥାନ୍ତା ?
Solution:
ଆମର ଦୈନନ୍ଦିନ ଜୀବନରେ ସଂଖ୍ୟା ପଢ଼ିବା । ଗଣନା କରିବା ଇତ୍ୟାଦିର ହିନ୍ଦୁ ସଂଖ୍ୟା ଏହା ସ୍ଥାନୀୟମାନ ପ୍ରଣାଳୀ ଉପରେ ଆଧାରିତ ।
ଠ କୁ ଗୋଟିଏ ଅଙ୍କ ଭାବେ ପ୍ରତ୍ୟେକ ସ୍ଥାନରେ ଏକ ଅଙ୍କ ଭାବେ ବ୍ୟବହାର ହେତୁ, ଏହି ପଦ୍ଧତିରେ ସଂଖ୍ୟା ଲେଖୁବାରେ କୌଣସି ଦ୍ଵନ୍ଦ ସୃଷ୍ଟି ହୁଏ ନାହିଁ ।

Question 4.
ପ୍ରାଚୀନ ଭାରତୀୟମାନେ ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ପାଇଁ ସମ୍ଭବତଃ 110 କୁ ଆଧାରରୂପେ ବ୍ୟବହାର କରୁଥିଲେ । କାରଣ ମଣିଷର 10ଟି ଆଙ୍ଗୁଠି ଅଛି ଏବଂ ଆମେ ଗଣନା କରିବା ପାଇଁ ଆମର ଆଙ୍ଗୁଠିକୁ ବ୍ୟବହାର କରିଥାଉ । କିନ୍ତୁ ଯଦି ଆମର କେବଳ ୫ଟି ଆଙ୍ଗୁଠି ଥାଆନ୍ତା ତେବେ କ’ଣ ହୋଇଥାନ୍ତା ? ସେ କ୍ଷେତ୍ରରେ ଆମେ ସଂଖ୍ୟା କିପରି ଲେଖୁଥାନ୍ତେ ? ଯଦି ଆମେ 10) ପରିବର୍ତ୍ତେ ୫ କୁ ଆଧାର ଭାବେ ବ୍ୟବହାର କରିଥାନ୍ତୁ, ତେବେ ହିନ୍ଦୁ ସଂଖ୍ୟା ସୂଚକଗୁଡ଼ିକ କିପରି ହୋଇଥାନ୍ତା ? ସେହିପରି 5 ଆଧାର ହୋଇଥିଲେ କ’ଣ ହୋଇଥାଆନ୍ତା ? ସେହିପରି 5 ଆଧାର ହୋଇଥିଲେ ହିନ୍ଦୁ ସଂଖ୍ୟା 25 କୁ 8 ଆଧାର ଏବଂ 5 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀ ବ୍ୟବହାର କରି ଲେଖୁବାକୁ ଚେଷ୍ଟାକର । ଏହାକୁ ତୁମେ 2 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀରେ ଲେଖୁରିବ କି ?

Solution:
8 ଆଧାର ପାଇଁ ସଂଖ୍ୟାଗୁଡ଼ିକ ହୋଇଥାନ୍ତି : 0, 1, 2, 3, 4, 5, 6, 7
ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ଯାଗୁଡ଼ିକ : 8⁰, 8¹, 8²,…….
5 ଆଧାର ପାଇଁ ସଂଖ୍ୟାଗୁଡ଼ିକ ହୋଇଥାନ୍ତି : 0, 1, 2, 3, 4
ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟାଗୁଡ଼ିକ : 5⁰, 5¹, 5², 5³, 5⁴
ଏବେ ଆମେ ହିନ୍ଦୁ ସଂଖ୍ୟା ତେଣୁ 25 କୁ 8 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଲେଖୁଲେ, 25 = 3 × 8¹ + 1 × 8⁰
25 କୁ 8 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟାରେ 3 1₈ ଲେଖାଯାଇପାରିବ ।
ଏହାକୁ ମଧ୍ୟ ଆମେ ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ 25 କୁ 5 ଆଧାର ବିଶିଷ୍ଟ ପ୍ରଣାଳୀରେ ଲେଖୁବା ।
25 = 1 × 5² + 0 × 5¹ + 0 × 5⁰, 25 = 100₅
ଏବେ 2 ଆଧାର ବିଶିଷ୍ଟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ, 25 = 1 × 2⁴ + 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
ତେଣୁ 25 = 11001₂

Class 8 Maths Chapter 3 MCQ Odia Medium

ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର।

Question 1.
ଆମେ ବ୍ୟବହାର କରୁଥିବା ମୌଖ୍କ ଓ ଲିଖ ସଂଖ୍ୟାର ସୃଷ୍ଟି ___________________ ଦେଶରେ ହୋଇଥିଲା?
Answer:
ଭାରତ

Question 2.
18 କୁ ସଙ୍କେତ ମାଧ୍ୟମରେ ପରିପ୍ରକାଶ କଲେ ___________________ ହେବ।
Answer:
XVIII

Question 3.
ଗୁମୁଲଗାଲ 6 ରୁ ବଡ଼ ଯେକୌଣସି ସଂଖ୍ୟାକୁ ___________________ କହୁଥିଲେ।
Answer:
ରାସ୍

Question 4.
ଯେଉଁ ସଂଖ୍ୟା ପାଇଁ ରୋମାନ୍ ପ୍ରଣାଳୀରେ ନୂଆ ସଙ୍କେତଗୁଡ଼ିକ ବ୍ୟବହାର ହୋଇଛି, ସେଗୁଡ଼ିକୁ ___________________ ସଂଖ୍ୟା କୁହାଯାଏ।
Answer:
ମାର୍ଗଦର୍ଶୀ ସଂଖ୍ୟା

Question 5.
ରୋମାନ୍ ପ୍ରଣାଳୀରେ 2367 ର ସଂଖ୍ୟା ସୂଚକ ହେଉଛି ___________________|
Answer:
MMCCCLXVII

Question 6.
715 କୁ ରୋମାନ୍ ପ୍ରଣାଳୀରେ ଲେଖୁଲେ ___________________ ହେବ।
Answer:
DCCXV

Question 7.
VXL ର ମାନ ___________________|
Answer:
250

Question 8.
ଏକ ଲିଖ୍ୟାତ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ସୂଚ।ଉଥ୍ବ| ସଂକେତରୁଡ଼ିକୁ ___________________ କୁହାଯାଏ।
Answer:
ସୂଚକ

Question 9.
ସାରା ବିଶ୍ଵରେ ବ୍ୟବହାର କରାଯାଉଥିବା ସଂଖ୍ୟା ପ୍ରଣାଳୀକୁ ___________________ କୁହାଯାଏ।
Answer:
ହିନ୍ଦୁ ସଂଖ୍ୟା ପ୍ରଣାଳୀ

Question 10.
302 କୁ ରୋମାନ୍ ପ୍ରଣାଳୀରେ ପରିପ୍ରକାଶ କଲେ ___________________ ହେବ।
Answer:
CCCII

ସଂକ୍ଷେପରେ ଉତ୍ତର ଲେଖ।

Question 1.
ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ରୋମାନ୍ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଲେଖ।
(a) 49
(b) 72
(c) 236
(d) 1769
Answer:
(a) 49 = XLIX (∵ 49 = 50 – 10 + 9)
(b) 72 = LXXII (∵ 72 = 50 + 10 + 10 + 2)
(c) 236 = CCXXXVI (∵ 236 = 200 + 30 + 5 + 1)
(d) 1769 = MDCCLXIX (∵ 1769 = 1000 + 500 + 200 + 50 + 10 + 9)

Question 2.
ନିମ୍ନଲିଖ୍ୟାତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଭାରତୀୟ ସଂଖ୍ୟା ପ୍ରଣାଳୀରେ ଲେଖ।
(a) XXXIX
(b) LXXXIV
(c) LI
(d) CXXXVII
(e) DCXXV
(f) CXCVIII
Answer:
(a) XXXIX = 30 + 9 = 39
(b) LXXXIV = 50 + 30 + 4 = 84
(c) LI = 50 + 1 = 51
(d) CXXXVII = 100 + 30 + 7 = 137
(e) DCXXV = 500 + 100 + 20 + 5 = 625
(f) CXCVIII = 100 + 90 + 8 = 198

Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 4 Data Handling and Presentation MCQ improves accuracy in objective exams.

MCQ on Data Handling and Presentation Class 6

Data Handling and Presentation MCQ Class 6

Class 6 Maths Data Handling and Presentation MCQ

Question 1.
The marks (out of 10) obtained by 28 students in a Mathematics test are listed as below:
8, 1, 2, 6, 5, 5, 5, 0, 1, 9, 7, 8, 0, 5, 8, 3, 0, 8, 10, 10, 3, 4, 8, 7, 8, 9, 2, 0
The number of students who obtained marks more than or equal to 5 is:
(a) 3
(b) 15
(c) 16
(d) 17
Solution:
(d) 17
The given marks can be arranged in ascending order as follows:
0, 0, 0, 0, 1,1,2, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 10, 10
Hence, the number of students who obtained marks more than or equal to 5 is 17.

Question 2.
The table below shows the marks obtained by 5 students in Science exam.

Name of Student	Attendance Marks	Assignment Marks	Theory (Marks)	Total marks
Abhijeet	5	10	45	60
Rajeev	4	8	58	70
Simran	5	6	38	49
Karuna	5	9	57	71
Jatin	4	10	65	79

Which student scored the highest marks in theory?
(a) Karima
(b) Jatin
(c) Abhijeet
(d) Simran
Solution:
(b) Jatin
From the table, we can see that the theory
column contains the highest marks of 65, which corresponds to Jatin. Hence, Jatin scored the highest marks in theory.

Question 3.
A dice is rolled 20 times. Following tally table shows which number comes up how many times.

Which of the number appeared maximum number of times?
(a) 1
(b) 2
(c) 3
(d) 6
Solution:
(a) 1
1 appears 7 times, which is maximum.

Question 4.
The following bar graph shows number of toys produced by a company during certain week:

The minimum number of toys produced on:
(a) Tuesday
(b) Monday
(c) Friday
(d) Saturday
Solution:
(d) Saturday
The bar graph shows that the minimum number of toys were produced on Saturday, with a total of 40.

Question 5.
The pictograph shows the number of bouquets sold by a flower shop in the past 4 days.


What is the difference between the maximum number of bouquets sold and the least number of bouquets sold?
(a) 5
(b) 21
(c) 15
(d) 33
Solution:
(b) 21
Maximum number of bouquets sold on thursday = 10 × 3 = 30
Least number of bouquets sold on Wednesday = 3 × 3 = 9
The difference between the maximum and least number of bouquets sold = 30 – 9 = 21

Question 6.
The following bar graph shows the number of sections in a class from class VI to class X:

The pair of classes in which the number of sections is same, is:
(a) VIII and X
(b) VIII and IX
(c) VII and IX
(d) IX and X
Solution:
(b) VIII and IX
The given bar graph shows that the number of sections is same in classes VIII and IX, equal to 4.

Question 7.
From the bar graph given in previous question, the total number of sections from class VI to class X, is:
(a) 19
(b) 20
(c) 22
(d) 25
Solution:
(c) 22
From the given bar graph,
Number of sections in class VI = 6
Number of sections in class VII = 5
Number of sections in class VIII = 4
Number of sections in class IX = 4
Number of sections in class X = 3
So, total number of sections from classes VI to X = 6 + 5 + 4 + 4 + 3 = 22

Question 8.
What are the advantages of using a bar graph over a pictograph?
(i) Bar graphs are more accurate for large values.
(ii) Bar graphs can be made fancy with pictures.
(iii) It is easier to compare different categories with bar graphs.
(iv) It can be more challenging to prepare a pictograph than a bar graph when the frequencies are not exact multiples of the scale
Choose the correct option from the following:
(a) Only (i) and (iv)
(b) Only (i) and (ii)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
Pictographs are a nice visual and suggestive way to represent data. They represent data through pictures of objects. It can be more challenging to prepare a pictograph when the amount of data is large or when the frequencies are not exact multiples of the scale or key. It is also easier to compare different categories of data with bar graphs.

Question 9.
A class has these shoe sizes (already in order):
3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7
(i) The shoe size range is 4.
(ii) There are 10 students with size 4.
(iii) Shoe size 4 is worn by more students than size 3.
(iv) Size 6 is the most common.
Choose the correct option from the following:
(a) Only (i) and (iv)
(b) Only (i) and (iii)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(b) Only (i) and (iii)
We have, size range = largest size – smallest size
= 7 – 3 = 4.
From the given list of shoe size, there are 8 students who wore shoes of size 4 while, and there are 3 students who wore shoes of size 3. Therefore, shoe size 4 is worn by more students than size 3.

Also, there are 10 students who wore shoes of size 5, while there are only 4 students who wore shoes of size 6. So, size 5 is the most common.

Question 10.
In which of the following situations data needs to be collected?
(i) Finding out the favourite fruit of 10 people.
(ii) Finding out the capital city of a state.
(iii) Counting the number of pets that the students own.
(iv) Finding out the total runs scored by Sachin Tendulkar in test cricket. ,
Choose the correct option from the following:
(a) Only (i) and (iii)
(b) Only (ii) and (iv)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(a) Only (i) and (iii)
We would need to ask each of the 10 people about their favourite fruit. So, this is an example of data collection.
The capital city of a particular state is a known fact, so there is no need to collect data.
We would need to ask each student about the number of pets they have, so this is an example of data collection.
Again, the total runs scored by Sachin Tendulkar in test cricket is a known fact, so there is no need to collect data.

Data Handling and Presentation Class 6 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): In a bar graph, all bars must be of equal width.
(R): Equal width ensures that the comparison is based only on the height of the bars.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Uniform width of bars is essential to not confuse width with quantity. Only height (or length) of the bar should represent value (frequency). Having equal widths supports accurate comparison.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): Bar graph with vertical bars is better to represent height of mountains.
(R): Vertical bars help us visually interpret increasing values as ‘growing upward’, like heights.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Heights naturally grow upwards. The vertical bars give an intuitive and realistic feel when comparing heights (like of mountains or buildings). The reason explains the assertion.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Data Handling and Presentation Class 6 Fill in the Blanks

Question 1.
Organising the data by arranging it in ________ or ______ order is called an array.
Solution: ascending, descending
Organising the data by arranging it in ascending or descending order is called an array.

Question 2.
Data obtained in the _____ form is called raw data.
Solution: original
Data obtained in the original form is called raw data.

Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 3 Number Play MCQ improves accuracy in objective exams.

MCQ on Number Play Class 6

Number Play MCQ Class 6

Class 6 Maths Number Play MCQ

Question 1.
How many supercells are there in the table below?

4532	1234	3165	3995
6721	8751	4987	1087
5421	3456	6134	5601
2136	4500	2180	3579

(a) 4
(b) 6
(c) 7
(d) 9
Solution:
(a) 4
We know, a cell is a supercell if the number in it is greater than the numbers in its neighbouring cells that are immediately to the left, right, top and bottom.

4532	1234	3165	3995
6721	8751	4987	1087
5421	3456	6134	5601
2136	4500	2180	3579

Thus, cell containing 3995 is a supercell as 3995 is greater than the numbers in its neighbouring cells i.e. 3165 and 1087.

Cell containing 8751 is a supercell as 8751 is greater than the numbers in its neighbouring cells i.e. 1234, 6721, 3456 and 4987.

Cell containing 6134 is a supercell as 6134 is greater than the numbers in its neighbouring cells i.e. 4987, 3456, 2180 and 5601.

Cell containing 4500 is a supercell as 4500 is greater than the numbers in its neighbouring cells i.e. 3456, 2136 and 2180. So, there are 4 supercells.

Question 2.
If there are 7 cells in a row, then the maximum number of supercells possible is:
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
(b) 4
Given, number of cells in a row, n = 7 (Odd number)
So, maximum number of supercells = \(\frac{n+1}{2}\)
= \(\frac{7+1}{2}=\frac{8}{2}\) = 4

Question 3.
Which of the following numbers should be filled in the empty cell such that there are 4 supercells in the table given below?

162	393	297
236		453
193	503	384

(a) 256
(b) 128
(c) 7
(d) 9
Solution:
(b) 128
In a grid, a cell is a supercell if the number in it is greater than the numbers in the neighbouring cells that are immediately to the left, right, top and bottom.
If 256 is filled in the empty cell, then there would be 3 supercells only.

162	393	297
236	256	453
193	503	384

If 128 is filled in the empty cell, then there would be 4 supercells.

162	393	297
236	128	453
193	503	384

If 421 is filled in the empty cell, then there would be 2 supercells only.

162	393	297
236	421	453
193	503	384

If 625 is filled in the empty cell, then there would be 1 supercell only.

162	393	297
236	625	453
193	503	384

Question 4.
Identify the numbers marked on the number line below.

What is the value of (x + y)?
(a) 10475
(b) 10480
(c) 10485
(d) 10490
Solution:
(d) 10490
There are 3 sub-divisions between 5240 and 5255.
Therefore, each sub-division represents
\(\frac{5255-5240}{3}=\frac{15}{3}\) = 5 numbers.
Thus, the labelled number line is shown below.

So, x = 5230 and y = 5260
Now, x + y = 5230 + 5260 = 10490

Question 5.
What is the highest sum of digits of a number between 35 and 45?
(a) 12
(b) 11
(c) 13
(d) 9
Solution:
(a) 12
Numbers between 35 and 45 are: 36, 37, 38, 39, 40, 41, 42, 43 and 44.

Numbers	Sum of digits
36	3 + 6 = 9
37	3 + 7 = 10
38	3 + 8 =11
39	3 + 9 = 12 (Highest)
40	4 + 0 = 4
41	4 + 1 = 5
42	4 + 2 = 6
43	4 + 3 = 7
44	4 + 4 = 8

Question 6.
In the number line pattern, if you are moving from 105 to 125 and then from 125 to 145, what will be the number after moving from 145 with the same pattern?
(a) 160
(b) 165
(c) 170
(d) 175
Solution:
(b) 165
First movement: From 105 to 125
Second movement: From 125 to 145
Next movement: From 145 to 145 + 20 = 165
So, the required number is 165.

Question 7.
Among the numbers 1 – 100, how many times will the digit ‘5’ occur?
(a) 19
(b) 20
(c) 21
(d) 22
Solution:
(b) 20
Numbers that contain digit ‘5’ are:
5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55(it has two 5s), 56, 57, 58, 59, 65, 75, 85 and 95
Thus, digit ‘5’ occurs 20 times.

Question 8.
Which of the following is a palindromic number?
(a) 1212
(b) 1331
(c) 4141
(d) 1009
Solution:
(b) 1331
We know, palindromic numbers read the same from left to right and right to left.
Thus, 1331 is a palindromic number.

Question 9.
What is the smallest sum of digits of a number from 116 to 125?
(a) 2
(b) 3
(c) 6
(d) 7
Solution:
(b) 3
Numbers from 116 to 125 are: 116, 117, 118, 119, 120, 121, 122, 123, 124 and 125.

Numbers	Sum of digits
116	1 + 1 + 6 = 8
117	1 + 1 + 7 = 9
118	1 + 1 + 8 = 10
119	1 + 1+ 9 =11
120	1 + 2 + 0 = 3 (smallest)
121	1 + 2 + 1 = 4
122	1 + 2 + 2 = 5
123	1 + 2 + 3 = 6
124	1 + 2 + 4 = 7
125	1 + 2 + 5 = 8

Question 10.
What is the sum of the smallest and largest 2-digit numbers with unique digits?
(a) 108
(b) 110
(c) 100
(d) 121
Solution:
(a) 108
Smallest 2-digit number with unique digits = 10
Largest 2-digit number with unique digits = 98
Required sum = 10 + 98 = 108

Question 11.
If you subtract the smallest 2-digit even number from smallest 3-digit odd number, what is the result?
(a) 91
(b) 81
(c) 101
(d) 110
Solution:
(a) 91
Smallest 2-digit even number = 10
Smallest 3-digit odd number = 101
Now, 101 – 10 = 91

Question 12.
A 24-hour digital clock is showing 14:41 now. How many minutes until the clock shows the next palindromic time?
(a) 60 minutes
(b) 70 minutes
(c) 80 minutes
(d) 90 minutes
Solution:
(b) 70 minutes
Time now = 14:41 and next palindrome time = 15:51
Thus, 15:51 – 14:41 = 1 hr 10 mins
=70 minutes
Hence, the clock shows the next palindromic time after 70 minutes.

Question 13.
The 5^th term in the Collatz sequence starting with 21 is:
(a) 8
(b) 16
(c) 10
(d) 17
Solution:
(a) 8
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
First term: 21
Second term: 3 × 21 + 1 = 64
[As 21 is an odd number.]
Third term: \(\frac{64}{2}\) = 32 [As 64 is an even number.]
Fourth term: \(\frac{32}{2}\) = 16
[As 32 is an even number.]
Fifth term: \(\frac{16}{2}\) = 8 [As 16 is an even number.]
Hence, the required term is 8.

Question 14.
The sum of 3^rd and 5^th terms in the Collatz sequence starting with 100 is:
(a) 28
(b) 63
(c) 34
(d) 41
Solution:
(b) 63
First term: 100
Second term: \(\frac{100}{2}\) = 50
[As 100 is an even number.]
Third term: \(\frac{50}{2}\) = 25
[As 50 is an even number.]
Fourth term: 3 × 25 + 1 = 76
[As 25 is an odd number.]
Fifth term: y = \(\frac{76}{2}\) [As 76 is an even number.]
Hence, the sum of 3^rd and 5^th terms = 25 + 38
= 63

Question 15.
Which of the following numbers are numbers in the collatz sequence starting with 18?
(i) 19
(ii) 52
(iii) 17
(iv) 35
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii)
Rule: If the number is even, divide by 2.
If the number is odd, multiply by 3 and add 1.
18 → 9 → 28 → 14 → 7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
From the given numbers, 17 and 52 are present in the Collatz sequence starting with 18.

Question 16.
Which of the following numbers are Supercells in the given grid?

109	62	573	432
101	673	809	84
572	961	274	340
209	173	114	200
381	524	468	93

(i) 673
(ii) 468
(iii) 809
(iv) 109
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(c) (iii) and (iv)
809 and 961 are greater than 673 in its neighbourh ood.
So, 673 is not a supercell.
524 is greater than 468 in its neighbourhood. So, 468 is not a supercell

There is no number greater than 809 in its neighbourhood, i.e. 573, 673, 84 and 274.
So, 809 is a supercell.
There is no number greater than 109 in its neighbourhood, i.e. 62 and 101.
So, 109 is a supercell.

Question 17.
Which of the following numbers are palindrome and have digit sum equal to 14?
(i) 545
(ii) 2552
(iii) 3434
(iv) 1771
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (i) and (ii)
545, 2552 and 1771 are palindromes but digit sum of 1771 is not 14.
1 + 7 + 7 + 1 = 16
3434 is not a palindrome as we get 4343 on reversing the digits, which is not same as 3434.

Number Play Class 6 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The coloured cell is die supercell in the given table.

15000

15500

16000

16500

(R): In a row, a cell is a supercell if the number in it is greater than the numbers in the neighbouring cells that are immediately to the left and right.
Solution:
(d) A is false but R is true.
We know, a cell is a supercell if the number in it is greater than the numbers in its neighbouring cells that are immediately to the left and right. Here, 16000 is greater than 15500 but smaller than 16500.
Thus, Assertion (A) is lalse, but Reason (R) is true.

Question 2.
(A): If 120 is the largest number in a grid, then cell containing 120 is a supercell.
(R): The cell having the largest number in a grid is always a supercell.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
If a number is the largest number in a grid, then it will be greater than all the numbers in its neighbouring cells.
Thus, the cell having the largest number in a grid is always a supercell.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): The cell containing 4500 is a supercell in the given table.

5100	4900	4600
4700	4500	5000
3900	4800	5300

(R): 4500 is smaller than 4900, 4700, 4800 and 5000.
Solution:
(d) A is false but R is true.
We know, a cell is a supercell if the number in it is greater than the numbers in the neighbouring cells that are immediately to the left, right, top and bottom.

As 4500 is smaller than 4900, 4700, 4800 and 5000, the cell containing 4500 is not a supercell. Thus, Assertion (A) is false, but Reason (R) is true.

Number Play Class 6 Fill in the Blanks

Question 1.
Two adjacent cells _______ be supercells.
Solution: cannot
Two adjacent cells cannot be supercells.

Question 2.
The cell having the ____ number in a grid is always a supercell.
Solution: largest
The cell having the largest number in a grid is always a supercell.

Question 3.
On the number line, a smaller number is always to the ______of a larger number.
Solution: left
On the number line, a smaller number is always to the left of a larger number.

Question 4.
The digits of number 7203 adds up to _______ .
Solution: 12
Sum of digits of 7203 is 7 + 2 + 0 + 3 = 12.

Question 5.
The total number of 4-digit numbers is __________ .
Solution: 9000
Smallest 4-digit number = 1000
Largest 4-digit number = 9999
The total number of 4-digit numbers = 9999 – 1000 + 1 = 9000

Question 6.
The largest palindromic number between 700 and 800 is _______ .
Solution: 797
The largest palindromic number between 700 and 800 is 797.

Question 7.
The next number in the Collatz sequence after 24 is __________ .
Solution: 12
Since 24 is an even number, the next number is \(\frac{24}{2}\) = 12.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 10 The Other Side of Zero Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 10 The Other Side of Zero Solutions

Ganita Prakash Class 6 Chapter 10 Solutions

Class 6 Maths Ganita Prakash Chapter 10 Solutions The Other Side of Zero

Question 1.
Evaluate these expressions:
(a) (+ 1) + (+ 4) = _______
(b) (+ 4) + (+ 1) = _______
(c) (+ 4) + (- 3) = _______
(d) (- 1) + (+ 2) = _______
(e) (- 1) + (+ 1) = _______
(f) 0 + (+ 2) = _______
(g) 0 + (- 2) = _______
Solutions:
(a) (+ 1) + (+ 4) = + 5
(b) (+ 4) + (+ 1) = + 5
(c) (+ 4) + (- 3) = + 1
(d) (- 1) + (+ 2) = + 1
(e) (- 1) + (+ 1) = 0
(f) 0 + (+ 2) = + 2
(g) 0 + (- 2) = – 2

Question 2.
Try to subtract: – 3 – (+ 5). How many zero pairs will you have to put in? What is the result?
Solution:
– 3 – (+ 5) = – 8
We want to take away 5 positive tokens when we have 3 negative tokens. So, we add 5 zero pairs. After removing the 5 positive tokens, we have 8 negative tokens left.

Question 3.
Suppose you start with 0 rupees in your bank account and then you have debits of ₹ 1, ₹ 2, ₹ 4, ₹ 8, ₹ 16, ₹ 32, ₹ 64 and ₹ 128 and then a single credit of ₹ 256. What is your bank account balance now?
Solution:
Total debits = ₹ 1 + ₹ 2 + ₹ 4 + ₹ 8 + ₹ 16 + ₹ 32 + ₹ 64 + ₹ 128 = ₹ 255
There is a single credit of ₹ 256.
∴ Final balance = Total credit – Total debit = ₹ 256 – ₹ 255 = ₹ 1

Question 4.
Looking at the geographical cross section, fill in the respective heights:

Solution:
A = + 1500 m
B = – 500 m
C = + 300 m
D = – 1200 m
E = + 1200 m
F = – 200 m
G = + 100 m

Question 5.
Leh in Ladakh gets very cold during winter. The following is a table of temperature readings taken during different times of the day and night in Leh on a day in November. Match the temperature with the appropriate time of the day and night.

Solution:
Usually, the coldest time of the day is very early in the morning, before the sun rises, and it starts to get warmer after sunrise, reaching the warmest in the afternoon. At night, the temperature starts to drop again.
So, we can match the temperatures like this:
– 4°C → 2:00 a.m. (This is the coldest time when it is still dark and freezing.)
– 2°C → 11:00 p.m. (It is night time, so it is also very cold, but not as cold as early morning.)
8°C → 11:00 a.m. (It is late morning, the sun is up and it is getting warmer.)
14°C → 2:00 p.m. (This is the afternoon and the warmest time of the day.)
This shows us how the temperature rises as the sun comes up and falls again after sunset.

Question 6.
Complete the grids to make the required border sum

Solution:

Question 7.
There are two dice whose faces have these numbers: – 1, 2, – 3, 4, – 5, 6. The smallest possible sum upon rolling these dice is – 10 = (- 5) + (- 5) and the largest possible sum is 12 = (6) + (6). Some numbers between (- 10) and (+ 12) are not possible to get by adding numbers on these two dice. Find those numbers.
Solution:
Let’s find the sums that are not possible when rolling these two dice.
The faces of the dice are:
– 1, 2, -3, 4, -5, and 6.
First, let’s list all possible sums:
The sum of two negative numbers:
(- 1) + (- 1) = -2; (- 1) + (- 3) = – 4; (- 1) + (- 5) = – 6
(- 3) + (- 3) = – 6; (- 3) + (- 5) = – 8 (- 5) + (- 5) = – 10
The sum of one negative and one positive number:
(-1) + 2 = 1; (-1)+ 4 = 3; (-1) + 6 = 5; (-3) + 2 = – 1; (-3)+ 4=1;
(- 3) + 6 = 3; (- 5) + 2 = – 3; (- 5) + 4 = – 1; (- 5) + 6 = 1
The sum of two positive numbers:
2 + 2 = 4; 2 + 4 = 6; 2 + 6 = 8; 4 + 4 = 8; 4 + 6=10; 6 + 6=12
Now, let’s list all the possible sums in ascending order:
– 10, -8, -6, -4, -3,-2,- 1, 1, 3, 4, 5, 6, 8, 10, 12
The sum of numbers between – 10 and 12 that are not possible to get are: – 9, – 7, – 5, 0, 2, 7, 9, 11 .

