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Students can use Class 7 Math Solution Odia Medium and Class 7 Maths Chapter 1 Question Answer Odia Medium ଆମ ଚାରିପଟେ ଥିବା ବଡ଼ ସଂଖ୍ୟା to check their answers after solving exercises.

7th Class Maths Chapter 1 Question Answer Odia Medium

Class 7 Maths Chapter 1 Odia Medium

1.1 ଏକ ଲକ୍ଷ ପ୍ରକାରର ବିହନ
Page No. (2 to 5)

Question 1.
ସଂରଚନାକୁ ଲକ୍ଷ୍ୟକରି ତଳେ ଦିଆଯାଇଥିବା କୋଠରିଗୁଡ଼ିକୁ ପୂରଣ କର । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

1,00,000 କୁ ‘ଏକ ଲକ୍ଷ’ କୁହାଯାଏ ।

Question 2.
ଖାଲିସ୍ଥାନ ପୂରଣ କରି ଲେଖ । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

Question 3.
ଯଦି ଦିନକୁ ଗୋଟିଏ ପ୍ରକାର ଚାଉଳ ଖାଆନ୍ତି, ତେବେ 100 ବର୍ଷରେ ଏକ ଲକ୍ଷ ପ୍ରକାର ଚାଉଳ ଖାଇପାରିବେ ନାହିଁ । ଯଦି ଆମେ ଦିନକୁ 2 ପ୍ରକାର ଚାଉଳ ଖାଇବା, ତେବେ କ’ଣ 100 ବର୍ଷରେ ଏକ ଲକ୍ଷ ପ୍ରକାରର ଚାଉଳ ଖାଇପାରିବା ?
Solution:
ଦତ୍ତ ଅଛି, ଯେ ଦିନକୁ ସେ 2 ପ୍ରକାର ଚାଉଳ ଖାଆନ୍ତି ।
1 ବର୍ଷ = 12 ମାସ = 365 ଦିନ ।
ଆମେ ଗୋଟିଏ ଦିନରେ ଖାଉ 2 ପ୍ରକାର ଚାଉଳ ।
ଗୋଟିଏ ବର୍ଷରେ ଖାଇବୁ = 2 × 365 = 730 ପ୍ରକାର ଚାଉଳ ।
100 ବର୍ଷରେ ଖାଇବୁ = 730 × 100 = 73,000 ପ୍ରକାର ଚାଉଳ ।
∴ ଆମେ 100 ବର୍ଷରେ 73,000 ପ୍ରକାର ଚାଉଳ ଖାଇବା, ଯାହା ଏକ ଲକ୍ଷ ନୁହେଁ ।

Question 4.
ଯଦି ଜଣେ ଦିନକୁ 3 ପ୍ରକାର ଚାଉଳ ଖାଏ, ତେବେ ସେ 100 ବର୍ଷ ଜୀବନ କାଳରେ ଏକ ଲକ୍ଷ ପ୍ରକାରର ଚାଉଳ ଖାଇପାରିବା କି ? ଆସ ନିର୍ଣ୍ଣୟ କରିବା ।
Solution:
ଦତ୍ତ ଅଛି ଯେ ଦିନକୁ ସେ 3 ପ୍ରକାର ଚାଉଳ ଖାଏ ।
1 ବର୍ଷ = 12 ମାସ = 365 ଦିନ ।
ଜଣେ ଗୋଟିଏ ଦିନରେ ଖାଏ 3 ପ୍ରକାର ଚାଉଳ ।
ଗୋଟିଏ ବର୍ଷରେ ଖାଏ = 3 × 365 = 1095 ପ୍ରକାର ଚାଉଳ ।
100 ବର୍ଷରେ ଖାଏ = 1095 × 100 = 1,09,5,00 ପ୍ରକାର ଚାଉଳ ।
∴ ସେ 100 ବର୍ଷରେ 1 ଲକ୍ଷ ପ୍ରକାରର ଚାଉଳ ଖାଇପାରିବ ।
> ରମେଶ କହିଲା, ‘‘ଆମେ ଜାଣିଲୁ ଯେ ଏକ ବର୍ଷରେ 365 ଦିନ ଅଛି (ଅଧ୍ବବର୍ଷକୁ ବାଦେଇ) । ଯଦି ଆମେ y ବର୍ଷ ବଞ୍ଚିବା, ଜୀବନ କାଳର ମୋଟ ଦିନ ସଂଖ୍ୟା ହେବ 365 × y ।

Question 5.
ତୁମେ y ପାଇଁ ଏକ ସଂଖ୍ୟା ନିଅ । ଏହି y ବର୍ଷର ଦିନଗୁଡ଼ିକର ସଂଖ୍ୟା ଏକ ଲକ୍ଷ ସହିତ କେତେ ନିକଟତର ଅଛି, ତୁମେ ବାଛିଥୁବା y ର ମାନ ପାଇଁ ନିର୍ଣ୍ଣୟ କର।
Solution:
ମନେକର ଆମେ 95 ବର୍ଷ ବଞ୍ଚିବୁ ଅର୍ଥାତ୍ y = 95 ।
ଆମ ଜୀବନସାରା ଦିନସଂଖ୍ୟା = 365 × 95 = 34,675 ଦିନ ।
1,00,000 ଓ 34,675ର ପାର୍ଥକ୍ୟ = 1,00,000 – 34,675 = 65,325

ନିଜେ କରି ଦେଖ 

Question 1.
2011 ଜନଗଣନା ଅନୁଯାୟୀ, ବରଗଡ଼ ସହରର ଜନସଂଖ୍ୟା ପ୍ରାୟ 83,651 ଥିଲା । ଏହା ଏକ ଲକ୍ଷରୁ କେତେ କମ୍ ?
Solution:
2011 ଜନଗଣନା ଅନୁଯାୟୀ ବରଗଡ଼ ସହରର ଜନସଂଖ୍ୟା = 83, 651
ଏହା 1 ଲକ୍ଷରୁ କମ୍ = 1,00,000 – 83,651 = 16,349
∴ ଏହା ଏକ ଲକ୍ଷରୁ 16,349 କମ୍ ।

Question 2.
2024 ମସିହାରେ ବରଗଡ଼ ସହରର ଆନୁମାନିକ ଜନସଂଖ୍ୟା 1,04,000 । ଏହା ଏକ ଲକ୍ଷଠାରୁ କେତେ ଅଧ୍ବକ ?
Solution:
2024 ଜନଗଣନା ଅନୁଯାୟୀ ବରଗଡ଼ ସହରର ଜନସଂଖ୍ୟା = 1,04,000
1 ଲକ୍ଷରୁ ଅଧିକ, 1,04,000 – 1,00,000 = 4,000
∴ ଏହା ଏକ ଲକ୍ଷରୁ 4,000) ବେଶୀ ।

Question 3.
2011 ରୁ 2024 ପର୍ଯ୍ୟନ୍ତ ବରଗଡ଼ ସହରର ଜନସଂଖ୍ୟା କେତେ ବୃଦ୍ଧି ପାଇଲା ?
Solution:
2011 ମସିହାରେ ଜନସଂଖ୍ୟା = 83,651
2024 ମସିହାରେ ଜନସଂଖ୍ୟା = 1,04,000
ଜନସଂଖ୍ୟା ବୃଦ୍ଧି ପାଇଲା = 1,04,000 – 83,651 = 20,349
∴ 2011 ରୁ 2024 ପର୍ଯ୍ୟନ୍ତ ବରଗଡ଼ ସହରର ଜନସଂଖ୍ୟା 20,349 ବୃଦ୍ଧି ପାଇଲା ।

ବଡ଼ସଂଖ୍ୟାକୁ ବୁଝିବା :

Question 1.
ଉଦାହରଣ : ସୋମୁର ଗୋଟିଏ ଦଶ (10) ମହଲା ବିଶିଷ୍ଟ ଘର ଅଛି । ସୋମୁର ଉଚ୍ଚତା 1 ମିଟର । ଯଦି ଏହି ଘରର ପ୍ରତ୍ୟେକ ମହଲା ସୋମୁ ଉଚ୍ଚତାର ପ୍ରାୟ 4 ଗୁଣ, ତେବେ ଘରର ଆନୁମାନିକ ଉଚ୍ଚତା କେତେ ?

Solution:
ସୋମୁର ଉଚ୍ଚତା 1 ମିଟର । ଘରର ପ୍ରତ୍ୟେକ ମହଲା ସୋମୁ ଉଚ୍ଚତାର ପ୍ରାୟ 4 ଗୁଣ ।
∴ ଘରର ପ୍ରତ୍ୟେକ ମହଲାର ଉଚ୍ଚତା = 4 × 1 = 4 ମିଟର ।
ସୋମୁ ଘରଟି 10 ମହଲା ବିଶିଷ୍ଟ । ସୋମୁ ଘରର ସମୁଦାୟ ଉଚ୍ଚତା = 10 × 4 = 40 ମିଟର ।
∴ ଘରର ଉଚ୍ଚତା 40 ମିଟର ।

Question 2.
ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍ କିମ୍ବା ସୋମୁର ଏହି ଦଶମହଲା ବିଶିଷ୍ଟ ଘରର ଉଚ୍ଚତା ମଧ୍ୟରୁ କେଉଁଟିର ଉଚ୍ଚତା ଅଧିକ ? କେତେ ମିଟର ଅଧୁକ ? ଆମେ ଜାଣିପାରିବା ଯେ, ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍‌ର ଉଚ୍ଚତା, ସେହି ଘରର ଉଚ୍ଚତାର 4 ଗୁଣ ପାଖାପାଖୁ ।

Solution:
ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍‌ର ଉଚ୍ଚତା = 180 ମିଟର ।
ସୋମୁର ଦଶମହଲା ବିଶିଷ୍ଟ ଘରର ଉଚ୍ଚତା = 40 ମିଟର ।
ଅଧ୍ବକ = 180 – 40 = 140 ମିଟର ।
∴ ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍‌ର ଉଚ୍ଚତା 140 ମିଟର ଅଧୂକ ।

Question 3.
ଖଣ୍ଡାଧାର ଜଳପ୍ରପାତ ସୋମୁର ଘରର ଉଚ୍ଚତା ଠାରୁ କେତେ ଅଧିକ ଉଚ୍ଚ ? …………… ମିଟର ।
Solution:
ଖଣ୍ଡାଧାର ଜଳପପ୍ରାତର ଉଚ୍ଚତା = 244 ମିଟର । ସୋମୁର ଘରର ଉଚ୍ଚତା = 40 ମିଟର ।
ଅଧିକ = 244 – 40 = 204 ମିଟର ।
∴ ଖଣ୍ଡାଧାର ଜଳପ୍ରପାତ ସୋମୁର ଘରର ଉଚ୍ଚତା ଠାରୁ 204 ମିଟର ଅଧୂକ ଉଚ୍ଚ ।

Question 4.
ସୋମୁର ଘର କେତେ ମହଲା ହେଲେ ଖଣ୍ଡାଧାର ଜଳପ୍ରପାତର ଉଚ୍ଚତାର ନିକଟତର ହେବ ?
Solution:
ଖଣ୍ଡାଧାର ଜଳପପ୍ରାତର ଉଚ୍ଚତା = 244 ମିଟର । ସୋମୁର ଗୋଟିଏ ମହଲାର ଉଚ୍ଚତା = 4 ମିଟର ।
244 ÷ 4 = 61 ମହଲା ।
∴ ସୋମୁର ଘର 61 ମହଲା ହେଲେ ଖଣ୍ଡାଧାର ଜଳପ୍ରପାତର ଉଚ୍ଚତାର ନିକଟତର ହେବ ।

ଏକ ଲକ୍ଷ ଗୋଟିଏ ବଡ଼ ସଂଖ୍ୟା କି ? :
Question 1.
ଏକ ଲକ୍ଷ ସଂଖ୍ୟାଟି ବଡ଼ କି ସାନ ? ଚିନ୍ତା କର ।
Solution:
ଏକ ଲକ୍ଷ ସଂଖ୍ୟାଟି ବଡ଼ କି ସାନ, ତାହା ତା’ର ପ୍ରସଙ୍ଗ ଉପରେ ନିର୍ଭର କରିଥାଏ । ଏକ ଲକ୍ଷ ଆନୁସଙ୍ଗିକ ଭାବରେ ଏକ ବଡ଼ ସଂଖ୍ୟା । ଯାହା ୧୦୦୦୦୦ ।
ମନେକର ଏକ ସହରର ଲୋକସଂଖ୍ୟା 1 ଲକ୍ଷ ।
ଏଠାରେ 1 ଲକ୍ଷ କହିଲେ ଏହା ଏକ ଛୋଟ ସହରକୁ ବୁଝାଏ ।
ମନେକର ଜଣେ ଲୋକ ମାସକୁ ୧ ଲକ୍ଷ ଟଙ୍କା ଦରମା ପାଉଛି । ଏହା ଏକ ବଡ଼ଭାବରେ ଗଣନା କରାଯିବ ।

Question 2.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାକୁ ଅକ୍ଷରରେ ଲେଖ ।
(a) 3,00,600
(b) 5,04,085
(c) 27,30,000
(d) 70,53,138
Solution:
(a) 3,00,600 – ତିନି ଲକ୍ଷ ଛଅଶହ ।
(b) 5,04,085 – ପାଞ୍ଚ ଲକ୍ଷ ଚାରି ହଜାର ପଞ୍ଚାଅଶୀ ।
(c) 27,30,000 – ସତେଇଶି ଲକ୍ଷ ତିରିଶି ହଜାର ।
(d) 70,53,138 – ସତୁରୀ ଲକ୍ଷ ତେପନ ହଜାର ଶହେ ଅଡ଼ତିରିଶି ।

Question 3.
ନିମ୍ନରେ କେତେକ ସଂଖ୍ୟାକୁ ଅକ୍ଷରରେ ଲେଖାଯାଇଛି । ପ୍ରତ୍ୟେକକୁ ଭାରତୀୟ ସ୍ଥାନୀୟମାନ ଲିଖନ ପ୍ରଣାଳୀରେ ଅଙ୍କରେ ଲେଖ ।
(a) ଏକ ଲକ୍ଷ ତେଇଶି ହଜାର ଚାରିଶହ ଛପନ
(b) ଚାରି ଲକ୍ଷ ସାତ ହଜାର ସାତଶହ ଚାରି
(c) ପଚାଶ ଲକ୍ଷ ପାଞ୍ଚ ହଜାର ପଚାଶ
(d) ଦଶ ଲକ୍ଷ ଦୁଇଶହ ପଞ୍ଚତିରିଶି
Solution:
(a) ଏକ ଲକ୍ଷ ତେଇଶି ହଜାର ଚାରିଶହ ଛପନ – 1, 23, 456
(b) ଚାରି ଲକ୍ଷ ସାତ ହଜାର ସାତଶହ ଚାରି – 4, 07, 704
(c) ପଚାଶ ଲକ୍ଷ ପାଞ୍ଚ ହଜାର ପଚାଶ – 50 05, 050
(d) ଦଶ ଲକ୍ଷ ଦୁଇଶହ ପଞ୍ଚତିରିଶି – 10, 00, 235

1.2 ଏକ ଦଶକର ରାଜ୍ୟ :
Page No. (5 to 8)

ଦଶକର ରାଜ୍ୟରେ ସ୍ଵତନ୍ତ୍ର ବଟନ୍ ଥିବା ସ୍ଵତନ୍ତ୍ର କାଲ୍‌କୁଲେଟର ଅଛି ।
Question 1.
‘ଚିନ୍ତାଶୀଳ ହଜାର’ କାଲ୍‌କୁଲେଟରରେ କେବଳ ‘ + 1000’ ବଟନ୍‌ ଅଛି । ଏହାକୁ କେତେଥର ଦବାଇଲେ, ନିମ୍ନରେ ଦିଆଯାଇଥ‌ିବା ସଂଖ୍ୟା ଦେଖାଇବ :

ଯେପରି
(a) ତିନି ହଜାର – 3 ଥର ।
(b) 10,000 – ………….. ଥର
(c) ତେପନ ହଜାର – …… ଥର
(d) 90,000 – ……………….. ଥର
(e) ଏକ ଲକ୍ଷ – …………….. ଥର
(f) …………… – 153 ଥର
(g) ଏକ ଲକ୍ଷ ହେବାକୁ କେତେ ହଜାର ଆବଶ୍ୟକ ?
Solution:
(a) ତିନି ହଜାର = \(\frac{3,000}{1000}\) = 3 ଥର ।
(b) 10,000 = \(\frac{10,000}{1000}\) = 10 ଥର ।
(c) ତେପନ ହଜାର = \(\frac{53,000}{1000}\) = 53 ଥର ।
(d) 90,000 = \(\frac{90,000}{1000}\) = 90 ଥର ।
(e) ଏକ ଲକ୍ଷ = \(\frac{1,00,000}{1000}\) = 100 ଥର ।
(f) 153 ଥର = 153 × 1000 = 1,53,000 ।
(g) 1,00,000 ÷ 1000 = 100 ଏକ ଲକ୍ଷ ହେବାକୁ 100) ହଜାର ଆବଶ୍ୟକ ।

Question 2.
‘କ୍ରାନ୍ତିକାରୀ ଦଶକ’ କାଲକୁଲେଟରରେ କେବଳ ‘+10’ ବଟନ୍‌ ଅଛି । ଏହାକୁ କେତେଥର ଦବାଇଲେ, ନିମ୍ନରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟା ଦର୍ଶାଇବ ?

ଯେପରି
(a) ପାଞ୍ଚ ଶହ ? 5 ଥର
(b) 780 ? ……… ଥର
(c) 1000 ? …….. ଥର
(d) 3700 ? ……… ଥର
(e) 10,000 ? ……. ଥର
(f) ଏକ ଲକ୍ଷ ? ……….. ଥର
(g) ……… ? 453 ଥର
Solution:
(a) ପାଞ୍ଚ ଶହ = \(\frac{500}{10}\) = 50 ଥର
(b) 780 = \(\frac{780}{10}\) = 78 ଥର
(c) 1000 = \(\frac{1000}{10}\) = 100 ଥର
(d) 3700 = \(\frac{3700}{10}\) = 370 ଥର
(e) 10,000 = \(\frac{10,000}{10}\) = 1000 ଥର
(f) ଏକ ଲକ୍ଷ = \(\frac{1,00,000}{10}\) = 10000 ଥର
(g) 453 × 10 = 4530 = 453 ଥର

Question 3.
‘ଉପଯୋଗୀ ଶତକ’ କାଲକୁଲେଟରରେ କେବଳ ‘+100’ ବଟନ୍‌ ଅଛି । ଏହାକୁ କେତେଥର ଦବାଇଲେ, ନିମ୍ନରେ ଦିଆଯାଇଥ‌ିବା ସଂଖ୍ୟା ଦର୍ଶାଇବ ?