Question 8.
This string has a total of 100 tokens arranged in a particular pattern. What is the value of the string?

Solution:
Total tokens =100
Total sets of 5 tokens = \(\frac{100}{5}\) = 20
Value of one set of 5 tokens = 3 + (- 2) = 1
Hence, total value of string = 1 × 20 = 20

InText Questions

Question 1.
Can there be a number less than 0? Can you think of any ways to have less than 0 of something?
Solution:
Yes, there can be numbers less than 0.
They are called negative numbers and written with a minus sign, like – 1, – 2, – 3, etc.
Temperature: In cold places, temperature can go below 0 °C, like – 5 °C or – 10 °C.
Money: If you have ₹ 10 and will have – ₹ 10. you spend ₹ 20, you

Question 2.
Connect the inverses by drawing lines.

Solution:
The inverses of + 5, – 7, – 8 and + 9 are – 5, + 7, + 8 and – 9 respectively.

Question 3.
Should we write – 3 <-4or-4 <-3?
Solution:
We should write – 4 < – 3, because on the number line, – 4 is to the left of – 3, which makes it smaller.
So, – 4 < -3 or -3 > -4.

Question 4.
Evaluate 15 – 5, 100 – 10 and 74 – 34.
Solution:
(i) What should we add to 5 to make 15? i.e.
5 + _______ = 15.
The missing number is 10. So, 15 – 5 = 10.

(ii) What should we add to 10 to make 100? i.e.
10 + ______ = 100.
The missing number is 90. So, 100 – 10 = 90.

(iii) What should we add to 34 to make 74? i.e.
34 + _______ = 74.
The missing number is 40. So, 74 – 34 = 40.

Question 5.
Use unmarked number lines to evaluate these expressions:
(a) _______ – 125 + (- 30) = _______
(b) _______ + 105 – (- 55) = _______
(c) _______ + 80 – (- 150) = _______
(d) _______ – 99 – (- 200) = _______
Solution:
(a) – 125 + (- 30) = – 155
We start at – 125 and move 30 steps left on the number line.

(b) + 105 – (- 55) = + 105 + ( + 55) = + 160
We start at + 105 and move 55 steps right on the number line.

(c) + 80 – (- 150) = + 80 + (+ 150) = + 230
We start at + 80 and move 150 steps right on the number line.

(d) -99-(-200) = -99 + 200 = + 101
We start at – 99 and move 200 steps right on the number line.

The Other Side of Zero Class 6 Extra Questions

The Other Side of Zero Class 6 Very Short Question Answer

Question 1.
In a game, Rishi scored + 10,-5, + 3,-8 and + 4 in five rounds. What is his total score?
Solution:
Given, the scores of Rishi in five rounds are +10, – 5, + 3, – 8 and + 4.
Total score of Rishi
= + 10 + (- 5) + (+ 3) + (- 8) + (+ 4)
= (+ 10 + 3 + 4) + (-5-8)
[Grouping positive and negative integers]
= (+ 17) + (- 13)
= (+ 17)-(+ 13)
[The number that is being added can be replaced by . its additive inverse and then subtracted.]
= + 4
∴ The total score of Rishi is + 4.

Question 2.
Arrange the following integers in ascending order:
9, -7, -4, 0, 3
Solution:
Given integers are: 9, – 7, – 4, 0, 3
For positive integers 3 and 9, we have 3 < 9.
For negative integers – 7 and – 4, we have – 7 < – 4.
We know that on the number line, the numbers right to 0 are greater than 0 and the numbers left to 0 are less than 0. Also, every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: – 7, -4, 0, 3, 9

Question 3.
A boat sailed 40 km to the east of a harbour and then 70 km to the west from there. How far from the harbour is the boat finally?
Solution:
The boat starts from point A at harbour. It sailed 40 km to the east to reach point B. From point B, it sailed 70 km to the west to reach point C, as shown in figure.

Now, the position of boat = (+ 40) + (- 70) = – 30
Hence, the boat is 30 km to the west of the harbour.

Question 4.
Write 5 distinct integers whose sum is 6.
Solution:
We know that the sum of an integer with its additive inverse is zero.
The additive inverse of – 1 is 1 and the additive inverse of – 2 is 2.
∴ 1 + (- 1) = 0 and 2 + (- 2) = 0
Now, [1 + (- 1)] + [2 + (- 2)] + 6 = 0 + 0 + 6 = 6
∴ 5 distinct integers whose sum is 6 are – 1, 1, -2, 2, 6.

Question 5.
Arrange the following integers in descending order:
-8, -3, 5, 1, -1
Solution:
Given integers are – 8, – 3, 5, 1 and – 1.
We know that a number is greater than every number that is to its left on the number line.
For positive integers 1 and 5, we have 5 > 1.
For negative integers -1,-3 and – 8, we have
– 1 > – 3 > – 8.
Also, every positive integer is greater than every negative integer.
Hence, the given integers in descending order:
5, 1, – 1, -3, -8

The Other Side of Zero Class 6 Short Question Answer

Question 1.
Complete the additions using tokens:
(i) (+ 4) + (- 8)
(ii) (- 3) + (+ 4)
(iii) (- 10) + ( + 6)
(iv) (+ 7) + (- 5)
Solution:
(i) (+ 4) + (- 8) = – 4

(ii) (- 3) + (+ 4) = + 1

(iii) (- 10) + (+ 6) = – 4

(iv) (+ 7) + (- 5) = + 2

Question 2.
Complete the subtractions using tokens:
(i) (+ 6) – (+ 3)
(ii) (+ 8) – (+ 7)
(iii) (- 9) – (- 5)
(iv) (- 7) – (- 4)
Solution:
(i) (+ 6) – (+ 3) = + 3

(ii) (+ 8) – (+ 7) = + 1

(iii) (- 9) – (- 5) = – 4

(iv) (- 7) – (- 4) = – 3

Question 3.
Complete the following subtractions using tokens:
(i) (+ 9) – (+ 2)
(ii) (+ 5) – (4- 3)
(iii) (- 8) – (- 3)
(iv) (- 6) – (- 1)
Solution:
(i) (+ 9) – (+ 2) = + 7

(ii) (+ 5) – (+ 3) = + 2

(iii) (- 8) – (- 3) = – 5

(iv) (- 6) – (- 1) = – 5

Question 4.
Complete the following subtractions using tokens:
(i) (+ 3) – (+ 8)
(ii) (+ 7) – (+ 9)
(iii) (+ 4) – (- 4)
(iv) (+ 6) – (- 5)
(v) (- 4) – (+ 3)
(vi) (- 6) – (+ 6)
Solution:
(i) (+ 3) – (+ 8) = – 5

(ii) (+ 7) – (+ 9)

(iii) (+ 4) – (- 4)

(iv) (+ 6) – (- 5)

(v) (- 4) – (+ 3)

(vi) (- 6) – (+ 6)

The Other Side of Zero Class 6 Long Question Answer

Question 1.
There are two dice whose faces have the numbers 1, – 2, 3, – 4, 5 and – 6. If these dice are rolled together, find the smallest possible sum of the numbers so obtained.
Solution:
Given, there are two dice whose faces have the numbers 1, – 2, 3, – 4, 5 and – 6.
To get the smallest possible sum, we need to add the smallest number from each die.
A number is always greater than any number to its left on the number line.
For positive integers 1, 3 and 5, we have 1 < 3 < 5.
For negative integers -2,-4 and – 6, we have -6 < -4 < -2.
We know that every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: -6, -4, -2, 1, 3, 5
The smallest number on each die is – 6.
Hence, smallest possible sum = (- 6) + (- 6) = – 12

Question 2.
Using the number line write the integer which is
(i) 3 more than – 1
(ii) 4 less than – 1
(iii) 4 more than – 9
Solution:
(i) To find the integer which is 3 more than – 1, we start at – 1 on the number line and move 3 steps forward. We reach 2. Hence, the required integer is 2.

(ii) To find the integer which is 4 less than – 1, we start at – 1 on the number line and move 4 steps backward. We reach – 5. Hence, the required integer is -5.

(iii) To find the integer which is 4 more than – 9, we start at – 9 on the number line and move 4 steps forward. We reach – 5. Hence, the required integer is – 5.

Question 3.
There are two dice whose faces have the numbers – 8, – 3, – 1, 2, 4 and 6. If these dice are rolled together, find the largest possible sum of the numbers so obtained.
Solution:
Given, there are two dice whose faces have the numbers – 8, – 3, – 1, 2, 4 and 6.
To get the largest possible sum, we need to add the largest number from each die.
We know that a number is greater than every number that is to its left on the number line.
For positive integers 2, 4 and 6, we have 2 < 4 < 6.
For negative integers -1,-3 and – 8, we have – 8 < – 3 < – 1.
We know that every positive integer is greater than every negative integer.
Hence, the given integers in ascending order: – 8, -3, – 1, 2, 4, 6
The largest number on each die is 6.
So, largest possible sum = (6) + (6) = 12

Question 4.
Complete the subtractions using tokens:
(i) (+ 4) – (+ 7)
(ii) (+ 5) – (+ 8)
(iii) (+ 5) – (- 3)
(iv) (+ 6) – (- 2)
(v) (- 4) – (+ 2)
Solution:
(i) (+ 4) – (+ 7) = – 3

[Adding 3 zero pairs]

(ii) (+ 5) – (+ 8) = – 3

[Adding 3 zero pairs]

(iii) (+ 5) – (- 3) = + 8

[Adding 3 zero pairs]

(iv) (+ 6) – (- 2) = + 8

[Adding 2 zero pairs]

(v) (- 4) – (+ 2) = – 6

[Adding 2 zero pairs]

Question 5.
Complete the grids to make the required border
(i)

-7
		4
	-3

Border sum = + 5

(ii)

	-12
-1
		7

Border sum = – 4

(iii)

3		11
-2		-3

Border sum = 0

(iv)

-15		5
10
		4

Border sum = 3
Solution:
We have to take that number which makes the sum of all the numbers in a row or column equal to the border sum. So, the complete grids are given below.
(i)

-7	3	9
-4		4
16	-3	-8

Border sum = + 5

(ii)

2	-12	6
-1		-17
-5	-6	7

Border sum = – 4

(iii)

3	-14	11
-1		-8
-2	5	-3

Border sum = 0

(iv)

-15	13	5
10		-6
8	-9	4

Border sum = 3

Question 6.
Complete the following additions using tokens:
(i) (+ 7) + (- 2)
(ii) (- 12) + (+ 4)
(iii) (- 9) + (+ 5)
(iv) (-7) + (+ 11)
(v) (- 3) + (+ 8)
(vi) (- 6) + (+ 12)
Solution:
(i) (+ 7) + (-2) = + 5

(ii) (- 12) + (+ 4) = – 8

(iii) (- 9) + (+ 5) = – 4

(iv) (-7) + (+ 11) = + 4

(v) (- 3) + (+ 8) = + 5

(vi) (- 6) + (+ 12) = + 6

Question 7.
Complete the grids to make the required border
(i)

	0
		5
-2
Border sum = – 2

(ii)

		6
-5
	-3
Border sum = + 3

(iii)

	0
3		5
	8
Border sum = 4

(iv)

3
-7		6
		-8
Border sum = 5

Solution:
We have to take that number which makes the sum of all the numbers in a row or column equal to the border sum. So, the complete grids are given below.
(i)

1	0	-3
-1		5
-2	4	-4

(ii)

5	-8	6
-5		-6
3	-3	3

(iii)

3	0	1
3		5
-2	8	-2

(iv)

3	-5	7
-7		6
-2	4	-8

The Other Side of Zero Class 6 Case Based Questions

Question 1.
In Shimla, the temperature in the morning was – 2°C. By afternoon, it rose by 5°C. In another city, Manali, the temperature was – 5°C in the morning and increased by 3°C in the afternoon.
Based on the above information, answer the following questions:
(i) What was the afternoon temperature in Shimla?
(ii) What was the afternoon temperature in Manali?
(iii) What is the difference in afternoon temperatures of both cities?
Solution:
(i) Given, morning temperature in Shimla = – 2 °C
Increase in temperature by afternoon = + 5 °C
To find afternoon temperature, we add the change to the morning temperature.
∴ Afternoon temperature in Shimla = (- 2 °C) + (+ 5°C) = + 3°C

(ii) Given, morning temperature in Manali = – 5°C
Increase in temperature by afternoon = + 3 °C
To find afternoon temperature, we add the change to the morning temperature.
∴ Afternoon temperature in Manali = (- 5 °C) + (+ 3 °C)= – 2 °C

(iii) Afternoon temperature in Shimla = + 3 °C
Afternoon temperature in Manali = – 2 °C
Difference in afternoon temperatures of both cities
= + 3 °C – (- 2 °C)
= + 3 °C + (+ 2 °C) = 5 °C

Question 2.
In a geographical cross-section, some points appear above sea level (marked with positive height values) and others lie below sea level (represented by negative values), using sea level defined as zero as the reference point.
Based on the given information, answer the following questions:

(i) Write the highest and lowest points in the given geographical cross section?
(ii) What is the difference between heights of the highest and lowest points?
(iii) Write the points A, B, …, G in a sequence of increasing order of heights.
OR
(iii) Write the points A, B, …, G in a sequence of decreasing order of heights.
Solution:
(i) E(500 m) is the highest point and C (- 400 m) is the lowest point.

(ii) Difference between heights of E and C
= 500 – (- 400) = 500 + (+ 400) = 900 m

(iii) Points in increasing order are as:
C(- 400 m), B(- 300 m), G(- 200 m), D(0 m), A(200 m), F(300 m), E(500 m)
OR
(iii) Points in decreasing order, are as:
E(500 m), F(300 m), A(200 m), D(0 m), G(- 200 m), B(- 300 m), C(- 400 m)

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 9 Symmetry Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 9 Symmetry Solutions

Ganita Prakash Class 6 Chapter 9 Solutions

Class 6 Maths Ganita Prakash Chapter 9 Solutions Symmetry

Question 1.
For each of the following figures, identify the line(s) of symmetry if it exists.

Solution:

Question 2.
Given the line(s) of symmetry, find the other hole(s).

Solution:

Question 3.
Find the lines of symmetry for the kolam below.

Solution:

Question 4.
Draw the following:
(a) A triangle with exactly one line of symmetry
(b) A triangle with exactly three lines of symmetry
(c) A triangle with no line of symmetry
Is it possible to draw a triangle with exactly two lines of symmetry?
Solution:
(a)

One line of symmetry (Isosceles Triangle)

(b)

Three lines of symmetry
(Equilateral Triangle)

(c)

No line of symmetry (Scalene Triangle)
No, it is not possible to draw a triangle with exactly two lines of symmetry.

Question 5.
Copy the following on a dot grid. For each figure draw two more lines to make a shape that has a line of symmetry.

Solution:

Question 6.
Color the sectors of the circle below so that the figure has i) 3 angles of symmetry, ii) 4 angles of symmetry, iii) what are the possible numbers of angles of symmetry you can obtain by coloring the sectors in different ways?

Solution:
(i) Three angles of symmetry

(ii) Four angles of symmetry

(iii)

For 2 angles of symmetry, the angles are: 180°, 360°
For 3 angles of symmetry, the angles are: 120°, 240°, 360° 1
For 4 angles of symmetry, the angles are: 90°, 180°, 270°, 360°
For 6 angles of symmetry, the angles are: 60°, 120°, 180°, 240°, 300°, 360°
For 12 angles of symmetry, the angles are: 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°, 330°, 360°

Question 7.
In a figure, 60° is an angle of symmetry. The figure has two angles of symmetry less than 60°. What is its smallest angle of symmetry?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry.

Let the smallest angle of symmetry be x. Then, the other angles of symmetry are 2x, 3x, 4x, 360°.

It is given that the figure has two angles of symmetry less than 60°. This means the third angle of symmetry is 60° itself.
∴ 3x = 60°
⇒ x = 20°
Hence, the smallest angle of symmetry is 20°.

Question 8.
How many lines of symmetry does the shape sequence, the Koch Snowflake sequence, have? Also, find numbers of angles of symmetry.
Solution:
The equilateral triangle has 3 lines of symmetry and 3 angles of symmetry.
In six-pointed star, lines of symmetry are 6 and the angle of symmetry is 6.
The lines of symmetry and angle of symmetry for the rest three figures are the same as six-pointed stars.

InText Questions

Question 1.
Is there any other way to fold the square so that the two halves overlap? How many lines of symmetry does the square shape have?
Solution:
Yes, there are other ways to fold a square so that the two halves overlap exactly. This can be done along its lines of symmetry.

A square has 4 lines of symmetry:
(i) A vertical line (through the midpoints of the top and bottom sides),
(ii) A horizontal line (through the midpoints of the left Square and right sides),
(iii) A diagonal from the top-left to bottom-right corner,
(iv) A diagonal from the top-right to bottom-left corner.
So, we can fold the square along any of these 4 lines, and the two halves will perfectly overlap.

Question 2.
We saw that the diagonal of a square is also a line of symmetry. Let us take a rectangle that is not a square. Is its diagonal a line of symmetry?
Solution:
No, the diagonal of a rectangle is not a line of symmetry. Folding along the diagonal does not produce two matching parts.

So, while a square has both its diagonals as lines of symmetry (because they divide the square into two mirror-image triangles), a non-square rectangle does not have this property.

Question 3.
Consider a figure with radial arms having exactly 7 angles of symmetry. What will be its smallest angle of symmetry? Is the number of degrees a whole number in this case? If not, express it as a mixed fraction.
Solution:
When a figure has rotational symmetry with 7 angles of symmetry, it means it can be rotated around a central point and still look the same 7 times in a full 360° turn.

To find the smallest angle of symmetry, divide 360° by the number of symmetrical positions:
Smallest angle of symmetry = \(\frac{360^{\circ}}{7}\) = \(\left(51 \frac{3}{7}\right)^0\) which is not a whole number.
Mixed fraction form: \(\left(51 \frac{3}{7}\right)^0\)

Symmetry Class 6 Extra Questions

Symmetry Class 6 Very Short Question Answer

Question 1.
In the below figure, line l is the line of symmetry. Complete the figure to be symmetric about line /.

Solution:
The complete figure is as shown below.

Question 2.
For each of the following figure, identify the line(s) of symmetry if it exists.

Solution:
The line(s) of symmetry in the given figures are as follows:

Question 3.
In the following figures if the dotted lines represent the lines of symmetry, find the other hole(s).

Solution:
The other hole(s) in the given figures is(are) as follows:

Question 4.
Observe each of the following sequences of paper folding and cutting. Draw the pattern obtained after unfolding the paper.

Solution:
The pattern obtained after unfolding the paper in each case are as follows:

Question 5.
In the following figure, identify the line of symmetry.

Solution:
We know that a line that cuts a plane figure into two parts that exactly overlap when folded along that line is called a line of symmetry or axis of symmetry of the figure.

In the given figure, we can see the figure is symmetrical about the line l.

Thus, the line of symmetry of the given figure is line l.

Question 6.
In a figure, 60° is the smallest angle of symmetry. What are the other angles of symmetry of this figure?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry.
So, the other angles of symmetry are
2 × 60° = 120°,
3 × 60° = 180°,
4 × 60° = 240°,
5 × 60° = 300° and
6 × 60° = 360°.

Question 7.
In a given figure, 45° is the smallest angle of symmetry. What are the other angles of symmetry?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry,

So, the other angles of symmetry are 2 × 45° = 90°, 3 × 45° = 135°, 4 × 45° = 180°, 5 × 45° = 225°, 6 × 45° = 270°, 7 × 45° = 315° and 8 × 45° = 360°.

Symmetry Class 6 Short Question Answer

Question 1.
Copy the following on square paper. Complete them so the dotted line is a line of symmetry.

Solution:
To complete a figure so that it becomes symmetric about dotted line, let us take the mirror image of the figure with respect to the dotted line. Therefore, the completed figures are as follows:

Question 2.
Each of the following figures shows a piece of paper with punched holes. Copy these figures onto a plain sheet of paper and draw the line of symmetry such that, when the paper is folded along this line, the holes on one side match exactly with the holes on the other side.

Solution:
The line(s) of symmetry in the given figures are as follows:

Question 3.
In each letter of the English alphabet given below, find the number of line(s) of symmetry.
(i) U
(ii) E
(iii) I
(iv) V
Solution:
We know that a line of symmetry cuts a plane figure into two equal but flipped figures.

Question 4.
For each of the following figure, identify the line(s) of symmetry if it exists.

Solution:
The line(s) of symmetry in the given figures are as follows:

Question 5.
In the following figures if the dotted lines represent the lines of symmetry, find the other hole(s).

Solution:
We know that a line that cuts a plane figure into two parts that exactly overlap when folded along that line is called a line of symmetry or axis of symmetry of the figure.

Therefore, the other hole(s) in the given figures is(are) as follows:

Question 6.
In a figure, 90° is an angle of symmetry. The figure has two angles of symmetry less than 90°. What is the smallest angle of symmetry?
Solution:
We know that the angles of symmetry of a figure having rotational symmetry about a point are multiples of the smallest angle of symmetry.

Let the smallest angle of symmetry be x. Then, the other angles of symmetry are 2x, 3x, 4x, …, 360°.

It is given that the figure has two angles of symmetry less than 90°. This means the third angle of symmetry is 90° itself.
∴ 3x = 90°
⇒ x = 30°
Hence, the smallest angle of symmetry is 30°.

Question 7.
How many angles of rotational symmetry does the given figure have and what are they?

Solution:
We notice that the figure returns to its original position only after a complete turn or a 360° rotation.

Therefore, the figure does not possess rotational symmetry, as 360° is the only angle at which it maps onto itself.

Hence, it has no angle of rotational symmetry.

Question 8.
Discuss the rotational symmetry of the given figure.

Solution:
We observe that when the given figure is rotated by 180° and 360°, it coincides exactly with its original position.

Thus, the given figure has rotational symmetry of order 2.

Question 9.
Discuss the rotational symmetry of the given figure.

Solution:
We observe that when the given figure is rotated by 120°, 240° and 360°, it coincides exactly with its original position.

Thus, the given figure has rotational symmetry of order 3.

Symmetry Class 6 Long Question Answer

Question 1.
Observe each of the following sequences of paper folding and cutting. Draw the pattern obtained after unfolding the paper.
(i)

(ii)

Solution:
The pattern obtained after unfolding the paper in each case are as follows:

Question 2.
Which of the following figures have rotational symmetry? Also, find the order of rotational symmetry.

Solution:
(i) We observe that the figure fits onto itself when we give it a half turn, i.e. when it is rotated through 180° as shown in given figure.

Thus, the figure has rotational symmetry of order 4.

(ii) We observe that the given figure fits onto itself once only when it is rotated through 360°, i.e. when it takes a full turn.

Thus, the figure has rotational symmetry of order 4.

(iii) We observe that the given figure fits onto itself once only when it is rotated through 360°, i.e. when it takes a full turn.

Thus, it does not have rotational symmetry.

Question 3.
Show that each of the letters H, I and N has a rotational symmetry of order 2. Also, mark the point of rotational symmetry in each case.
Solution:
Each of the letters H, I and N fits into itself when it is rotated through 180° and 360° about the marked point. Thus, each of the letters H, I and N has a
rotational symmetry of order 2.

Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 2 Lines and Angles MCQ improves accuracy in objective exams.

MCQ on Lines and Angles Class 6

Lines and Angles MCQ Class 6

Class 6 Maths Lines and Angles MCQ

Question 1.
A line segment has:
(a) No endpoint
(b) One endpoint
(c) Two endpoints
(d) Infinite endpoints
Solution:
(c) Two endpoints
We know that the shortest path from point A to point B (including A and B) is called the line segment AB. It is denoted by either \(\overline{A B}\) or \(\overline{B A}\). The points A and B are called the endpoints of the line segment \(\overline{A B}\).

Question 2.
Which of these represents a ray in real life?
(a) Beam of light from a torch
(b) A thread
(c) A compass needle
(d) A pencil
Solution:
(a) Beam of light from a torch
We know that in geometry, a ray starts at one point and goes on infinitely in one direction.

The light comes out from the head of the torch and travels in a straight line. This is like a ray in real life.
A thread has two ends. In geometry, this is more like a line segment, not a ray.

A compass needle rotates and points in a direction, but it has two ends, and it does not extend infinitely. It’s not fixed in one direction from one point.
A pencil is a physical object with two definite ends. lake a thread, it is a line segment, not a ray.

Question 3.
The shortest path between two points is:
(a) A ray
(b) A line segment
(c) A curve
(d) A line
Solution:
(b) A line segment
We know that the shortest path between two points (including both points) is called a line segment.

Question 4.
Can you draw a complete line on paper?
(a) Yes, only if the line is horizontal
(b) Yes, only if the line is vertical
(c) No
(d) Can’t say
Solution:
(c) No
We know that a line passing through two points A and B is written as \(\overleftrightarrow{A B}\). It extends infinitely in both directions. Therefore, a complete line cannot be drawn on paper.

Question 5.
An angle is formed by two:
(a) Points
(b) Rays with different starting points
(c) Rays with a common starting point
(d) Line segments
Solution:
(c) Rays with a common starting point
We know that an angle is formed by two rays having a common starting point.

Question 6.
Which of the following does not show angle formation by rotation?
(a) Opening a book
(b) Drawing a straight line
(c) Turning the hands of a clock
(d) Opening a pair of scissors
Solution:
(b) Drawing a straight line
Opening a book: Yes, it makes an angle between the two covers.
Drawing a straight line: No turning or rotation happens. It’s just a line.
Turning the hands of a clock: Yes, the hands rotate and form angles.
Opening a pair of scissors: Yes, the blades form an angle when opened.

Question 7.
Superimposition of angles means:
(a) Extending both arms
(b) Placing one angle over the other
(c) Comparing only vertices
(d) Measuring angles with a ruler
Solution:
(b) Placing one angle over the other
The word superimpose means to place one thing exactly on top of another so that you can compare them. So, superimposition of angles means placing one angle over another angle so that their vertices (corners) and one arm (side) coincide.

This helps us see whether the two angles are equal or not, without using a protractor.

Question 8.
In ∠POM, what is the vertex of the angle?
(a) P
(b) M
(c) O
(d) PO
Solution:
(c) O
In ∠POM, 0 is the vertex of the angle as it is written in the middle.

Question 9.
Superimposed angles are equal when:
(a) They look similar
(b) Their vertices match but arms don’t
(c) Both rays and vertex match exactly
(d) They are on the same side
Solution:
(c) Both rays and vertex match exactly
We know that when two angles are superimposed, and the common vertex and the two rays of both angles lie on each other, then the sizes of the angles are equal.

Question 10.
An angle greater than 90° but less than 180° is called:
(a) Right angle
(b) Acute angle
(c) Reflex angle
(d) Obtuse angle
Solution:
(d) Obtuse angle
We know that an angle greater than 90° but less than 180° is called obtuse angle.

Question 11.
An angle that measures exactly 180° is called a:
(a) Right angle
(b) Reflex angle
(c) Obtuse angle
(d) Straight angle
Solution:
(d) Straight angle
An angle that measures exactly 180° is called a straight angle.

Question 12.
Which of the following is a reflex angle?
(a) 85°
(b) 135°
(c) 180°
(d) 220°
Solution:
(d) 220°
An angle that is greater than 180° but less than 360° is known as a reflex angle. Thus, 220° is a reflex angle.

Question 13.
The markings, formed by repeated folding of a semi-circle, divide it into equal parts using the concept of:
(a) Angle scaling
(b) Rotation
(c) Angle bisector
(d) Circular radius
Solution:
(c) Angle bisector
When we fold a semi-circle repeatedly, each fold divides the angle into two equal parts. This is done using the concept of an angle bisector, which divides the angle into two equal halves.

Question 14.
What is the measure of the angle between the hour hand and the minute hand of a clock when it is 4 o’clock?
(a) 180°
(b) 120°
(c) 90°
(d) 60°
Solution:
(b) 120°
We know that the clock has 12 hours and a full circle is 360°.

So, each hour mark is \(\frac{360^{\circ}}{12}\) = 30 apart.
Thus, the angle increases by 30° for each hour.
At 4 o’clock, the hour hand is at 4 and the minute hand is at 12.
From the figure, we can see that there is a gap of 4 hour marks between the hour hand and the minute hand.
Therefore, angle between the hour hand and the minute hand at 4 o’clock = 4 × 30° = 120°

Question 15.
If there are 18 spokes in a motorcycle wheel, then the angle between two adjacent spokes is:
(a) 10°
(b) 12°
(c) 18°
(d) 20°
Solution:
(d) 20°
We know that a complete angle is equal to 360°. Given, there are 18 spokes in a motorcycle wheel.
Therefore, these 18 spokes divide the complete angle into 18 equal parts.
Thus, the angle between two adjacent spokes = \(\frac{360^{\circ}}{18}\) = 20°

Question 16.
Which of the following statements are true?
(i) A line segment has two endpoints.
(ii) A ray extends endlessly in both directions.
(iii) A line can be named using any two points on it.
(iv) A ray has one endpoint and goes on endlessly in one direction.
Choose the correct option from the following:
(a) Only (i) and (iv)
(b) Only (i) and (ii)
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
We know that
A line segment has two endpoints.
A line is named using any two points that lie on the line.
A ray has one endpoint and extends in one direction.
Thus, the statements (i), (iii) and (iv) are correct.

Question 17.
Which of the following are correct ways to name a given angle?

(i) ∠XYZ
(ii) ∠ZXY
(iii) ∠X
(iv) ∠YXZ
Choose the correct option from the following:
(a) Only (i) and (ii)
(b) Only (ii) and (iii)
(c) Only (iii) and (iv)
(d) (ii), (iiii) and (iv)
Solution:
(d) (ii), (iiii) and (iv)
To name an angle, we write the letter of one arm, followed by the letter of the vertex (always in the middle), and then the letter of the other arm.

For example, if O is the vertex and OA and OB are the arms or sides of the angle, then it can be written as ∠AOB or ∠BOA. If there is no other angle formed at the same vertex, we may simply name it ∠O.

Therefore, the given angle can be written as ∠YXZ, ∠ZXY or ∠X.
Thus, (ii), (iii) and (iv) are correct ways of writing the given angle.

Question 18.
While measuring an angle using a protractor, which of the following steps are correct?
(i) Place centre of protractor on the vertex of the angle.
(ii) Align one arm with the 0° mark.
(iii) Always use the outer scale.
(iv) Use subtraction if arms fall on different marks.
Choose the correct option from the following:
(a) Only (i) and (ii)
(b) (i), (ii) and (iv)
(c) Only (iii) and (iv)
(d) (ii), (iii) and (iv)
Solution:
(b) (i), (ii) and (iv)
While measuring an angle using a protractor, place the centre of the protractor on the vertex of the angle and align one arm with the 0° mark.

Now, depending on the orientation of the given angle, either the inner or outer scale of the protractor can he used.

If neither arm lies on the 0° mark, then subtraction can help to determine the actual measurement.
Thus, (i), (ii) and (iv) are correct.

Lines and Angles Class 6 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled .as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): Any two points determine a unique line.
(R): Through two distinct points, only one line can be drawn.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Through two distinct points, only one line can be drawn.

So, any two points determine a unique line.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): An angle is formed when two rays have a common starting point.
(R): The two rays are called the arms of the angle and the common point is called the vertex.
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
We know that an angle is formed by two rays having a common starting point.

Here is an angle formed by rays \(\overrightarrow{O P} \text { and } \overrightarrow{O M}\) where O is the common starting point.
The point 0 is called the vertex of the angle, and the rays \(\overrightarrow{O P} \text { and } \overrightarrow{O M}\) are called the arms of the angle.
Thus, both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Question 3.
(A): Superimposition helps us to compare the size of two angles accurately.
(R): By placing one angle over another with the same vertex, we can directly compare their
opening.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
The word superimpose means to place one thing exactly on top of another so that you can compare them.
So, superimposition of angles means placing one angle over another angle so that their vertices (corners) and one arm (side) coincide.
This helps us to compare the size of two angles accurately.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): Folding a semi-circle twice gives an angle of 90°.
(R): Each fold halves the angle, allowing us to mark standard angles like 45°, 90°, and 135°.
Solution:
(d) A is false but R is true.

We know that a semi-circle is half of a full circle, which is 180°. If we fold the semi-circle into 2 equal parts, each part is 90°, and if we fold it again, we get 45°. So, each fold halves the angle, allowing us to mark standard angles like 45°, 90°, and 135°.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 5.
(A): The angle between the hands of a clock at 3 o’clock is 90°.
(R): Each hour mark on a clock corresponds to a 30° rotation.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that the clock has 12 hours and a full circle is 360°.
So, each hour mark is \(\frac{360^{\circ}}{12}\) = 30° apart.

Thus, the angle increases by 30° for each hour.
At 3 o’clock, the hour hand is at 3 and the minute hand is at 12.

From the figure, we can see that there is a gap of 3 hour marks between the hour hand and the minute hand.

Therefore, angle between the hour hand and the minute hand at 3 o’clock = 3 x 30° = 90°

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Lines and Angles Class 6 Fill in the Blanks

Question 1.
A line has no endpoints and extends infinitely in ___________ directions.
Solution: both
We know that a line is a geometrical figure that is straight, has no width, and extends indefinitely in both directions.
Therefore, a line has no endpoints and extends infinitely in both directions.

Question 2.
If A and B are two points, the line segment between them is written as ____ .
Solution: \(\overline{A B}\) or \(\overline{B A}\)
We know that if A and B are two points, the line segment between them is written as \(\overline{A B}\) or \(\overline{B A}\).

Question 3.
The correct way to name an angle formed by points D, B and E in the given order, is ______ .
Solution: ∠DBE or ∠EBD
An angle is formed by two rays having a common starting point.
Given, the points are D, B and E.
So, the arms of the angle will be \(\overrightarrow{B D} \text { and } \overrightarrow{B E}\).

Thus, The correct way to name an angle formed by points D, B and E in the give order, is ∠DBE or ∠EBD.

Question 4.
The common starting point of the arms of an angle is called the ____ .
Solution: vertex
We know that the common starting point of the arms of an angle is called the vertex.

Question 5.
Any point outside the arms of an angle lies in its ______ .
Solution: exterior
We know that any point outside the arms of an angle lies in its exterior.

Question 6.
While comparing two angles using superimposition, their ____ and _____ must coincide.
Solution: vertices, one arm
While comparing two angles using superimposition, their vertices and one arm must coincide,

Question 7.
There are 360° in a _____ turn.
Solution: full
We know that there are 360° in a full turn.

Question 8.
When we fold a quarter circle in half again, the resulting angle is __________ .
Solution: 45°
We know that a full circle is 360°.
A quarter circle is 90°. If we fold that quarter circle i. e. 90° in half again, we get \(\frac{90^{\circ}}{2}\) = 45°.
So, when we fold a quarter circle in half again, the resulting angle is 45°.

Question 9.
Two times of 45° is a _____ angle.
Solution: right
Two times of 45° = 2 × 45° = 90°, which is a right angle.
Thus, two times of 45° is a right angle.

Question 10.
A line that divides an angle into two equal parts, is called _________ .
Solution: angle bisector
A line that divides an angle into two equal parts, is called angle bisector.

Students can use Class 8 Math Solution Odia Medium and Class 8 Maths Chapter 2 Question Answer Odia Medium ଘାତର ଖେଳ to check their answers after solving exercises.

8th Class Maths Chapter 2 Question Answer Odia Medium

Class 8 Maths Chapter 2 Odia Medium

2.1 ଘାତଖେଳର ଅଭିଜ୍ଞତା (Page No. 19)

Question 1.
ଖଣ୍ଡିଏ କାଗଜ ଫର୍ଦ ନିଅ। ଏହାକୁ ଗୋଟିଏ ଥର ଭାଙ୍ଗ। ପୁଣି ଏହାକୁ ବାରମ୍ବାର ଭାଙ୍ଗ। ଏହାକୁ ବାରମ୍ବାର କେତେଥର ଭାଙ୍ଗି ହେବ?
Solution:
କାଗଜର ଆରମ୍ଭ ବିନ୍ଦୁ (୦ ଭାଙ୍ଗ) : କାଗଜର ପ୍ରକୃତ ମୋଟେଇ 0.001 ସେ.ମି.।
ପ୍ରଥମ ଭାଙ୍ଗ : କାଗଜକୁ ପ୍ରଥମ ଭାଙ୍ଗରେ ଏହାର ମୋଟେଇ ଅଧାର ଦୁଇଗୁଣ।
ତେଣୁ ଏହା 0.001 ସେ.ମି. × 2 = 0.002 ସେ.ମି. ।
ଦ୍ଵିତୀୟ ଭାଙ୍ଗ : ଏହାକୁ ପୁଣି ଭାଙ୍ଗିଲେ ଏହା ହେବ = 0.001 ସେ.ମି. × 2 × 2 = (0.001 × 2²) = 0.004 ସେ.ମି.।
ଏହି କ୍ରମରେ ‘n’ ଥର ଭାଙ୍ଗିଲେ = 0,001 × 2ⁿ ହେବ।

Question 2.
ଗୋଟିଏ ଫର୍ଦ୍ଦ କାଗଜକୁ ତୁମ ଇଚ୍ଛାନୁସାରେ ଯେତେଥର ଚାହୁଁଛ ସେତେଥର ଭାଙ୍ଗ କରିପାରିବ। 30 ଭାଙ୍ଗ କରିବା ପରେ ଏହାର ମୋଟେଇ କେତେ ହେବ? ଅନୁମାନ କର।
Solution:
ଆମେ ଜାଣୁ କାଗଜର ମୋଟେଇ = 0.001 ସେ.ମି.।
ଏହାକୁ 30 ଥର ଭାଙ୍ଗିଲେ ଏହାର ମୋଟେଇ = 0.001 × 2³⁰ ସେ.ମି.।
= 0.001 × 1,073,741,824 ସେ.ମି.
= 1,073,741.824 ସେ.ମି.
= 10,737,41824 ମିଟର (1 ମିଟର = 100 ସେ.ମି.)
= 10.74 କି.ମି. (1 କି.ମି. = 1000 ମିଟର)
∴ 30 ଭାଙ୍ଗ କରିବା ପରେ କାଗଜର ମୋଟେଇ 10.74 କି.ମି. ହେବ।

Question 3.
ଗୋଟିଏ ଫର୍ଦ କାଗଜକୁ 46 ଭାଙ୍ଗ କରିବା ପରେ ଏହା କେତେ ମୋଟା ହେବ? ଧରନିଅ କାଗଜର ମୋଟେଇ 0.001 ସେ.ମି. ।
Solution:
ଆମେ ଜାଣୁ କାଗଜର ମୋଟେଇ = 0.001 ସେ.ମି.।
ଏହାକୁ 46 ଥର ଭାଙ୍ଗିଲେ ଏହାର ମୋଟେଇ = 0.001 × 2⁴⁶ ସେ.ମି.। = 703687,442 କି.ମି. ହେବ।

Page No. 20

Question 1.
ନିମ୍ନ ସାରଣୀରେ ପ୍ରତ୍ୟେକ ଭାଙ୍ଗପରେ କାଗଜ ମୋଟେଇର ତାଲିକା ଦିଆଯାଇଛି। ପ୍ରତ୍ୟେକ ଭାଙ୍ଗପରେ ଏହାର ମୋଟେଇ ଦୁଇଗୁଣ ହେଉଛି।

(≈ ଚିହ୍ନଟିକୁ ଆମେ ‘ପ୍ରାୟତଃ ସମାନ’ ବୋଲି ବ୍ୟବହାର କରିଥାଉ ।)
Solution:

2.2 ଘାତାଙ୍କୀୟ ସଂକେତ ଏବଂ ପ୍ରକ୍ରିୟା

Page No. 22

Question 1.
ଏକ କାଗଜ ଫର୍ଦ 10 ଥର ଭାଙ୍ଗିବା ପରେ ତାହାର ମୋଟେଇ ପରିପ୍ରକାଶ କିଭଳି ହେବ? ପ୍ରାରମ୍ଭିକ ମୋଟେଇ v ଅକ୍ଷର ସଂଖ୍ୟାଦ୍ଵାରା ସୂଚିତ କରାଯାଇଛି।
(i) 10v
(ii) 10 + v
(iii) 2 × 10 × v
(iv) 2¹⁶
(v) 2¹⁰ v
(vi) 10² v
Solution:
ପ୍ରତିଥର ଭାଙ୍ଗିବା ପରେ ଏହାର ମୋଟେଇ ଦୁଇଗୁଣ ବୃଦ୍ଧି ହେଲା।
∴ 1 ଥର ଭାଙ୍ଗ, ମୋଟେଇ = 2 × v
2 ଥର ଭାଙ୍ଗ, ମୋଟେଇ = 2² × v
ସେହିପରି 10 ଥର ଭାଙ୍ଗ, ମୋଟେଇ = 2¹⁰ × v
ତେଣୁ ଠିକ୍ ଉତ୍ତରଟି ହେଉଛି (v) 2¹⁰ × v

Question 2.
32400 ସଂଖ୍ୟାକୁ ଏହାର ମୌଳିକ ଗୁଣନୀୟକରେ ପ୍ରକାଶ କର ଏବଂ ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ସେମାନଙ୍କ ଘାତାଙ୍କ ରୂପରେ ଚିହ୍ନଟ କର।
Solution:

32400 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3 × 3 × 3
ଏହାର ଘାତାଙ୍କୀୟ ପରିପ୍ରକାଶଟି ହେବ = 32400 = 2⁴ × 5² × 3⁴

Question 3.
(-1)⁵ ଟି କ’ଣ? ଏହା ଧନାତ୍ମକ ନା ଋଣାତ୍ମକ? (-1)⁵⁶ ଟି କ’ଣ?
Solution:
(−1)⁵ = -1 (ଏହା ଋଣାତ୍ମକ) [(-1)^{ସଂଖ୍ୟ} ‍= -1 (ଋଣାତ୍ମକ)]
(-1)⁵⁶ = +1 (ଏହୀ ଧନାତ୍ମକ) [(-1)^{ଯୁଗ ସଂଖ୍ୟ} = 1 (ଧନାତ୍ମକ)]

Question 4.
(-2)⁴ = 16 କି? ଯାଞ୍ଚକର।
Solution:
(-2)⁴ = (-2) × (-2) × (-2) × (-2) = 4 × 4 = 16
∴ (-2)⁴ = 16, ଏହା ଠିକ୍।

ନିଜେ କରି ଦେଖ (Page No. 22)

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ପରିପ୍ରକାଶକୁ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖି :
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) а × а × а × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 = 6⁴
(ii) y × y = y²
(iii) b × b × b × b = b⁴
(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³
(v) 2 × 2 × a × a = 2² × a²
(vi) a × a × a × c × c × c × c × d = a³ × c⁴ × d¹

Question 2.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ ସେମାନଙ୍କର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ ରୂପରେ ପ୍ରକାଶ କର।
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:
(i) 648 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 3 × 3 × 3 × 3
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 2³ × 3⁴
(ii) 405 ର ମୌଳିକ ଗୁଣନୀୟକ = 3 × 3 × 3 × 3 × 5
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 3⁴ × 5
(iii) 540 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 3 × 3 × 3 × 5
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 2² × 3³ × 5
(iv) 3600 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
ଗୁଣନୀୟକଗୁଡ଼ିକର ଘାତର ଗୁଣଫଳ = 2⁴ × 3² × 5²

Question 3.
ନିମ୍ନଲିଖତ ପ୍ରତ୍ୟେକର ସାଂଖ୍ୟକ ମାନ ଲେଖ।
(i) 2 × 10³
(ii) 7² × 2³
(iii) 3 × 4⁴
(iv) (-3)² × (-5)²
(v) 3² × 10⁴
(vi) (-2)⁵ × (-10)⁶
Solution:
(i) 2 × 10³ = 2 × 10 × 10 × 10
= 2 × 1000
= 2000

(ii) 7² × 2³ = 7 × 7 × 2 × 2 × 2
= 49 × 8
= 392

(iii) 3 × 4⁴ = 3 × 4 × 4 × 4 × 4
= 3 × 256
= 768

(iv) (-3)² × (-5)² = -3 × -3 × -5 × -5
= 9 × 25
= 225

(v) 3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10
= 9 × 10,000
= 90,000

(vi) (-2)⁵ × (-10)⁶ = −2 × −2 × −2 × -2 × −2 × 10 × 10 × 10 × 10 × 10 × 10
= (-32) × 10,00,000
= -3,20,00,000

Page No. 23

Question 1.
ମୋଟ କେତୋଟି କୋଠରି ଥିଲା?

Solution:
ଦତ୍ତ ଚିତ୍ରରୁ ଆମେ ଦେଖୁଲୁ ଯେ କୋଠରି ସଂଖ୍ୟା ହେଉଛି 34 ।
3 × 3 × 3 × 3 = 243
27 × 3 = 81
81 × 3 = 243

Question 2.
ମୋଟ କେତୋଟି ହୀରା ଥୁଲା? ଉପରୋକ୍ତ ଗୁଣଫଳ ବ୍ୟବହାର କରି କେବଳ ଗୋଟିଏ ଗୁଣନଦ୍ଵାରା ଆମେ ଏହା ଜାଣିପାରିବା କି?
Solution:
ମୋଟ ହୀରାର ସଂଖ୍ୟା = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 3⁷
3⁷ = (3 × 3 × 3 × 3) × (3 × 3 × 3)
3 × 3 × 3 × 3 = 3⁴ = 243
3 × 3 × 3 = 3³
∴ 3⁴ × 3³ = 81 × 27 = 2187

Page No. 24

Question 1.
3⁷ କୁ 3² × 3⁵ ଭାବରେ ମଧ୍ୟ ଲେଖାଯାଇପାରିବ। ତୁମେ ଏହାର କାରଣ ଦର୍ଶାଇ ପାରିବ କି?
Solution:
ଏହାକୁ ସମାନ ଅକ୍ଷର-ସଂଖ୍ୟା ଥ‌ିବା ଘାତଗୁଡ଼ିକର ଗୁଣନ ଭାବରେ ସହଜରେ ବିସ୍ତାରିତ କରି ଏହାକୁ ଲେଖୁ ପାରିବା।

Question 2.
p⁴ × p⁶ ର ଗୁଣଫଳକୁ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖ।
Solution:
p⁴ × p⁶ = (p × p × p × p) × (p × p × p × p × p × p) = p¹⁰
ଏହାକୁ ଆମେ ସାଧାରଣ ରୂପରେ ନିମ୍ନମତେ ପ୍ରକାଶ କରିପାରିବା।
n^a × n^b = n^a+b, ଯେଉଁଠାରେ a ଓ b ଗଣନସଂଖ୍ୟା ଅଟନ୍ତି।

Question 3.
ଏହାକୁ ପର୍ଯ୍ୟବେକ୍ଷଣ କରି ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ଗଣନା କର।
(i) 2⁹
(ii) 5⁷
(iii) 4⁶
Solution:
(i) 2⁹ = 2³ × 2³ × 2³
= 8 × 8 × 8
= 512

(ii) 5⁷ = 5² × 5² × 5² × 5
= 25 × 25 × 25 × 5
= 625 × 125
= 78125

(iii) 4⁶ = 4² × 4² × 4²
= 16 × 16 × 16
= 256 × 16
= 4096

Question 4.
2¹⁰ ମଧ୍ୟ (2⁵)² ସହିତ ସମାନ କି? ଏହାକୁ ଏକ ଗୁଣଫଳ ଭାବରେ ଲେଖ।
Solution:
2¹⁰ = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)
= (2⁵) × (2⁵)
= (2⁵)², ଏହା ସମାନ।

Question 5.
ନିମ୍ନଲିଖୁତ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ଅତିକମ୍‌ରେ ଦୁଇଟି ଭିନ୍ନଭିନ୍ନ ଉପାୟରେ ଏକ ଘାତର ଘାତ ଭାବରେ ଲେଖ।
(i) 8⁶
(ii) 7¹⁵
(iii) 9¹⁴
(iv) 5⁸
Solution:

Page No. 25

Question 1.
ଏକ କୁହୁକ ପୋଖରୀ ମଝିରେ ଗୋଟିଏ ଗୋଲାପୀ ପଦ୍ମଫୁଲ ଅଛି। ଏହି ପୋଖରୀରେ ପ୍ରତିଦିନ ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ଦ୍ଵିଗୁଣିତ ହୁଏ। 30 ଦିନ ପରେ ପୋଖରାଟି ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ। କେଉଁଦିନ ପୋଖରୀଟି ଅଧାପୂର୍ଣ୍ଣ ଥିଲା?
Solution:
ଏହି ପୋଖରୀରେ ପ୍ରତିଦିନ ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ଦ୍ଵିଗୁଣିତ ହୁଏ।
30 ଦିନ ପରେ ପୋଖରୀଟି ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ।
30 ଦିନ ପରେ ପୋଖରୀଟି ଅଧାପୂର୍ଣ୍ଣ ଥିଲା।
ପ୍ରଥମ ଦିନ → 1 = 2⁰
ଦ୍ଵିତୀୟ ଦିନ → 2 = 2¹
ତୃତୀୟ ଦିନ → 2²
ଚତୁର୍ଥ ଦିନ → 2³
………………………….
………………………….
୨୯ ଦିନ → 2²⁸
୩୦ ଦିନ → 2²⁹

Question 2.
ଯଦି ପୋଖରୀଟି 30 ତମ ଦିନରେ ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ, ତେବେ 29 ତମ ଦିନରେ ଏହାର କେତେ ଅଂଶ ପଦ୍ମଫୁଲରେ ଭର୍ତ୍ତି ହୋଇଥିଲା?
Solution:
ଯଦି ପୋଖରୀଟି 30 ତମ ଦିନରେ ପଦ୍ମଫୁଲରେ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଭର୍ତ୍ତି ହୋଇଯାଏ, ତେବେ 29 ତମ ଦିନରେ ଏହାର ଅଧା ଅଂଶ ପଦ୍ମଫୁଲରେ ଭର୍ତ୍ତି ହୋଇଥିଲା।

Question 3.
ପଦ୍ମଫୁଲର ସଂଖ୍ୟା (ଘାତାଙ୍କୀୟ ରୂପରେ) ଲେଖ, ଯେତେବେଳେ ପୋଖରୀଟି–
(i) ସମ୍ପୂର୍ଣ୍ଣ ଭର୍ତ୍ତି ହୋଇଥିଲା।
(ii) ଅଧା ଭର୍ତ୍ତି ହୋଇଥିଲା।
Solution:
(i) 30 ତମ ଦିନରେ ପୋଖରୀଟି ସମ୍ପୂର୍ଣ୍ଣ ଭର୍ତ୍ତି ହୋଇଥିଲା = 2²⁹
(ii) 29 ତମ ଦିନରେ ପୋଖରୀଟି ଅଧା ଭର୍ତ୍ତି ହୋଇଥିଲା = 2²⁸
ଆଉ ଏକ ପୋଖରୀ ଅଛି ଯେଉଁଥରେ ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ପ୍ରତିଦିନ ତିନିଗୁଣ ହୁଏ । ଯେତେବେଳେ ଉଭୟ ପୋଖରୀରେ କୌଣସି ଫୁଲ ନଥୁଲା, ଶିବାନୀ ଦୁଇଗୁଣ ହେଉଥିବା ପୋଖରୀରେ ଗୋଟିଏ ପଦ୍ମଫୁଲ ରଖୁଲେ । 4 ଦିନପରେ ସେ ସେଠାରୁ ସମସ୍ତ ପଦ୍ମଫୁଲ ନେଇ ତିନିଗୁଣ ହେଉଥ‌ିବା ପୋଖରୀରେ ରଖିଲେ ।

Question 4.
4 ଦିନପରେ ତିନିଗୁଣ ହେଉଥ‌ିବା ପୋଖରୀରେ କେତୋଟି ପଦ୍ମଫୁଲ ଥୁବ?
Solution:
ପ୍ରଥମ 4 ଦିନପରେ, ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ହେଉଛି – 1 × 2 × 2 × 2 × 2 = 2⁴
ପରବର୍ତୀ 4 ଦିନପରେ, ପଦ୍ମଫୁଲର ସଂଖ୍ୟା ହେଉଛି – 2⁴ × 3 × 3 × 3 × 3 = 2⁴ × 3⁴

Question 5.
ଯଦି ଶିବାନୀ ଫୁଲଗୁଡ଼ିକୁ ପୋଖରୀରେ ରଖୁବାର କ୍ରମ ବଦଳାଇଥାନ୍ତେ, ତେବେ କେତୋଟି ପଦ୍ମଫୁଲ ରହିଥାନ୍ତା?
Solution:
1 × 3⁴ × 2⁴ = (3 × 3 × 3 × 3) × (2 × 2 × 2 × 2)

Question 6.
ଏହି ଗୁଣଫଳକୁ ଘାତାଙ୍କୀୟ ସଙ୍କେତ mn ଭାବରେ ପ୍ରକାଶ କରାଯାଇପାରିବ କି ? ଯେଉଁଠାରେ m ଏବଂ n ଗଣନ ସଂଖ୍ୟା ଅଟନ୍ତି ।
Solution:
ସଂଖ୍ୟାଗୁଡ଼ିକୁ ପୁନଃଗୋଷ୍ଠୀଭୁକ୍ତ କରାଗଲେ = (3 × 2) × (3 × 2) × (3 × 2) × (3 × 2)
= (3 × 2)⁴
= 6⁴

Question 7.
2⁵ × 5⁵ ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର।
Solution:
2⁵ × 5⁵ = (2 × 5)⁵
= 10⁵
= 1,00,000

Question 8.
\(\frac{10^4}{5^4}\) କୁ ସରଳକର ଏବଂ ଏହାକୁ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖ।
Solution:
\(\frac{10^4}{5^4}=\left(\frac{10}{5}\right)^4=(2)^4=2^4\) [∵ \(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{~b}^{\mathrm{m}}}=\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)^{\mathrm{m}}\)]

Page No. 26

Question 1.
ଉଦାହରଣ : ଇତୁ ପାଖରେ 4ଟି ପୋଷାକ ଓ 3ଟି ଟୋପି ଅଛି। ଇଡୁ କେତେ ପ୍ରକାରରେ ପୋଷାକ ଓ ଟୋପିକୁ ମିଶାଇ ପାରିବେ?