(a) ଚାରି ଶହ ? 4 ଥର
(b) 3,700 ? ………… ଥର
(c) 10,000 ? ………… ଥର
(d) ତେପନ ହଜାର ? …………… ଥର
(e) 90,0000 ? ………… ଥର
(f) 97,000 ? ………… ଥର
(g) 1,00,000 ? ………… ଥର
(h) ………… ? 582 ଥର
(i) ଦଶ ହଜାର ହେବାକୁ କେତେ ଶହ ଦରକାର ?
(j) ଏକ ଲକ୍ଷ ହେବାକୁ କେତେ ଶହ ଦରକାର ?
(k) ‘ ଉପଯୋଗୀ ଶତକ’ କାଲ୍‌କୁଲେଟର୍ କହୁଛି ‘‘ଏଠାରେ କିଛି ସଂଖ୍ୟା ଅଛି ଯାହା ‘କ୍ରାନ୍ତିକର ଦଶକ’ ଏବଂ ‘ଚିନ୍ତାଶୀଳ ହଜାର’ ଦେଖାଇ ପାରିବେ ନାହିଁ କିନ୍ତୁ ମୁଁ ପାରିବି ।’’ ଏହି ଭକ୍ତି ସତ୍ୟ କି ? ଚିନ୍ତା କର ଏବଂ ଅନୁସନ୍ଧାନ କର ।
Solution:
(a) ଚାରି ଶହ = \(\frac{400}{100}\) = 4 ଥର
(b) 3,700 = \(\frac{3700}{100}\) = 37 ଥର
(c) 10,000 = \(\frac{10,000}{100}\) = 100 ଥର
(d) ତେପନ ହଜାର = \(\frac{53,000}{100}\) = 530 ଥର
(e) 90,000 = \(\frac{90,000}{100}\) = 900 ଥର
(f) 97,000 = \(\frac{97,000}{100}\) = 970 ଥର
(g) 1,00,000 = \(\frac{1,00,000}{100}\) = 1000 ଥର
(h) \(\frac{58,200}{100}\) = 582 ଥର
(i) \(\frac{10,000}{100}\) = 100 ଥର ।
∴ ଦଶ ହଜାର ହେବାକୁ 100 ଶହ ଦରକାର ।
(j) \(\frac{1,00,000}{100}\) = = 1000 ଥର ।
∴ ଏକ ଲକ୍ଷ ହେବାକୁ 1000) ଶହ ଦରକାର ।
(k) ‘ଉପଯୋଗୀ ଶତକ’ କାଲ୍‌କୁଲେଟର୍ କହୁଛି ‘‘ଏଠାରେ କିଛି ସଂଖ୍ୟା ଅଛି ଯାହା ‘କ୍ରାନ୍ତିକର ଦଶକ’ ଏବଂ ‘ଚିନ୍ତାଶୀଳ ହଜାର’ ଦେଖାଇ ପାରିବେ ନାହିଁ କିନ୍ତୁ ମୁଁ ପାରିବି ।’’ ଏହି ଉକ୍ତି ସତ୍ୟ ।

Question 4.
‘କ୍ରିଏଟିଭ୍ ଚିଟ୍ଟି’ ଏକ ଭିନ୍ନ ପ୍ରକାରର କାଲକୁଲେଟର । ଏଥରେ ନିମ୍ନଲିଖ୍ତ ବଟନ୍‌ଗୁଡ଼ିକ ଅଛି : +1, +10, +100, +1000, +10000, +100000 ଏବଂ +1000000 । ଏହା ସବୁବେଳେ ଅନେକ ଉପାୟରେ କାମ କରିଥାଏ । ସଂଖ୍ୟା 321 ପାଇବାକୁ ହେଲେ, ଏହା +10କୁ 32 ଥର ଏବଂ +1 କୁ ଥରେ ଦବାଏ । ଏହା 321 ପାଇପାରିବ କି ? ବିକଳ୍ପ ଭାବରେ, ଏହା +100 କୁ ଦୁଇଥର ଏବଂ +10କୁ 12 ଥର ଏବଂ +1 କୁ ଥରେ ଦବାଇ ପାଇପାରିବ ।
321 କୁ ପାଇବା ପାଇଁ ଅତିବେଶିରେ + 100) ବଟନ୍‌କୁ କେତେଥର ଦବାଯାଇପାରିବ ?

Solution:
+10 ବଟନକୁ 32 ଥର ଦବାଇଲେ = 32 × 10 = 320
+1 ବଟନକୁ 1 ଥର ଦବାଇଲେ = 1
ବର୍ତ୍ତମାନ ଯୋଗ କଲେ, 320 + 1 = 321
∴ ହଁ 321 ପହଞ୍ଚିବ ।
ବିକଳ୍ପ ଭାବରେ, +100 ବଟନକୁ 2 ଥର ଚିପିଲେ = 2 × 100 = 200
+10 ବଟନକୁ 12 ଥର ଚିପିଲେ = 12 × 10 = 120
+1 ବଟନକୁ 1 ଥର ଟିପିଲେ = 1 × 1 = 1
ବର୍ତ୍ତମାନ ଯୋଗକଲେ, 200 + 120 + 1 = 321
ଆଉଥରେ ଏହା 321ରେ ପହଞ୍ଚିବ ।
+100 ବଟନକୁ 3 ଥର ଟିପିଲେ = 3 × 100 = 300
+10 ବଟନକୁ 2 ଥର ଟିପିଲେ = 2 × 10 = 20
+1 ବଟନକୁ 1 ଥର ଟିପିଲେ = 1 × 1 = 1
ବର୍ତ୍ତମାନ ଯୋଗକଲେ = 300 + 20 + 1 = 321
∴ 321 କୁ ପାଇବା ପାଇଁ ଅତିବେଶିରେ +100 ବଟନକୁ 3 ଥର ଦବାଯାଇପାରିବ ।

Question 5.
5072 ପାଇବା ପାଇଁ ଅନେକ ଉପାୟ ମଧ୍ୟରୁ ଦୁଇଟି ଉପାୟ ନିମ୍ନରେ ଦର୍ଶାଯାଇଛି ; ଏହି ଦୁଇଟି ଉପାୟକୁ ଏହିପରି ପରିପ୍ରକାଶ କରାଯାଇପାରିବ :
(a) 50 × 100) + (7 × 10) + (2 × 1) = 5072
(b) (3 × 1000) + (20 × 100) + (72 × 1) = 5072
• 5072 ପାଇବା ପାଇଁ ଏକ ଭିନ୍ନ ଉପାୟ ଖୋଜ ଏବଂ ଏହା ପାଇଁ ଏକ ପରିପ୍ରକାଶ ଲେଖ ।
Solution:
ଏଠାରେ 5 × 1000 + 0 × 100 + 7 × 10 + 2 × 1
= 5000 + 0 + 70 + 2
= 5072

ନିଜେ କରି ଦେଖ 

Question 1.
ନିଜେ କରି ଦେଖ । ତଳେ ଦିଆଯାଇଥିବା ପ୍ରତ୍ୟେକ ସଂଖ୍ଯାପାଇଁ, ବଟନ୍ ଦବାଇବା-ମାଧ୍ୟମରେ ସଂଖ୍ୟା ପରିପ୍ରକାଶ କରିବା ପାଇଁ ଅତି କମ୍‌ରେ ଦୁଇଟି ଭିନ୍ନ ଉପାୟ ଲେଖ । ‘କ୍ରିଏଟିଭ୍ ଚିଟ୍ଟି? ପରି ଚିନ୍ତାକର ଏବଂ ସୃଜନଶୀଳ ହୁଅ ।
(a) 8300
(b) 40629
(c) 56354
(d) 66666
(e) 367813
Solution:
(a) 8300
(i) (8 × 1000) + (3 × 100)
(ii) (83 × 1000)

(b) 40629
(i) (4 × 10,000) + (6 × 100) + (2 × 10) + (9 × 1)
(ii) (406 × 100) + (2 × 10) + (9 × 1)

(c) 56354
(i) (56 × 1000) + (35 × 10) + (4 × 1)
(ii) (5635 × 10) + (4 × 1)

(d) 66666
(i) (6 × 10000 + (66 × 100) + (66 × 1)
(ii) (66 × 1000) + (666 × 1)

(e) 367813
(i) (367 × 1000 + (813 × 1)
(ii) (3 × 100000 + (67 × 1000) + (8 × 100) + (13 × 1)

Question 2.
ତୁମପାଇଁ କ୍ରିଏଟିଭ୍ ଚିଟ୍ଟି କାଲକୁଲେଟରର କିଛି ପ୍ରଶ୍ନ :

(a) ତୁମକୁ ଠିକ୍ 30 ଥର ବଟନ୍ ଦବାଇବାକୁ ପଡ଼ିବ । ତୁମେ କେଉଁ ତିନିଅଙ୍କ ବିଶିଷ୍ଟ ବୃହତ୍ତମ ସଂଖ୍ୟା ତିଆରି କରିପାରିବ ? ତୁମେ କେଉଁ ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟାଟି ତିଆରି କରିପାରିବ ?
Solution:
ଏଠାରେ (9 × 100) + (8 × 10) + (13 × 1) = 993
ସମୁଦାୟ ବଟନ ଦବାଇବାକୁ ହେବ = 9 + 8 + 13 = 30 ଥର ।
∴ ଆମେ 30 ଥର ବଟନ୍ ଦବାଇଲାପରେ ତିନିଅଙ୍କ ବିଶିଷ୍ଟ ବୃହତ୍ତମ ସଂଖ୍ୟା ୨93 ତିଆରି କରିପାରିବା ।
ଏଠାରେ (8 × 10) + (22 × 1) = 80 + 22 = 102
ସମୁଦାୟ ବଟନ ଦବାଇବାକୁ ହେବ = 8 + 22 = 30
∴ ଆମେ 30 ଥର ବଟନ୍ ଦବାଇଲାପରେ ତିନିଅଙ୍କ ବିଶିଷ୍ଟ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟା 102 ତିଆରି କରିପାରିବା ।

(b) 25 ଥର ବଟନ୍ ଦବାଇ 997 ସଂଖ୍ୟାଟି ପାଇପାରିବା । ତୁମେ ଭିନ୍ନ ଭିନ୍ନ ଥର ବଟନ୍ ଦବାଇବା ମାଧ୍ୟମରେ
997 ସଂଖ୍ୟାଟି ତିଆରି କରିପାରିବ କି ?
Solution:
ଏଠାରେ (8 × 100) + (19 × 10) + (7 × 1) = 800 + 190 + 7 = 997
ସମୁଦାୟ ବଟନ୍ ଦବାଇବାକୁ ହେବ = 8 + 19 + 7 = 34
∴ ଆମେ 34 ଥର ବଟନ୍ ଦବାଇବା ମାଧ୍ୟମରେ ୨97 ସଂଖ୍ୟାଟି ତିଆରି କରିପାରିବା ।

Question 3.
ସିଷ୍ଟେମେଟିକ୍ ସିୱି ଏକ ଭିନ୍ନ ପ୍ରକାରର କାଲ୍‌କୁଲେଟର୍ । ଏଥରେ ନିମ୍ନଲିଖ୍ ବଟନ୍‌ଗୁଡ଼ିକ ଅଛି – +1, +10, +100, +1000, +10000, +100000 । ଏହାର ବଟନ୍‌କୁ ଯଥାସମ୍ଭବ କମ୍ ଥର ଦବାଇ ବିଭିନ୍ନ ସଂଖ୍ୟା ଗଠନ କର । ଆମେ କିପରି ଯଥାସମ୍ଭବ କମ୍ ଥର ବଟନ୍ ଦବାଇ ସଂଖ୍ୟାଗୁଡ଼ିକ ପାଇପାରିବା ?

(a) 5072
(b) 8300
Solution:
(a) ଏଠାରେ 5072 = (5 × 1000) + (7 × 10) + (2 × 1)
ସମୁଦାୟ ଥର ବଟନ୍ ଦବାଇବା ସଂଖ୍ୟା = 5 + 7 + 2 = 14
∴ 14 ଥର ବଟନ୍ ଦବାଇ ଆମେ 5072 ସଂଖ୍ୟାଟି ପାଇପାରିବା ।

(b) ଏଠାରେ 8300 = (8 × 1000) + (3 × 00) ସମୁଦା
ୟ ଥର ବଟନ୍ ଦବାଇବା ସଂଖ୍ୟା = 8 + 3 = 11
∴ 11 ଥର ବଟନ୍ ଦବାଇ ଆମେ 8300 ସଂଖ୍ୟାଟି ପାଇପାରିବା ।

• 23 ଥର ଠାରୁ କମ୍ ଥର ବଟନ୍ ଦବାଇ 5072 ପାଇବାର ଅନ୍ୟ କୌଣସି ଉପାୟ ଅଛି କି ? ସେଥ‌ିପାଇଁ ପରିପ୍ରକାଶଟି ଲେଖ ।
Solution:
ଏଠାରେ (5 × 1000) + (7 × 10) + (2 × 1)
ସମୁଦାୟ ଥର ବଟନ୍ ଦବାଇବା = 5 + 7 + 2 = 14
∴ 5072 ସଂଖ୍ୟାଟି ପାଇବା ପାଇଁ ଅତିକମ୍‌ରେ 14 ଥର ବଟନ୍ ଦବାଇବାକୁ ପଡ଼ିବ ।

ନିଜେ କରି ଦେଖ 

Question 1.
ପୂର୍ବ ଅଭ୍ୟାସରେ ଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକ ପାଇଁ ସର୍ବନିମ୍ନ ସଂଖ୍ୟକ ବଟନ୍ ଦବାଇ କିପରି ପ୍ରତ୍ୟେକ ସଂଖ୍ୟା ପାଇବ, ତାହା ଖୋଜ ଏବଂ ସେଗୁଡ଼ିକର ପରିପ୍ରକାଶ ଲେଖ ।
(a) 8300
(b) 40629
(c) 56354
(d) 66666
(e) 367813
Solution:
(a) 8300 = (8 × 1000) + (3 × 100)
∴ 8300 ସଂଖ୍ୟାଟି ପାଇପାରିବା = 8 + 3 = 11 ଥର ବଟନ୍ ଦବାଇବା ପରେ ।

(b) 40629 = (4 × 10,000 + (6 × 100) + (2 × 10) + (9 × 1)
∴ 40629 ସଂଖ୍ୟାଟି ପାଇପାରିବା = 4 + 6 + 2 + 9 = 21 ଥର ବଟନ୍ ଦବାଇବା ପରେ ।

(c) 56354 = (5 × 10,000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1)
∴ 56354 ସଂଖ୍ୟାଟି ପାଇପାରିବା = 5 + 6 + 3 + 5 + 4 = 23 ଥର ବଟନ୍ ଦବାଇବା ପରେ ।

(d) 66666 = (6 × 10,000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1)
∴ 66666 ସଂଖ୍ୟାଟି ପାଇପାରିବା = 6 + 6 + 6 + 6 + 6 = 30 ଥର ବଟନ୍ ଦବାଇବା ପରେ ।

(e) 367813 = (3 × 1,00,000) + (6 × 10,000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1)
∴ 367813 ସଂଖ୍ୟାଟି ପାଇପାରିବା = 3 + 6 + 7 + 8 + 1 + 3 = 28 ଥର ବଟନ୍ ଦବାଇବା ପରେ ।

Question 2.
ପ୍ରତ୍ୟେକ ସଂଖ୍ୟା ଏବଂ ଅନୁରୂପ ସର୍ବନିମ୍ନ ସଂଖ୍ୟକ ବଟନ୍ ଦବାଇବା ମଧ୍ଯରେ କୌଣସି ସମ୍ପର୍କ ଥୁବାର ତୁମେ ଦେଖିପାରୁଛ କି ?
Solution:
ସର୍ବନିମ୍ନ ସଂଖ୍ୟକ ବଟନ୍ ଦବାଇବା = ସଂଖ୍ୟାରେ ଥିବା ଅଙ୍କମାନଙ୍କର ସମଷ୍ଟି ।

Question 3.
ତୁମେ ଲକ୍ଷ୍ୟ କର ଯେ, ସର୍ବନିମ୍ନ ବଟନ୍ ଦବାଇବା ପରିପ୍ରକାଶଗୁଡ଼ିକ, ସଂଖ୍ୟାଗୁଡ଼ିକର ଭାରତୀୟ ସ୍ଥାନୀୟମାନ ଲିଖନ ପ୍ରଣାଳୀ ମଧ୍ୟ ଦେଇଥାଏ । ଏହାର କାରଣ ବିଷୟରେ ତୁମେ ଚିନ୍ତାକର ।
Solution:
ଭାରତୀୟ ସ୍ଥାନୀୟମାନ ଲିଖନ ପ୍ରଣାଳୀଦ୍ଵାରା ଆମେ ହଜାର, ଲକ୍ଷ ଓ କୋଟି ସ୍ଥାନୀୟମାନଙ୍କୁ ସହଜରେ ଦଳ କରି ଜାଣିଥାଉ ।

1.3 କୋଟି ବା କୋଟିରୁ ଅଧ୍ବକ :

Page No. 9

Question 1.
ଏକ ହଜାର ଲକ୍ଷରେ କେତୋଟି ଶୂନ ଅଛି ? ………………….
Solution:
ଏକ ହଜାର ଲକ୍ଷ = 10,00,00,000 ରେ 5 ଟି ଶୂନ ଅଛି ।

Question 2.
ଏକ ଶହ ହଜାରରେ କେତୋଟି ଶୂନ ଅଛି ?
Solution:
ଏକ ଶହ ହଜାର = 1,00,000 ରେ 5 ଟି ଶୂନ ଅଛି ।

9876501234 ସଂଖ୍ୟାଟିକୁ ପ୍ରଥମେ କମା ଦେଇ ସହଜରେ ପଢ଼ିହେବ ।
(a) 9,87,65,01,234 = ୨ ଅରବ 87 କୋଟି 65 ଲକ୍ଷ 1 ହଜାର 234 କିମ୍ବା 987 କୋଟି 65 ଲକ୍ଷ 1 ହଜାର 234 (ଭାରତୀୟ ଲିଖନ ପ୍ରଣାଳୀରେ)
(b) 9,87,65,01,234 = 9 ବିଲିୟନ୍ 876 ମିଲିୟନ୍ 501 ହଜାର 234 (ଆମେରିକୀୟ ଲିଖନ ପ୍ରଣାଳୀରେ)

ନିଜେ କରି ଦେଖ 

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଭାରତୀୟ ସ୍ଥାନୀୟମାନ ଅନୁସାରେ ପଢ଼ ଏବଂ ସେଗୁଡ଼ିକର ସଂଖ୍ୟା ନାମ ଉଭୟ ଭାରତୀୟ ଓ ଆମେରିକୀୟ ପ୍ରଣାଳୀରେ ଲେଖ ।
(a) 4050678
(b) 48121620
(c) 20022002
(d) 246813579
(e) 34500543
(f) 1020304050
Solution:
(a) ଭାରତୀୟ ପ୍ରଣାଳୀ : 40,50,678 = ଚାଳିଶି ଲକ୍ଷ ପଚାଶ ହଜାର ଛଅ ଶହ ଅଠସ୍ତରୀ ।
ଆମେରିକୀୟ ପ୍ରଣାଳୀ : 4,050,678 = ଚାରି ମିଲିୟନ୍ ପଚାଶ ହଜାର ଛଅ ଶହ ଅଠସ୍ତରୀ ।

(b) ଭାରତୀୟ ପ୍ରଣାଳୀ : 4,81,21,620 = ଚାରି କୋଟି ଏକାଅଶୀ ଲକ୍ଷ ଏକୋଇଶି ହଜାର ଛଅ ଶହ କୋଡ଼ିଏ ।
ଆମେରିକୀୟ ପ୍ରଣାଳୀ : 48,121,620 = ଅଡ଼ଚାଳିଶ ମିଲିୟନ୍ ଶହେ ଏକୋଇଶି ହଜାର ଛଅ ସହ କୋଡ଼ିଏ ।

(c) ଭାରତୀୟ ପ୍ରଣାଳୀ : 2,00,22,002 = ଦୁଇ କୋଟି ବାଇଶି ହଜାର ଦୁଇ ।
ଆମେରିକୀୟ ପ୍ରଣାଳୀ : 20, 022, 002 = କୋଡ଼ିଏ ମିଲିୟନ୍ ବାଇଶି ହଜାର ଦୁଇ ।

(d) ଭାରତୀୟ ପ୍ରଣାଳୀ : 24,68,13,579 = ଚବିଶି କୋଟି ଅଡ଼ଷଠୀ ଲକ୍ଷ ତେର ହଜାର ପାଞ୍ଚଶହ ଅଣାଅଶୀ ।
ଆମେରିକୀୟ ପ୍ରଣାଳୀ : 246,813,579 = ଦୁଇଶହ ଛୟାଳିଶି ମିଲିୟନ୍ ଆଠ ଶହ ତେର ହଜାର ପାଞ୍ଚଶହ ଅଣାଅଶୀ ।

(e) ଭାରତୀୟ ପ୍ରଣାଳୀ : 34,50,00,543 = ଚଉତିରିଶି କୋଟି ପଚାଶ ହଜାର ପାଞ୍ଚଶହ ତେୟାଳିଶି ।
ଆମେରିକୀୟ ପ୍ରଣାଳୀ : 345,000,543 = ତିନିଶହ ପଇଁଚାଳିଶି ମିଲିୟନ୍ ପାଞ୍ଚଶହ ତେୟାଳିଶି ।

(f) ଭାରତୀୟ ପ୍ରଣାଳୀ : 1,02,03,04,050 = ଏକ ଅରବ ଦୁଇ କୋଟି ତିନି ଲକ୍ଷ ଚାରି ହଜାର ପଚାଶ ।
ଆମେରିକୀୟ ପ୍ରଣାଳୀ : 1,020,304,050 = ଏକ ବିଲିୟନ୍ କୋଡ଼ିଏ ମିଲିୟନ୍ ତିନିଶହ ଚାରି ହଜାର ପଚାଶ ।