Solution:
ପ୍ରତ୍ୟେକ ଟୋପିପାଇଁ, ସେ 4ଟି ପୋଷାକ ମଧ୍ୟରୁ ଯେକୌଣସି ଗୋଟିଏ ବାଛିପାରିବେ। ତେଣୁ 3ଟି ଟୋପି ପାଇଁ 4 + 4 + 4 = 4 × 3 = 12 ସମାବେଶ ସମ୍ଭବ ଅଟେ। ଆମେ ଏହାକୁ ଅନ୍ୟ ଏକ ଉପାୟରେ ମଧ୍ୟ ଲକ୍ଷ୍ୟ କରିପାରିବା, ପ୍ରତ୍ୟେକ ପୋଷାକ ପାଇଁ ଇତୁ 3ଟି ଟୋପି ମଧ୍ୟରୁ ଯେକୌଣସି ଗୋଟିଏ ବାଛିପାରିବେ, ତେଣୁ 4ଟି ପୋଷାକ ପାଇଁ 3 + 3 + 3 + 3 = 3 × 4 = 12 ସମାବେଶ ସମ୍ଭବ।

Question 2.
ରକି ପାଖରେ 7ଟି ପୋଷାକ, 2ଟି ଟୋପି ଏବଂ 3 ଯୋଡ଼ା ଜୋତା ଅଛି । ରକି କେତେ ପ୍ରକାରରେ ଏଗୁଡ଼ିକୁ ପିନ୍ଧିପାରିବ?
Solution:
7 × 2 × 3 = 42 ପ୍ରକାର।
ରକି 42 ପ୍ରକାରରେ ଏଗୁଡ଼ିକୁ ପିନ୍ଧିପାରିବ।

Question 3.
ଇତୁ ଏବଂ ରକିଙ୍କୁ ଏକ ପୁରୁଣା ଷ୍ଟାମ୍ପ ଏବଂ ମୁଦ୍ରା ଥିବା ବାକ୍ସ ମିଳିଲା, ଯାହାକୁ କି ତାଙ୍କ ଜେଜେ ସଂଗ୍ରହ କରିଥିଲେ । ଏହାର ଚାବି 5 ଅଙ୍କ ବିଶିଷ୍ଟ ପାସୱାର୍ଡ଼ ଦ୍ଵାରା ସୁରକ୍ଷିତ ଥିଲା । ଏହାର ପାସ୍ୱାର୍ଡ଼ କାହାରିକୁ ଜଣା ନଥୁବାରୁ ସେମାନଙ୍କୁ ପ୍ରତ୍ୟେକ ପାସ୍ୱାର୍ଡ଼ଦ୍ୱାରା ଚେଷ୍ଟା କରିବା ବ୍ୟତୀତ ବାକ୍ସକୁ ଖୋଲିବା ପାଇଁ ଅନ୍ୟ କୌଣସି ବିକଳ୍ପ ଉପାୟ ନଥିଲା । ଦୁର୍ଭାଗ୍ୟ ଏହି ଯେ, ସବୁ ସମ୍ଭାବ୍ୟ ସମାବେଶଥର ଚେଷ୍ଟା କରିବା ପରେ ହିଁ ଶେଷ ପାସ୍ୱାର୍ଡ଼ରେ ଲିକ୍ ଖୋଲିଲା ; ସେମାନେ କେତୋଟି ପାସ୍ୱାର୍ଡ଼ ନେଇ ଚେଷ୍ଟା କରିଥିଲେ ?
Solution:
ପ୍ରତ୍ୟେକ ସଂଖ୍ୟା 0 ରୁ ୨ ପର୍ଯ୍ୟନ୍ତ ଯେକୌଣସି ସଂଖ୍ୟା ହୋଇପାରେ,
ପ୍ରତି ଅଙ୍କରେ 10ଟି ବିକଳ୍ପ ସମ୍ଭବ ହେବ ।
ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ପାସ୍ୱାର୍ଡ଼ ପାଇଁ = 10² = 100 ଟି ବିକଳ୍ପ ଅଛି ।
ପାଞ୍ଚ ଅଙ୍କ ବିଶିଷ୍ଟ ପାସ୍ୱାର୍ଡ଼ ପାଇଁ ମୋଟ ମିଶ୍ରଣ = 10⁵ = 1,00,000 ବିକଳ୍ପ ଅଛି ।

Question 4.
ମନେକର ଆମ ପାଖରେ 2 ଅଙ୍କବିଶିଷ୍ଟ ତାଲା ଅଛି ଏବଂ ଏଥ‌ିପାଇଁ କେତୋଟି ପାସ୍ୱାର୍ଡ଼ ସମ୍ଭବ ତାହା ଜାଣିବାକୁ ଆମେ ଚେଷ୍ଟା କରିବା ।
Solution:
ପ୍ରଥମ ଅଙ୍କ ପାଇଁ 10 ଟି ବିକଳ୍ପ ଅଛି (ଯଦି ) (ପ୍ରଥମ ଅଙ୍କ, ତେବେ 00, 01, 02, 03. 09) ସମ୍ଭବ । ଯଦି ।
ପ୍ରଥମ ଅଙ୍କ, ତେବେ 10, 11, 12, 13, …, 19 ସମ୍ଭବ ।
ତେଣୁ ଏକ 2 ଅଙ୍କ ବିଶିଷ୍ଟ ଲକ୍ ପାଇଁ ସମୁଦାୟ ସମାବେଶ ସଂଖ୍ୟା ହେଉଛି 10 × 10 = 100

Question 5.
ମନେକର ଆମ ପାଖରେ ୩ ଅଙ୍କ ବିଶିଷ୍ଟ ତାଲା ଅଛି ।
Solution:
ପୂର୍ବରୁ ଥିବା 100 (2 ଅଙ୍କ ବିଶିଷ୍ଟ) ପାସ୍ୱାର୍ଡ଼ ମଧ୍ୟରୁ ପ୍ରତ୍ୟେକ ଥର ତୃତୀୟ ଅଙ୍କ ପାଇଁ
100 × 10 = 1000ଟି ସମାବେଶ ଅଛି ।
ସେଗୁଡ଼ିକୁ ତାଲିକାଭୁକ୍ତ କଲେ = 000, 001, 002, ……, 997, 998, 999

Question 6.
କେତୋଟି 5 ଅଙ୍କ ବିଶିଷ୍ଟ ପାସ୍ୱାର୍ଡ଼ ସମ୍ଭବ?
Solution:
ପ୍ରତ୍ୟେକ ଅଙ୍କର 10 ଟି ପସନ୍ଦ ଅଛି । ତେଣୁ ଏକ 5 ଅଙ୍କ ବିଶିଷ୍ଟ ତାଲାରେ ଥିବ = 10 × 10 × 10 × 10 × 10 = 10⁵ = 1,00,000 ସଂଖ୍ୟକ ପାସ୍ୱାର୍ଡ଼ ।
ତେଣୁ 99,999 ପର୍ଯ୍ୟନ୍ତ 5 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଲେଖାଯାଇପାରିବ ।
00000, 00001, 00002, ……… 00010, 00011, ………. 00100, 00101, ………. 30456, ………. 99998, 99999

2.3 ଘାତର ଅପରପାର୍ଶ୍ବ

Page No. 27

Question 1.
2 ର ଘାତ ଆକାରରେ 2¹⁰⁰ ÷ 2²⁵ର ମାନ କେତେ ହେବ ?
Solution:
2¹⁰⁰ ÷ 2²⁵ = 2^100-25 = 2⁷⁵

Page No. 28

Question 1.
n କାହିଁକି 0 ହୋଇପାରିବ ନାହିଁ ?
Solution:
n ର ମୂଲ୍ୟ 0 ହେଲେ, ଏହାର ମୂଲ୍ୟ। ସଙ୍ଗେ ସମାନ।

Question 2.
2⁰ ମୂଲ୍ୟ କେତେ?
Solution:
ଆମେ 2⁰ କୁ ଏପରି ଭାବରେ ବ୍ୟାଖ୍ୟା କରିବା ଯେ ଉପରୋକ୍ତ ସାଧାରଣ ରୂପ ସତ୍ୟ ହେବ।
\(2^0=2^{4-4}=2^4 \div 2^4=\frac{2 \times 2 \times 2 \times 2}{2 \times 2 \times 2 \times 2}\) = 1

Page No. 29

Question 1.
ଆମେ \(10^3=\frac{1}{10^{-3}}\) ଲେଖୁପାରିବା କି?
Solution:
ହଁ, ଆମେ ଲେଖୁପାରିବା = \(\frac{1}{10^{-3}}=\frac{1}{1 / 10^3}=1 \div \frac{1}{10^3}=1 \times 10^3=10^3\)

Question 2.
ନିମ୍ନଲିଖୁତ ପଦଗୁଡ଼ିକର ସମତୁଲ୍ୟ ରୂପ ଲେଖ :
(i) 2^-4
(ii) 10^-5
(iii) (-7)^-2
(iv) (-5)^-3
(v) 10^-100
Solution:
(i) \(2^{-4}=\frac{1}{2^4}=\frac{1}{16}\)
(ii) \(10^{-5}=\frac{1}{10^5}=\frac{1}{1,00,000}\)
(iii) \((-7)^{-2}=\frac{1}{(-7)^2}=\frac{1}{49}\)
(iv) \((-5)^{-3}=\frac{1}{(-5)^3}=\frac{1}{-125}\)
(v) \(10^{-100}=\frac{1}{10^{100}}\)

Question 3.
ସରଳକର ଏବଂ ଉତ୍ତରଗୁଡ଼ିକୁ ଘାତଙ୍କୀୟ ରୂପରେ ଲେଖ :
(i) 2^-4 × 2⁷
(ii) 3² × 3^-5 × 3⁶
(iii) p³ × p^-10
(iv) 2⁴ × (-4)^-2
(v) 8^p × 8^q
Solution:
(i) \(2^{-4} \times 2^7=\frac{1}{2^4} \times 2^7=2^{7-4}=2^3\)
(∵ a^m ÷ aⁿ = a^m-n)

(ii) \(3^2 \times 3^{-5} \times 3^6=3^2 \times \frac{1}{3^5} \times 3^6=3^2 \times 3^6 \times \frac{1}{3^5}=3^{2+6-5}=3^{8-5}=3^3\)
(∵ a^m × aⁿ = a^m+n & a^m ÷ aⁿ = a^m-n)

(iii) \(p^3 \times p^{-10}=p^3 \times \frac{1}{p^{10}}=\frac{p^3}{p^{10}}=p^{3-10}=p^{-7}\)
(∵ a^m ÷ aⁿ = a^m-n)

(iv) \(2^4 \times(-4)^{-2}=2^4 \times \frac{1}{(-4)^2}=2^4 \times \frac{1}{16}=2^4 \times \frac{1}{2^4}=2^{4-4}=2^0=1\)
(∵ a^m ÷ aⁿ = a^m-n)

(v) 8^p × 8^q = 8^p+q (∵ a^m × aⁿ = a^m+n)

Page No. 30

Question 1.
16384 (= 4⁷) ସଂଖ୍ୟାଟି 1024 (= 4⁵) ଠାରୁ 16 (= 4²) ଗୁଣ ବଡ଼ କି?
Solution:
ହଁ, ଯେହେତୁ 4⁷ ÷ 4⁵ = 4²
ତେଣୁ 16384(= 4⁷) ସଂଖ୍ୟାଟି 1024 (= 4⁵) ଠାରୁ 16 (= 4²) ଗୁଣ ବଡ଼।

Question 2.
4^-2 ଠାରୁ 4² କେତେଗୁଣ ବଡ଼?
Solution:
4² = 16
⇒ 4^-2 = \(\frac{1}{4^2}\)
⇒ 4^-2 = \(\frac{1}{16}\)
⇒ \(\frac{4^2}{4^{-2}}\) = 4² × 4² = 4²⁺² = 4⁴ = 256
4^-2 ଠାରୁ 4² ହେଉଛି 256 (4⁴) ଗୁଣ ବଡ଼।

Question 3.
7ର ଘାତରେଖା ବ୍ୟବହାର କରି ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଲେଖ।

Solution:

2.4 10 ର ଘାତ

Page No. 30

Question 1.
ଉପରୋକ୍ତ ଉପାୟରେ ଏହି ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଲେଖ :
(i) 172
(ii) 5642
(iii) 6474
Solution:
(i) 172 = 1 × 100 + 7 × 10 + 2 × 1 = (1 × 10²) + (7 × 10¹) + 2 × 10⁰)
(ii) 5642 = 5 × 1000 + 6 × 100 + 4 × 10 + 2 × 1 = (5 × 10³) + (6 × 10²) + (4 × 10¹) + (2 × 10⁰)
(iii) 6474 = 6 × 1000 + 4 × 100 + 7 × 10 + 4 × 1 = (6 × 10³) + (4 × 10²) + (7 × 10¹) + (4 × 10⁰)

Question 2.
561,903 କୁ ଆମେ କିପରି ଲେଖିପାରିବା?
Solution:
561.903 = (5 × 100) + (6 × 10) + 1 + (9 × \(\frac {1}{10}\)) + (0 × \(\frac {1}{100}\)) + (3 × \(\frac {1}{1000}\))
10 ର ଘାତ ବ୍ୟବହାର କରି ଆମେ ଏହାକୁ ଲେଖୁବା :
561.903 = (5 × 10²) + (6 × 10¹) + (1 × 10⁰) + (9 × 10^-1) + (0 × 10^-2) + (3 × 10^-3)

Page No. 32

Question 1.
ଏହି ତିନୋଟି ଦୂରତା ମଧ୍ଯରୁ କେଉଁ ଦୂରତାଟି ସବୁଠାରୁ ସାନ?

Solution:
(i) 1.4335 × 10¹² ମିଟର = 14.335 × 10¹¹ ମିଟର।
(ii) 1.439 × 10¹² ମିଟର = 14.39 × 10¹¹ ମିଟର।
(iii) 1.496 × 10¹¹ ମିଟର।
ତୁଳନା କଲେ 14.39 × 10¹¹ ମିଟର > 14.335 × 10¹¹ ମିଟର > 1.496 × 10¹¹ ମିଟର।
∴ (iii) 1.496 × 10¹¹ ମିଟର ସବୁଠାରୁ ସାନ।
ସୂର୍ଯ୍ୟ ଓ ପୃଥ‌ିବୀ ମଧ୍ୟରେ ଦୂରତା = 1.496 × 10¹¹ ମିଟର।

Question 2.
ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ମାନକ ରୂପରେ ପ୍ରକାଶ କର।
(i) 59,583
(ii) 65,950
(iii) 34,30,000
(iv) 70,04,00,00,000
Solution:
(i) 59,583 = \(\frac {59853}{10000}\) × 10000 = 5.9853 × 10⁴
(ii) 65,950 = \(\frac {65950}{10000}\) × 10000 = 6.595 × 10⁴
(iii) 34,30,000 = \(\frac {3430000}{1000000}\) × 1000000 = 3.43 × 10⁶
(iv) 70,04,00,00,000 = \(\frac {70040000000}{10000000000}\) × 10000000000 = 7.004 × 10¹⁰

2.5 ତୁମେ କେବେ ଭାବିଛ କି?

Page No. 33

Question 1.
ଦାନ କରାଯାଉଥ‌ିବା ଗୁଡ଼ର ମୂଲ୍ୟ (ଟଙ୍କାରେ) କେତେ ହେବ ? ଦାନ କରାଯାଇଥିବା ଗହମର ମୂଲ୍ୟ (ଟଙ୍କାରେ) କେତେ ହେବ?
Solution:
ଗୁଡ଼ର ମୂଲ୍ୟ (ଟଙ୍କାରେ) = ରକିର ଓଜନ (କିଲୋଗ୍ରାମ୍‌ରେ) × 1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ।
ଗହମର ମୂଲ୍ୟ (ଟଙ୍କାରେ) = ଇତୁର ଓଜନ (କିଲୋଗ୍ରାମିଂରେ) × 1 କି.ଗ୍ରା. ଗହମର ମୂଲ୍ୟ।

Question 2.
ଅଜ୍ଞାତ ରାଶି ପାଇଁ ଆବଶ୍ୟକ ଏବଂ ଉପଯୁକ୍ତ ଆନୁମାନିକ ଆକଳନ କର ଏବଂ ଉତ୍ତର ନିର୍ଣ୍ଣୟ କର । ମନେରଖ ରକିକୁ 13 ବର୍ଷ ଏବଂ ଇତୁକୁ 11 ବର୍ଷ ।
Solution:
ମନେକର ରକିର ବୟସ = 13 ବର୍ଷ ଓ ଓଜନ = 45 କି.ଗ୍ରା.।
1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ = ଟ.70.00
ଦାନ କରାଯାଇଥିବା ଗୁଡ଼ର ମୂଲ୍ୟ = 45 × 70 ଟଙ୍କା = ଟ.3150.00 ଅଟେ।
ମନେକର ଇତୁର ବୟସ = 11 ବର୍ଷ ଓ ଓଜନ = 50 କି.ଗ୍ରା.।
1 କି.ଗ୍ରା. ଗହମର ମୂଲ୍ୟ = ଟ.50.00
ଦାନ କରାଯାଇଥ‌ିବା ଗହମର ମୂଲ୍ୟ = 50 × 50 ଟଙ୍କା = ଟ.2500.00 ଅଟେ।

Question 3.
ଅନୁମାନ : କୌଣସି ଗଣନା | ହିସାବ ନକରି, ଉତ୍ତର କ’ଣ ହୋଇପାରେ ତାହା ବିଷୟରେ ସ୍ଵତଃସ୍ଫୁର୍ଭ (ଶୀଘ୍ର ଅନୁମାନ କର)
Solution:
ରକିର ବୟସ = 13 ବର୍ଷ ।
ତେଣୁ ଆମେ ଅନୁମାନ କରିପାରିବା ଯେ ତା’ର ଓଜନ ପ୍ରାୟ 40ରୁ 50 କି.ଗ୍ରା. ମଧ୍ଯରେ ହୋଇପାରେ ।
ଯଦି ଆମେ ଧରିନେବା । କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ = 60 ଟଙ୍କା ।
ତେବେ 1 ଟଙ୍କିଆ ମୁଦ୍ରାର ସଂଖ୍ୟା = 45 × 60 = 2700

Page No. 34

Question 1.
ଆକଳନ ଏବଂ ଆନୁମାନିକ ଗଣନା :
(i) ଉତ୍ତର ପାଇବା ପାଇଁ ଆବଶ୍ୟକ ପରିମାଣଗୁଡ଼ିକ ମଧ୍ୟରେ ଥିବା ସମ୍ପର୍କ ବର୍ଣ୍ଣନା କର ।
(ii) ଯଦି ଆବଶ୍ୟକ ତଥ୍ୟ ଉପଲବ୍‌ଧ ନଥାଏ, ତେବେ ଉପଯୁକ୍ତ ଅନୁମାନ ଓ ଆକଳନ କର ।
(iii) ହିସାବ କର ଏବଂ ଉତ୍ତର ଖୋଜ (ଏବଂ ତୁମର ଅନୁମାନ, ଠିକ୍ ଉତ୍ତରରେ କେତେ ନିକଟତର ଥୁଲା ଦେଖ ।)
Solution:
(i) ଉତ୍ତର ପାଇବା ପାଇଁ ଆବଶ୍ୟକ ପରିମାଣଗୁଡ଼ିକ ମଧ୍ୟରେ ଥିବା ସମ୍ପର୍କ ବର୍ଣ୍ଣନା :
ରକିର ଓଜନ କେତେ । ଟଙ୍କା ମୁଦ୍ରା ମେଳ ଖାଉଛି ତାହା ଜାଣିବା ପାଇଁ ଆମକୁ ଚିହ୍ନଟ କରିବାକୁ ପଡ଼ିବ : ରକିର ଓଜନ (କିଲୋଗ୍ରାମରେ), 1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ ।
ଦାନର ମୂଲ୍ୟ = ଓଜନ କିଲୋଗ୍ରାମରେ × 1 କି.ଗ୍ରା.ଗୁଡ଼ର ମୂଲ୍ୟ।
ଆବଶ୍ୟକ 1 ଟଙ୍କିଆ ମୁଦ୍ରା ସଂଖ୍ୟା = ଦାନର ମୂଲ୍ୟ ÷ 1

(ii) ଯଦି ଆବଶ୍ୟକ ତଥ୍ୟ ଉପଲବ୍‌ଧ ନଥାଏ, ତେବେ ଉପଯୁକ୍ତ ଅନୁମାନ ଓ ଆକଳନ :
ଆସ ଯୁକ୍ତିଯୁକ୍ତ ଆକଳନ ବ୍ୟବହାର କରିବା :
ରକିର ଓଜନ ପ୍ରାୟ 45 କି.ଗ୍ରା. (ସାଧାରଣ 13 ବର୍ଷର ପିଲା)
1 କିଲୋଗ୍ରାମ୍ ଗୁଡ଼ର ଓଜନ ପ୍ରାୟ ଟ.60.00 (ହାରାହାରି ବଜାର ଦର)
45 କି.ଗ୍ରା ଗୁଡ଼ର ମୂଲ୍ୟ = 45 × 60 = ଟ.2700.00

(iii) ହିସାବ କର ଏବଂ ଉତ୍ତର ଖୋଜ : (ତୁମର ଅନୁମାନ, ଠିକ୍ ଉତ୍ତରର କେତେ ନିକଟତର ଥୁଲା ଦେଖ।)
1 ଟଙ୍କିଆ ମୁଦ୍ରା ଆବଶ୍ୟକ = 6.2700 ÷ ଟ.1 = 2700

Question 2.
ମୁଦ୍ରା ସଂଖ୍ୟା ଶହ ଶହ, ହଜାର ହଜାର, ଲକ୍ଷ ଲକ୍ଷ, କୋଟି କୋଟି କିମ୍ବା ତା’ଠାରୁ ଅଧିକ ହେବ କି ? ଅନୁମାନ କରି କୁହ।
Solution:
ଆମେ ଶୀଘ୍ର ବୁଦ୍ଧିମାନ କରି ଅନୁମାନ କରିବା।
ରକିର ଓଜନ ପ୍ରାୟ 45 କି.ଗ୍ରା. ଏବଂ 1 କି.ଗ୍ରା. ଗୁଡ଼ର ମୂଲ୍ୟ 60 ଟଙ୍କା
∴ ମୋଟ ମୂଲ୍ୟ = 45 × 60 ଟଙ୍କା = 2700 ଟଙ୍କା ।
ଯଦି ଆମେ 1 ଟଙ୍କିଆ ମୁଦ୍ରା ବ୍ୟବହାର କରୁ, ତେଣୁ 2700 ମୁଦ୍ରା ଆବଶ୍ୟକ,
ଯାହା ହଜାର ପରିସର ମଧ୍ୟରେ ପଡ଼ିଥାଏ ।
ତେଣୁ ହଜାର ମୁଦ୍ରା ଆବଶ୍ୟକ । ଲକ୍ଷ କିମ୍ବା କୋଟି ନୁହେଁ, କିନ୍ତୁ ଶହେରୁ ଅଧିକ।

Question 3.
ଅଜ୍ଞାତ ରାଶି ପାଇଁ ଆବଶ୍ୟକୀୟ ଅନୁମାନ ଏବଂ ଆକଳନ କରି ଉତ୍ତର ଖୋଜ ।
Solution:
ପିଲାମାନେ ଶିକ୍ଷକଙ୍କ ସହାୟତାରେ ନିଜେ କରିବେ ।

Question 4.
ଇଡ଼ୁ ପଚାରିଲା, ଯଦି ଆମେ 5 ଟଙ୍କିଆ ମୁଦ୍ରା କିମ୍ବା 10 ଟଙ୍କିଆ ନୋଟ୍ ବ୍ୟବହାର କରୁ, ଏହା କେତେ ଟଙ୍କା ହୋଇପାରେ?
Solution:
ମୋଟ ଦାନର ମୂଲ୍ୟ 1140 ଟଙ୍କା ।
5 ଟଙ୍କିଆ ମୁଦ୍ରାର ସଂଖ୍ୟା = 1140 ÷ 5 = 228
10 ଟଙ୍କିଆ ମୁଦ୍ରାର ସଂଖ୍ୟା = 1140 ÷ 10 = 114

Question 5.
ଇଡୁ କହିଲା, ‘‘ଯେତେବେଳେ ମୁଁ ବଡ଼ ହେବି, ପ୍ରତିବର୍ଷ ମୁଁ ମୋ ଓଜନର ନୋଟବୁକ୍ ଦାନ କରିବି ।’’ ରକି କହିଲା, “ଯେତେବେଳେ ମୁଁ ବଡ଼ ହେବି ପ୍ରତିବର୍ଷ ମୁଁ ମୋ ଓଜନର ଅନ୍ନଦାନ କରିବି ।’’ ଏହି ଦାନରୁ ବର୍ଷକୁ କେତେ ଲୋକ ଉପକୃତ ହେବେ ? ପୁନଶ୍ଚ ଖୋଜିବା ପୂର୍ବରୁ ପ୍ରଥମେ ଅନୁମାନ କର ।
Solution:
ଇତୁର ନୋଟବୁକ ଦାନ ପ୍ରତିବର୍ଷ 100ରୁ 200 ଛାତ୍ର ଉପକୃତ ହୋଇପାରନ୍ତି ।
ରକିର ଖାଦ୍ୟ ପ୍ରଦାନ : ପ୍ରାୟ ବାର୍ଷିକ 150 ରୁ 300
ଲୋକଙ୍କୁ ଖାଦ୍ୟ ଦେବାପାଇଁ ଯଥେଷ୍ଟ ଶସ୍ୟ କିମ୍ବା ଆନୁମାନିକ ଆକଳନ :
ଇତୁର ଓଜନ ପ୍ରାୟ 38 କି.ଗ୍ରା.।
1 ଟି ନୋଟବୁକ୍‌ର ମୂଲ୍ୟ ପ୍ରାୟ 20 ଟଙ୍କା ।
ରକିର ଓଜନ ପ୍ରାୟ 45 କି.ଗ୍ରା.।
ଦାନ କରାଯାଇଥିବା 1 କି.ଗ୍ରା. ଶସ୍ୟ କିମ୍ବା ଖାଦ୍ୟର ମୂଲ୍ୟ ପ୍ରାୟ 30 ଟଙ୍କା ।
ପ୍ରତିବ୍ୟକ୍ତି ପାଇଁ ଖାଦ୍ୟ ଖର୍ଚ୍ଚ ପ୍ରାୟ 15 ଟଙ୍କା ।
ଇତ୍ରୁର ନୋଟବୁକ୍ ଦାନ : ତାଙ୍କ ଓଜନ ମୂଲ୍ୟର ନୋଟ୍‌ବୁକ୍ ଦାନର ମୂଲ୍ୟ = 38 × 20 ଟଙ୍କା = 760 ଟଙ୍କା।
ଯଦି ପ୍ରତ୍ୟେକ ଛାତ୍ରଙ୍କୁ ଦୁଇଟି ନୋଟବୁକ୍‌ ଆବଶ୍ୟକ ହୁଏ,
ସାହାଯ୍ୟ ପାଇପାରୁଥିବା ଛାତ୍ରସଂଖ୍ୟା = 760 ÷ 40 = 19 ଜଣ।
∴ ପ୍ରତିବର୍ଷ 19 ଜଣ ଛାତ୍ର ନୋଟ୍‌ବୁକ୍‌ ପାଇପାରିବେ ।
ରକିର ଖାଦ୍ୟଦାନ : ତାଙ୍କ ଓଜନ ମୂଲ୍ୟର ଶସ୍ୟ କିମ୍ବା ଖାଦ୍ୟ ଦାନର ମୂଲ୍ୟ = 45 × 30 = 1350 ଟଙ୍କା।
ପ୍ରତ୍ୟେକ ଖାଦ୍ୟର ମୂଲ୍ୟ = 15 ଟଙ୍କା ।
ଖାଦ୍ୟ ଯୋଗାଇ ପାରୁଥୁବା ଲୋକଙ୍କ ସଂଖ୍ୟା = 1350 ÷ 15 = 90 ଜଣ।
∴ 90 ଜଣ ଲୋକ ଗୋଟିଏ ଥର ଖାଦ୍ୟ ପାଇବେ କିମ୍ବା ବର୍ଷକରେ 90 ରୁ କମ୍ ଏକାଧିକ ଥର ଖାଇପାରିବେ ।

Question 6.
ରକି ଓ ଇଡୁ ଆଉ କେହି କହୁଥିବାର ଶୁଣିଲେ – ଆମେ ଏହି ସ୍ଥାନରେ ପହଞ୍ଚିବା ପାଇଁ 400 କି.ମି. ପଦଯାତ୍ରା କଲୁ । ଆମେ ଆଜି ସକାଳୁ ପହଞ୍ଚିଲୁ । ସେମାନେ କେତେ ସମୟ ପୂର୍ବରୁ ଯାତ୍ରା ଆରମ୍ଭ କରିଥିବେ ?
Solution:
ଯଦି କେହିଜଣେ ପାଦରେ 400 କି.ମି. ଚାଲିଥାନ୍ତି, କେତେ ସମୟ ଲାଗିପାରେ ?
ଆସ ଅନୁମାନ କରିବା ଯେ, ସେମାନେ ପ୍ରତିଦିନ ପ୍ରାୟ 25 ରୁ 30 କି.ମି. ଚାଲେ ।
ଆନୁମାନିକ ଦିନ ସଂଖ୍ୟା = 400 ÷ 25 = 16 ଦିନ ।
କିମ୍ବା ତା’ର ଗତି ଓ ବିଶ୍ରାମ ଉପରେ ନିର୍ଭର କରି 13 ରୁ 20 ଦିନ ହୋଇପାରେ ।
ଆକଳନ ଓ ଗଣନା :
ଦୈନିକ ଅତିକ୍ରାନ୍ତ ଦୂରତା = 30 କି.ମି. (ଧରାଯାଉ)
ଅତିକ୍ରାନ୍ତ ଦୂରତା = 400 କି.ମି. ।
ଦିନସଂଖ୍ୟା = 400 ÷ 30 = 13.3 ଦିନ।
∴ ପ୍ରାୟତଃ 13 ରୁ 14 ଦିନ ।

Page No. 35

Question 1.
ଯଦି ଜଣେ ବ୍ୟକ୍ତି ନିରନ୍ତର ଚାଲିଥାଏ, ତେବେ ସେ ନିଜ ଜୀବନକାଳରେ କେତେଥର ପୃଥିବୀକୁ ପରିକ୍ରମା (ବିଶ୍ଵ ପରିକ୍ରମଣ) କରିପାରିବେ ? ପୃଥ‌ିବୀର ପରିଧ୍ଵକୁ 40,000 କି.ମି. ବୋଲି ଧରି ନିଆଯାଉ ।
Solution:
ଆନୁମାନିକ : ପୃଥ‌ିବୀର ପରିଧ୍ଵ = 40,000 କି.ମି. ।
ହାରାହାରି ଚାଲିବାର ଗତି ଘଣ୍ଟାପ୍ରତି 5 କି.ମି. ।
ପ୍ରତିଦିନ ସର୍ବାଧ‌ିକ ଚାଲିବା ପ୍ରାୟ = 8 ଘଣ୍ଟା ।
ଆସ ଅନୁମାନ କରିବା, 60 ବର୍ଷର (ବୟସ 15ରୁ 75 ମଧ୍ୟରେ)
ଗୋଟିଏ ଦିନରେ ଅତିକ୍ରାନ୍ତ ଦୂରତା = 8 × 5 = 40 କି.ମି. ।
ମୋଟ ଚାଲିଯାଇଥବା ଦିନ = 60 × 365 = 21,900 ଦିନ ।
ତା’ର ଜୀବନକାଳରେ ମୋଟ ଚାଲିଥିବା ଦୂରତା = 21,900 × 40 = 8,76,000 କି.ମି.।
∴ ମୋଟ ପୃଥ‌ିବୀ ପରିକ୍ରମା = 8,76,000 ÷ 40,000 = 21.9 ଥର ।

Page No. 36

Question 1.
3,84,400 କି.ମି.ରେ କେତେ 20 ସେ.ମି. ଅଛି ?