Question 2.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଭାରତୀୟ ସ୍ଥାନୀୟମାନ ପଦ୍ଧତି ଅନୁସାରେ ଲେଖ ।
(a) ଏକ କୋଟି ଏକ ଲକ୍ଷ ଏକ ହଜାର ଦଶ
(b) ଏକ ବିଲିୟନ୍ ଏକ ମିଲିୟନ ଏକ ହଜାର ଏକ
(c) ଦଶ କୋଟି କୋଡ଼ିଏ ଲକ୍ଷ ତିରିଶ ହଜାର ଚାଳିଶ
(d) ନଅ ବିଲିୟନ୍ ଅଶୀ ମିଲିୟନ ସାତ ଶହ ହଜାର ଛଅ ଶହ
Solution:
(a) ଏକ କୋଟି ଏକ ଲକ୍ଷ ଏକ ହଜାର ଦଶ = 1,01,01,010
(b) ଏକ ବିଲିୟନ୍ ଏକ ମିଲିୟନ ଏକ ହଜାର ଏକ = 1,00,10,01,001
(c) ଦଶ କୋଟି କୋଡ଼ିଏ ଲକ୍ଷ ତିରିଶ ହଜାର ଚାଳିଶ = 10,20,30,040
(d) ନଅ ବିଲିୟନ୍ ଅଶୀ ମିଲିୟନ ସାତ ଶହ ହଜାର ଛଅ ଶହ = 9,080,700,600

Question 3.
ତୁଳନା କର ଏବଂ ‘<‘, ‘>’ କିମ୍ବା ‘=’ ଲେଖ ।
(a) 30 ହଜାର ……………… 3 ଲକ୍ଷ
(b) 500 ଲକ୍ଷ ……….. 5 ମିଲିୟନ୍
(c) 800 ହଜାର ………….. 8 ମିଲିୟନ୍
(d) 640 ହଜାର ………….. 30 ବିଲିୟନ୍
Solution:
(a) 30 ହଜାର ( < ) 3 ଲକ୍ଷ ∴ 30,000 < 3,00,000
(b) 500 ଲକ୍ଷ (> ) 5 ମିଲିୟନ୍ ∴ 5,00,00,000 > 5,000,000
(c) 800 ହଜାର ( < ) 8 ମିଲିୟନ୍ ∴ 8,00,000 < 8,000,000
(d) 640 କୋଟି ( < ) 60 ବିଲିୟନ୍ ∴ 6,40,00,00,000 < 60,000,000,00

1.4 ଠିକ୍ ଏବଂ ଆନୁମାନିକ ମାନ :

Page No. (10 to 12)

Question 1.
ସେହି ପରିସ୍ଥିତିଗୁଡ଼ିକୁ ଦଳଭିତରେ ବିଚାର କର ଏବଂ ଆଲୋଚନା କର ଯେଉଁଠାରେ (କ) ଆସନ୍ନମାନଟି ପ୍ରକୃତ ମାନଠାରୁ ଅଧିକ ହେବ । (ଖ) ଆସନ୍ନମାନଟି ପ୍ରକୃତମାନଠାରୁ କମ୍ ହେବ (ଗ) ଆସନ୍ନମାନଟି ପ୍ରକୃତ ମାନଠାରୁ ଅଧିକ ବା କମ୍ ହେବ, (ଘ) ଠିକ୍ ସଂଖ୍ୟା ଆବଶ୍ୟକ ହେବ ।
Solution:
(କ) ନିକଟତମ ଦଶକ ଆସନ୍ନମାନ ନିର୍ଣ୍ଣୟ :
1 ରୁ 4 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକ 10 ଅପେକ୍ଷା ‘0’ ର ନିକଟତମ ।
ତେଣୁ 1, 2, 3, 4 ଆସନ୍ନମାନ ‘0’ ଅଟେ ।
6 ରୁ 9 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକ ‘0’ ଅପେକ୍ଷା ’10’ ର ନିକଟତମ ।
ତେଣୁ 6, 7, 8, 9 ଆସନ୍ନମାନ 10 ଅଟେ ।
5 ସଂଖ୍ୟାଟି 0 ଓ 10 ଠାରୁ ସମଦୂରବର୍ତ୍ତୀ ।
ତେଣୁ 5, ସଂଖ୍ୟାଟିର ଆସନ୍ନମାନ 10 ଅଟେ ।
ତେଣୁ 12, 25 ଓ 17 ସଂଖ୍ୟାଗୁଡ଼ିକର ଆସନ୍ନମାନ ଯଥାକ୍ରମେ 10, 30 ଓ 20 ।

(ଖ) ନିକଟତମ ଶତକ ଆସନ୍ନମାନ ନିର୍ଣ୍ଣୟ :
1 ରୁ 49 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକ 100 ଅପେକ୍ଷା ‘0’ ନିକଟତମ ।
ତେଣୁ ଏହାର ଆସନ୍ନମାନ = 0
ସେହିପରି 51 ରୁ 99 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକର ଆସନ୍ନମାନ = 100 ।
ସେହିପରି 212, 375 ଓ 495 ର ନିକଟତମ ଆସନ୍ନମାନ ଯଥାକ୍ରମେ 200, 400 ଓ 500 ।

(ଗ) ନିକଟତମ ହଜାରର ଆସନ୍ନମାନ ନିର୍ଣ୍ଣୟ :
1 ରୁ 499 ସଂଖ୍ୟାଗୁଡ଼ିକ 1000 ଅପେକ୍ଷା ‘0’ ର ନିକଟତମ ।
ତେଣୁ ଏହାର ଆସନ୍ନମାନ = 0
କିନ୍ତୁ 501 ଠାରୁ 999 ପର୍ଯ୍ୟନ୍ତ ସଂଖ୍ୟାଗୁଡ଼ିକର ଆସନ୍ନମାନ = 1000

ମନେରଖ :
– ନିକଟତମ ଦଶକର ଆସନ୍ନମାନ ଏକ ଅଙ୍କ ବିଶିଷ୍ଟ । ଯଦି ଏକକ ଘରର ଅଙ୍କ 5 ଠାରୁ କମ୍ ହେବ, ତେବେ ଏହି ସଂଖ୍ୟାମାନଙ୍କର ଆସନ୍ନମାନରେ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଟି 0 ହେବ ।
– ନିକଟତମ ଶତକର ଆସନ୍ନମାନରେ ଯଦି ଦଶକ ଘରର ଅଙ୍କଟି 5 କିମ୍ବା ତାଠାରୁ ବଡ଼ ତେବେ ଏହାର ଆସନ୍ନମାନ ଏକକ ଓ ଦଶକ ଅଙ୍କ ପ୍ରତ୍ୟେକ ) ଓ ଶତକ ଅଙ୍କ । ଅଧିକ ହେବ । ଉଦାହରଣ 153 → 200

ନିକଟତମ ପଡ଼ୋଶୀ :
Question 2.
ବଡ଼ ସଂଖ୍ୟାଗୁଡ଼ିକ ସହିତ ନିକଟତମ ହଜାର, ଲକ୍ଷ କିମ୍ବା କୋଟି ଜାଣିବା ଉପଯୋଗୀ ଅଟେ । (ଆସନ୍ତମାନ) ଲେଖ ।
ଉଦାହରଣ : ସଂଖ୍ୟା 6,72,85,183 ର ନିକଟତମ ସଂଖ୍ୟା
Solution:

ନିକଟତମ ହଜାର	6,7,2,85,000	ନିକଟତମ 10 ହଜାର	6,72,90,000
ନିକଟତମ ଲକ୍ଷ	6,73,00,000	ନିକଟତମ 10 ଲକ୍ଷ	6,70,00,000
ନିକଟତମ କୋଟି	7,00,00,000

Question 3.
ନିମ୍ନ ସଂଖ୍ୟାଗୁଡ଼ିକର ପାଞ୍ଚୋଟି ନିକଟତମ ସଂଖ୍ୟା (ଆସନମାନ) ଲେଖ ।
(a) 3,87,69,957
(b) 29,05,32,481
Solution:
(a) 3,87,69,957
ଏଠାରେ ନିକଟତମ ହଜାର 3,87,70,000
ନିକଟତମ 10 ହଜାର 3,87,70,000
ନିକଟତମ ଲକ୍ଷ 3,88,00,000
ନିକଟତମ 10 ଲକ୍ଷ 3,90,00,000
ନିକଟତମ କୋଟି 4,00,00,000

(b) 29,05,32,481
ନିକଟତମ ହଜାର 29,05,32,000
ନିକଟତମ 10 ହଜାର 29,05,30,000
ନିକଟତମ ଲକ୍ଷ 29,05,00,000
ନିକଟତମ 10 ଲକ୍ଷ 29,10,00,000
ନିକଟତମ କୋଟି 29,00,00,000

Question 4.
ତୁମ ପାଖରେ ଏକ ସଂଖ୍ୟା ଅଛି ଯାହାର ନିକଟବର୍ତୀ ଏହିପରି ପାଞ୍ଚୋଟି ସଂଖ୍ୟା 5,00,00,000 ଅଟେ । ସଂଖ୍ୟାଟି କେତେ ? ଏହିପରି କେତୋଟି ସଂଖ୍ୟା ଅଛି ?
Solution:
4,99,99,500 ଓ 5,00,00,000 (ପ୍ରଥମ ଓ ଶେଷ ସଂଖ୍ୟାକୁ ବାଦ୍ ଦେଇ) ମଧ୍ୟବର୍ତ୍ତୀ ଏକ ସଂଖ୍ୟା, ଯାହାର ନିକଟବର୍ତ୍ତୀ ଏହିପରି ପାଞ୍ଚଟି ସଂଖ୍ୟାର ଆସନ୍ନମାନ 5,00,00,000 । ଏହିପରି 1000)ଟି ସଂଖ୍ୟା ଅଛି ।

1. 4,63,128 + 4,19,682
ନମିତା : ‘‘ଯୋଗଫଳ ପ୍ରାୟ 8,00,000 ଏବଂ 8,00,000 ରୁ ଅଧିକ ’’
ରମେଶ : ‘‘ଯୋଗଫଳ ପ୍ରାୟ 9,00,000 ଏବଂ 9,00,000 ରୁ କମ୍ ।”
ନମିତା ଏବଂ ରମେଶ ସରଳ ପରିପ୍ରକାଶର ମାନ ଆକଳନ କରିଛନ୍ତି ।

(a) ଏହି ଆକଳନଗୁଡ଼ିକ ଠିକ୍ ଅଛି କି ? କାହାର ଆକଳନ ଯୋଗଫଳର ନିକଟତର ?
Solution:
ଏହି ଆକଳନଗୁଡ଼ିକ ଠିକ୍ ଅଛି । ରମେଶର ଆକଳନ ଯୋଗଫଳର ନିକଟତର ।

(b) ଯୋଗଫଳ 8,50,000 ରୁ ଅଧିକ ହେବ କିମ୍ବା 8,50,000 ରୁ କମ୍ ହେବ ? ତୁମେ ଏହା କାହିଁକି ଭାବୁଛ ?
Solution:
ହଁ, ଏହାର ଯୋଗଫଳ 8,50,000 ଠାରୁ ବଡ଼ ହୋଇପାରେ । କାରଣ ଯଦି ଆମେ ଏହାର ବାମରୁ 2ଟି ଅଙ୍କ ଯୋଗ କରିବା ଯୋଗଫଳ 87 ହେବ, ଯାହା 85 ଠାରୁ ବଡ଼ ।

(c) ଯୋଗଫଳ 8,83,128 ରୁ ଅଧିକ ହେବ କିମ୍ବା 8,83,128 ରୁ କମ୍ ହେବ ? ତୁମେ ଏପରି କାହିଁକି ଭାବୁଛ ?\
Solution:
ଯୋଗଫଳ 8,83,128ରୁ କମ୍ ହେବ ।

(d) 4,63,128 + 4,19,682ର ଠିକ୍ ମୂଲ୍ୟ କେତେ ହେବ ?
Solution:
4,63,128 + 4,19,682 = 8,82,810 ହେବ ।

2. 14,63,128 – 4,90,020
ନମିତା : ‘‘ପାର୍ଥକ୍ୟ ପାଖାପାଖୁ 10 ଲକ୍ଷ ଏବଂ 10 ଲକ୍ଷରୁ କମ୍ ।”
ରମେଶ : ‘‘ପାର୍ଥକ୍ୟ ପାଖାପାଖୁ 9 ଲକ୍ଷ ଏବଂ 9 ଲକ୍ଷରୁ ବେଶୀ ।’’

(a) ଏହି ଆକଳନଗୁଡ଼ିକ ଠିକ୍ ଅଛି କି ? କାହାର ଆକଳନ ପାର୍ଥକ୍ୟର ନିକଟତର ?
Solution:
ଏହାର ପ୍ରକୃତ ପାର୍ଥକ୍ୟ = 14,63,128 – 4,90,020 = 9,73,108
ପାଖାପାଖୁ ପାର୍ଥକ୍ୟ = 15,00,000 – 5,00,000 = 10,00,000
∴ ନମିତାର ଆକଳନ ପ୍ରକୃତ ପାର୍ଥକ୍ୟର ନିକଟତମ ।

(b) ପାର୍ଥକ୍ୟ 9,50,000 ରୁ ଅଧିକ ହେବ କିମ୍ବା 9,50,000 ରୁ କମ୍ ହେବ । ତୁମେ ଏପରି କାହିଁକି ଭାବୁଛ ?
Solution:
ପ୍ରକୃତ ପାର୍ଥକ୍ୟ = 9,73,108 । ଏହା 9,50,000 ଠାରୁ ବଡ଼ ।
ବର୍ତ୍ତମାନ 14,63,128 ଓ 4,90,020ର ଆକଳନ ଆସନ୍ନ ନିକଟତମ 10 ହଜାର ଘର ।
ଆମେ ପାଇବା 14,60,000 ଓ 4,90,000
∴ ପାର୍ଥକ୍ୟ = 14,60,000 – 4,90,000 = 9,70,000
ଏହା 9,50,000 ଠାରୁ ବଡ଼ ।

(c) ପାର୍ଥକ୍ୟ 9,63,128 ରୁ ଅଧିକ ହେବ କିମ୍ବା 9,63,128 ରୁ କମ୍ ହେବ କି ? ତୁମେ ଏପରି କାହିଁକି ଭାବୁଛ ?
Solution:
ପ୍ରକୃତ ପାର୍ଥକ୍ୟ = 9,73,108 । ଏହା 9,63,128 ଠାରୁ ବଡ଼ ।
∴ ପାର୍ଥକ୍ୟ = 9,,73,108 – 9,63,128 = 9,98୦
ଏହା ପ୍ରକୃତ ପାର୍ଥକ୍ୟଠାରୁ ଦୂର ।

(d) 14,63,128 – 4,90,020 ର ଠିକ୍ ମୂଲ୍ୟ ଆଳକନ କର ।
Solution:
ପ୍ରକୃତ ମୂଲ୍ୟ = 14,63,128 – 4,90,000 = 9,73,108

ସହରମାନଙ୍କର ଜନସଂଖ୍ୟା

ଉପରେ ଦିଆଯାଇଥବା ସାରଣୀର ତଥ୍ୟ ଅନୁଯାୟୀ ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।
Question 1.
ଉପରୋକ୍ତ ତଥ୍ୟ ବିଷୟରେ ତୁମ ଶ୍ରେଣୀରେ ଆଲୋଚନା କର ।
Solution:
ଅଧିକାଂଶ ସହରର ଲୋକସଂଖ୍ୟା ଉଲ୍ଲେଖନୀୟ ଭାବରେ ବୃଦ୍ଧି ପାଉଥିବାର ଦେଖାଯାଉଛି । ବେଙ୍ଗାଲୁରୁ, ହାଇଦ୍ରାବାଦ, ସୁରତ ଇତ୍ୟାଦି ସହରର ଲୋକସଂଖ୍ୟା ଦୁଇଗୁଣ ଲେଖାଏଁ ବୃଦ୍ଧି ପାଇଛି । କୋଲକାତା ପରି ଅଳ୍ପ କେତେକ ସହରର ଲୋକସଂଖ୍ୟା ଅଳ୍ପ ପରିମାଣରରେ ବୃଦ୍ଧି ଘଟିଛି ।

Question 2.
ଉପରୋକ୍ତ ସାରଣୀ ପାଇଁ ଏକ ଉପଯୁକ୍ତ ଶୀର୍ଷକ କ’ଣ ହେବ ?
Solution:
2001 ଓ 2011ରେ ପ୍ରମୁଖ ଭାରତୀୟ ସହରର ଜନସଂଖ୍ୟା ।

Question 3.
2011 ମସିହାରେ ଭୁବନେଶ୍ଵର ସହରର ଜନସଂଖ୍ୟା କେତେ ? ଆନୁମାନିକ ଭାବରେ, 2001 ତୁଳନାରେ ଏହା କେତେ ବୃଦ୍ଧି ପାଇଛି ?
Solution:
2011 ରେ ଭୁବନେଶ୍ଵରର ଜନସଂଖ୍ୟା = 8,85,363
2001 ରେ ଭୁବନେଶ୍ଵରର ଜନସଂଖ୍ୟା ଥିଲା = 6,58,220
ଜନସଂଖ୍ୟାର ବୃଦ୍ଧି = 8,85,363 – 6,58,220 = 2,27,143
∴ ଆନୁମାନିକ ଭାବରେ 2001 ତୁଳନାରେ ଏହା 2.2 ଲକ୍ଷ ବୃଦ୍ଧି ପାଉଛି

Question 4.
2001 ଏବଂ 2011 ମଧ୍ୟରେ କେଉଁ ସହରର ଜନସଂଖ୍ୟା ସର୍ବାଧକ ବୃଦ୍ଧି ପାଇଛି ?
Solution:
କେତେକ ସହରର ଜନସଂଖ୍ୟା ବୃଦ୍ଧି ପାଇଛି ।
ହାଇଦ୍ରାବାଦ = 68,09,970 – 36,37,483 = 31,72,487
ନୂଆଦିଲ୍ଲୀ = 1,10,07,835 – 98,79,172 = 11,28,663
ବେଙ୍ଗାଲୁରୁ = 84,25,970 – 43,01,326 = 41,24,644
ଅହମ୍ମଦାବାଦ = 55,70,585 – 35,20,085 = 20,25,500

Question 5.
କେଉଁ କେଉଁ ସହରର ଜନସଂଖ୍ୟା 2001 ତୁଳନାରେ ପ୍ରାୟ ଦୁଇଗୁଣ ହୋଇଛି ?
Solution:
ହାଇଦ୍ରାବାଦର ଜନସଂଖ୍ୟା 2001 ତୁଳନାରେ ପ୍ରାୟ ଦୁଇଗୁଣ ହୋଇଛି ।

Question 6.
ଭୁବନେଶ୍ଵରର ଜନସଂଖ୍ୟାକୁ କେଉଁ ସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନକଲେ ମୁମ୍ବାଇ ଜନସଂଖ୍ୟାର ପାଖାପାଖୁ ହେବ ?
Solution:
ଭୁବନେଶ୍ୱରର ଜନସଂଖ୍ୟା (2001) = 6,58,220
ମୁମ୍ବାଇର ଜନସଂଖ୍ୟା (2001) = 1,19,78,450
ଗୁଣ୍ଯ = 1,19,78,450 ÷ 6,58,220 = 18.2
∴ ଭୁବନେଶ୍ଵରର ଜନସଂଖ୍ୟାକୁ 18 ଦ୍ଵାରା ଗୁଣନ କଲେ ମୁମ୍ବାଇ ସହରର ଜନସଂଖ୍ୟା ପାଖାପାଖୁ ହେବ ।

1.5 ଗୁଣନରେ ସଂରଚନା :

Page No. (13 to 15)

• ଏକ ଗୁଣନ ସମ୍ବନ୍ଧୀୟ ଖେଳରେ ଦୁଇଜଣ ପିଲା ଗୋଟିଏ ସଂଖ୍ୟାକୁ 10,100, 1000, ….ଇତ୍ୟାଦି ଦ୍ଵାରା ଗୁଣିବା ପାଇଁ ଏକ ଆକର୍ଷଣୀୟ କୌଶଳ ପାଇପାରିଛନ୍ତି ।

ସଂକ୍ଷିପ୍ତ ପ୍ରଣାଳୀରେ ଗୁଣନ
Question 1.
ଗୋଟିଏ ପିଲା 116 × 5 ର ଗୁଣଫଳକୁ ନିମ୍ନ ପ୍ରକ୍ରିୟାରେ ନିଶ୍ଚୟ କଲା ।
Solution:
116 × 5 = = 58 × 10 = 580

Question 2.
ଅନ୍ୟ ଜଣକ ନିମ୍ନ ପ୍ରକ୍ରିୟାରେ 824 × 25 ର ଗୁଣଫଳ ନିଶ୍ଚୟ କଲା ।
Solution:
824 × 25 = = 20,600

Question 3.
ଗୁଣନ ଓ ଭାଗକ୍ରିୟାଦ୍ଵାରା ଗୋଟିଏ ସଂଖ୍ୟାକୁ 5 ଦ୍ଵାରା ଗୁଣିବା ଅର୍ଥ, ସେହି ସଂଖ୍ୟାକୁ 2 ଦ୍ଵାରା ଭାଗକରି 10 ଦ୍ଵାରା ଗୁଣିବା ସହ ସମାନ କି ?
Solution:
ହଁ, ସମାନ ହେବ । କାରଣ \(\frac{1}{2}\) = \(\frac{10 \div 2}{2 \div 2}\) = 5

ନିଜେ କରି ଦେଖ ।

Question 1.
ନିମ୍ନଲିଖତ ଗୁଣନଗୁଡ଼ିକର ଗୁଣଫଳ ଶୀଘ୍ର ନିଷ୍କ୍ରିୟ କରିବାର ଉପାୟ ଲେଖ ।
(a) 2 × 1768 × 50
(b) 72 × 125 (ସୂଚନା : 125 = \(\frac{1000}{8}\) )
(c) 125 × 40 × 8 × 25
Solution:
(a) 2 × 1768 × 50 = 2 × 50 × 1768 = 100 × 1768 = 176800

(b) 72 × 125 = 72 × \(\frac{1000}{8}\) = 9 × 1000 = 9000

(c) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25 = 1000 × 1000 = 1000000

Question 2.
କମ୍ ସମୟରେ ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର ।
(a) 25 × 12 =
(b) 25 × 240 =
(c) 250 × 120 =
(d) 2500 × 12 =
(e) ……………. × ………………. = 120000000
Solution:
(a) 25 × 12 = \(\frac{100}{4}\) × 12 = 100 × 3 = 300
(b) 25 × 240 = \(\frac{100}{4}\) × 240 = 100 × 60 = 6000
(c) 250 × 120 = \(\frac{1000}{4}\) × 120 = 1000 × 30 = 30000
(d) 2500 × 12 = \(\frac{10,000}{4}\) × 12 = 10,000 × 3 = 30,000
(e) 25000 × 4800 = \(\frac{1,00,000}{4}\) × 4800 = 1,00,000 × 1200 = 120000000

ଗୁଣଫଳ କେତେ ବଡ଼ ?

Question 1.
ତଳେ ଦିଆଯାଇଥ‌ିବା କୋଠରିଗୁଡ଼ିକରେ ଗୁଣନଗୁଡ଼ିକ ମଜାଦାର ସଂରଚନା ସୃଷ୍ଟି କରୁଛନ୍ତି । ସେହି ସଂରଚନାକୁ ଖୋଜିବା ପାଇଁ ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର । ଲକ୍ଷ୍ୟ କରିଥିବା ସଂରଚନାକୁ ଆଧାର କରି ଗୁଣନଗୁଡ଼ିକୁ ଆଗକୁ ବଢ଼ାଅ ।

Solution:

Question 2.
ପ୍ରତ୍ୟେକ ସ୍ଥଳରେ ଗୁଣନ କରାଯାଇଥିବା ଦୁଇଟି ସଂଖ୍ୟା ଓ ସେମାନଙ୍କ ଗୁଣଫଳର ଅଙ୍କଗୁଡ଼ିକର ସଂଖ୍ୟାକୁ ଲକ୍ଷ୍ୟ କର । ଗୁଣନ କରାଯାଉଥିବା ସଂଖ୍ୟା ଓ ସେମାନଙ୍କର ଗୁଣଫଳର ଅଙ୍କମାନଙ୍କ ସଂଖ୍ୟା ମଧ୍ୟରେ କିଛି ସମ୍ପର୍କ ରହିଛି କି ?
Solution:
ହଁ, ସମ୍ପର୍କ ରହିଛି । ଆମେ ଯେତେବେଳେ ନିମ୍ନ ଅଙ୍କ ସଂଖ୍ୟାର ଗୁଣନ କରୁ,
1- ଅଙ୍କ ସଂଖ୍ୟା × 1-ଅଙ୍କ ସଂଖ୍ୟା = 1 ଓ 2 ଅଙ୍କ ।
2- ଅଙ୍କ ସଂଖ୍ୟା × 2-ଅଙ୍କ ସଂଖ୍ୟା = 3 ଓ 4 ଅଙ୍କ ।
3- ଅଙ୍କ ସଂଖ୍ୟା × 3-ଅଙ୍କ ସଂଖ୍ୟା = 5 ଓ 6 ଅଙ୍କ ।
4- ଅଙ୍କ ସଂଖ୍ୟା × 4-ଅଙ୍କ ସଂଖ୍ୟା = 7 ଓ 8 ଅଙ୍କ ।

Question 3.
• ଦୁଇଟି 2-ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଗୁଣପଳ କେବଳ 3-ଅଙ୍କ କିମ୍ବା 4-ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହୋଇପାରିବ । ଏହା କ’ଣ ଠିକ୍ ?
Solution:
ଏହା ଠିକ୍ ।
କାରଣ, ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ସାନ ସଂଖ୍ୟାର ଗୁଣଫଳ = 10 × 10 = 100 (3 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା)
ଦୁଇଟି ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ବୃହତ୍ତମ ସଂଖ୍ୟାର ଗୁଣଫଳ = 99 × 99 = 9801 (4 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା)

Question 4.
ଏହି ଉକ୍ତିଟି ଠିକ୍ ନା ନୁହେଁ ଜାଣିବାପାଇଁ ଆମେ 2 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ସହିତ ସମସ୍ତ ସମ୍ଭାବ୍ୟ ଗୁଣନ ଚେଷ୍ଟା କରିବା ଉଚିତ୍ କି ? କିମ୍ବା ଏହାକୁ ପ୍ରମାଣିତ କରିବା ପାଇଁ ଅନ୍ୟ କିଛି ଭଲ ଉପାୟ ଅଛି କି ?

‘‘ଆମେ ଦୁଇଟି 2-ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଗୁଣଫଳର ଅଙ୍କଗୁଡ଼ିକର ସଂଖ୍ୟା ବିଷୟରେ ଜାଣିବାକୁ ଚାହୁଁଛି । ସେହିପରି ସବୁଠାରୁ ଛୋଟ ଗୁଣଫଳ ଜାଣିବା ପାଇଁ ମୁଁ 10 × 10 ନେଲି । ତେଣୁ ଅନ୍ୟ ସମସ୍ତ ଗୁଣଫଳ 100 ରୁ ଅଧ‌ିକ ହେବ । ସେହିପରି ବୃହତ୍ତମ ଗୁଣଫଳ ଜାଣିବା ପାଇଁ ମୁଁ 3-ଅଙ୍କ ବିଶିଷ୍ଟ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟା (100) × 100) କୁ ଗୁଣନ କରି ଗୁଣଫଳ 10,000 ପାଇଲି । ତେଣୁ ସମସ୍ତ ଦୁଇଟି 2- ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାମାନଙ୍କର ଗୁଣଫଳ 10,000ରୁ କମ୍ ହେବ ।’’
Solution:
ନା, ସମସ୍ତ ସମ୍ଭାବ୍ୟ ଗୁଣଫଳର ଆବଶ୍ୟକତା ନାହିଁ ।
10 × 10 ଓ 99 × 99 ରେ କମ୍ ଓ ଅଧିକ ଅଙ୍କର ସମ୍ଭାବନା ଥାଏ ।

Question 5.
ଗୋଟିଏ ତିନିଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ସହ ଅନ୍ୟ ଏକ 3- ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଗୁଣନ କଲେ ଗୁଣଫଳ ଚାରିଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହେବ କି ?
Solution:
ନା, ଦୁଇଟି ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ ସାନ ସଂଖ୍ୟାର ଗୁଣଫଳ = 100 × 100 = 10,000 (5 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା) ଦୁଇଟି ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ ବଡ଼ ସଂଖ୍ୟାର ଗୁଣଫଳ = 999 × 999 = 9,98,001 (6 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା) ତେଣୁ ଦୁଇଟି 3 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାକୁ ଗୁଣନ କଲେ ଗୁଣଫଳ 4 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହୋଇପାରିବ ନାହିଁ ।

Question 6.
ଗୋଟିଏ ଚାରି 4- ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ସହ ଅନ୍ୟ ଏକ 2-ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଗୁଣନ କଲେ ଗୁଣଫଳ 5-ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହେବ କି ?
Solution:
ଏହା ସମ୍ଭବ ।
4 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ସହ 2 ଅଙ୍କ ବିଶିଷ୍ଟ ସାନ ସଂଖ୍ୟାର ଗୁଣଫଳ
= 1000 × 10 = 10,000 (5-ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା)
ଗୋଟିଏ 4 ଅଙ୍କ ବିଶିଷ୍ଟ ବଡ଼ ସଂଖ୍ୟା ସହ 2 ଅଙ୍କ ବିଶିଷ୍ଟ ବଡ଼ ସଂଖ୍ୟାର ଗୁଣଫଳ
= 9999 × 99 = 9,89,901 (6 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା) ।

Question 7.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ବିଭିନ୍ନ ଅଙ୍କବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଗୁଣନ ପ୍ରକ୍ରିୟାକୁ ଲକ୍ଷ୍ୟ କର । ତୁମେ କିଛି ସଂରଚନାଥୁବାର ଲକ୍ଷ୍ୟ କରୁଛ କି ? ଉକ୍ତ ସଂରଚନା ଅନ୍ୟ ସଂଖ୍ୟାଗୁଡ଼ିକ ପାଇଁ ମଧ୍ୟ ପ୍ରଯୁଜ୍ୟ କି, ଚେଷ୍ଟା କରି ଦେଖ । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

ବଡ଼ସଂଖ୍ୟା ସମ୍ପର୍କରେ ମଜାଦାର ତଥ୍ୟ :
ବଡ଼ ସଂଖ୍ୟାଗୁଡ଼ିକର ଗୁଣନ ଓ ହରଣ କରିବା ସମୟରେ କିଛି ରୋଚକ ତଥ୍ୟ ଉପଲବ୍‌ଧୁ ହୋଇଥାଏ । ଏହାର ଲିଖନ ପ୍ରଣାଳୀରେ ଲେଖାଯାଇଥାଏ । ଗୁଣଫଳ ଓ ଭାଗଫଳକୁ ଆମେରିକୀୟ ସଂଖ୍ୟା
ଗୁଣଫଳ ନିଷ୍କ୍ରିୟ : ଗୁଣଫଳ ବିଷୟରେ ରୋଚକ ତଥ୍ୟ ପାଇବା ପାଇଁ ଗୁଣନ କରିବା :

Question 1.
କିମ୍ବଦନ୍ତୀ ଅନୁସାରେ 1250 × 380 ହେଉଛି ପୁରନ୍ଦରଦାସଙ୍କ ରଚିତ କୀର୍ତ୍ତନ ସଂଖ୍ୟା । ଏହାର ଗୁଣଫଳ କେତେ?
Solution:
କୀର୍ତ୍ତନ ସଂଖ୍ୟା = 1250 × 380 = (125 × 10) × (38 × 10)
= 125 × 38 × 10 × 10 = 4750 × 100 = 475000 = 4,75,000

Question 2.
2100 × 70,000 ହେଉଛି, ପୃଥ‌ିବୀ ଓ ସୂର୍ଯ୍ୟ ମଧ୍ୟରେ ଥ‌ିବା ଦୂରତା କିଲୋମିଟରରେ ଆନୁମାନିକ ମୂଲ୍ୟ ।
Solution:
ପୃଥ‌ିବୀ ଓ ସୂର୍ଯ୍ୟ ମଧ୍ୟରେ ଦୂରତା = 2100 × 70,000 = (21 × 1,00) × (7 × 10,000)
= 21 × 7 × 1,00 × 10,000 = 147 × 10,00,000 = 147000000.
= 14,70,00,000 କିଲୋମିଟର ।

Question 3.
6400 × 62,500 ହେଉଛି ଆମାଜନ୍ ନଦୀ ପ୍ରତି ସେକେଣ୍ଡରେ ଆଟଲାଣ୍ଟିକ୍ ମହାସାଗରକୁ ଛାଡ଼ୁଥୁବା ପାଣିର ପରିମାଣ (ଲିଟରରେ) ।
ପାଣିର ପରିମାଣ = 6400 × 62,500 = (64 × 100) × (625 × 100)
= 64 × 625 × 100 × 100 = 40,000 × 10,000
= 40,00,00,000 ଲିଟର ।

Question 4.
ଭାଗଫଳ ନିଶ୍ଚୟ : ହରଣ ବିଷୟରେ ରୋଚକ ତଥ୍ୟ ଜାଣିବା ପାଇଁ ଦିଆଯାଇଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଭାଗ କରିବା । 13,95,000 ÷ 150 ହେଉଛି, ପୃଥିବୀର ଦୀର୍ଘତମ ଏକକ ଟ୍ରେନ୍ ଯାତ୍ରାର ଦୂରତା( କି.ମି.ରେ) ।
Solution:
ଯାତ୍ରାର ଦୂରତା = \(\frac{13,95,000}{150}\) = \(\frac{1395}{15}\) × \(\frac{1000}{10}\) = 93 × 100 = 9300 କି.ମି. ।

Question 5.
ନୀଳତିମିର ଓଜନ 10,50,00,000 ÷ 700 କିଲୋଗ୍ରାମର ଅଧିକ ହୋଇପାରେ ।
Solution:
ନୀଳତିମିର ଓଜନ = \(\frac{10,50,00,000}{700}\) = \(\frac{105}{7}\) × \(\frac{10,00,000}{100}\) = 15 × 10000 = 1,50,000 କିଲୋଗ୍ରାମ୍ ।

Question 6.
52,00,00,00,000 ÷ 130 ଟନ୍ ହେଉଛି 2021 ମସିହାରେ ବିଶ୍ଵରେ ନିର୍ଗତ ହୋଇଥିବା ପ୍ଲାଷ୍ଟିକ୍ ବର୍ଜ୍ୟବସ୍ତୁର ଓଜନ ।
Solution:
ବର୍ଜ୍ୟବସ୍ତୁର ଓଜନ = \(\frac{52,00,00,00,000}{130}\) = 40,00,00,000 କିଲୋଗ୍ରାମ୍ ।

1.6 ତୁମେ କେବେ ଭାବିଛ କି ?

Page No. (19 to 21)

Question 1.
ଟାଇଟାନିକ୍ ଜାହାଜରେ 2500 ଜଣ ଯାତ୍ରୀ ବସିପାରନ୍ତି । ତେବେ ଭୁବନେଶ୍ଵର ସହରର ସମସ୍ତ ଲୋକ ସେହିପରି 400ଟି ଜାହାଜରେ ବସିପାରିବେ କି ?

Solution:
ଗୋଟିଏ ଟାଇଟାନିକ୍ ଜାହାଜରେ ବସିପାରନ୍ତି = 2500 ଜଣ ଯାତ୍ରୀ ।
400 ଟି ଜାହାଜରେ ବସିପାରିବେ = 2500 × 400 = 10,00,000 ଜଣ ଯାତ୍ରୀ ।
ଭୁବନେଶ୍ୱର ସହରର ଜନସଂଖ୍ୟା = ୫ ଲକ୍ଷ 85 ହଜାରରୁ ଅଧିକ (2011 ଜନଗଣନା ଅନୁଯାୟୀ) = 8,85,000
∴ ଭୁବନେଶ୍ଵର ସହରର ସମସ୍ତ ଲୋକ ଜାହାଜରେ ବସିପାରିବେ ।

Question 2.
ନମିତା ପଚାରିଲା, ‘ଯଦି ସେ ପ୍ରତିଦିନ 100 କି.ମି. ଯାତ୍ରା କରିବ, ତେବେ ସେ 10 ବର୍ଷରେ ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିପାରିବ କି ? (ପୃଥ‌ିବୀ ଓ ଚନ୍ଦ୍ର ମଧ୍ୟରେ ଦୂରତା 3,84,400 କି.ମି.)
• ସେ ଏକ ବର୍ଷରେ କେତେ ଦୂରତା ଅତିକ୍ରମ କରିଥାନ୍ତା ?
• ସେ 10 ବର୍ଷରେ କେତେ ଦୂରତା ଯାଇପାରିଥାନ୍ତା ?
Solution:
ପୃଥ‌ିବୀ ଓ ଚନ୍ଦ୍ର ମଧ୍ୟରେ ଦୂରତା = 3,84,400 କି.ମି. ।
1 ଦିନରେ ଯାତ୍ରା କରେ = 100 କି.ମି. ।
1 ବର୍ଷ = 365 ଦିନ ।
। ବର୍ଷରେ ଯାତ୍ରା କରିବ = 100 × 365 = 36,500 କି.ମି. ।
ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିବାକୁ ସମୟ ଲାଗିବ = 3,84,400 ÷ 36500 = 10.5 ବର୍ଷ ।
∴ ସେ 10.5 ବର୍ଷରେ ଚନ୍ଦ୍ରରେ ପହଞ୍ଚିପାରିବ ।
1 ବର୍ଷରେ ସେ ଯାତ୍ରା କରେ = 36,500 କି.ମି. ।
10 ବର୍ଷରେ ସେ ଯାତ୍ରା କରିବ = 36,500 × 10 = 3,65,000 କି.ମି. ।
∴ ସେ 10 ବର୍ଷରେ 3,65,000 କି.ମି. ଯାତ୍ରା କରିବ ।

Question 3.
ଯଦି ତୁମେ ଦିନକୁ 1000 କି.ମି. ଯାତ୍ରା କର, ତେବେ ତୁମେ ଜୀବନକାଳ ମଧ୍ୟରେ ସୂର୍ଯ୍ୟ ପାଖରେ ପହଞ୍ଚି ପାରିବ କି ? ଚିନ୍ତାକରି ଉତ୍ତର ଦିଅ । (ପୂର୍ବ ଅଭ୍ୟାସରେ ପୃଥ‌ିବୀ ଓ ସୂର୍ଯ୍ୟ ମଧ୍ୟରେ ଦୂରତା ଦିଆଯାଇଛି)
Solution:
ଛାତ୍ରଛାତ୍ରୀମାନେ ଅଭ୍ୟାସ କରନ୍ତୁ ।

Question 4.
ଉପଯୁକ୍ତ କଳ୍ପନା କରି ନିମ୍ନ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।
(a) ଯଦି ଗୋଟିଏ କାଗଜ ଫର୍ଦ୍ଦର ଓଜନ 5 ଗ୍ରାମ୍ ହୁଏ, ତେବେ ଏକାଥରକେ ଏକ ଲକ୍ଷ କାଗଜ ଫର୍ଦ୍ଦକୁ ତୁମେ ଉଠାଇ ପାରିବ କି ?
Solution:
ଗୋଟିଏ କାଗଜ ଫର୍ଦ୍ଦର ଓଜନ = 5 ଗ୍ରାମ୍ ।
1 ଲକ୍ଷ କାଗଜ ଫର୍ଦର ଓଜନ = 1,00,000 × 5 = 5,00,000 ଗ୍ରାମ୍ ।
1 କି.ଗ୍ରା. = 1000 ଗ୍ରାମ୍ । 5,00,000 ଗ୍ରାମ୍ = 5,00,000 ÷ 1000 = 500 କି.ଗ୍ରା. ।
∴ ଜଣେ ସାଧାରଣ ଲୋକ 500 କି.ଗ୍ରା. ଏକାଥରକେ ଉଠାଇ ପାରିବ ନାହିଁ ।

(b) ଯଦି ସାରା ବିଶ୍ଵରେ ପ୍ରତି ମିନିଟ୍‌ରେ 250 ଟି ଶିଶୁ ଜନ୍ମ ହୁଅନ୍ତି, ତେବେ ଗୋଟିଏ ଦିନରେ ଏକ ମିଲିୟନ୍ ଶିଶୁ ଜନ୍ମ ହୋଇପାରିବେ କି ?
Solution:
1 ଘଣ୍ଟା = 60 ମିନିଟ୍, 1 ଦିନ = 24 ଘଣ୍ଟା ।
1 ମିନିଟ୍‌ରେ ଜନ୍ମ ନିଅନ୍ତି = 250 ଶିଶୁ ।
1 ଘଣ୍ଟାରେ ଜନ୍ମ ନିଅନ୍ତି = 250 × 60 = 15,000 ଶିଶୁ ।
1 ଦିନରେ ଜନ୍ମ ନିଅନ୍ତି = 15,000 × 24 = 3,60,000 ଶିଶୁ ।
1 ମିଲିୟନ୍ = 10 ଲକ୍ଷ ।
3,60,000 ସଂଖ୍ୟାଟି 1 ମିଲିୟନ୍ (10) ଲକ୍ଷ)ରୁ ସାନ ।
∴ ଗୋଟିଏ ଦିନରେ ଏକ ମିଲିୟନ୍ ଶିଶୁ ଜନ୍ମ ହୋଇପାରିବେ ନାହିଁ ।