Solution:
ଯଦି ଆମେ ହିସାବ କରୁ, ତେବେ ଫଳାଫଳ 1,92,20,00,000 ପାହାଚ ମିଳିବ । ଏହା 192 କୋଟି 20 ଲକ୍ଷ ପାହାଚ କିମ୍ବା 1 ବିଲିୟନ 992 ମିଲିୟନ୍ ପାହାଚ ସହ ସମାନ । ପ୍ରତ୍ୟେକ ପଦକ୍ଷେପ ପରେ ପୃଥ‌ିବୀଠାରୁ ଦୂରତାରେ ନିର୍ଦ୍ଦିଷ୍ଟ ବୃଦ୍ଧି (ପ୍ରତ୍ୟେକ ପଦକ୍ଷେପ 20) ସେ.ମି. ବୃଦ୍ଧି)କୁ କୈକ ବୃଦ୍ଧି କୁହାଯାଏ ।

Question 2.
ତୁମେ ରେଖକ ବୃଦ୍ଧି ଏବଂ ଘାତାଙ୍କୀୟ ବୃଦ୍ଧିର କିଛି ଉଦାହରଣ ଦେଇପାରିବ କି ?
Solution:
ନିଜେ ଅଭ୍ୟାସ କର ।

Page No. 38-39

Question 1.
ବିଶ୍ଵ ଜନସଂଖ୍ୟା ପ୍ରାୟ 8 × 10⁹ ଓ ଆଫ୍ରିକୀୟ ହାତୀ ସଂଖ୍ୟା ହେଉଛି 4 × 10⁵ ଆମେ କହିପାରିବା କି, ପାଇଁ ପ୍ରାୟ 20,000 ଲୋକ ଅଛନ୍ତି ।
Solution:
ବିଶ୍ଵ ଜନସଂଖ୍ୟା ପ୍ରାୟ = 8 × 10⁹ = 8,00,00,00,000
ଆଫ୍ରିକୀୟ ହାତୀ ସଂଖ୍ୟା = 4 × 10⁵ = 4,00,000
∴ ପ୍ରତ୍ୟେକ ହାତୀ ପାଇଁ ଲୋକ = \(\frac {8,00,00,00,000}{4,00,000}\) = 20,000

Question 2.
ବୈଜ୍ଞାନିକ ସଂକେତ ବ୍ୟବହାର କରି ଗଣନା କର ଏବଂ ଉତ୍ତର ଲେଖ।
(i) ବିଶ୍ଵରେ ମନୁଷ୍ୟ ଓ ପିମ୍ପୁଡ଼ିମାନଙ୍କ ସଂଖ୍ୟାର ଅନୁପାତ କେତେ?
(ii) ଯଦି ଷ୍ଟଲିଂ ପକ୍ଷୀମାନଙ୍କର ଏକ ଦଳରେ 10,000 ପକ୍ଷୀ ଥାଆନ୍ତି, ତେବେ ବିଶ୍ଵରେ କେତେ ଦଳ ପକ୍ଷୀ ଥାଇପାରନ୍ତି?
(iii) ଯଦି ଗୋଟିଏ ଗଛରେ 10⁴ ସଂଖ୍ୟକ ପତ୍ର ଥାଏ, ତେବେ ପୃଥ‌ିବୀର ସମସ୍ତ ଗଛରେ ଥିବା ପତ୍ର ସଂଖ୍ୟା କେତେ ନିର୍ଣ୍ଣୟ କର।
(iv) ଯଦି ତୁମେ କାଗଜ–ଫର୍ଦଗୁଡ଼ିକୁ ଉପରକୁ ଉପର ଥାକ କରି ରଖୁବ, ତେବେ ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିବା ପାଇଁ ତୁମକୁ କେତେ ଫର୍ଦ୍ଦ କାଗଜ’ ଆବଶ୍ୟକ ହେବ?
Solution:
(i) ବିଶ୍ଵରେ ପିମ୍ପୁଡ଼ିମାନଙ୍କ ସଂଖ୍ୟା = 20 କ୍ୱାଡ୍ରିଲିୟନ୍ = 2 × 10¹⁶ ପ୍ରାୟ ।
ବିଶ୍ୱରେ ମନୁଷ୍ୟମାନଙ୍କ ସଂଖ୍ୟା = 8 ବିଲିୟନ୍ = 8 × 10⁹ ପ୍ରାୟ ।
ବିଶ୍ଵରେ ମନୁଷ୍ୟ ଓ ପିମ୍ପୁଡ଼ିମାନଙ୍କ ସଂଖ୍ୟାର ଅନୁପାତ = \(\frac{2 \times 10^{16}}{8 \times 10^9}\)
= 0.25 × 107
= 2.5 × 106

(ii) ଷ୍ଟଲିଂ ପକ୍ଷୀସଂଖ୍ୟା = 3.1 × 10⁸ ପ୍ରାୟ ।
ଷ୍ଟଲିଂ ପକ୍ଷୀମାନଙ୍କର ଗୋଟିଏ ଦଳରେ ପକ୍ଷୀସଂଖ୍ୟା = 1 × 10⁴
ବିଶ୍ଵରେ ଥ‌ିବା ଷ୍ଟଲିଂ ପକ୍ଷୀମାନଙ୍କର ଦଳସଂଖ୍ୟା = \(\frac{3.1 \times 10^8}{1 \times 10^4}\) = 3.1 × 10⁴

(iii) ବିଶ୍ଵର ଗଛସଂଖ୍ୟା ପ୍ରାୟ = 3 × 10¹²
ଗୋଟିଏ ଗଛର ପତ୍ର ସଂଖ୍ୟା ପ୍ରାୟ = 1 × 10⁴
ବିଶ୍ଵର ସମୁଦାୟ ଗଛର ପତ୍ର ସଂଖ୍ୟା = 3 × 10¹² × 1 × 10⁴ = 3 × 10¹⁶

(iv) ଚନ୍ଦ୍ରର ଦୂରତା = 3.84 × 10⁸ ମିଟର।
ଗୋଟିଏ କାଗଜର ମୋଟେଇ = 1 × 10^-4 ମିଟର।
ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିବା ପାଇଁ କାଗଜ ଦରକାର = \(\frac{3.84 \times 10^8}{1 \times 10^{-4}}\) = 3.84 × 10¹²

Page No. 39

Question 1.
‘‘ତୁମର ବୟସ କେତେ ?” ଇଡୁ ପଚାରିଲା :
‘‘ମୁଁ କିଛି ସପ୍ତାହ ପୂର୍ବରୁ 13 ବର୍ଷ ପୂରଣ କରିଛି’’ ରକି ଉତ୍ତର ଦେଲା ।
‘‘ତୁମର ବୟସ କେତେ ?” ଇଡୁ ପୁଣି ପଚାରିଲା ।
‘‘ଆଜି ମୋତେ 4840 ଦିନ’’ ରକି କହିଲା ।
‘‘ତୁମକୁ କେତେ ବୟସ’’ ଇଡୁ ଆଉ ଥରେ ପଚାରିଲା,
‘‘ମୋର ବୟସ ————– ଘଣ୍ଟା !’’ ରକି କହିଲା ।
Solution:
1 ବର୍ଷ = 365 ଦିନ ।
4840 ଦିନ = \(\frac {4840}{365}\) ବର୍ଷ = 13.26 ବର୍ଷ |
1 ଦିନ = 24 ଘଣ୍ଟା ।
4840 ଦିନ = 24 × 4840 = 116160 ଘଣ୍ଟା
∴ ରକିର ବୟସ 13,26 ବର୍ଷ ଓ 116160 ଘଣ୍ଟା ।

Question 2.
ଇତୁ ‘‘ଆଜି ମୋର ବୟସ 4070 ଦିନ । ତୁମେ ମୋର ଜନ୍ମତାରିଖ ଖୋଜି ପାରିବ କି ?’’
Solution:
ନିଜେ କର ।

Question 3.
ଯଦି ତୁମେ ଏକ ନିୟୁତ ସେକେଣ୍ଡ ବଞ୍ଚୁଛ, ତେବେ ତୁମକୁ କେତେ ବୟସ ହେବ?
Solution:
ମିନିଟ୍ = 60 ସେକେଣ୍ଡ ।
1 ଘଣ୍ଟା = 60 ମିନିଟ୍ = 60 × 60 ସେକେଣ୍ଡ = 3600 ସେକେଣ୍ଡ ।
1 ଦିନ = 24 ଘଣ୍ଟା = 24 × 3600 = 86,400 ସେକେଣ୍ଡ ।
ତୁମେ ବଞ୍ଚ = 1 ନିୟୁତ ସେକେଣ୍ଡ = 10,00,000 ସେକେଣ୍ଡ ।
ଦିନ ସଂଖ୍ୟା = \(\frac {10,00,000}{86,400}\) = 11.57 ଦିନ।

Page No. 40

Question 1.
10⁵ ସେକେଣ୍ଡ = 1.16 ଦିନ ଏବଂ 10⁶ ସେକେଣ୍ଡ = 11.57 ଦିନ । କିଛି ଘଟଣା ବିଷୟରେ ଚିନ୍ତାକର ଯାହାର ସମୟ (i) 10⁵ ସେକେଣ୍ଡ ଏବଂ (ii) 10⁶ ସେକେଣ୍ଡର କ୍ରମରେ ଅଛନ୍ତି, ସେଗୁଡ଼ିକୁ ବୈଜ୍ଞାନିକ ସଂକେତରେ ଲେଖ ।
Solution:
(i) 10⁵ ସେକେଣ୍ଡର ଘଟଣାଗୁଡ଼ିକ (ପ୍ରାୟ 1.16 ଦିନ)
(କ) ବହୁସ୍ତରୀୟ ସାଇକେଲ ଚାଳନା ଦୌଡ଼ ପର୍ଯ୍ୟାୟ (ଟୁର ଡି ଫ୍ରାନ୍ସର ଗୋଟିଏ ପର୍ଯ୍ୟାୟ ପରି)
ସମୟ ବ୍ୟବଧାନ : 1 × 10⁵ ସେକେଣ୍ଡ ।
(ଖ) ଏକ କ୍ଷୁଦ୍ର ଚଳଚ୍ଚିତ୍ର ବା ଡକୁମେଣ୍ଟାରୀ ଫିଲ୍ମ କରିବା ।
ସୁଟିଂ କାର୍ଯ୍ୟସୂଚୀ : 1.2 × 10⁵ ସେକେଣ୍ଡ ।

(ii) 10⁶ ସେକେଣ୍ଡର ଘଟଣାଗୁଡ଼ିକ (ପ୍ରାୟ 11.57 ଦିନ)
(କ) ବଡ଼ ଉତ୍ସବର ଅବଧୂ (ଯଥା- କୁମ୍ଭମେଳା କିମ୍ବା ଅଲମ୍ପିକ୍ସ ଉଦ୍‌ଘାଟନୀ ଇଭେଣ୍ଟ)
ସମ୍ପୂର୍ଣ୍ଣ ଉତ୍ସବ ଅବଧୂ ପ୍ରାୟ 1 × 10⁶ ସେକେଣ୍ଡ ।
(ଖ) କୁକୁଡ଼ା ଅଣ୍ଡାର ଉଷୁମାଇବା ସମୟ (ଛୁଆ ହେବା ପର୍ଯ୍ୟନ୍ତ) ପ୍ରାୟ 1.2 × 10⁶ ସେକେଣ୍ଡ।

Page No. 42

Question 1.
ବୈଜ୍ଞାନିକ ସଂକେତ ବ୍ୟବହାର କରି ଗଣନ କର ଏବଂ ଉତ୍ତର ଲେଖ।
(i) ଯଦି ପ୍ରତି ସେକେଣ୍ଡରେ ଗୋଟିଏ ତାରା ଗଣାଯାଏ, ତେବେ ବ୍ରହ୍ମାଣ୍ଡର ସମସ୍ତ ତାରା ଗଣନା କରିବା ପାଇଁ କେତେ ସମୟ ଲାଗିବ? ବୈଜ୍ଞାନିକ ‘ସଂକେତ ବ୍ୟବହାର କରି ସେକେଣ୍ଡ ଏକକରେ ଉତ୍ତର ଦିଅ।
(ii) ଯଦି ଜଣେ ବ୍ୟକ୍ତି 10 ସେକେଣ୍ଡରେ ଏକ ଗ୍ଲାସ ପାଣି (200 ମି.ଲି.) ପିଇପାରନ୍ତି, ତେବେ ପୃଥ‌ିବୀରେ ଥିବା ସବୁ ପାଣି ପିଇ ଶେଷ କରିବାକୁ ତାଙ୍କୁ କେତେ ସମୟ ଲାଗିବ?
Solution:
(i) ପର୍ଯ୍ୟବେକ୍ଷଣ ଯୋଗ୍ୟ ସ୍ଥାନରେ ଥିବା ତାରାଗୁଡ଼ିକର ଆନୁମାନିକ ସଂଖ୍ୟା :
ବ୍ରହ୍ମାଣ୍ଡରେ ତାରା ସଂଖ୍ୟା : 1 × 10²⁴ (1 ସେକ୍ସଟିଲିୟନ୍)
ଯଦି ଆମେ ପ୍ରତି ସେକେଣ୍ଡରେ ଗୋଟିଏ ତାରା ଗଣନା କରନ୍ତି,
ତେବେ ଆବଶ୍ୟକ ସମୟ = 1 × 10²⁴ ସେକେଣ୍ଡ ।
(ii) ପୃଥିବୀରେ ଥିବା ମୋଟ ଜଳର ଆୟତନ ହେଉଛି 1.38 × 10⁹ କି.ମି.³
ଘନ କି.ମି.କୁ ମିଲିଲିଟରରେ ପରିଣତ କଲେ = 1.386 × 10⁹ × 10¹⁵ ମି.ଲି.। = 1.386 × 10²⁴ ମି.ଲି.।
ପ୍ରତ୍ୟେକ ଗ୍ଲାସ୍ଟର କ୍ଷମତା = 200 ମି.ଲି.।
ଏକ ଗ୍ଲାସ୍ ପାଣି ପିଇବାକୁ ସମୟ ଲାଗେ = 10 ସେକେଣ୍ଡ ।
ସମୁଦାୟ ଗ୍ଲାସ୍ ସଂଖ୍ୟା = 6.93 × 10²¹
∴ ସବୁ ପାଣି ପିଇ ଶେଷ କରିବାକୁ ସମୟ ଲାଗିବ = 6.93 × 10²¹ × 10 = 6.93 × 10²² ସେକେଣ୍ଡ।

2.5 ଇତିହାସ ପୃଷ୍ଠାରୁ

ନିଜେ କରି ଦେଖ : (Page No. 44-45)

Question 1.
2²²⁴ ÷ 4³² ମାନର ଏକକ ସ୍ଥାନରେ ଥ‌ିବା ଅଙ୍କଟି କେତେ? (ସୂଚନା 4 = 2²)
Solution:
2²²⁴ ÷ 4³² = 2²²⁴ ÷ (2²)³² = 2²²⁴ ÷ 2⁶⁴ = 2^(224-64) = 2¹⁶⁰
[∵ (a^m)ⁿ = a^mn]
2 ଘାତର ଏକକ ଅଙ୍କ ପ୍ରତି ଚାରି ଘାତରେ ଏକ ପୁନରାବୃତ୍ତି ଢାଞ୍ଚା ଅନୁସରଣ କରେ ।
2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16 (ଏକକ ଅଙ୍କ 6),
2⁵ = 32 (ଏକକ ଅଙ୍କ 2), 2⁶ = 64 (ଏକକ ଅଙ୍କ 4)
2ର ଘାତ ହେଉଛି 160। ତେଣୁ = \(\frac {160}{4}\) = 40
ଯେତେବେଳେ ଭାଗଶେଷ 0, ଚକ୍ରରେ ଏକକ ସ୍ଥାନରେ ଥିବା ଶେଷ ଅଙ୍କଟି 6 ହେବ।
∴ 2²²⁴ ÷ 4³² ମାନର ଏକକ ସ୍ଥାନରେ ଥିବା ଅଙ୍କଟି 6।

Question 2.
ଗୋଟିଏ ପାତ୍ରରେ 5ଟି ବୋତଲ ଅଛି । ପ୍ରତିଦିନ ଗୋଟିଏ ନୂଆପାତ୍ର ଅଣାଯାଉଥାଏ । 40 ଦିନପରେ ସେଥିରେ କେତେ ନୂଆ ବୋତଲ ଥିବ?
Solution:
ଗୋଟିଏ ପାତ୍ରରେ 5ଟି ବୋତଲ ଅଛି ।
40 ଦିନ ପର୍ଯ୍ୟନ୍ତ ପ୍ରତିଦିନ ଗୋଟିଏ ନୂଆପାତ୍ର ଅଣାଯାଉଥାଏ ।
40 ଦିନପରେ ବୋତଲ ସଂଖ୍ୟା = 40ଟି ପାତ୍ର × 5ଟି ବୋତଲ = 200 ବୋତଲ ।

Question 3.
ନିମ୍ନରେ ଦିଆଯାଇଥବା ସଂଖ୍ୟାକୁ ଦୁଇ କିମ୍ବା ଅଧିକ ଘାତାଙ୍କର ଗୁଣଫଳ ଭାବରେ ତିନୋଟି ଭିନ୍ନଭିନ୍ନ ଉପାୟରେ ଲେଖ । ଘାତାଙ୍କଗୁଡ଼ିକ ଯେକୌଣସି ପୂର୍ବସଂଖ୍ୟା ହୋଇପାରିବ ।
(i) 64³
(ii) 192⁸
(iii) 32^-5
Solution:

ବିକଳ୍ପ ଉତ୍ତର :
ଆମେ ଜାଣିଛେ (a^m)ⁿ = a^mn ଓ a^m+n = a^m × aⁿ
(i) 64³ = (8²)³ = 8^2×3 = 8² × 8³
64³ = (4³)³ = 4^3×3 = 4³ × 4³

(ii) 192⁸ = (64 × 3)⁸ = (8² × 3¹)⁸ = 8^2×8 × 3⁸ = 8¹⁶ × 3⁸
192⁸ = (64 × 3)⁸ = (4³ × 3¹)⁸ = 4²⁴ × 3⁸
192⁸ = (64 × 3)⁸ = (2⁶ × 3¹)⁸ = 2⁴⁸ × 3⁸

(iii) 32^-5 = (2⁵)^-5 = 2^5×(-5)
(32)^-5 = (2⁵)^-5 = 2^-25 = 2^(-5-10-10) = 2^-5 × 2^-10 × 2^-10
(32)^-5 = (2⁵)^-5 = 2^-25 = 2^(10-35) = 2¹⁰ × 2^-35

Question 4.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ପ୍ରତ୍ୟେକ ଉକ୍ତିକୁ ପରୀକ୍ଷା କର ଏବଂ କେଉଁଗୁଡ଼ିକ ‘ସର୍ବଦା ସତ୍ୟ’, ‘ବେଳେବେଳେ ସତ୍ୟ’ କିମ୍ବା ‘ଆଦୌ’ ସତ୍ୟ ନୁହେଁ? ନିର୍ଣ୍ଣୟ କର । ତୁମର ଯୁକ୍ତିକୁ ବର୍ଣ୍ଣନା କର ।
(i) ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
(ii) ଚତୁର୍ଥ ଘାତାଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
(iii) ଗୋଟିଏ ସଂଖ୍ୟାର ପଞ୍ଚମ ଘାତାଙ୍କ ସେହି ସଂଖ୍ୟାର ଘନଦ୍ଵାରା ବିଭାଜ୍ୟ ।
(iv) ଦୁଇଟି ଘନସଂଖ୍ୟାର ଗୁଣଫଳ ଏକ ଘନସଂଖ୍ୟା ଅଟେ ।
(v) q⁴⁶ ସଂଖ୍ୟାଟି ଉଭୟ ଚତୁର୍ଥ ଘାତାଙ୍କ ଓ ଷଷ୍ଠ ଘାତାଙ୍କ ଅଟେ । (q ଏକ ମୌଳିକ ସଂଖ୍ୟା ଅଟେ)।
Solution:
(i) ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
ଉଦାହରଣ : 64 = 4³ = 8² ହେଉଛି ଗୋଟିଏ ଘନ ଏବଂ ବର୍ଗ ସଂଖ୍ୟା ।
ଯେପରିକି 8 = 2³ ହେଉଛି ଗୋଟିଏ ଘନସଂଖ୍ୟା କିନ୍ତୁ ବର୍ଗସଂଖ୍ୟା ନୁହେଁ ଏବଂ
9 = 3² ହେଉଛି ଗୋଟିଏ ବର୍ଗସଂଖା କିନ୍ତୁ ଘନସଂଖ୍ୟା ନୁହେଁ ।
ତେଣୁ ଏହି ଉକ୍ତିଟି ବେଳେବେଳେ ସତ୍ୟ ।

(ii) ଚତୁର୍ଥ ଘାତାଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟ ବର୍ଗ ସଂଖ୍ୟା ଅଟନ୍ତି ।
ଏହି ଉକ୍ତିଟି ସର୍ବଦା ସତ୍ୟ ।
ଉଦାହରଣ : ଏକ ଚତୁର୍ଥ ଘାତକୁ x⁴ ଲେଖାଯାଇପାରିବ ।
ଏହାକୁ (x²)² ଭାବରେ ପୁନଃ ଲେଖାଯାଇପାରେ, ଯାହା x²ର ବର୍ଗ ।

(iii) ଗୋଟିଏ ସଂଖ୍ୟାର ପଞ୍ଚମ ଘାତାଙ୍କ ସେହି ସଂଖ୍ୟାର ଘନଦ୍ଵାରା ବିଭାଜ୍ୟ।
ମନେକର ସଂଖ୍ୟାଟି = n
ଏହାର ପଞ୍ଚମ ଘାତ = n⁵ ଏବଂ ଘନ = n³
ଯେହେତୁ n⁵ = n³ × n²
∴ n⁵ ଯେକୌଣସି ପୂର୍ବସଂଖ୍ୟା ପାଇଁ n³ ଦ୍ଵାରା ବିଭାଜ୍ୟ।
ତେଣୁ ଏହି ଉକ୍ତିଟି ସର୍ବଦା ସତ୍ୟ।

(iv) ଦୁଇଟି ଘନସଂଖ୍ୟାର ଗୁଣଫଳ ଏକ ଘନସଂଖ୍ୟା ଅଟେ ।
ମନେକର ଘନସଂଖ୍ୟାଦ୍ଵୟ a³ ଓ b³
ସେମାନଙ୍କର ଗୁଣଫଳ = a³ × b³ = (a × b)³
ଯେହେତୁ ଗୁଣଫଳ ଅନ୍ୟ ସଂଖ୍ୟା (a × b)ର ଘନ ଭାବରେ ପ୍ରକାଶ କରାଯାଇପାରେ,
ଏହା ସର୍ବଦା ଗୋଟିଏ ଘନସଂଖ୍ୟା ହେବ।
ତେଣୁ ଉକ୍ତିଟି ସର୍ବଦା ସତ୍ୟ ଅଟେ।

(v) q⁴⁶ ସଂଖ୍ୟାଟି ଉଭୟ ଚତୁର୍ଥ ଘାତାଙ୍କ ଓ ଷଷ୍ଠ ଘାତାଙ୍କ ଅଟେ । (q ଏକ ମୌଳିକ ସଂଖ୍ୟା ଅଟେ) ।
q⁴⁶ କୁ ଚତୁର୍ଥ ଘାତରେ ପରିଣତ କରିବା ପାଇଁ 46 ସଂଖ୍ୟାଟି 4 ଦ୍ଵାରା ବିଭାଜ୍ୟ ହେବ, ଯାହା ସମ୍ଭବ ନୁହେଁ ।
q⁴⁶ କୁ ଷଷ୍ଠ ଘାତରେ ପରିଣତ କରିବା ପାଇଁ 46 ସଂଖ୍ୟାଟି 6 ଦ୍ଵାରା ବିଭାଜ୍ୟ ହେବ, ଯାହା ସମ୍ଭବ ନୁହେଁ ।
∴ q⁴⁶ ସଂଖ୍ୟାଟି ଉଭୟ ଚତୁର୍ଥ ଘାତାଙ୍କ ଓ ଷଷ୍ଠ ଘାତାଙ୍କ ହୋଇପାରିବ ନାହିଁ । ତେଣୁ ଉକ୍ତିଟି ସତ୍ୟ ନୁହେଁ ।

Question 5.
ନିମ୍ନ ରାଶିଗୁଡ଼ିକୁ ସରଳ କର ଏବଂ ଘାତାଙ୍କୀୟ ରୂପରେ ଲେଖ :
(i) 10^-2 × 10^-5
(ii) 5⁷ ÷ 5⁴
(iii) 9^-7 ÷ 9⁴
(iv) (13^-2)^-3
(v) m⁵n¹²(mn)⁹
Solution:
(i) 10^-2 × 10^-5 = 10^-2-5 = 10^-7 [∵ a^m × aⁿ = a^m+n]
(ii) 5⁷ ÷ 5⁴ = 5^7-4 = 5³ [∵ a^m ÷ aⁿ = a^m-n]
(iii) 9^-7 ÷ 9⁴ = 9^-7-4 = 9^-11 [∵ a^m ÷ aⁿ = a^m-n]
(iv) (13^-2)^-3 = (13)^-2×(-3) = (13)⁶ [∵ (a^m)ⁿ = a^mn]
(v) m⁵n¹²(mn)⁹ = m⁵n¹²m⁹n⁹ = m⁵⁺⁹ . n¹²⁺⁹ = m¹⁴n²¹ [∵ (a^m)ⁿ = a^mn, a^m × aⁿ = a^m+n]

Question 6.
ଯଦି 12² = 144, ତେବେ ନିମ୍ନଲିଖୂତ ପରିପ୍ରକାଶଗୁଡ଼ିକର ମାନ କେତେ ହେବ?
(i) (1.2)²
(ii) (0.12)²
(iii) (0.012)²
(iv) 120²
Solution:
(i) (1.2)² = \(\left(\frac{12}{10}\right)^2=\frac{144}{100}\) = 1.44
(ii) (0.12)² = \(\left(\frac{12}{100}\right)^2=\frac{144}{10000}\) = 0.0144
(iii) (0.012)² = \(\left(\frac{12}{1000}\right)^2=\frac{144}{10,00,000}\) = 0.000144
(iv) 120² = (12 × 10)² = 144 × 100 = 14400

Question 7.
ସମାନ ମୂଲ୍ୟବିଶିଷ୍ଟ ସଂଖ୍ୟାଗୁଡ଼ିକ ଗୋଲ ବୁଲାଅ-
2⁴ × 3⁶, 6⁴ × 3², 6¹⁰, 18² × 6², 6²⁴
Solution:

Question 8.
ନିମ୍ନଲିଖୁତ ପ୍ରତ୍ୟେକ ସ୍ଥଳରେ ବୃହତ୍ତର ସଂଖ୍ୟାଟିକୁ ଚିହ୍ନଟ କର ।
(i) 4³ କିମ୍ବା 3⁴
(ii) 2⁸ କିମ୍ବା 8²
(iii) 100² କିମ୍ବା 2¹⁰⁰
Solution:
(i) 4³ = 6⁴, 3⁴ = 81
⇒ 81 > 64
∴ 3⁴ > 4³

(ii) 2⁸ = 256, 8² = 64
⇒ 256 > 64
∴ 2⁸ > 8²

(iii) 100² = 10000, 2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰
1024 ସଂଖ୍ୟାଟି 100 ଠାରୁ ବଡ଼,
ଏହାର ଘାତ 10 ବୃଦ୍ଧି ହେଲେ, 10000 ଠାରୁ ବଡ଼ ହେବ ।
∴ 2¹⁰⁰ > 100²

Question 9.
ଏକ ଡାଏରୀ ଫାର୍ମ ବର୍ଷକୁ 8.5 ବିଲିୟନ ପ୍ୟାକେଟ୍ କ୍ଷୀର ଉତ୍ପାଦନ କରିବାକୁ ଯୋଜନା କରୁଛି । ସେମାନେ ପ୍ରତ୍ୟେକ ପ୍ୟାକେଟ୍‌ ପାଇଁ ଏକ ଅନନ୍ୟ ID (ପରିଚୟ ପତ୍ର) କୋର୍ଡ଼ ଚାହୁଁଛନ୍ତି । ଯଦି ସେମାନେ 0 – 9 ଅଙ୍କଗୁଡ଼ିକୁ ବ୍ୟବହାର କରିବାକୁ ସ୍ଥିର କରନ୍ତି, ତେବେ କୋଡ଼ରେ କେତୋଟି ଅଙ୍କ ରହିବ?
Solution:
ଗୋଟିଏ ବର୍ଷରେ ଉତ୍ପାଦିତ ମୋଟ ପ୍ୟାକେଟ୍ ସଂଖ୍ୟା = 8.5 ବିଲିୟନ୍ ।
ଆବଶ୍ୟକ କୋଡ଼୍ ସଂଖ୍ୟା = 8.5 × 10⁹ କୋଡ଼୍ ।
0 ରୁ 9 ପର୍ଯ୍ୟନ୍ତ ଅଙ୍କଗୁଡ଼ିକ ବ୍ୟବହାର କରି ID କୋଡ଼ପାଇଁ ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାର 10ଟି ବିକଳ୍ପ ଅଛି ।
n ସଂଖ୍ୟା ବିଶିଷ୍ଟ କୋଡ଼୍ ତିଆରି କରିବାକୁ ସମ୍ଭାବ୍ୟ ମୋଟ କୋଡ୍ ସଂଖ୍ୟା = 10ⁿ
∴ 10ⁿ ≥ 8.5 × 10⁹
10⁹ = 1,00,00,00,000
10ⁿ ≥ 8.5 × 10⁹ର ସର୍ବନିମ୍ନ ମୂଲ୍ୟ ସନ୍ତୁଷ୍ଟ କରିବାକୁ nର ମୂଲ୍ୟ = 10
∴ କୋଡ଼ର ଅଙ୍କସଂଖ୍ୟା 10¹⁰