(c) ତୁମେ 1 ମିଲିୟନ୍ ମୁଦ୍ରା, ଗୋଟିଏ ଦିନରେ ଗଣିପାରିବ କି ? (ମନେକର, ତୁମେ 1 ସେକେଣ୍ଡରେ 1ଟି ମୁଦ୍ରା ଗଣିପାରୁଛ)
Solution:
1 ଘଣ୍ଟା = 60 ମିନିଟ୍, 1 ମିନିଟ୍ = 60 ସେକେଣ୍ଡ, 1 ଦିନ = 24 ଘଣ୍ଟା ।
1 ସେକେଣ୍ଡରେ ଗଣେ = 1 ମୁଦ୍ରା
1 ମିନିଟ୍‌ରେ ଗଣିବ = 1 × 60 = 60 ମୁଦ୍ରା |
1 ଘଣ୍ଟାରେ ଗଣିବ = 60 × 60 = 3600 ମୁଦ୍ରା ।
1 ଦିନରେ ଗଣିବ = 3600 × 24 = 86,400 ମୁଦ୍ରା ।
1 ମିଲିୟନ୍ = 10 ଲକ୍ଷ ।
86,400 ସଂଖ୍ୟାଟି । ମିଲିୟନ୍ (10 ଲକ୍ଷ)ରୁ ସାନ ।
∴ ଗୋଟିଏ ଦିନରେ ସେ ଏକ ମିଲିୟନ୍ ମୁଦ୍ରା ଗଣିପାରିବ ନାହିଁ ।

ନିଜେ କରି ଦେଖ :

Question 1.
0 ରୁ 9 ପର୍ଯ୍ୟନ୍ତ ଅଙ୍କଗୁଡ଼ିକ କେବଳ ଥରେ ମାତ୍ର ବ୍ୟବହାର କରି (ପ୍ରଥମ ଅଙ୍କ ‘0’ ହୋଇପାରିବ ନାହିଁ) ଗୋଟିଏ 10 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଗଠନ କର ଓ ଲେଖ, ଯାହା–
(a) 5ର ବୃହତ୍ତମ ଗୁଣିତକ
(b) କ୍ଷୁଦ୍ରତମ ଯୁଗ୍ମ ସଂଖ୍ୟା
Solution:
(a) 0 ରୁ 9 ପର୍ଯ୍ୟନ୍ତ ଅଙ୍କଗୁଡ଼ିକୁ ଥରେ ମାତ୍ର ବ୍ୟବହାର କରି ବଡ଼ ସଂଖ୍ୟା = 9876543210
ସଂଖ୍ୟାର ଡାହାଣପଟେ 0 କିମ୍ବା 5 ଥିଲେ ସଂଖ୍ୟାଟି 5 ଦ୍ଵାରା ବିଭାଜ୍ୟ ହେବ ।
ଏଣୁ 9876543210 ସଂଖ୍ୟାଟି 5 ଦ୍ଵାରା ବିଭାଜ୍ୟ ।
∴ 5 ର ବୃହତ୍ତମ ଗୁଣିତକ ସଂଖ୍ୟାଟି ହେଲା 98765432

(b) ସାନରୁ ବଡ଼ କ୍ରମରେ ଲେଖୁଲେ ସଂଖ୍ୟାଟି ହେବ = 0123456789
ଆମେ ଜାଣିଛେ ସଂଖ୍ୟାଟି 0 ରୁ ଆରମ୍ଭ ହୋଇପାରିବ ନାହିଁ,
ତେଣୁ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟାଟି ହେବ = 10223456789
କିନ୍ତୁ ସଂଖ୍ୟାଟି ଯୁଗ୍ମ ସଂଖ୍ୟା ନୁହେଁ ।
ସଂଖ୍ୟାଟିର ଡାହାଣପଟ ସଂଖ୍ୟାଦୁଇଟିକୁ ଅଦଳ ବଦଳ କଲେ ସଂଖ୍ୟାଟି ଯୁଗ୍ମ ସଂଖ୍ୟା ହେବ ।
∴ କ୍ଷୁଦ୍ରତମ ଯୁଗ୍ମ ସଂଖ୍ୟାଟି = 1023456798

Question 2.
10,30,285 ସଂଖ୍ୟାକୁ ଅକ୍ଷରରେ ‘‘ଦଶ ଲକ୍ଷ ତିରିଶ ହଜାର ଦୁଇ ଶହ ପଞ୍ଚାଅଶୀ’’ ଲେଖାଯାଏ, ଯେଉଁଥରେ 18 ଟି ଅକ୍ଷର ଅଛି । ସର୍ବାଧ‌ିକ ସଂଖ୍ୟକ ଅକ୍ଷର ଥିବା ଗୋଟିଏ 7 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଲେଖ ।
Solution:
10,30,285 = ଦଶ ଲକ୍ଷ ତିରିଶ ହଜାର ଦୁଇ ଶହ ପଞ୍ଚାଅଶୀ (ଅକ୍ଷର ସଂଖ୍ୟା = 18)
ଗୋଟିଏ ସାତ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହେଉଛି 77,77,777
77,77,777 = ସତସ୍ତରୀ ଲକ୍ଷ ସତସ୍ତରୀ ହଜାର ସାତ ଶହ ସତସ୍ତରୀ (ଅକ୍ଷର ସଂଖ୍ୟା = 21)

Question 3.
ଏକ 9 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଲେଖ ଯେଉଁଠାରେ ଯେକୌଣସି ଦୁଇଟି ଅଙ୍କର ସ୍ଥାନ ପରିବର୍ତ୍ତନ କଲେ ଏକ ବଡ଼ ସଂଖ୍ୟା ମିଳିବ । ଏହିପରି କେତୋଟି ସଂଖ୍ୟା ଅଛି ?
Solution:
ମନେକର ଏକ 9 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା = 123456789
ଯେକୌଣସି ଦୁଇଟି ଅଙ୍କର ସ୍ଥାନ ପରିବର୍ତ୍ତନ କଲେ ମିଳୁଥିବା ବଡ଼ ସଂଖ୍ୟାଟି = 213456789

Question 4.
12345123451234512345 ସଂଖ୍ୟାରୁ 10ଟି ଅଙ୍କ କାଟିଦିଅ, ଯାହାଦ୍ଵାରା ଅବଶିଷ୍ଟ ସଂଖ୍ୟାଟି ଯଥାସମ୍ଭବ ବଡ଼ ସଂଖ୍ୟା ହେବ ?
Solution:
1. ଆମର 20 ଅଙ୍କ ଅଛି ଏବଂ ଆମକୁ 10 ଅଙ୍କ କରିବାକୁ ପଡ଼ିବ, ଯାହାଫଳରେ ଆମ ପାଖରେ 10 ଅଙ୍କ ରହିବ । ପରିମାଣ ସ୍ଵରୂପ ସଂଖ୍ୟାକୁ ସର୍ବାଧ‌ିକ କରିବା ପାଇଁ, ଆମେ ବାମ ଅଙ୍କକୁ ବଡ଼ କରିବାକୁ ଚାହୁଁଛୁ । ମୂଳସଂଖ୍ୟା = 12 । ପ୍ରଥମ କିଛି ଅଙ୍କ ଛୋଟ । ପ୍ରଥମ ସ୍ଥାନରେ ଏକ ବଡ଼ ଅଙ୍କ ପାଇବା ପାଇଁ ଆମେ ପ୍ରାରମ୍ଭିକ 12345 କୁ କାଟି ପାରିବା ।

2. ପରବର୍ତୀ ଉପରକୁ ଅଙ୍କଗୁଡ଼ିକୁ ପ୍ରଥମ ପାଞ୍ଚଟି ଅଙ୍କ କାଟିବା ପରେ ଅଛି । 1234512345 ଏବେ ପ୍ରଥମ ଅଙ୍କ 1 । ଯଦି ଆମେ ପ୍ରଥମ ଚାରୋଟି ଅଙ୍କ କାଟିବା, ତେବେ ଅବଶିଷ୍ଟ ସଂଖ୍ୟା 5 ସହିତ ଆରମ୍ଭ ହେବ । ଏହା 1 ଅପେକ୍ଷା ଭଲ ।

3. ଏବେ ଆମର ଆରମ୍ଭ ସଂଖ୍ୟା = 5। ଆମେ ଏ ପର୍ଯ୍ୟନ୍ତ 4 ଅଙ୍କ କାଟି ସାରିଛେ । 5 ଅଙ୍କ କାଟିବାକୁ ପଡ଼ିବ । ବାକିସଂଖ୍ୟା ହେଉଛି 5123451234512345 । ଆମେ ଚାହୁଁଛୁ ଯେ ପରବର୍ତ୍ତୀ ଅଙ୍କ ଯଥାସମ୍ଭବ ବଡ଼ ହେଉ । ଆମେ 5 ପରେ 1234 କୁ କାଟିବା, ଯାହା ଫଳରେ ଆମ ପାଖରେ 551234512345 ରହିବ । ଆମେ ଏବେ 4 + 4 = 8 ଅଙ୍କ କାଟି ସାରିଛୁ ।

4. ଆମକୁ ଆଉ 2ଟି ଅଙ୍କ କାଟିବାକୁ ପଡ଼ିବ । ସଂଖ୍ୟାକୁ ଯଥାସମ୍ଭବ ବଳ୍କ କରିବା ପାଇଁ ଆମକୁ 1 କିମ୍ବା 2 କାଟିବା ଉଚିତ । ତେଣୁ 1234 ତା’ପରେ 1234 ଏବଂ 1 ଓ 2 କାଟିବା ପରେ ଅବଶିଷ୍ଟ ସଂଖ୍ୟା ହେଉଛି 5534512345

Question 5.
‘Zero’ ଓ ‘One’ ଉଭୟ ଶବ୍ଦରେ ‘e’ ଏବଂ ‘o’ ଅକ୍ଷର ଅଛି । ‘One’ ଏବଂ ‘Two’ ଉଭୟ ଶବ୍ଦରେ ‘o’ ଅକ୍ଷର ଅଛି ଏବଂ ‘two’ ଏବଂ ‘three’ ଉଭୟ ଶବ୍ଦରେ ‘t’ ଅକ୍ଷର ଅଛି । କୌଣସି ଇଂରାଜୀ ଅକ୍ଷର ସମାନ ନଥାଇ ଦୁଇଟି କ୍ରମିକ ସଂଖ୍ୟା (ଇଂରାଜୀ ବନାନ) ନିର୍ଣ୍ଣୟ କରିବା ପାଇଁ ତୁମକୁ କେତେ ପର୍ଯ୍ୟନ୍ତ ଗଣିବାକୁ ପଡ଼ିବ ?
Solution:
ଏଠାରେ ଶବ୍ଦ ‘one’ ଏବଂ ‘two’ରେ ‘o’ ଅକ୍ଷର ଅଛି । two ଏବଂ three ଉଭୟ ଶବ୍ଦରେ ‘t’ ଅକ୍ଷର ଅଛି ।
three ଏବଂ four ଉଭୟ ଶବ୍ଦରେ ‘r’ ଅକ୍ଷର ଅଛି । four ଏବଂ five ଉଭୟ ଶବ୍ଦରେ ‘f’ ଅକ୍ଷର ଅଛି ।
five ଏବଂ six ଉଭୟ ଶବ୍ଦରେ ‘i’ ଅକ୍ଷର ଅଛି । six ଏବଂ seven ଉଭୟ ଶବ୍ଦରେ ‘s’ ଅକ୍ଷର ଅଛି ।
seven ଏବଂ eight ଉଭୟ ଶବ୍ଦରେ ‘e’ ଅକ୍ଷର ଅଛି ।
eight ଏବଂ nine ଉଭୟ ଶବ୍ଦରେ ‘e’,’n’ ଅକ୍ଷର ଅଛି ।
nine ଏବଂ ten ଉଭୟ ଶବ୍ଦରେ ‘n’, ‘e’ ଅକ୍ଷର ଅଛି ।
ଏହା ଦର୍ଶାଉଛି ଯେ, ସମସ୍ତ କ୍ରମାଗତ ସଂଖ୍ୟାରେ ଅତିକମ୍‌ରେ ଗୋଟିଏ ସାଧାରଣ ସଂଖ୍ୟା ଅଛି ।

Question 6.
ମନେକର ତୁମେ ସମସ୍ତ ସଂଖ୍ୟା 1, 2, 3, 4, 5 ……9, 10, 11 ………………. ଲେଖୁଛି । ତୁମେ ଲେଖୁଥ‌ିବା ଦଶମ ଅଙ୍କଟି ହେଉଛି ‘1’ ଏବଂ ଏକାଦଶ ଅଙ୍କଟି ହେଉଛି ‘0’, ଯାହା ସଂଖ୍ୟା 10 ର ଏକ ଅଂଶ ।
(a) ସେହିପରି ଲେଖୁ ଚାଲିଲେ, 1000 ତମ ଅଙ୍କଟି କ’ଣ ହେବ ? ଏହି ଅକଟି କେଉଁ ସଂଖ୍ୟାରେ ଥବ ?
(b) କେଉଁ ସଂଖ୍ୟାରେ ନିୟୁତତମ ଅଙ୍କ ରହିଥ‌ିବ ?
(c) ତୁମେ କେତେବେଳେ ଅଙ୍କ ‘+10,000’ ଏବଂ ‘+100’ ବଟନ୍ ଅଛି ।
Solution:
(a) ଏଠାରେ 1 ରୁ 9 ପର୍ଯ୍ୟନ୍ତ ଅଙ୍କ ହେଉଛି 9ଟି ଅଙ୍କ ।
10 ରୁ 99 ପର୍ଯ୍ୟନ୍ତ ଅଙ୍କ = 90 × 2 = 180 ଟି ଅଙ୍କ ।
1 ରୁ 99 ପର୍ଯ୍ୟନ୍ତ ମୋଟ ଅଙ୍କ = 180 + 9 = 189 ଟି ଅଙ୍କ ।
1000 ତମ ଅଙ୍କ ପାଇବାକୁ ବାକିଥିବା ଅଙ୍କ = 1000 – 189 = 811
ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା = \(\frac{811}{3}\) = 270 ପଡ଼ି 1 ବଳିଲା ।
ପ୍ରଥମ 3 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହେଉଛି 100 ।
270 ତମ 3 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଟି = 100 + 270 – 1 = 369 ।
ପରବର୍ତୀ ସଂଖ୍ୟାଟି ହେଉଛି 370 ।
ତେଣୁ 370 ର ପ୍ରଥମ ଅଙ୍କଟି 3 ଯାହା 1000 ତମ ଅଙ୍କଟି ହେବ ।

(b)

• ଏକ ଅଙ୍କ ବିଶିଷ୍ଟ (1 – 9) ସଂଖ୍ୟା = 9 ଟି ଅଙ୍କ ।
• ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ( 10 – 99) ସଂଖ୍ୟା = 90 × 2 = 180ଟି ଅଙ୍କ ।
  ମୋଟ ଅଙ୍କ = 9 + 180 = 189 ଟି ଅଙ୍କ ।
• ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ (100 – 999) ସଂଖ୍ୟା = 900 × 3 = 2700ଟି ଅଙ୍କ ।
  ମୋଟ ଅଙ୍କ = 189 + 27000 = 2889 ଟି ଅଙ୍କ ।
• ଚାରି ଅଙ୍କ ବିଶିଷ୍ଟ (1000 – 9999) ସଂଖ୍ୟା = 9000 × 4 = 36000ଟି ଅଙ୍କ ।
  ମୋଟ ଅଙ୍କ = 2889 + 36000 = 38,889 ଟି ଅଳ୍ପ ।
• ପାଞ୍ଚ ଅଙ୍କ ବିଶିଷ୍ଟ (1000 – 99999) ସଂଖ୍ୟା = 90000 × 5 = 4,50,000ଟି ଅଙ୍କ ।
  ମୋଟ ଅଙ୍କ = 4,50,000 + 38,889 = 4,88,889 ଟି ଅଙ୍କ ।
• ଏବେ, ଆମକୁ 6 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ଆବଶ୍ୟକ ।
  1,000,000 – 4,88,889 = 5,11,111 ଟି ଅଳ୍ପ ବାକି ।
  ପ୍ରତ୍ୟେକ 6 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାରେ ଅଙ୍କ ସଂଖ୍ୟା = 6
  511111 ÷ 6 = 85,185 ପଡ଼ି 1 ବଳିଲା ।
  ତେଣୁ ନିୟୁତତମ ଅଙ୍କଟି ହେଉଛି 85,185, 6 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ପ୍ରଥମ ଅଙ୍କ ।
  ତେଣୁ ପ୍ରଥମ 6 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା = 1,00,000 + 85,185 – 1 = 1,85,185
• ନିୟୁତ ତମ ସଂଖ୍ୟା ଏହି ସଂଖ୍ୟା 1,85,185ରେ ଅବସ୍ଥିତ ।

(c)

• ଏକ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା (1 – 9) କେବଳ ଥରେ ଲେଖାଏଁ
  ‘5’ ଆସୁଥିବା ମୋଟ ସଂଖ୍ୟା = 1 ଥର ।
• ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ (10 – 99) ସଂଖ୍ୟା ମଧ୍ୟରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 19 ଥର ।
  ମୋଟ ସଂଖ୍ୟା = 1 + 19 = 20 ଥର ।
• ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ (100 – 999) ସଂଖ୍ୟା ମଧ୍ୟରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 280 ଥର ।
  ମୋଟ ସଂଖ୍ୟା = 280 + 20 = 300
• ଚାରି ଅଙ୍କ ବିଶିଷ୍ଟ (1000 – 9999) ସଂଖ୍ୟା ମଧ୍ୟରେ ‘5’ ଆସୁଥ‌ିବା ସଂଖ୍ୟା = 3700 ଥର ।
  ମୋଟ ସଂଖ୍ୟା = 300 + 3700 = 4000
• ସଂଖ୍ୟା 1000 ରୁ ଆରମ୍ଭ କରି 5000 ତମ ପାଇଁ ଆମକୁ ଆବଶ୍ୟକ 5000 – 4000 = 1000ଟି ‘5’ ଯାହା 10,001 – 10,999 ମଦ୍ୟରେ ଆସିବ ।
• (10,000 – 10,999) ମଧ୍ୟରେ ଏକ ଅଙ୍କରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା 100 (ଉଦାହରଣ – 10,005, 10,015, ……..)
  ଦଶକ ଅଙ୍କରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 100
  (10,500 – 10,599) ଶତକ ଅଙ୍କରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 100
  ଏବେ ମୋଟ ସଂଖ୍ୟା (ଯଥା – 10,500 – 10,599) 4000 + 300 = 4300
  (11,000 – 11,999) ମଧ୍ୟରେ
  ଏକକ ଅଙ୍କରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 100
  ଦଶକ ଅଙ୍କରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 100
  ଶତକ ଅଙ୍କରେ ‘5’ ଆସୁଥିବା ସଂଖ୍ୟା = 100
  ମୋଟ ସଂଖ୍ୟା = 4300 + 300 = 4600
  (12,000 – 12,999) ସଂଖ୍ୟା ମଧ୍ୟରେ ‘5’ ଆସୁଥ‌ିବା ସଂଖ୍ୟା = 4600 + 300 = 4900
  (13,000 – 13,99) ସଂଖ୍ୟା ମଧ୍ୟରେ = 20 ଥର ମୋଟ = 4920
  13,100 – 13, 199 ସଂଖ୍ୟା ମଧ୍ୟରେ = 20 ଥର ମୋଟ = 4940
  13,200 – 13, 299 ସଂଖ୍ୟା ମଧ୍ୟରେ = 20 ଥର ମୋଟ = 4960
  13,300 – 13, 399 ସଂଖ୍ୟା ମଧ୍ୟରେ = 20 ଥର ମୋଟ = 4980
  13,400 – 13, 495 ସଂଖ୍ୟା ମଧ୍ଯରେ = 20 ଥର ମୋଟ = 5000

Question 7.
ଗୋଟିଏ କାଲକୁଲେଟର୍‌ର କେବଳ ‘+10,000’ ଏବଂ ‘+100’ ବଟନ୍ ଅଛି । ନିମ୍ନଲିଖତ ସଂଖ୍ୟାଗୁଡ଼ିକ ପାଇଁ କେତେଥର କେଉଁ ବଟନ୍ ଦବାଉଛ, ତା’ର ଏକ ପରିପ୍ରକାଶ ଲେଖ ।
(a) 20,800
(b) 92,100
(c) 1,20,500
(d) 65,30,000
(e) 70,25,700
Solution:
(a) 20,800 (2 × 10,000) + (8 × 100) ବଟନ ଦବାଇବା = 2 + 8 = 10 ଥର ।
(b) 92,100 = (9 × 10,000) + (21 × 100) ବଟନ ଦବାଇବା = 9 + 21 = 30 ଥର ।
(c) 1,20,500 (12 × 10,000) + (5 × 100) ବଟନ ଦବାଇବା = 12 + 5 = 17 ଥର ।
(d) 65,30,000 (653 × 10,000) ବଟନ ଦବାଇବା = 653 ଥର ।
(e) 70,25,700 = (702 × 10,000) + (57 × 100) ବଟନ ଦବାଇବା = 702 + 57 = 759 ଥର ।

Question 8.
କେତେ ଲକ୍ଷରେ ଏକ ବିଲିୟନ୍ ହୁଏ ?
Solution:
ଆମେ ଜାଣିଛେ 1 ଲକ୍ଷ = 1,00,000
1 ବିଲିୟନ୍ = 1,000,000,000
1,000,000,000 ÷ 1,00,000 = 10,000
∴ 10,000 ଲକ୍ଷରେ ଏକ ବିଲିୟନ୍ ହୁଏ ।

Question 9.
1 ରୁ 9 ଥିବା ଦୁଇ ସେଟ୍ ସଂଖ୍ୟା କାର୍ଡ଼ ତୁମକୁ ଦିଆଯାଇଛି । ତଳେ ଦିଆଯାଇଥିବା ପ୍ରତ୍ୟେକ କୋଠରୀରେ ସେଥୁରୁ ଏପରି ଗୋଟିଏ ଲେଖାଏଁ କାର୍ଡ଼ ରଖ ଯେପରି –
(a) ସୃଷ୍ଟି ହୋଇଥବା ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ ସର୍ବାଧ‌ିକ ହେବ ।
(b) ଗଠିତ ସଂଖ୍ୟାଦ୍ଵୟର ସମ୍ଭାବ୍ୟ ସର୍ବନିମ୍ନ ବିୟୋଗଫଳ ହେବ ।

Solution:
(a) ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ ସର୍ବାଧ‌ିକ ହେବ,
ଅର୍ଥାତ୍ 7 ଅଙ୍କ ବିଶିଷ୍ଟ ଓ 5 ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ବୃହତ୍ତମ ହେବ ।
7 ଅଙ୍କ ବିଶିଷ୍ଟ ବୃହତ୍ତମ ସଂଖ୍ୟା = 98,76,543
5 ଅଙ୍କ ବିଶିଷ୍ଟ ବୃହତ୍ତମ ସଂଖ୍ୟା = 98.765
ଯୋଗକଲେ = 98,76,543 + 98,765 = 99,75,308

(b) ଗଠିତ ସଂଖ୍ୟାଦ୍ଵୟର ସମ୍ଭାବ୍ୟ ବିୟୋଗଫଳ ସର୍ବନିମ୍ନ ହେବ; ଅର୍ଥାତ୍ 7 ଅଙ୍କ ବିଶିଷ୍ଟ ଗଠିତ ସଂଖ୍ୟା କ୍ଷୁଦ୍ରତମ ହେବ ଓ ପାଞ୍ଚ ଅଙ୍କ ବିଶିଷ୍ଟ ଗଠିତ ସଂଖ୍ୟା ବୃହତ୍ତମ ହେବ ।
7 ଅଙ୍କ ବିଶିଷ୍ଟ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟା =12, 34, 567
5 ଅଙ୍କ ବିଶିଷ୍ଟ ବୃହତ୍ତମ ସଂଖ୍ୟା = 98,765
ବିୟୋଗ କଲେ = 12,34,567 – 98,765 = 11,35,802

Question 10.
ତୁମକୁ କିଛି ସଂଖ୍ୟା କାର୍ଡ଼ ଦିଆଯାଇଛି । 4000, 13000, 300, 70000, 150000, 20, 5 । କାର୍ଡ଼ଗୁଡ଼ିକୁ ବ୍ୟବହାର କରି ତୁମେ ଚାହୁଁଥୁବା ଗାଣିତିକ ପ୍ରକ୍ରିୟା ଅନୁସାରେ ନିମ୍ନରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକର ନିକଟତର ହୁଅ । (ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସଂଖ୍ୟା ତିଆରି କରିବା ପାଇଁ ପ୍ରତ୍ୟେକ କାର୍ଡ଼କୁ କେବଳ ଥରେ ବ୍ୟବହୃତ କରିପାରିବା ।)

ଯେପରି –
(a) 1,10,000 4000 × (20 + 5) + 13000 = 1,13,000
(b) 2,00,000
(c) 5,80,000
(d) 12,45,000
(e) 20,90,800
Solution:
(a) 1,10,000 = 70,000 + (4,000 × 5) + 13,000
= 70,000 + 20,000 + 13,000 = 1,03,000

(b) 2,00,000 = 1,50,000 + 70,000 – 4,000 × 5
= 1,50,000 + 70,000 – 20,000 = 2,00,000

(c) 5,80,000 = 70,000 × 5 + 1,50,000 + 4,000 × 20
= 3,50,000 + 1,50,000 + 80,000 = 5,80,000

(d) 12,45,000 = 70,000 × 20 – 1,50,000 – 4,000 – 300 × 3
= 14,00,000 – 1,50,000 – 4,000 – 1,500 = 12,44,500

(e) 20,90,800 = 13,000 × 300 – 70,000 (20 + 5) – 1,50,000 + 4,000
= 39,00,000 – 17,50,000 – 1,50,00 + 4,000 = 20,04,000

Question 11.
‘ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍’ ଉଚ୍ଚତା ବିଶିଷ୍ଟ ଏକ ମୁଦ୍ରା ସ୍ତମ୍ଭ ସୃଷ୍ଟି କରିବା ପାଇଁ କେତୋଟି ମୁଦ୍ରା ଆବଶ୍ୟକ ହେବ ? ମନେକର ପ୍ରତ୍ୟେକ ମୁଦ୍ରାର ମୋଟେଇ 1 ମି.ମି. ।
Solution:
ଗୋଟିଏ ମୁଦ୍ରାର ମୋଟେଇ = 1 ମି.ମି. ।
1 ମିଟର = 1000 ମି.ମି. ।
ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍‌ର ଉଚ୍ଚତା = 180 ମିଟର = 180 × 1000 = 1,80,000 ମି.ମି. ।
1 ମି.ମି. = 1ଟି ମୁଦ୍ରା ।
1,80,000 ମି.ମି. = 1,80,000 ÷ 1 1,80,000ଟି ମୁଦ୍ରା ।
∴ ‘ଷ୍ଟାଚ୍ୟୁ ଅଫ୍ ୟୁନିଟ୍’ ଉଚ୍ଚତା ବିଶିଷ୍ଟ ଏକ ମୁଦ୍ରା ସ୍ତମ୍ଭ ସୃଷ୍ଟି କରିବା ପାଇଁ 1,80,000ଟି ମୁଦ୍ରା ଆବଶ୍ୟକ ହେବ ।

Question 12.
ଧୂସର ମୁଣ୍ଡଥିବା ଏକ ପ୍ରକାର ସାମୁଦ୍ରିକ ପକ୍ଷୀ ‘ଆଲ୍‌ବାଟ୍ରୋସ୍’ ଗୁଡ଼ିକର ପ୍ରାୟ 7 ଫୁଟ ଓସାରର ଡେଣା ଥାଏ । ସେମାନେ ସମୁଦ୍ରରେ ଏକ ସ୍ଥାନରୁ ଅନ୍ୟ ସ୍ଥାନକୁ ସ୍ଥାନାନ୍ତର କରିବାରେ ଜଣାଶୁଣା । ଆଲ୍‌ବାଟ୍ରୋସ୍‌ଗୁଡ଼ିକ ଦିନକୁ ପ୍ରାୟ 900 – 1000 କି.ମି. ପର୍ଯ୍ୟନ୍ତ ଯାଇପାରନ୍ତି । ରେକର୍ଡ଼ ହୋଇଥବା ଦୀର୍ଘତମ ଏକକ ଯାତ୍ରା ହେଉଛି ପ୍ରାୟ 12,000 କି.ମି. । ତେବେ ଏପରି ଯାତ୍ରା କରିଲେ ପ୍ରଶାନ୍ତ ମହାସାଗର ଅତିକ୍ରମ କରିବାପାଇଁ ପ୍ରାୟ କେତେଦିନ ଲାଗିବ ?
Solution:
ସମୁଦାୟ ଅତିକ୍ରମ କରିବାକୁ ପଡ଼ିବ = 12,000 କି.ମି. ।
ପକ୍ଷୀଟି ଗୋଟିଏ ଦିନରେ ଯାଏ 900 – 1000 କି.ମି. ।
ଯଦି ପକ୍ଷୀଟି ଗୋଟିଏ ଦିନରେ ୨୦୦ କି.ମି. ଯାଏ ତେବେ 12,000 କି.ମି. ଅତିକ୍ରମ କରିବାକୁ ଲାଗିବ = 12,000 ÷ 900 = 13.3 ଦିନ ।
ଯଦି ପକ୍ଷୀଟି ଗୋଟିଏ ଦିନରେ 1000 କି.ମି. ଯାଏ ତେବେ 12,000 କି.ମି. ଅତିକ୍ରମ କରିବାକୁ ଲାଗିବ = 12,000 ÷ 1000 = 12 ଦିନ ।
∴ ପକ୍ଷୀଟିକୁ ପ୍ରଶାନ୍ତ ମହାସାଗର ଅତିକ୍ରମ କରିବାପାଇଁ ପ୍ରାୟ 12 – 14 ଦିନ ସମୟ ଲାଗିବ ।

Question 13.
ଫ୍ଲେମିଙ୍ଗା ପକ୍ଷୀ ପ୍ରତିବର୍ଷ ଶୀତଋତୁରେ ସାଇବେରିଆ ଅଞ୍ଚଳରୁ ଚିଲିକାକୁ ଆସିଥାନ୍ତି । ତାକୁ ସାଇବେରିଆରୁ ଚିଲିକାକୁ ଅବିରତ ଯାତ୍ରା କରିବାରେ 13,560 କି.ମି. ପଥ ଅତିକ୍ରମ କରିବାକୁ ପଡ଼ିଥାଏ । ଏହାର ଯାତ୍ରା 13 ଅକ୍ଟୋବର, 2022 ଠାରୁ ଆରମ୍ଭ ହୋଇ 11 ଦିନ ପର୍ଯ୍ୟନ୍ତ ଚାଲିଲା । ତେବେ ଏହାର ପ୍ରତିଦିନ ଅତିକ୍ରମ କରିଥିବା ଦୂରତା ଏବଂ ପ୍ରତି ଘଣ୍ଟାରେ ଅତିକ୍ରମ କରିଥୁବା ଦୂରତା ନିଶ୍ଚୟ କର ।

Solution:
ପକ୍ଷୀଟି 11 ଦିନରେ ଅତିକ୍ରମ କରେ 13,560 କି.ମି. ।
ଗୋଟିଏ ଦିନରେ ଅତିକ୍ରମ କରେ = 13,560 ÷ 11 = 1233 କି.ମି. ।
1 ଦିନ = 24 ଘଣ୍ଟା ।
24 ଘଣ୍ଟାରେ ଯାଏ = 1233 କି.ମି. ।
1 ଘଣ୍ଟାରେ ଯିବ = 1233 ÷ 24 = 51.36 କି.ମି. ବା 51 କି.ମି. । ।
∴ ତେବେ ପକ୍ଷୀଟି ପ୍ରତିଦିନ 1233 କି.ମି. ଓ ପ୍ରତି ଘଣ୍ଟାରେ 51 କି.ମି. ଦୂରତା ଅତିକ୍ରମ କରିବ ।

Question 14.
ଶଙ୍ଖଚିଲ ପକ୍ଷୀ ଭୂମିସ୍ତରରୁ 4500 – 6000 ମିଟର ପର୍ଯ୍ୟନ୍ତ ଉଡ଼ିପାରନ୍ତି । ଦେଓମାଳି ପର୍ବତ ହେଉଛି 1672 ମିଟର ଉଚ୍ଚ । ଉଡ଼ାଜାହାଜ 10,000 – 12,800 ମିଟର ଉଚ୍ଚତା ପର୍ଯ୍ୟନ୍ତ ଉଡ଼ିପାରେ । ସୋମୁଙ୍କ କୋଠା ତୁଳନାରେ ଏହି ଉଚ୍ଚତାଗୁଡ଼ିକ କେତେ ଗୁଣ ବଡ଼ ?

Solution:
ସୋମୁଙ୍କ ନୋଠାଘରର ଉଚ୍ଚତା = 40 ମିଟର ।
ଶଙ୍ଖଚିଲ ଭୂମିସ୍ତରରୁ 4500 – 6000 ମିଟର ଉଡ଼ିପାରେ ।
ସର୍ବନିମ୍ନ ଦୂରତାକୁ ହିସାବକୁ ନେଲେ = 4500 ÷ 40 = 112.5 ଗୁଣ ।
ସର୍ବୋଚ୍ଚ ଦୂରତାକୁ ହିସାବକୁ ନେଲେ = 6000 ÷ 40 = 150 ଗୁଣ ।
∴ ସୋମୁଙ୍କ କୋଠାଠାରୁ ଶଙ୍ଖଚିଲର ଉଡ଼ିବା ପ୍ରାୟ 112 ରୁ 150 ଗୁଣ ଅଧିକ ।
ଦେଓମାଳି ପର୍ବତର ଉଚ୍ଚତା = 1672 ମିଟର ।
1672 ÷ 40 = 41.8 ବା 42 ଗୁଣ ।
∴ ଦେଓମାଳି ପର୍ବତ ସୋମୁର କୋଠାଠାରୁ 42 ଗୁଣ ବଡ଼ ।
ଉଡ଼ାଜାହାଜ 10,000 – 12,800 ମିଟର ଉଚ୍ଚତା ପର୍ଯ୍ୟନ୍ତ ଉଡ଼ିପାରେ ।
ସର୍ବନିମ୍ନ ଦୂରତାକୁ ହିସାବକୁ ନେଲେ = 10,000 ÷ 40 = 250 ଗୁଣ ।
ସର୍ବୋଚ୍ଚ ଦୂରତାକୁ ହିସାବକୁ ନେଲେ = 12,800 ÷ 40 = 320 ଗୁଣ ।
∴ ସୋମୁଙ୍କ କୋଠା ଅପେକ୍ଷା ପ୍ରାୟ 250 – 320 ଗୁଣ ଅଧ‌ିକ ଉଚ୍ଚରେ ଉଡ଼ାଜାହାଜଟି ଉଡୁଥଲା ।

Students can use Class 8 Math Solution Odia Medium and Class 8 Maths Chapter 1 Question Answer Odia Medium ବର୍ଗ ଓ ଘନ to check their answers after solving exercises.

8th Class Maths Chapter 1 Question Answer Odia Medium

Class 8 Maths Chapter 1 Odia Medium

Page No. 1

Question 1.
ପ୍ରକ୍ରିୟା ଆରମ୍ଭ ହେବା ପୂର୍ବରୁ ଖୋଇନାମ ଜାଣିପାରିଥିଲେ ଯେ ଶେଷରେ କେଉଁ ଲକରଗୁଡ଼ିକ ଖୋଲା ରହିବ । ସେ କିପରି ଜାଣିପାରିଲେ?
Solution:
ଯଦି କୌଣସି ଲକରକୁ ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଥର ଖୋଲାଯାଏ ବନ୍ଦ କରାଯାଏ, ତେବେ ଏହା ଖୋଲା କିମ୍ବା ବନ୍ଦ ରହିବ । ଏକ ଲକରକୁ ବନ୍ଦ ବା ଖୋଲାଯିବାର ସଂଖ୍ୟା ଲକର ନମ୍ବରର ଗୁଣନୀୟକର ସଂଖ୍ୟା ସହିତ ସମାନ । ମନେକର ଲକରକୁ 6 ଥର ପାଇଁ, ପ୍ରଥମ ବ୍ୟକ୍ତି ଖୋଲନ୍ତି, ଦ୍ଵିତୀୟ ବ୍ୟକ୍ତି ଏହାକୁ ବନ୍ଦ କରନ୍ତି, ତୃତୀୟ ବ୍ୟକ୍ତି ଏହାକୁ ଖୋଲନ୍ତି ଏବଂ ଷଷ୍ଠ ବ୍ୟକ୍ତି ଏହାକୁ ବନ୍ଦ କରନ୍ତି । 1, 2, 3 ଓ 6 ପ୍ରତ୍ୟେକ ରେ ଗୋଟିଏ ଲେଖାଏଁ ଗୁଣନୀୟକ । ଯଦି ଗୁଣନୀୟକ ସଂଖ୍ୟା ଯୁଗ୍ମ, ତେବେ ଲକରକୁ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟକ ଲୋକଙ୍କଦ୍ୱାରା ଖୋଲା ବା ବନ୍ଦ କରିହେବ ଏବଂ ଶେଷରେ ଏହା ବନ୍ଦ ରହିବ।
ଗୋଟିଏ ସଂଖ୍ୟାର ପ୍ରତ୍ୟେକ ଗୁର୍ଣନୀୟକର ଏକ ‘ସହଭାଗୀ ଗୁଣନୀୟକ’ ଥାଏ, ଯେଉଁ ଦୁଇଟିର ଗୁଣଫଳ ସଂଖ୍ୟାଟି ସହିତ ସମାନ ହୋଇଥାଏ ।
6 = 1 × 6 = 2 × 3
ଗୁଣନୀୟକଗୁଡ଼ିକ ହେଲେ : 1, 2, 3 ଏବଂ 6।

Question 2.
ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାର ଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଥାଏ କି?
Solution:
ନା, ପ୍ରତ୍ୟେକ ସଂଖ୍ୟାର ଯୁଗ୍ମ ସଂଖ୍ୟାକ ଗୁଣନୀୟକ ନ ଥାଏ । ଉଦାହରଣ : 1, 4, 9
1 = 1 × 1
ଏକମାତ୍ର ଗୁଣନୀୟକ
ହେଉଛି ।
4 = 1 × 4 = 2 × 2
ଗୁଣନୀୟକଗୁଡ଼ିକ ହେଲେ
1, 2 ଏବଂ 4
9 = 1 × 9 = 3 × 3
ଗୁଣନୀୟକଗୁଡ଼ିକ ହେଲେ
1, 3 ଏବଂ 9
କେତେକ କ୍ଷେତ୍ରରେ, 2 × 2 ଭଳି ଯୋଡ଼ିରେ ଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକ ସମାନ।

Page No. 2

Question 1.
ଉପରୋକ୍ତ ତଥ୍ୟକୁ ଆଧାର କରି ତୁମେ ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଥିବା ଅଧିକ କିଛି ସଂଖ୍ୟା ପାଇବ କି?

Solution:
ହଁ । 36ରେ ଏକ ଗୁଣନୀୟକ ଯୋଡ଼ି ହେଉଛି 6 × 6 ଅଛି, ଯେଉଁଠାରେ ଉଭୟ ସଂଖ୍ୟା 6 ଅଟେ । ଯଦି 6 ବ୍ୟତୀତ 3 ରେ ପ୍ରତ୍ୟେକ ଗୁଣନୀୟକର ଏକ ଭିନ୍ନ ସହଭାଗୀ ଗୁଣନୀୟକ ଅଛି, ତେବେ କହିବା 36 ରେ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଅଛି ।
ନିମ୍ନଲିଖୂତ ସମସ୍ତ ସଂଖ୍ୟାର ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଅଛି : 1 × 1, 2 × 2, 3 × 3, 4 × 4,…..
କୌଣସି ସଂଖ୍ୟାକୁ ସେହି ସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣିଲେ, ଗୁଣଫଳକୁ ଉକ୍ତ ସଂଖ୍ୟାର ବର୍ଗ ବା ଏକ ‘ବର୍ଗସଂଖ୍ୟା’ କୁହାଯାଏ।
କେବଳ ବର୍ଗସଂଖ୍ୟାଗୁଡ଼ିକର ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଥାଏ, କାରଣ ସେମାନଙ୍କର ପ୍ରତ୍ୟେକର ଏକ ଗୁଣନୀୟକ ଥାଏ, ଯାହାର ବର୍ଗ ସେହି ସଂଖ୍ୟା ସହ ସମାନ ହୋଇଥାଏ । ତେଣୁ, ଯେଉଁ ଲକରର ସଂଖ୍ୟା ଏକ ବର୍ଗ ସଂଖ୍ୟା, ତାହା ଖୋଲା ରହିବ ।

Page No. 3

Question 1.
ଖୋଲାଥିବା ଲକର ସଂଖ୍ୟାଗୁଡ଼ିକ ଲେଖ।
Solution:
1, 4, 9,…………….