Question 10.
64 ଏକ ବର୍ଗ ସଂଖ୍ୟା (8²) ଏବଂ ଘନସଂଖ୍ୟା (4³) ସହ ସମାନ । ଏହିପରି ଅନ୍ୟ କୌଣସି ସଂଖ୍ୟା ଅଛି କି ଯାହା ଉଭୟ ବର୍ଗ ଓ ଘନ ଅଟେ ? ସାଧାରଣ ଭାବେ ଏହିପରି ସଂଖ୍ଯାଗୁଡ଼ିକୁ ପ୍ରକାଶ କରିବାର କୌଣସି ଉପାୟ ଅଛି କି?
Solution:

ହଁ ଏହିପରି ଅନେକ ସଂଖ୍ୟା ଅଛି ଯାହା ଉଭୟ ବର୍ଗ ଓ ଘନ ଅଟେ।
ସାଧାରଣ ନିୟମ : ଏକ ସଂଖ୍ୟା ଉଭୟ ପୂର୍ବବର୍ଗ ଏବଂ ପୂର୍ବଘନ ସଂଖ୍ୟା ହେବ, ଯଦି ଏହା ଏକ ପୂର୍ଣ୍ଣ ଷଷ୍ଠ ଘାତ ହୁଏ ।
ଅର୍ଥାତ୍ କୌଣସି ପୂର୍ବସଂଖ୍ୟା x ପାଇଁ ଏହାକୁ x⁶ ଭାବରେ ଲେଖାଯାଇପାରିବ।
ତେବେ ଏକ ସଂଖ୍ୟା ଉଭୟ ପୂର୍ବବର୍ଗ ଓ ପୂର୍ଣ୍ଣ ଘନ ହେବ।

Question 11.
ଏକ ଡିଜିଟାଲ ଲକର୍‌ରେ 5 ଅକ୍ଷର ବିଶିଷ୍ଟ ଆଲଫାନ୍ୟୁମେରିକ୍ (ଏଥିରେ ଉଭୟ ଅଙ୍କ ଏବଂ ଅକ୍ଷର ରହିପାରିବ) ପାସ୍କୋର୍ଡ଼ ଅଛି, କୋର୍ଡ଼ଗୁଡ଼ିକର କିଛି ଉଦାହରଣ ହେଉଛି G89PO, 38098, BRJKW ଏବଂ 003AZ । ଏହିପରି କେତୋଟି କୋର୍ଡ଼ ସମ୍ଭବ?
Solution:
ଅକ୍ଷର ସଂଖ୍ୟା (A – Z) = 26
ଅଙ୍କ ସଂଖ୍ୟା (0 – 9) = 10
ତେଣୁ ପ୍ରତ୍ୟେକ ଅକ୍ଷର 36ଟି ଆଲ୍‌ଫାନ୍ୟୁମେରିକ୍ ଅକ୍ଷର ମଧ୍ୟରୁ ଯେକୌଣସି ହୋଇପାରେ ।
ଏହା ସହିତ ପ୍ରତ୍ୟେକ 5ଟି ସ୍ଥାନ ପାଇଁ 36ଟି ବିକଳ୍ପ ଅଛି ।
ମୋଟ୍ କୋଡ଼୍ = 36⁵ = 6,04,66,176
ତେଣୁ, 5 ଅକ୍ଷର ବିଶିଷ୍ଟ 6,04,66,176ଟି ଆଲଫାନ୍ୟୁମେରିକ୍ ପାସ୍‌ର୍ଡ଼ ସମ୍ଭବ ।

Question 12.
ସମଗ୍ର ବିଶ୍ଵରେ ମେଣ୍ଢାମାନଙ୍କ ସଂଖ୍ୟା 2024 ମସିହା ପ୍ରାୟ 10⁹ ଏବଂ ଛେଳିମାନଙ୍କର ସଂଖ୍ୟା ପ୍ରାୟ ସମାନ । ମେଣ୍ଢା ଓ ଛେଳିମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା ନିମ୍ନୋକ୍ତ ବିକଳ୍ପଗୁଡ଼ିକ ମଧ୍ୟରୁ କେଉଁଟି ?
(i) 20⁹
(ii) 10¹¹
(iii) 10¹⁰
(iv) 10¹⁸
(v) 2 × 10⁹
(vi) 10⁹ + 10⁹
Solution:
ମେଣ୍ଢାମାନଙ୍କ ସଂଖ୍ୟା = 10⁹
ଛେଳିମାନଙ୍କ ସଂଖ୍ୟା = 10⁹
ମେଣ୍ଢା ଓ ଛେଳିମାନଙ୍କର ମୋଟ ସଂଖ୍ୟା = 10⁹ + 10⁹ = 2 × 10⁹
∴ (v) ବିକଳ୍ପଟି ଠିକ୍ ।

Question 13.
ହିସାବ କର ଏବଂ ବୈଜ୍ଞାନିକ ସଂକେତରେ ଉତ୍ତର ଲେଖ ।
(i) ଯଦି ବିଶ୍ଵର ପ୍ରତ୍ୟେକ ବ୍ୟକ୍ତିଙ୍କ ପାଖରେ 30 ଖଣ୍ଡ ପୋଷାକ ଥାଏ, ତେବେ ମୋଟ ପୋଷାକ ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର।
(ii) ବିଶ୍ଵରେ ପ୍ରାୟ 100 ନିୟୁତ ମହୁଫେଣା ଅଛି; ପ୍ରତ୍ୟେକ ଫେଣାରେ ପ୍ରାୟ 50,000 ମହୁମାଛି ଥାଆନ୍ତି, ତେବେ ମହୁମାଛି ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର।
(iii) ମାନବ ଶରୀରରେ ପ୍ରାୟ 38 ଟ୍ରିଲିୟନ୍ ବ୍ୟାକ୍ଟେରିଆ କୋଷ ଅଛି, ପୃଥୁବୀରେ ଥିବା ସମସ୍ତ ମଣିଷଙ୍କ ଶରୀରରେ ଥିବା ବ୍ୟାକ୍ଟେରିଆ ସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର।
(iv) ଜୀବନକାଳରେ ଖାଇବାରେ ବିତାଇଥବା ମୋଟ ସମୟକୁ ସେକେଣ୍ଡ ଏକକରେ ନିର୍ଣ୍ଣୟ କର।
Solution:
(i) ଆମେ ଜାଣୁ ଯେ, ବିଶ୍ଵର ଜନସଂଖ୍ୟା = 8.2 ବିଲିୟନ୍ = 8.2 × 10⁹
ପ୍ରତ୍ୟେକ ବ୍ୟକ୍ତିଙ୍କ ପାଖରେ ଥିବା ପୋଷାକ ସଂଖ୍ୟା = 30 ଖଣ୍ଡ ।
∴ ବିଶ୍ଵର ମୋଟ ପୋଷାକ ସଂଖ୍ୟା = 8.2 × 10⁹ × 30
= 246 × 10⁹
= 2.46 × 10¹¹ ଟି ପୋଷାକ।

(ii) ବିଶ୍ଵର ମହୁଫେଣା ସଂଖ୍ୟା = 100 ନିୟୁତ = 1 × 10⁸
ପ୍ରତ୍ୟେକ ଫେଣାରେ ମୁହମାଛି ଥାଆନ୍ତି = 5,000 = 5 × 10⁴
ମୋଟ ମହୁମାଛି ସଂଖ୍ୟା = 1 × 10⁸ × 5 × 10⁴ = 5 × 10¹² ମହୁମାଛି ।

(iii) ମାନବ ଶରୀରରେ ବ୍ୟାକ୍ଟେରିଆ କୋଷ ସଂଖ୍ୟା = 38 ଟ୍ରିଲିୟନ୍ = 3.8 × 10¹³
ପୃଥ‌ିବୀର ଜନସଂଖ୍ୟା = 8.2 ବିଲିୟନ୍ = 8.2 × 10⁹
ପୃଥ‌ିବୀରେ ଥ‌ିବା ସମସ୍ତ ମଣିଷଙ୍କ ଶରୀରରେ ଥିବା ବ୍ୟାକ୍ଟେରିଆ ସଂଖ୍ୟା = 3.8 × 10¹³ × 8.2 × 10⁹
= 31.16 × 10¹³⁺⁹
= 31.16 × 10²²

(iv) ମନେକର ଦୈନିକ ଖାଇବାରେ ବିତାଉଥବା ହାରାହାରି ସମୟକୁ 1.5 ଘଣ୍ଟା ଧରାଯାଉ ।
ସେକେଣ୍ଡରେ ପ୍ରକାଶ କଲେ = 1.5 × 60 × 60 = 5400 ସେକେଣ୍ଡ = 5.4 × 10³ ସେକେଣ୍ଡ ।
ଜଣେ ବ୍ୟକ୍ତିର ଜୀବନକାଳ ସମୟ ଆନୁମାନିକ = 70 ବର୍ଷ ।
ଜୀବନକାଳ ସମୟକୁ ସେକେଣ୍ଡରେ ପ୍ରକାଶ କଲେ = 70 × 365 × 24 × 60 × 60
= 161,148,960,000
= 1.61 × 10¹¹
ଖାଇବାରେ ବିତାଇଥବା ମୋଟ ସମୟ = 5.4 × 10³ × 1.61 × 10¹¹
= 8.694 × 10³⁺¹¹
= 8.7 × 10¹⁴ ସେକେଣ୍ଡ।

Question 14.
1 ଅରବ / 1 ବିଲିୟନ ସେକେଣ୍ଡ ପୂର୍ବରୁ ତାରିଖ କେତେ ଥିଲା?
Solution:
1 ଅରବ = 10⁹
ମିନିଟ୍ = \(\frac{1,000,000,000}{60}\)
ଘଣ୍ଟା = \(\frac{1,000,000,000}{60 \times 60}\)
ଦିନ = \(\frac{1,000,000,000}{60 \times 60 \times 24}\)
ବର୍ଷ = \(\frac{1,000,000,000}{365 \times 60 \times 60 \times 24}\)
ଆମେ କ୍ୟାଲେଣ୍ଡର ଦେଖୁଲେ ଏହା ପ୍ରାୟ 31 ବର୍ଷ, 8 ମାସ ଓ 15 ଦିନ ହେବ ।
ମନେକର ଆଜିର ଦିନ 1 ଜାନୁୟାରୀ 2026
ଆଜିର ଦିନର 31 ବର୍ଷ, 8 ମାସ ଓ 15 ଦିନ ପୂର୍ବେ ଥୁଲା = 24 ଏପ୍ରିଲ୍ 1994

Class 8 Maths Chapter 2 MCQ Odia Medium

ସମ୍ଭାବ୍ୟ ଚାରୋଟି ଉତ୍ତର ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛି ଲେଖ ।

Question 1.
2⁴ ର ମୂଲ୍ୟ କେତେ?
(a) 8
(b) 4
(c) 16
(d) 32
Answer:
(c) 16

Question 2.
10^-10 ର ମୂଲ୍ୟ କେତେ?
(a) \(\frac{1}{(10)^{10}}\)
(b) \(\frac{1}{(100)^{10}}\)
(c) (100)¹⁰
(d) 1000
Answer:
(a) \(\frac{1}{(10)^{10}}\)

Question 3.
5² × 5^-2 ର ମୂଲ୍ୟ କେତେ ?
(a) 1
(b) 5
(c) 5⁴
(d) 5^-4
Answer:
(a) 1

Question 4.
(-9)³ × (-9)⁸ ର ମୂଲ୍ୟ କେତେ?
(a) (-9)⁵
(b) (-9)^-5
(c) (-9)¹¹
(d) (-9)^-11
Answer:
(c) (-9)¹¹

Question 5.
625 କୁ ଘାତ ରାଶିରେ ପ୍ରକାଶ କଲେ କେତେ ହେବ ?
(a) 5³
(b) 5²
(c) 5⁴
(d) 5⁵
Answer:
(c) 5⁴

Question 6.
2.08 × 10^-8 ର ସାଧାରଣ ରୂପ କେତେ?
(a) 0.0000208
(b) 0.000028
(c) 0.000208
(d) 0.0002080
Answer:
(a) 0.0000208

Question 7.
ଯଦି 3 × 10² ଟି ଗଛ ଅଛି ଓ ପ୍ରତ୍ୟେକ ଗଛରେ 10⁴ ଟି ପତ୍ର ଅଛି, ତେବେ ମୋଟ ପତ୍ର ସଂଖ୍ୟା କେତେ ?
(a) 3 × 10¹⁶
(b) 3 × 10⁸
(c) 30 × 10¹²
(d) 10¹⁸
Answer:
(a) 3 × 10¹⁶

Question 8.
5⁰ ର ମୂଲ୍ୟ କେତେ?
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
(a) 1

Question 9.
1 କୋଟିକୁ 10 ର ଘାତରେ ପ୍ରକାଶ କଲେ କେତେ ହେବ?
(a) 10⁵
(b) 10⁶
(c) 10⁷
(d) 10⁸
Answer:
(c) 10⁷

Question 10.
(2^-1 × 3^-1)^-1 ର ମୂଲ୍ୟ କେତେ?
(a) 4
(b) 5
(c) 6
(d) 1
Answer:
(c) 6

Question 11.
3⁴ × 3³ ÷ 3⁵ = _________
(a) -9
(b) \(\frac {1}{9}\)
(c) 3⁴
(d) 9
Answer:
(d) 9

Question 12.
3⁹ × 3⁵ ÷ 9⁷ = ?
(a) 2
(b) 3
(c) 5
(d) 1
Answer:
(d) 1

Question 13.
1000 ଯେଉଁ ଆଧାରର ତୃତୀୟ ଘାତ, ସେହି ଆଧାରର nତମ ଘାତ କେତେ ?
(a) n
(b) 10ⁿ
(c) 10n
(d) n¹⁰
Answer:
(b) 10ⁿ

Question 14.
(-2)ⁿ = -512 ଏଠାରେ n ଏକ-
(a) ଗଣନ ସଂଖ୍ୟା
(b) ପୂର୍ଣ୍ଣସଂଖ୍ୟା
(c) ପରିମେୟ ସଂଖ୍ୟା
(d) ବାସ୍ତବ ସଂଖ୍ୟା
Answer:
(a) ଗଣନ ସଂଖ୍ୟା

Question 15.
8³ × 4² ÷ 16² = ___________
(a) 2
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 16.
256 ଯେଉଁ ଆଧାରର ଚତୁର୍ଥ ଘାତ ତା’ର ଘନ କେତେ ?
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(d) 64

Question 17.
625କୁ ଘାତ ରାଶି ରେ ପରି ଣତ କଲେ _________ ହେବ ।
(a) 25²
(b) 5³
(c) 5⁴
(d) କେଉଁଟି ନୁହେଁ
Answer:
(c) 5⁴

Question 18.
(-8)³ ର ମାନ _________
(a) 512
(b) -512
(c) 64
(d) -64
Answer:
(b) -512

Question 19.
10 ର ଚତୁର୍ଥ ଘାତ = _________
(a) 10
(b) 1000
(c) 10000
(d) କେଉଁଟି ନୁହେଁ
Answer:
(c) 10000

Question 20.
5ର _________ ଘାତ 625
(a) 2
(b) 4
(c) 3
(d) 5
Answer:
(b) 4

Question 21.
5 ଆଧାରର ଚତୁର୍ଥ ଘାତ _________ ଆଧାରର ଦ୍ବିତୀୟ ଘାତ ସହ ସମାନ ।
(a) 5
(b) 15
(c) 25
(d) 20
Answer:
(c) 25

Question 22.
256 ଯେଉଁ ଆଧାରର ଚତୁର୍ଥ ଘାତ, ତାହାର 2ୟ ଘାତ _________
(a) 4
(b) 8
(c) 32
(d) 16
Answer:
(d) 16

Question 23.
(4² × 4³) ÷ 4⁵ କୁ ସରଳ କଲେ _________ ହେବ ।
(a) 4⁰
(b) 1
(c) 4¹
(d) କେଉଁଟି ନୁହେଁ
Answer:
(b) 1

Question 24.
(64)³ କୁ ମୌଳିକ ଆଧାର ବିଶିଷ୍ଟ ଘାତରାଶିରେ ପ୍ରକାଶ କଲେ _________ ହେବ ।
(a) 4³
(b) 4⁹
(c) 2⁶
(d) 2¹⁸
Answer:
(d) 2¹⁸

Question 25.
(125)^m-1 = _________
(a) 5^3m-3
(b) 5^3m-1
(c) 5^m-3
(d) 5^m-1
Answer:
(a) 5^3m-3

Question 26.
କେଉଁ କ୍ଷେତ୍ରରେ n ଏକ ଗଣନ ସଂଖ୍ୟା ହେବ?
(a) 5ⁿ = 100
(b) 4ⁿ = 1024
(c) 5ⁿ = 1250
(d) \(\left(\frac{2}{3}\right)^n=\frac{32}{15}\)
Answer:
(b) 4ⁿ = 1024

ସଂକ୍ଷେପରେ ଉତ୍ତର ଲେଖ।

Question 1.
ନିମ୍ନ ଘାତରାଶିମାନଙ୍କର ଆଧାର ଓ ଘାତାଙ୍କ ଦର୍ଶାଇ ମାନ ନିର୍ଣ୍ଣୟ କର ।
(i) (1)¹⁵
(ii) (-1)¹¹
(iii) (-1)¹⁸
(iv) (9)⁵
(v) (-2)⁵
(vi) \(\left(\frac{1}{2}\right)^6\)
(vii) \(\left(\frac{2}{3}\right)^5\)
(viii) (5 × 2)⁴
(ix) (10)⁷
(x) (-10)⁵
Answer:
(i) (1)¹⁵ ରେ ଆଧାର 1 ଓ ଘାତାଙ୍କ 15 ଅଟେ ।
(1)¹⁵ ର ମାନ = 1 × 1 × 1 × 1 × ….× 1 (15 ଥର) = 1

(ii) (-1)¹¹ ରେ ଆଧାର -1 ଓ ଘାତାଙ୍କ 11
(-1)¹¹ ର ମାନ = -1 [∵ (−1)^m = -1, ଯେଉଁଠାରେ m ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା]

(iii) (-1)¹⁸ ରେ ଆଧାର -1 ଓ ଘାତାଙ୍କ 18
(-1)¹⁸ ର ମାନ = 1 [∵ (-1)^m = 1, ଯେଉଁଠାରେ m ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା)

(iv) (9)⁵ ରେ ଆଧାର 9 ଓ ଘାତାଙ୍କ 5
(9)⁵ = 9 × 9 × 9 × 9 × 9 = 59049
∴ (9)⁵ ର ମାନ = 59049

(v) (-2)⁵ ରେ ଆଧାର -2 ଓ ଘାତାଙ୍କ 5।
(-2)⁵ = (-2) × (-2) × (-2) × (-2) × (-2) = -32
∴ (-2)⁵ ର ମାନ = -32

(vi) \(\left(\frac{1}{2}\right)^6\) ରେ ଆଧାର \(\frac {1}{2}\) ଓ ଘାତାଙ୍କ 6।
\(\left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{64}\)
∴ \(\left(\frac{1}{2}\right)^6\) ର ମାନ = \(\frac {1}{64}\)

(vii) \(\left(\frac{2}{3}\right)^5\) ରେ ଆଧାର \(\frac {2}{3}\) ଓ ଘାତାଙ୍କ 5।
\(\left(\frac{2}{3}\right)^5=\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{32}{243}\)
∴ \(\left(\frac{2}{3}\right)^5\) ର ମାନ = \(\frac {32}{243}\)

(viii) (5 × 2)⁴ = (10)⁴
= 10 × 10 × 10 × 10
= 10000
(5 × 2)⁴ ରେ ଆଧାର 10 ଓ ଘାତାଙ୍କ 4
∴ ମାନ = 10000

(ix) (10)⁷ ରେ ଆଧାର 10 ଓ ଘାତାଙ୍କ 7।
(10)⁷ = 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10000000
∴ (10)⁷ ର ମାନ = 10000000

(x) (-10)⁵ ରେ ଆଧାର -10 ଓ ଘାତାଙ୍କ 5
(-10)⁵ = (-10)(-10)(-10)(-10)(-10) = -100000
∴ (-10)⁵ ର ମାନ = -100000

Question 2.
\(\frac{2^3 \times 3^4}{3 \times 2^5}\) କୁ ସରଳ କର।
Answer:
\(\frac{2^3 \times 3^4}{3 \times 2^5}=\left(\frac{2^3}{2^5}\right) \times\left(\frac{3^4}{3}\right)=\frac{1}{2^{5-3}} \times 3^{4-1}\) = \(\frac{1}{2^2} \times 3^3=\frac{27}{4}\)

Question 3.
ନିମ୍ନଲିଖିତ ରାଶିଗୁଡ଼ିକୁ ଏକ ଆଧାର ବିଶିଷ୍ଟ ଘାତ ରାଶି ରୂପେ ପ୍ରକାଶ କର।
(i) (9)³ × (27)⁴
(ii) (8)³ × (-4)⁴
(iii) (7)⁸ × (-7)⁵
Answer:
(i) (9)³ × (27)⁴ = (3²)³ × (3³)⁴
= 3^2×3 × 3^3×4 [∵ (a^m)ⁿ = a^m×n]
= 3⁶ × 3¹²
= 3⁶⁺¹²
= 3¹⁸

(ii) (8)³ × (-4)⁴ = 8³ × (-1 × 4)⁴
= 8³ × (-1)⁴ × 4⁴ [∵ (ab)^m = a^m b^m]
= (-1)⁴ × 8³ × 4⁴
= 1 × (2³)³ × (2²)⁴
= 2⁹ × 2⁸ [∵ (-1)⁴ = 1]
= 2⁹⁺⁸
= 2¹⁷

(iii) (7)⁸ × (-7)⁵
= 7⁸ × (-1 × 7)⁵ [∵ (ab)^m = a^m . b^m]
= 7⁸ × (-1)⁵ × 7⁵
= (-1)⁵ × 7⁸ × 7⁵
= -1 × 7⁸⁺⁵
= -1 × 7¹³
= (-1)¹³ × 7¹³
= (-1 × 7)¹³
= (-7)¹³

Question 4.
ମୌଳିକ ଆଧାର ବିଶିଷ୍ଟ ଘାତରାଶିରେ ପ୍ରକାଶ କର।
(i) (9)⁷
(ii) (125)^m-1
Answer:
(i) (9)⁷ = (3²)⁷ = 3^2×7 = 3¹⁴
[∵ 9 ର ଏକ ମୌଳିକ ଗୁଣନୀୟକ 3]

(ii) (125)^m-1 = (5³)^m-1 = 5^3(m-1) = 5^3m-3
[∵ 125 ର ଏକ ମୌଳିକ ଗୁଣନୀୟକ 5]

Question 5.
ମାନ ନିର୍ଣ୍ଣୟ କର ।
(i) (3¹¹ × 4⁵) ÷ (4⁴ × 3⁶)
(ii) (4³ × 4² × 4) ÷ (2⁴ × 2³ × 2²)
Answer:
(i) (3¹¹ × 4⁵) ÷ (4⁴ × 3⁶)
= \(\frac{3^{11} \times 4^5}{4^4 \times 3^6}\)
= \(\frac{3^{11}}{3^6} \times \frac{4^5}{4^4}\)
= 3^11-6 × 4^5-4
= 3⁵ × 4
= 243 × 4
= 972

(ii) (4³ × 4⁵ × 4) ÷ (2⁴ x 2³ x 2²)
= 4³⁺²⁺¹ ÷ 2⁴⁺³⁺²
= 4⁶ ÷ 2⁹
= (2²)⁶ ÷ 2⁹
= 2¹² ÷ 2⁹
= 2^12-9
= 2³
= 8

Question 6.
ନିମ୍ନ ଘାତରାଶି ବିଶିଷ୍ଟ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ସରଳ କର।
(i) \(\left(\frac{2}{9}\right)^5 \div\left(-\frac{2}{9}\right)^4\)
(ii) \(\left(\frac{1}{25}\right)^4 \div 5^4\)
(iii) \(\frac{3^8 \times a^5}{27 \times a^2}\) (a ≠ 0)
(iv) (4² × 4³) ÷ 4⁵
(v) \(\left(\frac{-2}{3}\right)^9 \div\left(\frac{2}{3}\right)^7\)
(vi) \(\left\{\left(\frac{1}{2}\right)^3\right\}^2 \div\left(\frac{1}{4}\right)^3\)
Answer:

Question 7.
ନିମ୍ନଲିଖୂତ ରାଶିଗୁଡ଼ିକୁ ଏକ ଆଧାର ବିଶିଷ୍ଟ ଘାତ ରାଶି ରୂପେ ପ୍ରକାଶ କର।
(i) \(\frac{7^4}{3^4}\)
(ii) 3⁹ ÷ 4⁹
(iii) \(\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)^7+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^3\)
(iv) \(\left(\frac{a}{b}\right)^4 \div\left(-\frac{b}{a}\right)^3\)
Answer:

Question 8.
ସରଳ କର
(i) (2² × 2)³
(ii) (ab)⁵ × a³ × b²
(iii) \(\left(\frac{a}{b}\right)^7 \times a^6 \times b^5 \times\left(\frac{b}{a}\right)^6\)
(iv) 3⁹ × 3⁵ ÷ 9⁷
(v) \(\left(\frac{2}{3}\right)^5 \div\left(\frac{2}{3}\right)^8 \times\left(\frac{2}{3}\right)^3\)
Answer:

Go through BSE Odisha Class 8 Science Solutions Chapter 11 Keeping Time with the Skies Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 11 Question Answer

Class 8 Science Ch 11 Keeping Time with the Skies Question Answer

Class 8 Science Chapter 11 Keeping Time with the Skies Question Answer

Probe and Ponder Questions

Question 1.
Have you ever seen the Moon during the day? Why do you think it is sometimes visible when the sun is up?
Answer:
Yes, the Moon can often be seen during the day, especially in its waxing and waning phases. This happens because the Moon reflects sunlight and during certain phases it is high enough in the sky while the Sun is also above the horizon, so the bright part of the Moon is visible to us.

Question 2.
Imagine you lived on the Moon instead of Earth. What would you mean by a day, a month or a year?
Answer:

• A day on the Moon (from one sunrise to the next) would be about 29.5 Earth days.
• A month might be defined by one complete orbit around Earth (which is also about 27.3 Earth days).
• A year is the time the Earth takes to complete one orbit around the Sun, which you could still observed from the Moon in about 365 days.

Question 3.
What would happen if Earth had two moons instead of one? How would that change the night sky?
Answer:
If Earth had two moons:

• The night sky would be brighter and more dynantic.
• There could be more frequent eclipses.
• The gravitational pull on Earth would be different, possibly affecting tides.
• The two moons might cross paths, creating fascinating views or even risks of collision over long periods.

Question 4.
If we didn’t have clocks or calendars, how else could we measure time?
Answer:
We could measure time by:

• Observing the position of the Sun (sunrise, noon, sunset).
• Using the phases of the Moon to count months.
• Tracking stars and constellations that change with the seasons.
• Using natural events, like plant flowering or animal behaviour to mark the passage of time.

Question 5.
Share your questions …………
Answer:
Here are some fun questions from the chapter.

• Why does the Moon change shapes?
• How do festivals link to Moon phases?
• Can we see statellites at night?
• Why do some calendars add extra days?

InText Questions

Question 1.
Why does the illuminated portion of the Moon seen from the Earth decrease when it appears closer to the Sun?(Page 174)
Answer:
The Moon does not have its own light; it shines because it reflects sunlight that falls on it. When the Moon appears closer to the Sun in the sky, the side of the Moon that is illuminated by the Sun faces away from the Earth. As a result, from Earth we can see only a smaller part of the bright portion and most of the Moon’s surface visible to us is dark. That is why the illuminated portion of the Moon seen from Earth decreases when it comes closer to the Sun – it appears as a cresent Moon and finally becomes invisible on the new Moon day.

Question 2.
So, changing phases of the Moon is a natural periodic event, with a cycle of almost a month, which can also be used for time keeping. Yes, along with the natural periodic events of day and night and the changing seasons about which we learnt earlier. But how are these periodic events used for keeping time? (Page 178)
Answer:
The natural periodic events – like day and night, phases of the Moon and changing seasons – helps measure time because they repeat regularly after fixed intervals.