Question 2.
ଖୋଇନାମ୍ ସଙ୍ଗେସଙ୍ଗେ ଏହି 10ଟି ଲକରରୁ ଶବ୍ଦ ସୂଚନା ସଂଗ୍ରହ କଲେ ଓ ବୁଝିପାରିଲେ।
ଠିକ୍ ଦୁଇଥର ଛୁଆଁ ଯାଇଥିବା ପ୍ରଥମ ପାଞ୍ଚୋଟି ଲକର ସଂଖ୍ୟାର ନାମ କୁହ ।
କେଉଁଗୁଡ଼ିକ ଏହି ପାଞ୍ଚୋଟି ଲକର?
Solution:
ପାଞ୍ଚଟି ଲକର୍ ସଂଖ୍ୟା ହେଲା 2, 3, 5, 7 ଓ 11 । ଯେଉଁ ଲକରଗୁଡ଼ିକ କେବଳ ଦୁଇଥର ଖୋଲା ବା ବନ୍ଦ ହୋଇଛି ତାହା ହେଉଛି ମୌଳିକ ସଂଖ୍ୟା, କାରଣ ପ୍ରତ୍ୟେକ ମୌଳିକସଂଖ୍ୟାର ଗୁଣନୀୟକ ହେଉଛି 1 ଓ ସେହି ସଂଖ୍ୟା । ତେଣୁ କୋଡ୍ ହେଉଛି 2-3-5-7-11।

1.1 ବର୍ଗ ସଂଖ୍ୟା (Square Numbers)

Page No. 4

Question 1.
ପ୍ରଥମ 30ଟି ସ୍ଵାଭାବିକ ସଂଖ୍ୟାର ବର୍ଗ ନିର୍ଣ୍ଣୟ କରି ନିମ୍ନ ସାରଣୀଟିକୁ ପୂରଣ କର । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

12 = 1	112 = 121	212 = 441
22 = 4	122 = 144	222 = 484
32 = 9	132 = 169	232 = 529
42 = 16	142 = 196	242 = 576
52 = 25	152 = 225	252 = 625
62 = 36	162 = 256	262 = 676
72 = 49	172 = 289	272 = 729
82 = 64	182 = 324	282 = 784
92 = 81	192 = 361	292 = 841
102 = 100	202 = 400	302 = 900

Question 2.
ତୁମେ ଏଠାରେ କେଉଁ ପ୍ରକାରର ସଂରଚନା ଲକ୍ଷ୍ୟ କରୁଅଛ? ଅନ୍ୟମାନଙ୍କ ସହିତ ଆଲୋଚନା କର ଓ ଅନୁଧାରଣ କର।
Solution:
ଏହି ସଂରଚନାରୁ ଆମେ ଜାଣିଲୁ, ଦୁଇଟି ପାଖାପାଖୁ ସଂଖ୍ୟାର ଯୋଗଫଳ ଗୋଟିଏ ବର୍ଗ ସଂଖ୍ୟା।
ତା’ର ପରବର୍ତୀ ପୂର୍ବ ବର୍ଗସଂଖ୍ୟା ଦୁଇଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଯୋଗଫଳ।
ଉଦାହରଣ : 1² = 1
2² = 1 + 3 = 4
3² = 1 + 3 + 5 = 9
4² = 1 + 3 + 5 + 7 = 16
5² = 16 + 9 = 25
6² = 25 + 11 = 36
7² = 36 + 13 = 49
………………………………….
………………………………….
………………………………….
28² = 729 + 55 = 784
29² = 784 + 57 = 841
30² = 841 + 59 = 900

Question 3.
ଏହି ସଂଖ୍ୟାଗୁଡ଼ିକର ଏକକ ସ୍ଥାନରେ କେଉଁସବୁ ଅଙ୍କ ଅଛି?
Solution:
ଏହି ସଂଖ୍ୟାଗୁଡ଼ିକର ଏକକ ସ୍ଥାନରେ 0, 1, 4, 5, 6 କିମ୍ବା 9 ଅଙ୍କଗୁଡ଼ିକ ରହିଥାଏ । 2, 3, 7 କିମ୍ବା 8 କୌଣସି ସଂଖ୍ୟାର ଏକକ ସ୍ଥାନ ଅଙ୍କ ହୋଇ ନ ଥାଏ ।

Question 4.
ଏକକ ସ୍ଥାନର 0, 1, 4, 5, 6 କିମ୍ବା 9 ଥିଲେ, ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ସଂଖ୍ୟାଟି ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ହୋଇଥାଏ କି ?
Solution:
ନା, ଏକକ ସ୍ଥାନରେ 0, 1, 4, 6 କିମ୍ବା ୨ ଥୁଲେ ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ସଂଖ୍ୟାଟି ପୂର୍ବବର୍ଗ ହୋଇପାରିବ ନାହିଁ । ଉଦାହରଣ 16 ଓ 36 ସଂଖ୍ୟାଦ୍ଵୟର ଏକକ ସ୍ଥାନରେ 6 ଅଛି ଓ ସଂଖ୍ୟାଦ୍ୱୟ ପୂର୍ଣ୍ଣବର୍ଗ ସଂଖ୍ୟା । ମାତ୍ର 26 ଓ 46 ର ଏକକ ସ୍ଥାନରେ 6 ଥିଲେ ମଧ୍ୟ ସଂଖ୍ୟାଦ୍ଵୟ ପୂର୍ବବର୍ଗ ନୁହଁନ୍ତି ।

Question 5.
5ଟି ସଂଖ୍ୟା ଲେଖ, ଯାହାର ଏକକ ଅଙ୍କକୁ ଦେଖ୍ ନିର୍ଣ୍ଣୟ କରିପାରିବ ଯେ ସେମାନେ ବର୍ଗ ସଂଖ୍ୟା ନୁହନ୍ତି?
Solution:
5 ଟି ସଂଖ୍ୟା 12, 32, 153, 4508, ଓ 9057
ଏଗୁଡ଼ିକର ଏକକ ଘରେ 2, 3, 7, 8 ଥିଲେ ଏମାନେ ବର୍ଗ ସଂଖ୍ୟା ନୁହଁନ୍ତି ।
ଯଦି ଗୋଟିଏ ସଂଖ୍ୟାର ଏକକ ସ୍ଥାନରେ 1 କିମ୍ବା 9 ଥାଏ, ତେବେ ତାହାର ବର୍ଗସଂଖ୍ୟାର ଏକକ ଘର 1 ହୋଇଥାଏ।
6 ରେ ଶେଷ ହେଉଥ‌ିବା ବର୍ଗ ସଂଖ୍ୟାଗୁଡ଼ିକ : 16 = 4², 36 = 6², 196 = 14², 256 = 16², 576 = 24² ଏବଂ 676 = 26²

Page No. 5

Question 1.
ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟାମାନଙ୍କ ମଧ୍ୟରୁ କେଉଁଗୁଡ଼ିକର ଏକକ ସ୍ଥାନରେ 6 ରହିବ?
(i) 38²
(ii) 34²
(iii) 46²
(iv) 56²
(v) 74²
(vi) 82²
Solution:
ଆମେ ଜାଣିଛେ, ଯଦି ଗୋଟିଏ ସଂଖ୍ୟାର ଏକକ ସ୍ଥାନର ଅଙ୍କଟି 4 ଓ 6 ଥାଏ, ତେବେ ତାହାର ବର୍ଗସଂଖ୍ୟାର ଏକକ ଘରେ 6 ରହିବ।
ତେଣୁ (ii) 34² (iii) 46² (iv) 56² (v) 74² ସଂଖ୍ୟାଗୁଡ଼ିକର ବର୍ଗସଂଖ୍ୟାର ଏକକ ସ୍ଥାନରେ 6 ରହିବ।

Question 2.
ପୂର୍ବରୁ ତୁମେ ପୂରଣ କରିଥିବା ସାରଣୀରୁ ସଂଖ୍ୟା ଏବଂ ସେମାନଙ୍କର ବର୍ଗଗୁଡ଼ିକୁ ଲକ୍ଷ୍ୟକରି ଏହିପରି ଅଧିକ ସଂରଚନା ଖୋଜ।
Solution:
ଆମେ ପୂରଣ କରିଥିବା ସାରଣୀରୁ ସଂଖ୍ୟା ଏବଂ ସେମାନଙ୍କର ବର୍ଗଗୁଡ଼ିକୁ ଲକ୍ଷ୍ୟକରି ଜାଣିଲୁ ଯେ-
(1) ଯଦି କୌଣସି ସଂଖ୍ୟାର ଏକକ ଘରେ 3 ଓ 7 ଥାଏ, ତେବେ ତାହାର ବର୍ଗସଂଖ୍ୟାର ଏକକ ଘରେ 9 ରହିବ।
(2) ଯଦି କୌଣସି ସଂଖ୍ୟାର ଏକକ ଘରେ 2 ଓ ୫ ଥାଏ, ତେବେ ତାହାର ବର୍ଗସଂଖ୍ୟାର ଏକକ ଘରେ 4 ରହିବ।
(3) ଯଦି କୌଣସି ସଂଖ୍ୟାର ଏକକ ଘରେ 5 ଥାଏ, ତେବେ ତାହାର ବର୍ଗସଂଖ୍ୟାର ଏକକ ଘରେ 5 ରହିବ।
ନିମ୍ନଲିଖୁତ ସଂଖ୍ୟା ଓ ସେମାନଙ୍କର ବର୍ଗଗୁଡ଼ିକୁ ବିଚାର କର।

Question 3.
ଯଦି ଗୋଟିଏ ସଂଖ୍ୟାର ଶେଷରେ 3ଟି ଶୂନ ଥାଏ, ତେବେ ତା’ର ବର୍ଗର ଶେଷରେ କେତୋଟି ଶୂନ ରହିବ?
Solution:
ଯଦି ଗୋଟିଏ ସଂଖ୍ୟାର ଶେଷରେ 3ଟି ଶୂନ ଥାଏ, ତେବେ ତା’ର ବର୍ଗର ଶେଷରେ ଟି ଶୂନ ରହିବ ।
ଉଦାହରଣ – 1000² = 1000000

Question 4.
ଏକ ସଂଖ୍ୟାର ଶେଷରେ ଥ‌ିବା ଶୂନମାନଙ୍କର ସଂଖ୍ୟା ଏବଂ ତାହାର ବର୍ଗର ଶେଷରେ ଥିବା ଶୂନମାନଙ୍କର ସଂଖ୍ୟା ବିଷୟରେ ତୁମେ କ’ଣ ଲକ୍ଷ୍ୟ କରୁଛ? ଏପରି ସବୁବେଳେ ହୋଇଥାଏ କି? ଆମେ କହିପାରିବା କି ବର୍ଗଗୁଡ଼ିକର ଶେଷରେ କେବଳ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟକ ଶୂନ ରହିବ?
Solution:
ଆମେ ଜାଣିଛୁ ଯେ, ଗୋଟିଏ ସଂଖ୍ୟାର ଶେଷରେ ଯେତୋଟି ଶୂନ ଥିବ ତା’ର ବର୍ଗର ଶେଷରେ ତା’ର ଦୁଇଗୁଣ ଶୂନ ରହିବ। ହଁ, ଏହା ସବୁବେଳେ ସମ୍ଭବ। ଆମେ କହିପାରିବା ଯେ ବର୍ଗଗୁଡ଼ିକର ଶେଷରେ କେବଳ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟକ ଶୂନ ରହିବ।

Question 5.
ଗୋଟିଏ ସଂଖ୍ୟା ତା’ର ବର୍ଗ ମଧ୍ୟରେ ଥିବା ସମ୍ପର୍କ ବିଷୟରେ ତୁମେ କ’ଣ କହିପାରିବ?
Solution:
ଆମେ କହିପାରିବା ଯେ ଗୋଟିଏ ସଂଖ୍ୟାକୁ ସେହି ସଂଖ୍ୟା ସହିତ ଗୁଣନ କଲେ ଆମେ ସେହି ସଂଖ୍ୟାର ବର୍ଗ ପାଇପାରିବା।

Question 6.
କ୍ରମିକ ବର୍ଗସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟରେ ଥିବା ପାର୍ଥକ୍ୟଗୁଡ଼ିକୁ ଖୋଜିବା । ତୁମେ କ’ଣ ଲଧ୍ୟ କରୁଛ?
Solution:
ଦୁଇଟି କ୍ରମିକ ବର୍ଗ ସଂଖ୍ୟା ମଧ୍ୟରେ ପାର୍ଥକ୍ୟ-
4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, 25 – 16 = 9

1 = 1 = 1²
1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
1 + 3 + 5 + 7 + 9 = 25 = 5²
1 + 3 + 5 + 7 + 9 + 11 = 36 = 6²

Page No. 6

Question 1.
ଏହି ସଂରଚନା ଆଧାରରେ 36² ନିର୍ଣ୍ଣୟ କର । 35² = 1225।
Solution:
ପ୍ରଶ୍ନରୁ ଆମେ ଜାଣୁଛୁ ଯେ, 1225 ହେଉଛି ପ୍ରଥମ 35ଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଯୋଗଫଳ।
36² ର ମୂଲ୍ୟ ଜାଣିବା ପାଇଁ ଆମକୁ 1225 ରେ 36 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟାକୁ ଯୋଗ କରିବାକୁ ପଡ଼ିବ।
ଅର୍ଥାତ୍ 1225 + 71 = 1296 (36 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 71)।

Question 2.
ତୁମେ 36 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା କିପରି ପାଇବ?
Solution:
ପ୍ରଥମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 1 – 1 = 1
ଦ୍ଵିତୀୟ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 2 – 1 = 3
ତୃତୀୟ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 3 – 1 = 5
ଚତୁର୍ଥ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 4 – 1 = 7
ପଞ୍ଚମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 5 – 1 = 9
ଷଷ୍ଠ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 6 – 1 = 11
ଏହି କ୍ରମଟିକୁ ଆଗକୁ ବଢ଼ାଇଲେ ଆମେ ପାଇବା 36 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 36 – 1 = 71

Question 3.
n ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା କେତେ ହେବ?
Solution:
n ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2n – 1 (‘n’ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା)
ତେଣୁ 36 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 36 – 1 = 72 – 1 = 71

Question 4.
ଏପରି ଗୋଟିଏ ସଂଖ୍ୟା ନେବା ଯାହା ଗୋଟିଏ ବର୍ଗ ସଂଖ୍ୟା ହୋଇ ନଥୁବ।
Solution:
ମନେକର ସଂଖ୍ୟାଟି = 38, ଏଥୁରୁ 1 ରୁ ଆରମ୍ଭ କରି କ୍ରମାଗତ ଭାବେ ଅଯୁଗ୍ମ ସଂଖ୍ୟାକୁ ବିୟୋଗ କଲେ,
38 – 1 = 37, 37 – 3 = 34, 34 – 5 = 29, 29 – 7 = 22, 22 – 9 = 13, 13 – 11 = 2, 2 – 13 = -11
ଏଥୁରୁ ଆମେ ଜାଣିଲେ, 38 କୁ 1 ରୁ ଆରମ୍ଭ କରି କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ଭାବେ ପ୍ରକାଶ କରିପାରିବା ନାହିଁ । ଯଦି ଗୋଟିଏ ଗଣନ ସଂଖ୍ୟାକୁ ।
ରୁ ଆରମ୍ଭ କରି କ୍ରମିକ ଅଯୁଗ୍ମ ସ୍ଵାଭାବିକ ସଂଖ୍ୟାର ଯୋଗଫଳ ଭାବେ ପ୍ରକାଶ କରାଯାଇପାରିବ ନାହିଁ, ତେବେ ତାହା ଏକ ପୂର୍ବବର୍ଗ ହୋଇପାରିବ ନାହିଁ ।

Page No. 7

Question 1.
1 ରୁ 100 ମଧ୍ୟରେ କେତୋଟି ବର୍ଗ ସଂଖ୍ୟା ଅଛି ? 101 ରୁ 200 ମଧ୍ଯରେ କେତୋଟି ଅଛି ? ତୁମେ ପୂର୍ବରୁ ପୂରଣ କରିଥବା ବର୍ଗଗୁଡ଼ିକର ସାରଣୀ ବ୍ୟବହାର କରି ପ୍ରତ୍ୟେକ 100ର ସଂଭାଗରେ କେତୋଟି ବର୍ଗସଂଖ୍ୟା ଅଛି ଗଣି ଲେଖ । 1000 ରୁ କମ୍ ସବୁଠାରୁ ବଡ଼ ବର୍ଗ ସଂଖ୍ୟାଟି କିଏ?

Solution:

1000 ରୁ କମ୍ ସବୁଠାରୁ ବଡ଼ ବର୍ଗ ସଂଖ୍ୟାଟି = 961

Question 2.
ତୁମେ ତ୍ରିଭୁଜାକାର ସଂଖ୍ୟା ଏବଂ ବର୍ଗ ସଂଖ୍ୟା ଆଗକୁ ବଢ଼ାଅ ଓ ପରବର୍ତ୍ତୀ ସଂଖ୍ୟା ମଧ୍ୟରେ କିଛି ସମ୍ପର୍କ ଲକ୍ଷ୍ୟ କରିପାରୁଛ କି? ଏହି ସଂରଚନାକୁ କେତେ ହେବ ଚିତ୍ରରେ ଦର୍ଶାଅ।

Solution:

Question 3.
ଗୋଟିଏ ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ 49 ବର୍ଗ ସେ.ମି.। ଏହାର ବାହୁର ଦୈର୍ଘ୍ୟ କେତେ?
Solution:
ଆମେ ଜାଣିଛୁ 7 × 7 = 49, କିମ୍ବା 7² = 49
ତେଣୁ 49 ବର୍ଗ ସେ.ମି. କ୍ଷେତ୍ରଫଳ ବିଶିଷ୍ଟ ଏକ ବର୍ଗକ୍ଷେତ୍ରର ପ୍ରତ୍ୟେକ ବାହୁର ଦୈର୍ଘ୍ୟ 7 ସେ.ମି.। ଆମେ 7 କୁ 49 ର ବର୍ଗମୂଳ କହୁ ।

Page No. 8

Question 1.
64 ର ବର୍ଗମୂଳ କେତେ?
Solution:
ଆମେ ଜାଣିଛୁ 8 × 8 = 64 ।
ତେଣୁ 8 ହେଉଛି 64 ର ବର୍ଗମୂଳ ।
ସେହିପରି (–8) × (–8) = 64
(-8)² = 64
∴ 8² = 64 ଓ (-8)² = 64
∴ 64 ର ବର୍ଗମୂଳଗୁଡ଼ିକ ହେଉଛି +8 ଓ –8।

Question 2.
ମନେକର, 576 କିମ୍ବା 327 ଭଳି ସଂଖ୍ୟା, ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା କି ନୁହେଁ କିପରି ଜାଣିବା ? ଯଦି ଏହା ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା, ତେବେ ଆମେ ଏହାର ବର୍ଗମୂଳ କିପରି ନିର୍ଣ୍ଣୟ କରିବା?
Solution:
576 ର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକ ନେଲେ,
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)
ଏଠାରେ ସମାନ ସଂଖ୍ୟାର ଚାରିଯୋଡ଼ି ଅଟେ । ଏଠାରେ କୌଣସି ମୌଳିକ ଗୁଣନୀୟକ ବଳକା ନାହିଁ । ତେଣୁ 576 ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ।
∴ \(\sqrt{576}=\sqrt{2^2 \times 2^2 \times 2^2 \times 5^2}\)
= 2 × 2 × 2 × 3
= 24
ସେହିପରି, 327 = 3 × 109
ଏହା ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ।

Question 3.
ଆମେ ସମସ୍ତ ବର୍ଗ ସଂଖ୍ୟାଗୁଡ଼ିକୁ କ୍ରମରେ ତାଲିକାଭୁକ୍ତ କରିପାରିବା ଏବଂ 576 ସେମାନଙ୍କ ମଧ୍ୟରେ ଅଛି କି ନାହିଁ ତାହା ଜାଣିପାରିବା ।
Solution:
20² = 400, 21² = 441, 22² = 484, 23² = 529, 24² = 576
ମାତ୍ର ଏହି ପ୍ରକ୍ରିୟା ବଡ଼ ସଂଖ୍ୟା ପାଇଁ ପ୍ରଯୁଜ୍ୟ ହେବ ନାହିଁ ।