• Day and Night: The rotation of the Earth on its axis causes day and night. One complete rotation takes 24 hours which is used to define one day.
• Phases of the Moon: The Moon revolves around the Earth and completes one full cycle of phases (from one full Moon to the next) in about 29.5 days, which forms the basis of a month.
• Changing Seasons: The Earth revolves around the Sun once in about 365 days. This regular change in the position of the Earth causes seasons and gives us a year.
• Thus, these periodic motions of celestial bodies – the Earth and the Moon acts as natural clocks and calendars that humans have used for keeping time since ancient times.

Question 3.
Why do most Indian festivals fall on different dates every year? (Page 183)
Answer:
Many Indian festivals are linked to the phases of the Moon and hence are based on either lunar or luni-solar calendars, not the regular Gregorian solar calendar that we use daily. A lunar month is about 29.5 days and a lunar year (12 lunar months) has about 354 days, which is 11 days shorter than the solar year of 365 days.

Because of this difference, festivals like Diwali, Holi, Eid-ul-Fitr and Buddha Purnima which follow the Moon’s phases appear on different Gregorian calendar dates each year. To adjust this difference, the luni-solar calendars sometimes add an extra month called Adhika Maasa (extra month) every few years which helps keep the festivals in the same season.

Question 4.
When I look at the night sky in early evening, I see some moving stars. What are they? Is their motion also periodic? (Page 185)
Answer:
The moving ‘stars’ seen in the night sky are actually artificial satellites, not real stars. These are man-made objects launched into space that revolve around the Earth. They reflect sunlight which is why they appear as small moving points of light in the sky. Yes, their motion is periodic. Each satellite orbits the Earth in a fixed path and takes about 100 minutes to complete one round. Artificial satellites are used for communication, weather forecasting, navigation, disaster management and scientific research.

Keeping Time with the Skies Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 187-188)

Question 1.
State whether the following statements are True or False.
(i) We can only see that part of the Moon which reflects sunlight towards us.
(ii) The shadow of Earth blocks sunlight from reaching the Moon, causing phases.
(iii) Calendars are based on various astronomical cycles which repeat predictably.
(iv) The Moon can only be seen at night.
Answer:
(i) True: We can only see the part of the Moon that reflects sunlight towards Earth.
(ii) False: The Earth’s shadow causes lunar eclipses, not the regular phases of the Moon.
(iii) True: Calendars are based on repeating astronomical events like day-night, Moon phases, and seasons.
(iv) False: The Moon can also be seen during the daytime, depending on its phase and position in the sky.

Question 2.
Amol was born on the 6th of May on a full Moon day. Does his birthday fall on the full Moon day every year? Explain your answer.
Answer:
No, Amol’s birthday does not fall on a full Moon Day every year. This is because the Moon’s phases follow a lunar cycle of about 29.5 days, while the calendar year follows the solar cycle of about 365 days. So, the date of the full Moon changes each year in the Gregorian calendar.

Question 3.
Name two things that are incorrect in the figure.

Answer:
Two incorrect things are :

• Stars are shown near the Moon during the daytime, which is incorrect because stars are not visible in the daytime sky.
• The Moon’s dark part is shaded incorrectly to show a phase. The shadow in the figure suggests it’s caused by Earth’s shadow, which is not true for regular Moon phases; they are caused by the Moon’s position relative to the Earth and Sun, not a shadow.

Question 4.
Look at the pictures of the Moon in the figure, and answer the following questions.

(i) Write the correct panel number corresponding to the phases of the Moon shown in the pictures above.

(ii) List the picture labels of the phases of the Moon that are never seen from Earth. Hint: You can use your observations from Activity 11.1 or Figure as reference.
Answer:
(i)

Picture label	Phase of Moon
C	Three days after New Moon
E	Full Moon
F	Three days after the Full Moon
A	A week after the Full Moon
B	Day of New Moon

(ii) Picture B (New Moon phase) is never seen from Earth because the illuminated side of the Moon is facing away from us.

Question 5.
Malini saw the Moon overhead in the sky at sunset.
(i) Draw the phase of the Moon that Malini saw.
(ii) Is the Moon in the waxing or the waning phase?
Answer:
(i) At sunset, the Moon is overhead only during the first quarter (a week after New Moon), when the right half is illuminated. So, we need to draw a half Moon (right half bright, left half dark).
(ii) Waxing phase (because it occurs after New Moon and the bright part is increasing).

Question 6.
Ravi said, “I saw a crescent Moon, and it was rising in the East when the Sun was setting.” Kaushalya said, “Once I saw the gibbous Moon during the afternoon in the East.” Who out of the two is telling the truth?
Answer:
Kaushalya is telling the truth because gibbous Moons can be seen in the East during the afternoon. Ravi’s statement is incorrect because Crescent Moons do not rise in the East at sunset. A crescent Moon appears just after the New Moon (waxing crescent) or just before the New Moon (waning crescent). Waxing crescent is visible after sunset in the western sky, not rising in the east, and waning crescent rises just before sunrise, not at sunset.

Question 7.
Scientific studies show that the Moon is getting farther away from the Earth and slower in its revolution. Will luni-solar calendars need an intercalary month more often or less often?
Answer:
Luni-solar calendars will need an intercalary month more often as the Moon moves farther and slower, and it takes longer to complete a cycle. So, a lunar year becomes even shorter compared to the solar year.

Question 8.
A total of 37 full Moons happen during 3 years in a solar calendar. Show that at least two of the 37 full moons must happen during the same month of the solar calendar.
Answer:
Yes, at least two full Moons must happen in the same solar month.

• A solar calendar has 12 months × 3 years =36 months.
• 37 full Moons in 36 months, at least one month must have 2 full Moons.

Question 9.
On a particular night, Vaishali saw the Moon in the sky from sunset to sunrise. What phase of the Moon would she have noticed?
Answer:
As the Moon is visible all night long only on a Full Moon, it is a Full Moon.

Question 10.
If we stopped having leap years, in approximately how many years would the Indian Independence Day happen in winter?
Answer:
One leap year adds ∼1 day every 4 years.
Without leap years, the calendar shifts by 1 day every 4 years.
There are roughly 183 days between 15 August (monsoon) and winter (mid-February).
183 days × 4=732 years
In approximately 730-732 years, 15 August would occur in winter.

Question 11.
What is the purpose of launching artificial satellites?
Answer:
Artificial satellites are launched for:

• Communication
• Navigation
• Weather monitoring -Scientific research
• Disaster management
• Earth observation

Question 12.
On which periodic phenomenon are the following measures of time based :
(i) day
(ii) month
(iii) year
Answer:
(i) Day → Earth’s rotation
(ii) Month → Moon’s revolution (phases of the Moon)
(iii) Year → Earth’s revolution around the Sun.

Class 8 Science Chapter 11 Question Answer

Activity 1

Activity 1.

Let us explore
Aim: To observe the shape of the moon from full moon day to new moon day.

Procedure:
1. Observe the Moon at sunrise in the western direction starting from the first day after the full Moon.
2. Construct a table same as Table -1 in your notebook. Complete the following details:

• Date
• When you saw the Moon (at sunrise or sunset)?
• Shade the corresponding Circle with pencil to show the bright portion of the Moon as shown in figure.

Table 1: Documenting changes in the Moon’s appearance

Answer:

Observations:

1. From the second day onwards also document the following:

• Is the size of bright portion of the Moon increasing or decreasing from the previous day.
• Yes, there is a change in shape of the moon everyday. Bright portion goes on decreasing.
• Whether the Moon appears closer to or farther from the Sun in the sky than the day before.

2. After about 15 days, you may not be able to see the Moon at sunrise or sunset. For the next 15 days, carry out this activity at sunset.

Inferences:

• Moon appear different each day.
• No, after 15 days, you may not be able to see the moon at sunrise or sunset.
• After the full moon, the bright face of the moon goes on decreasing every night. By another fifteen days again new moon is formed.
• The crescent moon goes on increasing everyday, till on the fifteenth day (from the new moon), the full face of the moon is visible.
• Each morning at sunrise, moon appears closer to the sun’s position in the sky.

Activity 2.

Let us explore
Aim: To show the Moon’s phases with a simple model.
Materials Required: A soft ball, a stick, a torch or lamp as the sun, your own head as the Earth.

Procedure:

• Take a small soft ball and insert a stick into it [(Figure (a)]. This represents the Moon.
• Move to a dark open place (at night), and ask your teacher or guardian to shine a torchlight towards you from about 3 m which will represent that light is coming from the Sun or stand near an electric lamp. Your head represents the Earth.
• Hold the ball in your hand, slightly above your head.
• Shine the torch toward the ball to represent sunlight.
• As you turn in a circle, the ball (“Moon”) shows a changing illuminated portion to your eyes.

Observations:

What does the model show

• Full Moon: When the ball is held opposite the lamp (behind you compared to the Sun), the side facing you is fully lit-just like a full Moon.
• New Moon: When the ball is held between your head and the lamp (towards the Sun), you see only the dark side-like a new Moon.
• Crescent and Gibbous Phases: Turning the ball slowly, the visible portion transitions. Sometimes you see a crescent (less than half lit), other times gibbous (more than half lit).
• The line between the bright and dark parts is always curved-this matches what we see in the real Moon.

Inference:
The Science behind the phasses

• Half illuminated, Half Dark: At every moment, half the Moon is lit by sunlight and half is in darkness.
• Moon’s Revolution: As the Moon revolves around the Earth, the angle between Earth, Moon and Sun changes. The part of the Moon we see as bright changes accordingly.

Activity 3.

Let us measure a day!

Aim: To find the duration of a day by observing the length of the shadow.
Materials Required: A 1m stick.

Procedure:

• Choose a small flat and open area in a ground which receives sunlight during the day. Fix a 1 m stick vertically in it as shown in figure.
• Let us start observing the shadow at 11:00 a.m. Every minute, mark a dot on the ground at the tip of the stick’s shadow. Keep marking dots until around 1:10 p.m.
• Identify when the shadow was shortest and find out its time by counting the number of dots. Record this time in Table. Repeat this experiment for the next few days.
• Now calculate the duration of the solar day. This can be done by finding a difference in time on two consecutive days as shown in Table.

Observations:
Table1: Finding the duration of a solar day

Answer:

Date	Time of shortest shadow (hh:mm)	Duration of day (hh:mm)
22 March 2025 23 March 2025 24 March 2025 25 March 2025 26 March 2025	12:20 12:20 12:19 12:19 12:18	…………… 24:00 23:59 23:59 24:00

1. The shortest shadow during the day marks when the sun is at highest point in the sky (noon).
2. Measuring from one day’s noon to the next day gives the length of a day.
3. The average solar day i.e. average duration of the day is about 24 hours.

Inference:
The Month and the Moon

• The phases of the Moon create another natural cycleone complete phase cycle (from full Moon to next full Moon) takes about 29.5 days (approximately one month).
• This lunar cycle is the basis for measuring a month. (see figure)

The Year and the Seasons

• One year is the duration for Earth to make a full revolution around the Sun, which is about 365 \(\frac{1}{4}\) days.
• The repetition of seasons (spring, summer, autumn, winter) marks the annual cycle.
• The Earth undergoes one cycle of seasons during this time, which can be used to define a solar year (see figure).

Activity 4.

Let us identify
Aim: To observe artificial satellites.
Materials Required: A telescope.

Procedure:

• Go to a location that has a clear view of sky; along with an adult. There should not be any obstruction of trees or tall buildings.
• To identify satellites in the sky, look for a small, bright, continuously moving dot in the sky, typically before sunrise or after sunset.

Observations:

• Can be seen without a telescope or with binoculars.
• Satellite-tracking mobile apps or websites help identify visible satellites in your location and when they will be passing above you in the sky.

Keeping Time with the Skies Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
How does the Moon’s position in the sky change each day?
Answer:
The Moon’s position shifts slightly eastward each day, so it is not in the same place at the same time. This is why its rise and set times also change daily.

Question 2.
Why does the Moon sometimes appear in the daylight?
Answer:
The Moon can rise before sunset, sometimes in the afternoon. In such cases, it is visible in the sky while the Sun is still up.

Question 3.
What does the simple ball-and-lamp model of the Moon help us understand?
Answer:
The model shows how sunlight falls on the Moon and creates different phases. By turning with the ball, we can see changes in the illuminated portion, similar to what happens in reality.

Question 4.
Why is the line between the bright and dark portions of the Moon always curved ?
Answer:
The Moon is spherical, so the dividing line between sunlight and shadow is curved. This curve is visible from Earth during all phases.

Question 5.
Why are Moon phases not caused by Earth’s shadow ?
Answer:
Moon phases happen because we see varying parts of its sunlit side as it orbits. Earth’s shadow only causes a lunar eclipse, which is rare.

Question 6.
Why don’t we have eclipses every full Moon or new Moon?
Answer:
The Moon’s orbit is tilted compared to Earth’s orbit. This tilt means the Sun, Earth and Moon usually don’t line up perfectly.

Long Answer Type Questions

Question 1.
Explain the waxing and waning periods of the Moon and how they form a monthly cycle.
Answer:

• The waxing period is when the bright portion of the Moon increases, starting from the new Moon and becoming full in about two weeks. The waning period is when the bright portion decreases, starting from the full Moon and becoming new Moon in about two weeks.
• Together, waxing and waning make a repeating cycle every month. This cycle takes about 29.5 days from one full Moon to the next.
• These changes occur because of the Moon’s revolution around Earth and the changing angle between the Sun, Earth and Moon.

Question 2.
Explain how the motion of the Sun in the sky helps in measuring a day.
Answer:

• The Sun appears to rise in the east, move across the sky and set in the west because Earth rotates on its axis. The highest position of the Sun in the sky, when shadows are shortest, marks noon.
• The time from one noon to the next is called a mean solar day. This length is about 24 hours, which is the basic unit for measuring time in days. This daily cycle has been used since ancient times to keep track of time.

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow :

Anita kept a Moon observation diary for a week. On Day 1, she saw a thin crescent on the right side. By Day 4, half of the Moon was visible. By Day 7, almost the full Moon was visible.

(a) Which phase did Anita observe on Day 1?
Answer:
Waxing Crescent

(b) What is the name of the phase on Day 4?
Answer:
First Quarter

(c) Was the Moon waxing or waning during these days?
Answer:
Waxing

Picture Based Questions

I. Look at the pictures and answer the following questions:

(a) Identify the pictures (i) and (ii).
Answer:
(i) Warli painting
(ii) Dhokra Brass Sculpture

(b) Write the name of painting given below and the state from given where it is associated ?
Answer:

Madhubani painting. It is a famous painting associated to Bihar.

Keeping Time with the Skies Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
Phases of the moon occur because:
(a) We can see only that part of the Moon which reflects light towards us.
(b) Our distance from the Moon keeps changing.
(c) The shadow of the earth covers only a part of the Moon’s surface.
(d) The thickness of the Moon’s atmosphere is not constant.
Answer:
(a) We can see only that part of the Moon which reflects light towards us.

Question 2.
What do they call the Moon when it’s more than half lit but not full ?
(a) Crescent
(b) Gibbous
(c) Quarter
(d) Eclipse Moon
Answer:
(b) Gibbous

Question 3.
What is the sequence of phases starting from New Moon?
(a) New Moon → Full Moon → Waxing → Crescent → First quarter
(b) New Moon → Waxing Crescent → First Quarter → Full Moon
(c) Full Moon → Waxing Gibbous → New Moon
(d) First quarter → Full Moon → New Moon
Answer:
(b) New Moon → Waxing Crescent → First Quarter → Full Moon

Question 4.
What causes the phases of the Moon ?
(a) The Earth’s rotation
(b) The Moon’s rotation
(c) The Moon’s orbit around the Earth
(d) The Sun’s movement
Answer:
(c) The Moon’s orbit around the Earth

Question 5.
How long does one complete cycle of Moon phases take?
(a) 7 days
(b) 15 days
(c) 29.5 days
(d) 365 days
Answer:
(c) 29.5 days

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.

(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): The Moon appears to change shape throughout the month.
Reason (R): The Earth casts different shadows on the Moon each night.
Answer:
(c) Assertion (A) is correct but the Reason (R) is wrong.

2. Assertion (A): The full Moon rises at sunset and sets at sunrise.
Reason (R): The full Moon is directly opposite the sun in the sky.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

Fill in the blanks

1. The Moon does not have its own light, it………… the sunlight.
Answer:
reflects

2. The changing shapes of the Moon that we see from Earth are called ………….
Answer:
Phases

3. A full cycle of the Moon’s phases takes about ……….. days.
Answer:
29.5

4. Waxing Moon is best seen at ………….
Answer:
Sunset

5. Waning Moon is best seen at ………….
Answer:
sunrise.

True or False

1. A lunar eclipse occurs when the Moon comes between the Earth and the Sun.
Answer:
False

2. The Full Moon appears once every lunar cycle.
Answer:
True

3. A Moon’s surface is smooth and shiny.
Answer:
False

4. We can sometimes see the Moon during the daytime.
Answer:
True

5. The Moon always shows the same side to the Earth.
Answer:
True

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 8 Playing with Constructions Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 8 Playing with Constructions Solutions

Ganita Prakash Class 6 Chapter 8 Solutions

Class 6 Maths Ganita Prakash Chapter 8 Solutions Playing with Constructions

Question 1.
Draw the rectangle and four squares configuration on a dot paper.

What did you do to recreate this figure so that the four squares are placed symmetrically around the rectangle? Discuss with your classmates.
Solution:
Draw a rectangle using four dots and draw four squares to make sure all . the squares are equal in size and placed around the rectangle in symmetry.

Question 2.
Take a central line of a different length and try to draw the wave on it.
Solution:

Step 1: Draw a central line AB = 10 cm.
Step 2: Since half of AB is 5 cm, mark a point X on AB such that AX = XB = 5 cm.
Step 3: Mark a point C on AX such that AC = CX =2.5 cm. Mark another point D on XB such that-.. . XD = DB = 2.5 cm.
Step 4: With C as centre and radius equal to AC, draw a semicircle above the line AB, Again, with D as centre and radius equal to BD, draw a semicircle below the line AB.
Step 5: The resultant figure is the required wavy wave with central line of length 10 cm.

InText Questions

Question 1.
Is it possible to construct a 4-sided figure in which all the angles are equal to 90° but opposite sides are not equal?
Solution:
Step 1: Draw a line segment AB = 7 cm.
Step 2: At A and B draw perpendiculars with the help of a protractor.
Step 3: fake two points C and D on the two perpendiculars such that AD = 4 cm and BC = 3 cm.
Step 4: Join DC.
Since the opposite sides AD and BC are not equal, it is found that neither ∠D nor ∠C is 90°.
Hence, we conclude that it is not possible to draw a four-sided figure with all angles equal to 90°, when opposite sides are not equal.

Question 2.
Construct a rectangle in which one of the diagonals divides the opposite angles into 45° and 45°. What do you observe about the sides?
Solution:
Step 1: Draw a horizontal line segment AB (say 5 cm). This will be one side of the rectangle.
Step 2: At point A, use a protractor to measure and draw AX at 45° angle.
Step 3: At point B, measure and draw a 90° angle. Draw a line segment BC extending from B at this angle meeting AX at C.
Step 4: At points A and C, draw a 90° angle which meets at the point D.

Thus, ABCD is the required rectangle.
We observe that all sides are equal. Hence, ABCD is a square.

Playing with Constructions Class 6 Extra Questions

Playing with Constructions Class 6 Very Short Question Answer

Question 1.
Construct a circle with the radius 4.5 cm.
Solution:
Steps of construction of a circle with radius 4.5 cm are as follows:
Step 1: Using a ruler, open the compass to a radius of 4.5 cm.
Step 2: Mark a point O on the paper. Place the pointed end of the compass on point O.
Step 3: Rotate the compass around point 0 to draw a circle ensuring that the pencil remains in contact with the paper at all times.

Question 2.
Construct a square with the side 4 cm.
Solution:

Steps of construction of a square with side 4 cm are as follows:
Step 1: Draw a line segment AB = 4 cm using a ruler.
Step 2: At points A and B, construct two perpendiculars to AB using a protractor.
Step 3: From points A and B, mark points D and C on the perpendiculars respectively such that AD = BC = 4 cm.
Step 4: Join CD to complete the square ABCD.

Question 3.
Construct a rectangle with the side lengths 4 cm and 7 cm.
Solution:

Steps of construction of a rectangle with side lengths 4 cm and 7 cm are as follows:
Step 1: Draw a line segment AB = 7 cm using a ruler.
Step 2: At points A and B, draw perpendiculars to AB using a protractor.
Step 3: From points A and B, mark points D and C on the perpendiculars respectively such that AD = BC = 4 cm.
Step 4: Join CD to complete the rectangle ABCD.

Playing with Constructions Class 6 Short Question Answer

Question 1.
Construct a circle with the following radius:
(i) 4 cm
(ii) 5 cm
(iii) 6 cm
Solution:
(i) Step by step construction of a circle with radius 4 cm:

Step 1: Using a ruler, open the compass to a radius of 4 cm.
Step 2: Mark a point 0 on the paper. Place the pointed end of the compass on point 0.
Step 3: Rotate the compass around point 0 to draw a circle ensuring that the pencil remains in contact with the paper at all times.

(ii) Follow the same steps for radius 5 cm as in (i).
(iii) Follow the same steps for radius 6 cm as in (i).

Question 2.
Recreate the following falling squares:

Solution:
In the figure, we have three identical squares of side length 3 cm.

Step 1: Construct a square ABCD with each side measuring 3 cm.
Step 2: Produce AD to a new point G and CD to a new point E such that DC = DE = 3 cm.
Step 3: Construct square DEFG.
Step 4: Produce EE to a new point J and GF to a A new point H such that FJ = FH = 3 cm.
Step 5: Construct square FHIJ.

Question 3.
Recreate the following falling squares:

Solution:
In the figure, we have three squares of side lengths 5 cm, 4 cm and 3 cm respectively.

Step 1: Construct a square ABCD with each side measuring 5 cm.
Step 2: Produce AD to a new point G and CD to a new point E such that DG = DE = 4 cm.
Step 3: Construct square DEFG.
Step 4: Produce EF to a new point J and GF to a new point H such that FJ = FH = 3 cm.
Step 5: Construct square FHIJ.

Question 4.
Recreate the following combinations of square and hole:

Centre of the hole is the same as the centre of the square containing it.
Solution:
Side length of square is not given. You can take any side length of your choice, say 5 cm.

Step 1: Construct a square ABCD of side 5 cm.
Step 2: Draw diagonals AC and BD. Let O be their point of intersection. Point O is the centre of square ABCD.
Step 3: Using a compass, draw a circle with centre O and radius less than half of the side length of the square ABCD.
The figure so obtained is the required square with a hole.

Question 5.
Construct a rectangle which can be divided into 2 identical squares.
Solution:
A rectangle that can be divided into two identical squares must have one side equal to twice of the other.

Let’s take 3 cm and 6 cm as the side lengths of the rectangle.
Steps of construction:
Step 1: Draw a line segment AB = 6 cm using a ruler, (this will be the longer side of the rectangle).
Step 2: Divide AB into two equal parts. Mark the midpoint M, so that TM = MB = 3 cm.
Step 3: At points A, M and B, draw perpendiculars to AB (using a protractor) long enough to mark the height of the rectangle i.e. 3 cm.
Step 4: Mark points D, N and C on the perpendiculars respectively such that
AD = MN = BC = 3 cm.
Step 5: Join CD to complete the rectangle ABCD.
Here, the rectangle ABCD is divided into two squares (AMND and MBCN) each with the side length of 3 cm.

Question 6.
Recreate the shadings as shown in the figure:

Choose measurements of your choice.
Note: The larger 4 sided- figure is a square and so are the smaller ones.
Solution:
Let the side length of the smaller squares be 2 cm.
Then, the side length of the larger square = 4 × 2 = 8 cm.
Steps of construction are as follows:

Step 1: Construct a square PQRS of side length 8 cm.
Step 2: Using a ruler, mark points at distances 2 cm on sides PQ, QR, RS and SP of square PQRS.
Step 3: Draw horizontal and vertical lines through the marked points to get smaller squares with side equal to 2 cm as shown in the figure.
Step 4: Draw diagonals of the smaller squares.
Step 5: In small squares with side equal to 2 cm, draw vertical lines in the portions below the diagonals as shown in the figure.

Question 7.
Construct a rectangle in which one of its sides is 5 cm and the length of a diagonal is 7.5 cm.
Solution:
Steps of constructions are as follows:
Step 1: Draw a line segment AB of length 5 cm.
Step 2: Using protractor, draw perpendicular BX to AB through B.

Step 3: Using compass, draw an arc with centre A and radius 7.5 cm.
Let arc and BX intersect at point C.
Step 4: Using protractor, draw perpendicular AY to AB through A, and perpendicular CZ to BC through C.
Let AY and CZ intersect at point D. ABCD is the required rectangle.

Question 8.
The distance between points A and B is 7 cm. Now, mark the points that are:

(i) 3 cm away from both A and B
(ii) 4.5 cm away from both A and B
Solution:
(i) Step 1: Using a compass, draw a circle with centre A and radius 3 cm.
Step 2: Using a compass, draw a circle with centre B and radius 3 cm.

Since the two circles do not intersect, there is no point that is exattly 3 cm away from both point A and point B.

(ii) Step 1: Using a compass, mark arcs (with centre A and radius 4.5 cm) above and below AB.
Step 2: Using a compass, mark arcs (with centre B and radius 4.5 cm) above and below AB such that they cut the arcs constructed in step 1.
Let above arcs intersect at point P and below arcs intersect at point Q.
P and Q are at distance of 4.5 cm from points A and B.

Question 9.
Construct a square with the following side length:
(i) 3 cm
(ii) 4.5 cm
(iii) 5.5 cm
Solution:
(i) Steps of construction of a square with side 3 cm are as follows:

Step 1: Draw a line segment AB = 3 cm using a ruler.
Step 2: At points A and B, construct two perpendiculars using a protractor.
Step 3: From points A and B, mark points D and C on the perpendiculars respectively such that AD = BC = 3 cm.
Step 4: Join CD to complete the square ABCD.

(ii) Follow the same steps for square with side 4.5 m as in (i).

(iii) Follow the same steps for square with side 5.5 cm as in (i).

Question 10.
Construct a rectangle of the following side lengths:
(i) 3 cm and 5 cm
(ii) 4 cm and 6.5 cm
(iii) 5.5 cm and 7.5 cm
Solution:
(i) Steps of construction of a rectangle with side lengths 3 cm and 5 cm are as follows:

Step 1: Draw a line segment AB = 5 cm using a ruler.
Step 2: At points A and B, draw perpendiculars to AB using a protractor.
Step 3: Mark points D and C on both the perpendiculars respectively such that AD = BC = 3 cm.
Step 4: Join CD to complete the rectangle ABCD.

(ii) Follow the same steps for rectangle with side lengths 4 cm and 6.5 cm as in (i).

(iii) Follow the same steps for rectangle with side lengths 5.5 cm and 7.5 cm as in (i).

Question 11.
Construct a square within a rectangle such that the centre of the square is the same as the centre of the rectangle. The sides of the rectangle are 7 cm and 4 cm.
Solution:
As square lies inside the rectangle, the side of the square is equal to the smaller side of the rectangle.
So, side of the square = 4 cm
Steps of construction:
Step 1: Draw a rectangle ABCD such that AB = 7 cm and BC = 4 cm.

Step 2: With the help of ruler, mark P, Q, R and S as the mid-points of sides AB, BC, CD and DA.
Step 3: Join PR and QS. Their point of intersection 0 is the centre of the rectangle.
Step 4: With P as centre and OP as radius, mark arcs on both sides ofP intersecting AB at E and F respectively.
Step 5: Similarly, with R as centre and OR (= OP) as radius, mark arcs on both sides of R intersecting CD at G and H respectively. Then, join HE and GF.
Thus, EFGH is the required square whose centre is same as the centre of rectangle ABCD.

Question 12.
Recreate the following combinations of square and hole:

Centre of the hole is the same as the centre of the square containing it.
Solution:
Side length of square is not given. You can take any side length of your choice, say 12 cm.

Step 1: Construct a square ABCD of side 12 cm.
Step 2: Using a ruler, mark points P, Q, R and S as the mid-points of AB, BC, CD and DA respectively.
Join PR and QS. Let 0 be their point of intersection.
Step 3: Draw diagonals AO and PS. Let X be their point of intersection. Point X is the centre of square APOS.
Step 4: Using a compass, draw a circle with centre X and radius less than half of the side length of the square APOS.
Step 5: Draw circles in squares PBQO, OQCR and ORDS (same as step 4). Name their centres as Y, Z and W respectively.

Playing with Constructions Class 6 Long Question Answer

Question 1.
Recreate the following combinations of square and curves:

Note: All the 4 arcs bulge uniformly from each side.
Solution:
Side length of square is not given. We can take any side length of our choice, say 6 cm.
Steps of constructions are as follows:
Step 1: Construct a square ABCD of side 6 cm.
Step 2: Using a ruler, mark points P, Q, R and S as the mid-points of AB, BC, CD and DA respectively.
Step 3: Join PR and produce it in both directions. Also, join QS and produce it in both directions.
Step 4: Mark points X,Y, Z and W such that

XP = YQ = ZR = WS ≥ \(\frac{A B}{2}\)
Step 5: Using compass, draw an arc with X as centre and AX as radius such that it passes through A and B.
Similarly, draw arcs with centres Y, Z and W and radius YB, ZC and DW respectively.
The figure so obtained is the required square with curves.

NOTE: If we take XP = YQ = ZR = WS < \(\frac{A B}{2}\), then adjacent arcs will intersect at two points. But that is not the case.

Question 2.
A rectangular orchard of size 20 m × 15 m is to be planned using the square and diagonal planting systems. Each tree requires 5 m spacing. The planner uses only a compass, ruler and protractor to design accurate layouts based on geometric principles.

Based on the above information, answer the following questions:
i) How many trees can be planted using the square system, with 5 in spacing on a 20m × 13 in held?
(ii) Two trees are 6 m apart. How can you find a point that is 4.5 m away from both? Explain using compass method.
Solution:
(i) Given, field dimensions = 20 m × 15 m

Along the 20 m side (length), trees can be planted at the positions 0 m, 5 m, 10 m, 15 m, and 20 m. That is, there are a total of five positions along the 20 m side.
Therefore, the total number of columns of trees = 5.

Along the 15 m side (breadth), trees can be planted at the positions 0 m, 5 m, 10 m, and 15 m. That is, there are a total of four positions along the 15 m side.

Therefore, the total number of rows of trees = 4.

Hence, total trees = 5 × 4 = 20 trees

(ii) Given, two trees are 6 metres apart, we can find a point that is 4.5 metres away from both by following these steps:

Step 1: Mark two points A and B, 6 metres apart and join them using ruler.
Step 2: Set your compass to 4.5 cm.
Step 3: With A as the centre, draw two arcs, one above and one below the line segment AB.
Step 4: With B as centre, draw arcs of the same radius to intersect the previous arcs.
Step 5: Label the intersection points as P and 0.
Hence, points P and Q are both exactly 4.5 m away from A and B.

Question 3.
Recreate the following House:

Note: All the lines forming the border of the house are of length 4 cm.
Solution:

Steps of construction are as follows:
Step 1: Draw a line segment AB = 4 cm.
Step 2: Construct perpendiculars AD and BC to AB such that BC = AD = 4 cm.
Step 3: With D as centre draw an arc of length 4 cm. Also, draw an arc of same length with C as centre meeting previously drawn arc at point P.
Step 4: Join CP and PD, and with P as centre and CP as radius draw an arc joining C and D. The figure so formed represents a house whose all sides are of length 4 cm.
Step 5: Mark points E and F on AB such that
AE = BF = \(\left(\frac{4-1}{2}\right)\) cm = \(\frac{3}{2}\) cm = 1.5 cm.
Step 6: Draw perpendiculars EH and EC on AB at E and F respectively such that EH = FG = 2 cm. Join GH.

Regular revision with Class 6 Maths MCQ with Answers and Ganita Prakash Class 6 Maths Chapter 1 Patterns in Mathematics MCQ improves accuracy in objective exams.

MCQ on Patterns in Mathematics Class 6

Patterns in Mathematics MCQ Class 6

Class 6 Maths Patterns in Mathematics MCQ

Question 1.
Which of the following is a triangular number?
(a) 8
(b) 11
(c) 18
(d) 21
Solution:
(d) 21
We know, the number sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28 … .
From the given numbers, 21 is present in the number sequence of triangular numbers.
Hence, 21 is a triangular number.

Question 2.
Which of the following is not a cube number?
(a) 8
(b) 27
(c) 125
(d) 169
Solution:
(d) 169
We know, the number sequence of cubes is 1, 8, 27, 64, 125, 216, … .
From the given numbers, 169 is not present in the number sequence of cubes.
Hence, 169 is not a cube.

Question 3.
Which of the following is not a power of 3?
(a) 9
(b) 63
(c) 81
(d) 243
Solution:
(b) 243
We know, the number sequence of powers of 3 is 1, 3, 9, 27, 81, 243, 729 … .
From the given numbers, 63 is not present in the number sequence of powers of 3.
Hence, 63 is not a power of 3.

Question 4.
The pattern in the sequence 1, 4, 9, 16, 25, … is:
(a) n^th term = 2n, where n = 1, 2, 3, 4, …
(b) n^th term = n³, where n = 1, 2, 3, 4, …
(c) n^th term = \(\frac{n}{2}\), where n = 1, 2, 3, 4, …
(d) n^th term = n², where n = 1, 2, 3, 4, …
Solution:
(d) n^th term = n², where n = 1, 2, 3, 4, …
Given sequence is 1, 4, 9, 16, 25, ….
First term = 1 = 1 × 1 = 1²
Second term = 4 = 2 × 2 = 2²
Third term = 9 = 3 × 3 = 3²
Fourth term = 16 = 4 × 4 = 4²
Fifth term = 25 = 5 × 5 = 5²
Therefore, the sequence follows the pattern:
n^th term = n~, where n = 1,2, 3, 4, …

Question 5.
The pattern in the sequence 1, 3, 6, 10, 15, 21, 28, … is:
(a) n^th term = Sum of first n counting numbers + 1, where n = 1, 2, 3, 4, …
(b) n^th term = Sum of first n counting numbers, where n = 1, 2, 3, 4, …
(c) n^th term = Sum of first n counting numbers × 2, where n = 1, 2, 3, 4, …
(d) n^th term = Sum of first n counting numbers ÷ 2, where n = 1, 2, 3, 4, …
Solution:
(b) n^th term = Sum of first n counting numbers, where n = 1, 2, 3, 4, …
Given sequence is 1, 3, 6, 10, 15, 21, 28, … .
First term = 1 (First counting number)
Second term = 3 = 1 + 2 (Sum of first 2 . counting numbers)
Third term = 6 = 1 + 2 + 3 (Sum of first 3 counting numbers)
Fourth term = 10 = 1 + 2 + 3 + 4 (Sum of first 4 counting numbers)
Fifth term = 15 = 1 + 2 + 3 + 4 + 5 (Sum of first 5 counting numbers)
Therefore, the sequence follows the pattern:
n^th term = Sum of first n counting numbers, where n = 1, 2, 3, 4, …

Question 6.
Which of the following sequence is represented by dots forming a square?
(a) 1, 4, 9, 16, 25, …
(b) 1, 3, 5, 7, 9, …
(c) 1, 2, 4, 8, 16, …
(d) 1, 3, 6, 10, 15, …
Solution:
(a) 1, 4, 9, 16, 25, …
We know, the square numbers 1, 4, 9, 16, 25,… are represented by dots forming the squares.

Question 7.
The sequence 1, 8, 27, 64, 125, … is visualised using:
(a) Squares
(b) Circles
(c) Triangles
(d) Cubes
Solution:
(d) Cubes
The given sequence 1, 8, 27, 64, 125, … is asequence of cubes.
We know, the sequence of cubes can be visualised using cubes.

Question 8.
What is the next number in the sequence: 1, 7, 19, 37,…?
(a) 50
(b) 61
(c) 63
(d) 65
Solution:
(b) 61
The given sequence is 1, 7, 19, 37, …, which is the sequence of hexagonal numbers.
The rule followed in the sequence of hexagonal numbers is
1^st term = 1, n^th number (term) = Preceding term + 6 × (n – 1); n = 2, 3, 4, …
∴ Next term, i.e. fifth term = 37 + 6 v (5 – 1) = 37 + 6 × 4 = 61

Question 9.
Triangular numbers can be visualised by arranging dots in the form of:
(a) Rectangles
(b) Squares
(c) Triangles
(d) Hexagons
Solution:
(c) Triangles
We know, the triangular numbers 1, 3, 6, 10, 15, … can be visualised by arranging dots in the form of triangles.

Question 10.
How many dots are there in the 4^th triangular number?
(a) 6
(b) 10
(c) 15
(d) 21
Solution:
(b) 10
We know, the sequence of triangular numbers is 1, 3, 6, 10, 15, 21, 28, 36, 45, … .
So, the 4^th triangular number is 10.
Thus, there are 10 dots in the triangle which represent the 4^th triangular number.

Question 11.
How many dots are there in the 4th hexagonal number?
(a) 19
(b) 37
(c) 61
(d) 91
Solution:
(b) 37
We know, the sequence of hexagonal numbers is 1, 7, 19, 37, 61, 91, ……..
So, the 4^th hexagonal number is 37.

Thus, there are 37 dots in the hexagon which represent the 4^th hexagonal number.

Question 12.
The sum 1 + 7 + 19 + 37 gives which type of number?
(a) Triangular
(b) Virahanka
(c) Square
(d) Hexagonal
Solution:
(c) Square
1 + 7 + 19 + 37 = 64, which is a square number.

Question 13.
A polygon having 6 sides is known as:
(a) Quadrilateral
(b) Pentagon
(c) Hexagon
(d) Heptagon
Solution:
(c) Hexagon
Hexagon has 6 sides.

Question 14.
The geometric pattern 3, 12, 48, 192, 768, …, represents:
(a) The number of sides in the sequence of Koch Snowflakes
(b) The number of line segments in the sequence of Complete Graphs
(c) The number of vertices in the sequence of Regular Polygons
(d) The number of stacked triangles in the sequence of Stacked Triangles
Solution:
(a) The number of sides in the sequence of Koch Snowflakes
The geometric pattern 3, 12, 48, 192, 768, … represents the number of sides in the Koch snowflakes sequence.

Question 15.
The number of line segments in the sequence of complete graphs form a sequence of:
(a) even numbers
(b) virahanka numbers
(c) powers of 2
(d) triangular numbers
Solution:
(d) triangular numbers
The number of line segments in a complete graph follows the sequence: 0, 1, 3, 6, 10, 15…
Therefore, complete graphs sequence is related to triangular number sequence, i.e. i, 3, 6, 10,15 … .

Question 16.
The number of stacked triangles in the stacked triangles sequence form a sequence of:
(a) triangular numbers
(b) square numbers
(c) powers of 2
(d) hexagonal numbers
Solution:
(b) square numbers
We know,

Number of stacked triangles:
1 = 1² 4 = 2² 9 = 3²
16 = 4² 25 = 5²
Thus, the number of stacked triangles in the stacked triangles sequence form a sequence of square numbers.

Question 17.
Which of the following statements is/are true regarding square numbers?
(i) The sum of the first n odd numbers gives a square number.
(ii) Adding counting numbers up from 1 to a number, and then back down again to 1, also gives a square number.
(iii) Adding up consecutive powers of 2, yields square number.
(iv) Multiplying triangular numbers by 6 and adding 1 to each term gives square numbers.
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (i) and (ii) only
(c) (iii) and (iv) only
(d) (i), (iii) and (iv)
Solution:
(b) (i) and (ii) only
We have, the sum of the first two odd numbers = 1 + 3 = 4 = 2²
The sum of the first three odd numbers = 1 + 3 + 5 = 9 = 3²
The sum of the first four odd numbers =1 + 3 + 5 + 7 = 16 = 4² and so on.
Hence, the sum of the first n odd numbers gives a square number.
So, statement (i) is correct.

Also, adding counting numbers up and then down gives a square number.
For example, 1 + 2 + 1 = 4 = 2² and 1 + 2 + 3 + 2 + 1 = 9 = 3² and so on.
So, statement (ii) is also correct.

Now, adding consecutive powers of 2 does not give square numbers. The sequence powers of 2 is given as 1,2, 4, 8, 16, 32, …
Adding two consecutive numbers at a time, we get 3 (1 + 2 ), 6 (2 + 4), etc., which are not square numbers.
So, statement (iii) is incorrect.

We know that if we multiply triangular numbers by 6 and then add 1, we get a new number sequence: 7, 19, 37, 61 and so on. The new sequence is the sequence of hexagonal numbers starting with 7.
So, statement (iv) is also incorrect.

Question 18.
Which of the following statements is/are true?
(i) The regular polygons sequence is related to counting numbers sequence starting with 3.
(ii) The complete graphs sequence is related to triangular numbers sequence, where numbers are given by \(\frac{n(n-1)}{2}\), n = 1, 2, 3, …, i.e. 0, 1, 3, 6, 10, 15 and so on.
(iii) The stacked squares sequence is related to square numbers sequence, i.e. 1, 4, 9, 16, 25 and so on.
(iv) The stacked triangles sequence is related to the square numbers sequence, i.e. 1, 4, 9, 16, 25 and so on.
Choose the correct option from the following:
(a) (i) and (ii) only
(b) (ii) and (iv) only
(c) (i) (ii) and (iii) only
(d) (i), (ii), (iii) and (iv)
Solution:
(d) (i), (ii), (iii) and (iv)
All the statements are correct.

Patterns in Mathematics Class 6 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): All is sequence (1, 1, 1, 1, …) gives counting numbers when added up successively.
(R): Adding 1 repeatedly increases the total by 1 each time.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, all is sequence is 1, 1, 1, 1, …..
Here, 1 = 1 = 1
1 + 1 = 2 = 1 + 1
1 + 1 + 1 = 3 = 2 + 1
1 + 1 + 1 + 1 = 4 = 3 + 1
So, all is sequence (1, 1, 1, 1, ….) gives the counting numbers when added up successively and adding 1 repeatedly increases the total by 1 each time.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The pattern 1, 2, 4, 8, 16, … represents the powers of 2.
(R): Each number in this sequence is obtained by adding 2 to the previous term.
Solution:
(c) A is true but R is false.
We know, the number sequence of powers of 2 is 1, 2, 4, 8, 16, 32, 64, … .
First term = 1
Second term = 2 = 1 × 2 (First term × 2)
Third term = 4 = 2 × 2 (Second term × 2)
Fourth term = 8 = 4 × 2 (Third term × 2)
Fifth term = 16 = 8 × 2 (Fourth term × 2)
So, we can say that each term of the number sequence of powers of 2 is obtained by multiplying 2 to the previous term.
Thus, Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): Square numbers can be obtained by adding up the odd numbers.
(R): The sum 1 + 3 + 5 + 7 + 9 = 25, which is a perfect square.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, the sequence of odd numbers is 1, 3, 5, 7, 9, … .
And, the sequence of squares is 1, 4, 9, 16, 25, ….
Now, 1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
So, the square numbers can be obtained by adding up the odd numbers.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): The hexagonal number sequence starts with 1 and continues with 7, 19, 37, 61, and so on.
(R): Each number in the sequence adds a growing number of layers of dots in a hexagonal form.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, the sequence of hexagonal numbers is 1, 7, 19, 37, 61, 91, … .
Pictorial representation of hexagonal numbers is given below:

Here, we observe that each number in the sequence adds a growing number of layers of dots in a hexagonal form.Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Patterns in Mathematics Class 6 Fill in the Blanks

Question 1.
The next term of the sequence 1, 16, 49, 100, 169,… is _______.
Solution: 256
Given sequence is 1, 16, 49, 100, 169, ….
First term = 1 = (3 × 0 + 1)²
Second term = 16 = (3 × 1 + 1)²
Third term = 49 = (3 × 2 + 1)²
Fourth term = 100 = (3 × 3 + 1)²
Fifth term = 169 = (3 × 4 + 1)²
∴ Next term, i.e. sixth term = (3 × 5 + 1)²
= (16)² = 256
Hence, the next term of the sequence 1, 16, 49, 100, 169, … is 256.

Question 2.
A square number sequence can be visualised as dots arranged in shape of a _______ .
Solution: square
We know, a square number sequence can be visualised as dots arranged in the shape of a square.

Question 3.
In the triangular number sequence, the 5^th number is ________ .
Solution: 15
We know, the triangular number sequence is 1,3, 6, 10, 15, 21, 28,… .
So, in the triangular number sequence, the 5^th number is 15.

Question 4.
The difference between consecutive square numbers forms an _______ sequence.
Solution: odd number
The difference between consecutive square numbers forms an odd number sequence: 3, 5, 7, 9, 11, … i.e. 4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, and so on.

Question 5.
The pictorial representation of the sequence 1, 8, 27, 64, … is based on _______ .
Solution: cubes
We know, the sequence 1, 8, 27, 64, … is the sequence of cubes.
Thus, the pictorial representation of the sequence 1,8, 27, 64, … is based on cubes.

Question 6.
A polygon having 8 sides is known as __________ .
Solution: Octagon
A polygon having 8 sides is known as Octagon.

Question 7.
The difference between the number of sides of a nonagon and a pentagon is _________ .
Solution: 4
A pentagon has 5 sides whereas a nonagon has 9 sides.
∴ The difference between the number of sides of a nonagon and a pentagon is 9 – 5 = 4.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 7 Finding the Unknown MCQ improves accuracy in objective exams.

MCQ on Finding the Unknown Class 7

Finding the Unknown MCQ Class 7

Class 7 Maths Finding the Unknown MCQ

Question 1.
A weighing scale is balanced when two identical boxes and a 4-unit weight are placed on one pan, and one identical box with a 12-unit weight is placed on the other pan. What is the weight of one box?
(a) 6
(b) 4
(c) 8
(d) 12
Solution:
(c) 8
Let the weight of one box be x units. Then,
Total weight on one pan = (2x + 4) units
Total weight on other pan = (x + 12) units
As weighing scale is balanced, weights on boththe pans are equal.
∴ 2x + 4 = x + 12
⇒ 2x – x = 12 – 4 ⇒ x = 8
Hence, the weight of one box is 8 units.

Question 2.
A sequence of arrangements uses 4 sticks in the first step and increases by 3 sticks in every next step. Which expression gives the number of sticks in step n?
(a) 3n
(b) 3n + 1
(c) 3n + 2
(d) 3n – 1
Solution:
(b) 3n + 1
Number of sticks in:
1^st step = 4
2^nd step = 4 + 3 × 1
3^rd step = 4 + 3 × 2
4^th step = 4 + 3 × 3
.
.
.
.
.
n^th step = 4 + 3 × (n – 1) = 4 + 3n – 3
= 3n + 1
Hence, the number of sticks in step n is 3n + 1.

Question 3.
The value of m so that 0.6 m – (-0.8 m + 0.4) = 0.2 – 0.5 m, is:
(a) \(\frac{7}{19}\)
(b) \(\frac{6}{19}\)
(c) \(\frac{9}{19}\)
(d) \(\frac{8}{19}\)
Solution:
(b) \(\frac{6}{19}\)
Given, 0.6m – (- 0.8 m + 0.4) = 0.2 – 0.5m
⇒ 0.6m + 0.8m – 0.4 = 0.2 – 0.5m
⇒ 1.4m – 0.4 = 0.2 – 0.5m
⇒ 1.4m + 0.5m = 0.2 + 0.4
⇒ 1.9m = 0.6
⇒ m = \(\frac{0.6}{1.9}\) ⇒ m = \(\frac{6}{19}\)

Question 4.
Which of the following is correct about x = x?
(a) The equation has no solution.
(b) The equation has exactly one solution.
(c) The equation has infinitely many solutions.
(d) The equation is incorrect.
Solution:
(c) The equation has infinitely many solutions.
We have, x = x
It means it is true (LHS = RHS) for all values of the variable. Hence, it has infinitely many solutions.

Question 5.
It is given that \(\frac{45}{56} \times \frac{14}{9} \times \frac{7}{3} \times 5=\frac{22050}{1512}\), then the value of \(\frac{45}{56} \times \frac{14}{9} \times 5\) is:
(a) \(\frac{3150}{504}\)
(b) \(\frac{2850}{564}\)
(c) \(\frac{3150}{564}\)
(d) \(\frac{2850}{504}\)
Solution:
(a) \(\frac{3150}{504}\)
Given, \(\frac{45}{56} \times \frac{14}{9} \times \frac{7}{3} \times 5=\frac{22050}{1512}\)
⇒ \(\frac{45}{56} \times \frac{14}{9} \times 5=\frac{22050}{1512} \times \frac{3}{7}=\frac{3150}{504}\)

Question 6.
If 3x – \(\frac{4}{5}=\frac{5 x}{3}+\frac{16}{5}\), then the value of x which satisfies the given equation is:
(a) Natural number greater than 7
(b) Natural number between 5 and 7
(c) Natural number less than 2
(d) Natural number between 2 and 4
Solution:
(d) Natural number between 2 and 4
Given, \(3 x-\frac{4}{5}=\frac{5 x}{3}+\frac{16}{5}\)
⇒ \(3 x-\frac{5 x}{3}=\frac{16}{5}+\frac{4}{5}\) ⇒ \(\frac{9 x-5 x}{3}=\frac{16+4}{5}\)
⇒ \(\frac{4 x}{3}=\frac{20}{5}\) ⇒ \(\frac{4 x}{3}=4\)
⇒ x = \(\frac{4 \times 3}{4}\) ⇒ x = 3

Question 7.
The sum of two consecutive even numbers is 70. The greater one is:
(a) 32
(b) 34
(c) 36
(d) 38
Solution:
(c) 36
Let x and x + 2 be two consecutive even numbers.
According to question, we have
x + (x + 2) = 70
⇒ 2x + 2 = 70 ⇒ 2x = 70 – 2
⇒ 2x = 68 ⇒ x = \(\frac{68}{2}\) = 34
So, smaller even number = 34 and greater even number = 34 + 2 = 36

Question 8.
Rajini has three balls, A, B and C. A is twice as heavy as B, B weighs one third of C. The heaviest ball is:
(a) A
(b) B
(c) C
(d) A and C
Solution:
(c) C
Let the weight of ball C be x units. Then,
Weights of ball B and ball A are \(\frac{x}{3}\) units and \(\frac{2x}{3}\) units respectively. Clearly, x > \(\frac{2x}{3}\) > \(\frac{x}{3}\)
[As weight cannot be negative, x > 0]
Thus, the heaviest ball is C.

Question 9.
Which of the following equations represents the statement “twice of a number is 25 more than one-third of the number”?
(a) \(2 x=\frac{x}{3}+25\)
(b) \(2 x+25=\frac{x}{3}\)
(c) \(2 x=\frac{25+x}{3}\)
(d) \(3 x=25+\frac{x}{2}\)
Solution:
(a) \(2 x=\frac{x}{3}+25\)
Let the number be x. Then,
Twice of the number = 2x and one-third of the number = \(\frac{x}{3}\)
According to question, we have 2x = \(\frac{x}{3}\) + 25

Question 10.
Adding 7 to the thrice of a whole number gives 34. The whole number is:
(a) 13
(b) 9
(c) 14
(d) 11
Solution:
(b) 9
Let the whole number be x. Then, thrice of the whole number = 3x
According to question, we have
3x + 7 = 34 ⇒ 3x = 34 – 7
⇒ 3x = 27 ⇒ x = \(\frac{27}{3}\) = 9

Question 11.
Which of the following equations have t = – 3 as a solution?
(i) 2t + 7 = 1
(ii) 4(t + 5) = 8
(iii) 2t + \(2\left(t+\frac{1}{4}\right)=\frac{-5}{2}\) + t
(iv) -2 = \(\frac{3}{5}\) – 7t
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (i) and (ii)
(i) LHS = 2t + 7 = 2 × (-3) + 7 = -6 + 7
= 1 = RHS
∴ t = – 3 is the solution of equation 2t + 7 = 1.

(ii) LHS = 4(t + 5) = 4(-3 + 5) = 4 × 2 = 8 = RHS
∴ t = – 3 is the solution of equation
4 (t + 5) = 8.

(iii) LHS = 2t + 2\(\left(t+\frac{1}{4}\right)\)
= 2 x (-3) + 2 \(\left(-3+\frac{1}{4}\right)\)
= -6 + \(2\left(\frac{-11}{4}\right)\) = -6 – \(\frac{11}{2}\)
= \(-\frac{23}{2}\)
RHS = \(\frac{3}{5}-7 t=\frac{3}{5}-7 \times(-3)\)
= \(\frac{3}{5}+21=\frac{3+105}{5}=\frac{108}{5}\) ≠ -2 = LHS

Question 12.
“48 is divided into 2 parts such that three times the greater part is 10 less than four times the smaller part”.
Identify the correct statements about the given situation.
(i) The smaller part is 16.
(ii) The smaller part is 22.
(iii) The greater part is 26.
(iv) The greater part is 32.
Choose the correct option from the following:
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Solution:
(c) (ii) and (iii)
Let the greater part be x. Then, the smaller part = 48 – x
According to given situation, we have
3x = 4(48 – x) – 10 ⇒ 3x = 4 × 48 – 4x – 10
⇒ 3x = 192 – 4x – 10 ⇒ 3x + 4x = 192 – 10
⇒ 7x = 182 ⇒ x = \(\frac{182}{7}\) = 26
So, the greater part is 26 and the smaller part is 48 – 26 = 22.
Thus, statements (ii) and (iii) are correct.

Question 13.
The sum of three consecutive odd numbers is 63.
Identify the correct statements about the given situation.
(i) Smallest odd number is 17.
(ii) Largest odd number is 23.
(iii) Smallest odd number is 19.
(iv) Largest odd number is 21.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii)
We know, difference between two consecutive odd numbers is 2.
Let x, x + 2 and x + 4 be the three consecutive odd numbers.
∴ x + (x + 2) + (x + 4) = 63
[Given, sum of three consecutive odd numbers is 63.]
⇒ 3x + 6 = 63 ⇒ 3x = 63 – 6
⇒ 3x = 57 ⇒ x = \(\frac{57}{3}\) = 19
So, the three consecutive odd numbers are 19, 21 and 23.
Thus, the smallest odd number is 19 and the largest odd number is 23.

Finding the Unknown Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1
(A): Only addition can be done on both sides of an equation, not subtraction.
(R): An equation remains balanced when we perform the same operation on both sides.
Solution:
(d) A is false but R is true.
An equation remains balanced when we perform the same operation (addition, subtraction, multiplication, division) on both sides.
∴ Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): Two complementary angles differ by 8° can be represented by equation x – (90° – x) = 8°.
(R): Sum of complementary angles is 180°.
Solution:
(c) A is true but R is false.
We know, sum of complementary angles is 90°.
Given, two complementary angles differ by 8°.
Let x be one of the two complementary angles. Then, the other angle = 90° – x
∴ x – (90° – x) = 8° is the required equation.
Thus, Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): For the given triangle, y + 2 (y + 15°) = 180°.
(R): In an isosceles triangle, angles opposite to equal sides are equal.

Solution:
(a) Both A and R are true and R is the correct explanation of A.
The given triangle is isosceles.
We know, angles opposite to equal sides of an isosceles triangle are equal.
∴ First angle = y, second angle = third angle
= y + 15°
Now, y + (y + 15°) + (y + 15°) = 180°
[The sum of all the angles of a triangle is 180°.]
⇒ y + 2(y + 15°) = 180°
∴ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Finding the Unknown Class 7 Fill in the Blanks

Question 1.
If the solution is known, the number of equations that can be built are ________ .
Solution: infinite
If the solution is known, the number of equations that can be built are infinite.

Question 2.
Fill in the blanks with integers:
(i) 11 (11 + ____ ) – 98 = 100
(ii) – 3 × (-9 + _____) = 42
Solution: 7, -5
(i) Let y be the required value to fill the blank.
∴ 11(11 + y) – 98 = 100
⇒ 11(11 + y) = 100 + 98
[Adding 98 to both sides]
⇒ 11(11 + y) = 198
⇒ (11 + y) = 198 ÷ 11
[Dividing both sides by 11]
⇒ 11 + y= 18
⇒ y = 18 – 11 = 7
So, 11(11 + 7) – 98 = 100

(ii) Let y be the required value to fill the blank.
∴ -3 × (- 9 + y) = 42
⇒ (- 9 + y) = 42 ÷ (-3)
[Dividing both sides by – 3]
⇒ – 9 + y = -14
⇒ y = – 14 + 9 [Adding 9 to both sides]
⇒ y = – 5
So, – 3 × (- 9 + (- 5)) = 42

Question 3.
In an equation, there is always an ________ sign.
Solution: equals
In an equation, there is always an equals sign.

Question 4.
The value of the variable which satisfies a equation is called a ________ to the equation.
Solution: solution
The value of the variable which satisfies an equation is called a solution to the equation.

Question 5.
The equation representing “one-fifth of a number is 3 more than one-sixth of the same number” is \(\frac{x}{5}\) = ______ .
Solution: \(\frac{x}{6}\) + 3
Let x be the number. Then,
One-fifth of the number = \(\frac{x}{5}\) and one-sixth of the number = \(\frac{x}{6}\)
\(\frac{x}{5}=\frac{x}{6}\) + 3
The equation representing “one-fifth of a number is 3 more than one-sixth of the same number” is \(\frac{x}{5}=\frac{x}{6}\) + 3.

Question 6.
If \(\frac{x}{7}\) of a number is 60. the number is _______ .
Solution: 105
Let the number be x.
∴ \(\frac{4}{7}\) × x = 60
⇒ x = 60 × \(\frac{7}{4}\) = 15 × 7 = 105
If \(\frac{4}{7}\) of a number is 60, the number is 105.