Question 4.
1 ରୁ ଆରମ୍ଭ କରି କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ଭାବେ ପ୍ରତ୍ୟେକ ବର୍ଗ ସଂଖ୍ୟାକୁ ପ୍ରକାଶ କରିପାରିବା। ଉଦାହରଣ : √81
Solution:
ମନେକର √81
81 – 1 = 80, 80 – 3 = 77, 77 – 5 = 72, 72 – 7 = 65, 65 – 9 = 56, 56 – 11 = 45, 45 – 13 = 32, 32 – 15 = 17, 17 – 17 = 0
81 ରୁ ଆମେ 1 ରୁ ଆରମ୍ଭ କରି କ୍ରମାଗତ ଭାବେ ୨ ଥର ଅଯୁଗ୍ମ ସଂଖ୍ୟା ବିୟୋଗ କରିବା ପରେ ଆମେ 0 ପାଇଲୁ ।
ତେଣୁ √81 = 9

Page No. 9

Question 1.
ଗୋଟିଏ ପୂର୍ବ ସଂଖ୍ୟାକୁ ତା’ ନିଜ ସହିତ ଗୁଣନ କଲେ ଏକ ପୂର୍ଷ ବର୍ଗ ସଂଖ୍ୟା ମିଳିଥାଏ । ଗୋଟିଏ ସଂଖ୍ୟାକୁ ଏହାର ମୌଳିକ ଗୁଣନୀୟକମାନଙ୍କର ଗୁଣଫଳ ଭାବରେ ପ୍ରକାଶ କଲେ, ଏହା ପୂର୍ବବର୍ଗ କି ନୁହେଁ, ଜାଣିବା ସହଜ ହେବ କି?
Solution:
ହଁ, ଯଦି ଆମେ ସଂଖ୍ୟାଟିର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ସମାନ ଯୋଡ଼ି ସଂଖ୍ୟାରେ ଭାଗକରିବା, ତେବେ ପ୍ରତ୍ୟେକ ଭାଗର ମୌଳିକ ଗୁଣନୀୟକମାନଙ୍କର ଗୁଣଫଳ ହଁ, ସଂଖ୍ୟାଟିର ବର୍ଗମୂଳ ହୋଇଥାଏ । ଏହି ପ୍ରକ୍ରିୟାଟି ସହଜ ହୋଇଥାଏ ।

Question 2.
324 ଏକ ପୂର୍ଣ୍ଣ ବର୍ଗ କି?
Solution:
324 = 2 × 2 × 3 × 3 × 3 × 3
ଏହାକୁ ନିମ୍ନମତେ ଭାଗ ଭାଗ କରିପାରିବା ।
324 = (2 × 3 × 3) × (2 × 3 × 3) = (2 × 3 × 3)² = 18²
ଆମେ ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ଯୋଡ଼ିରେ ଲେଖୁଲେ,
\(\sqrt{324}=\sqrt{2 \times 2 \times 3 \times 3 \times 3 \times 3}=\sqrt{(2 \times 3 \times 3)^2}\) = 18
∴ 324 ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା।

Question 3.
156 ଗୋଟିଏ ପୂର୍ବବର୍ଗ କି?
Solution:
156 କୁ ମୌଳିକ ଗୁଣନୀୟକମାନଙ୍କର ଗୁଣଫଳରେ ପ୍ରକାଶ କଲେ,
156 = 2 × 2 × 3 × 13
ଆମେ ଏହି ଗୁଣନୀୟକଗୁଡ଼ିକୁ ଯୋଡ଼ିରେ ପ୍ରକାଶ କରିପାରିବା ନାହିଁ ।
ତେଣୁ, 156 ଗୋଟିଏ ପୂର୍ବବର୍ଗ ନୁହେଁ ।

Question 4.
ମୌଳିକ ଗୁଣନୀୟକ ନିଷ୍କ୍ରିୟ କରି 1156 ଏବଂ 2800 ପୂର୍ବବର୍ଗ କି ନୁହେଁ ପରୀକ୍ଷା କର।
Solution:
(a) 1156 କୁ ମୌଳିକ ଗୁଣନୀୟକମାନଙ୍କର ଗୁଣଫଳରେ ପ୍ରକାଶ କଲେ,
1156 = 2 × 2 × 17 × 17 = (2 × 17)²
ଆମେ ଏହି ଗୁଣନୀୟକଗୁଡ଼ିକୁ ଯୋଡ଼ିରେ ପ୍ରକାଶ କରିପାରିବା
ତେଣୁ 1156 ଗୋଟିଏ ପୂର୍ବବର୍ଗ।

(b) 2800 କୁ ମୌଳିକ ଗୁଣନୀୟକମାନଙ୍କର ଗୁଣଫଳରେ ପ୍ରକାଶ କଲେ, ଆମେ ପାଇବା
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
ଆମେ ଏହି ଗୁଣନୀୟକଗୁଡ଼ିକୁ ଯୋଡ଼ିରେ ପ୍ରକାଶ କରିପାରିବା ନାହିଁ ।
ତେଣୁ, 2800 ଗୋଟିଏ ପୂର୍ଣବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ ।
(i) 1600 = (40)² ଓ 2500 = (50)² ମଧ୍ୟରେ √1936 ଅଛି, ତେଣୁ 40 < 1936 < 50
(ii) 1936ରେ ଶେଷ ଅଙ୍କ 6 । ତେଣୁ ବର୍ଗମୂଳର ଶେଷ ଅଙ୍କ 4 କିମ୍ବା 6 ହେବ । ଏହା 44 କିମ୍ବା 46 ହୋଇପାରେ।
(iii) ଯଦି ଆମେ 452 ର ମୂଲ୍ୟ ଜାଣିବା, ତେବେ ଆମେ ଏହାକୁ 40 – 50 ସଂଭାଗରେ ରଖୁବା, ଯାହା ମଧ୍ଯ 40 – 45 ବା 45 – 50 ର ସଂଭାଗ ହୋଇପାରେ ।
ଆମେ ଲେଖୁପାରିବା
45² = (40 + 5) (40 + 5)
= 40² + 2 × 40 × 5 + 5²
= 1600 + 400 + 25
= 2025
(iv) 2025 > 1936, ତେଣୁ 40 < √1936 < 45
(v) ଏଥୁରୁ ଆମେ ଜାଣିପାରିବା √1936 = 44

Question 5.
ସୋନୁ ଓ ବିଜୁ ଏକ କେଳ ଖେଳନ୍ତି । ଜଣେ ଗୋଟିଏ ସଂଖ୍ୟା କହିଲେ ଅନ୍ୟ ଜଣେ ତା’ର ବର୍ଗମୂଳ କହି ଉତ୍ତର ଦିଏ । ସୋନୁ 25 କହି ଖେଳ ଆରମ୍ଭ କଲା ଓ ବିଜୁ ସାଙ୍ଗେସାଙ୍ଗେ ଉତ୍ତର 5 ଦେଲା । ତା’ପରେ ବିଜୁ କହିଲା 81, ସୋନୁ ଉତ୍ତର ଦେଲା ୨ । ସୋନୁ 250 କହିବା ପର୍ଯ୍ୟନ୍ତ ଖେଳ ଚାଲିଥିଲା । 250 ଗୋଟିଏ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନ ହୋଇଥିବାରୁ ବିଜୁ ଏହାର ଉତ୍ତର ଦେଇପାରିଲା ନାହିଁ ।
କେଉଁଟି 250ର ର୍ବଗମୂଳର ନିକଟତମ ସଂଖ୍ୟା
Solution:
ଏଥିପାଇଁ 250ର ବର୍ଗମୂଳ କେତେ ହେବ, ତାହା ଆକଳନ କରିବାକୁ ପଡ଼ିବ।
ଆମେ ଜାଣିଛୁ ଯେ, 100 < 250 < 400 ଏବଂ √100 = 10 ଓ √400 = 20
ତେଣୁ, 10 < √250 < 20
ତଥାପି ଆମେ ସେହି ସଂଖ୍ୟାର ନିକଟତର ହୋଇନାହୁଁ, ଯାହାର ବର୍ଗ 250।
ଆମେ ଜାଣୁ ମେ 15² = 225 ଏବଂ 16² = 256।
ତେଣୁ 15 < √250 < 16
225 ତୁଳନାରେ 250 ର ଅଧ୍ଵକ ନିକଟତର ହେଉଛି 256,
ତେଣୁ, 250 ର ମୂଲ୍ୟ ପ୍ରାୟ 16, ଯଦିଓ ଏହା 16 ଠାରୁ ସାମାନ୍ୟ କମ୍।

Question 6.
ଅଞ୍ଛଳ ପାଖରେ 125 ବର୍ଗ ସେ.ମି. କ୍ଷେତ୍ରଫଳ ବିଶିଷ୍ଟ ଖଣ୍ଡିଏ ବର୍ଗାକୃତି କପଡ଼ା ଅଛି । ଏଥରୁ 15 ସେ.ମି. ଦୈର୍ଘ୍ୟର ବର୍ଗାକୃତି କପଡ଼ା କଟାଯାଇପାରିବ କି ? ଯଦି ନୁହେଁ, ସେ ଜାଣିବାକୁ ଚାହାନ୍ତି ଯେ, ଏହି କପଡ଼ା ଖଣ୍ଡରୁ କେଉଁ ସର୍ବାଧ‌ିକ ଆକୃତିର ରୁମାଲ କାଟିହେବ, ଯାହାର ପାର୍ଶ୍ଵର ଲମ୍ବ ପୂର୍ବସଂଖ୍ୟା ହେଉଥ‌ିବ?
Solution:
125 ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ ।
ନିକଟତମ ପୂର୍ଣ୍ଣବର୍ଗ ସଂଖ୍ୟାଗୁଡ଼ିକ ହେଲେ 11² = 121 ଓ 12² = 144
ତେଣୁ ଏହି କପଡ଼ାରୁ ଅତି ବେଶୀରେ 11 ସେ.ମି. ଦୈର୍ଘ୍ୟର ବର୍ଗାକୃତି ରୁମାଲଟିଏ ପ୍ରସ୍ତୁତ କରାଯାଇପାରିବ।

ନିଜେ କରି ଦେଖ (Page No. 10-11)

Question 1.
ନିମ୍ନଲିଖ୍ୟାତ ସଂଖ୍ୟାଗୁଡ଼ିକ ମଧ୍ୟରୁ କେଉଁଗୁଡ଼ିକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Solution:
(i) 2032 ସଂଖ୍ୟାର ଏକକ ଘରେ 2 ଥିବାରୁ ସଂଖ୍ୟାଟି ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ।
(ii) 2018 ସଂଖ୍ୟାର ଏକକ ଘରେ 8 ଥିବାରୁ ସଂଖ୍ୟାଟି ପୂର୍ଣବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ।
(iii) 1027 ସଂଖ୍ୟାର ଏକକ ଘରେ 7 ଥିବାରୁ ସଂଖ୍ୟାଟି ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ।
(iv) 1089 ସଂଖ୍ୟାର ଏକକ ଘରେ 9 ଥିବାରୁ ସଂଖ୍ୟାଟି ଗୋଟିଏ ପୂର୍ଣବର୍ଗ ସଂଖ୍ୟା ଅଟେ।

Question 2.
64², 108², 292² ଓ 36² ମଧ୍ୟରୁ କେଉଁଟିର ଶେଷ ଅଙ୍କ 4 ହେବ?
Solution:
(i) 64 ସଂଖ୍ୟାର ଏକକ ଘର ସଂଖ୍ୟା = 4
∴ 4² = 4 × 4 = 16 (ଶେଷ ଅଙ୍କଟି = 6)
(ii) 108 ସଂଖ୍ୟାର ଏକକ ଘର ସଂଖ୍ୟା = 8
∴ 8² = 8 × 8 = 64 (ଶେଷ ଅଙ୍କଟି = 4)
(iii) 292 ସଂଖ୍ୟାର ଏକକ ଘର ସଂଖ୍ୟା = 2
∴ 2² = 2 × 2 = 4 (ଶେଷ ଅଙ୍କଟି = 4)
(iv) 36 ସଂଖ୍ୟାର ଏକକ ଘର ସଂଖ୍ୟା = 6
∴ 6² = 6 × 6 = 36 (ଶେଷ ଅଙ୍କଟି = 6)
ତେଣୁ, ଯେଉଁ ସଂଖ୍ୟାଗୁଡ଼ିକର ବର୍ଗ 4ରେ ଶେଷ ହୁଏ, ସେଗୁଡ଼ିକ ହେଉଛି 108² ଓ 209²

Question 3.
125² = 15625, ତେବେ 126² ର ମାନ କେତେ ହେବ?
(i) 15625 + 126
(ii) 15625 + 26²
(iii) 15625 + 253
(iv) 15625 + 251
(v) 15625 + 51²
Solution:
ଏଠାରେ 126² = (125 + 1)²
= (125)² + 2 × 125 × 1 + (1)² [∵ ଯେହେତୁ (a + b)² = a² + 2ab + b²]
= 15625 + 250 + 1
= 15625 + 251
ତେଣୁ, (iv) ଉତ୍ତରଟି ଠିକ୍।

Question 4.
441 ବର୍ଗମିଟର କ୍ଷେତ୍ରଫଳ ବିଶିଷ୍ଟ ଗୋଟିଏ ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ କେତେ?
Solution:
ବର୍ଗକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = ବାହୁ × ବାହୁ = 441
⇒ ବାହୁ² = 441 ବର୍ଗ ମିଟର
⇒ ବାହୁ = √441 ବର୍ଗମିଟର।
441 = (3 × 3) × (7 × 7)
⇒ √441 = 3 × 7 = 21 ମିଟର।
∴ ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ = 21 ମିଟର।

Question 5.
4, 9 ଓ 10 ଦ୍ଵାରା ବିଭାଜ୍ୟ କ୍ଷୁଦ୍ରତମ ବର୍ଗସଂଖ୍ୟାଟି କେତେ?
Solution:
4, 9 ଓ 10 ଦ୍ଵାରା ବିଭାଜ୍ୟ କ୍ଷୁଦ୍ରତମ ବର୍ଗସଂଖ୍ୟାଟି ପାଇବାକୁ ହେଲେ
4, 9 ଓ 10 ର ଲ.ସା.ଗୁ. ନିର୍ଣ୍ଣୟ କରିବାକୁ ହେବ।

ଲ.ସା.ଗୁ. = 2 × 2 × 3 × 3 × 5 = 180
180 ର ମୌଳିକ ଗୁଣନୀୟକ ଗୁଡ଼ିକ = (2 × 2) × (3 × 3) × 5
ଯେହେତୁ 5 ସଂଖ୍ୟାଟି ଯୋଡ଼ିରେ ନାହିଁ, 180 ସଂଖ୍ୟାଟି ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ ।
ପୂର୍ଣ୍ଣ ବର୍ଗସଂଖ୍ୟାଟି ପାଇବାକୁ ହେଲେ, 180 କୁ 5 ଦ୍ଵାରା ଗୁଣନ କରିବାକୁ ହେବ = 180 × 5 = 900
∴ 4, 9 ଓ 10 ଦ୍ଵାରା ବିଭାଜ୍ୟ କ୍ଷୁଦ୍ରତମ ବର୍ଗସଂଖ୍ୟାଟି 900।

Question 6.
କେଉଁ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟାଦ୍ଵାରା 9408କୁ ଗୁଣନ କଲେ, ଗୁଣଫଳ ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ହେବ? ଗୁଣଫଳର ବର୍ଗମୂଳ ନିର୍ଣ୍ଣୟ କର ।
Solution:
9408 = (2 × 2) × (2 × 2) × (2 × 2) × 3 × (7 × 7)
9408 ସଂଖ୍ୟାର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ଯୋଡ଼ିରେ ସଜାଇଲେ 3 ବଳିପଡ଼ିବ।
9408 କୁ 3 ଦ୍ଵାରା ଗୁଣନ କଲେ ଆମେ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ପାଇବା
ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା = 9408 × 3 = 28224

ବର୍ତ୍ତମାନ \(\sqrt{28224}=\sqrt{(2 \times 2) \times(2 \times 2) \times(3 \times 3) \times(3 \times 3) \times(7 \times 7)}\)
= 2 × 2 × 3 × 3 × 7
= 252
∴ ସଂଖ୍ୟାଟି 252।

Question 7.
ନିମ୍ନଲିଖତ ସଂଖ୍ୟାଗୁଡ଼ିକର ବର୍ଗ ମଧ୍ୟରେ କେତୋଟି ସଂଖ୍ୟା ରହିବ?
(i) 16 ଓ 17
(ii) 99 ଓ 100
Solution:
(i) 16² ଓ 17² ମଧ୍ୟରେ ଥିବା ସଂଖ୍ୟା = 2 × 16 = 32
(ii) 99² ଓ 100² ମଧ୍ୟରେ ଥିବା ସଂଖ୍ୟା = 2 × 99 = 198

Question 8.
ନିମ୍ନଲିଖୁତ ସଂରଚନାରେ ଖାଲିଥିବା ସ୍ଥାନରେ ଠିକ୍ ସଂଖ୍ୟା ଲେଖୁ ପୂରଣ କର।
(i) 1² + 2² + 2² = 3²
(ii) 2² + 3² + 6² = 7²
(iii) 3² + 4² + 12² = 13²
(iv) 4² + 5² + 20² = (____)²
(v) 9² + 10² + (____)² = (____)²
Solution:
(i) 1² + 2² + 2² = 3²
(ii) 2² + 3² + 6² = 7²
(iii) 3² + 4² + 12² = 13²
(iv) 4² + 5² + 20² = (21)²
(v) 9² + 10² + (90)² = (91)²

Question 9.
ପାର୍ଶ୍ଵରେ ଦିଆଯାଇଥ‌ିବା ଚିତ୍ରରେ କେତୋଟି ଛୋଟ ବର୍ଗ ଚିତ୍ର ଅଛି? ଭକ୍ତ ସଂଖ୍ୟାର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ଲେଖ।

Solution:
ଗୋଟିଏ ଧାଡ଼ିରେ ଥୁବା ବର୍ଗଚିତ୍ର ସଂଖ୍ୟା = 9
ଗୋଟିଏ ସ୍ତମ୍ଭରେ ଥିବା ବର୍ଗଚିତ୍ର ସଂଖ୍ୟା = 9
ଗୋଟିଏ ବଡ଼ ବର୍ଗଚିତ୍ର ମଧ୍ୟରେ ଥ‌ିବା ଛୋଟ ବର୍ଗଚିତ୍ର = 5 × 5 = 25
ସମୁଦାୟ ଛୋଟ ବର୍ଗଚିତ୍ର = 9 × 9 × 25 = 2025
2025 ର ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକ = 3 × 3 × 3 × 3 × 5 × 5 = 45²

1.2 ଘନ ସଂଖ୍ୟା (Cubic Numbers)

Page No. 11

Question 1.
ଉଦାହରଣ : 1 ସେ.ମି. ବାହୁ ବିଶିଷ୍ଟ କେତୋଟି ସମଘନକୁ ନେଲେ 3 ସେ.ମି. ବାହୁ ବିଶିଷ୍ଟ ଗୋଟିଏ ସମଘନ ପ୍ରସ୍ତୁତ ହୋଇପାରିବ?
Solution:
1 ସେ.ମି. ବାହୁ ବିଶିଷ୍ଟ 3ଟି ସମଘନକୁ ନେଲେ 3 ସେ.ମି. ବାହୁ ବିଶିଷ୍ଟ ଗୋଟିଏ ସମଘନ ପ୍ରସ୍ତୁତ ହୋଇପାରିବ।

Question 2.
9 ଏକ ଘନ ସଂଖ୍ୟା ହେବ କି?

Solution:
ଆମେ ଜାଣିଛେ, 2 × 2 × 2 = 8 ଏବଂ 3 × 3 × 3 = 27
ଏଥୁରୁ ଜଣାପଡୁଛି, ୨ ଗୋଟିଏ ଘନ ସଂଖ୍ୟା ନୁହେଁ ବା
10 ଠାରୁ 26 ମଧ୍ଯରେ କୌଣସି ଘନ ସଂଖ୍ୟା ନାହିଁ ।

Question 3.
4 ଏକକ ଦୈର୍ଘ୍ୟ ବିଶିଷ୍ଟ ଗୋଟିଏ ସମଘନରେ କେତୋଟି ଏକକ ସମଘନ ରହିବ?

Solution:
ଏଥିରେ 64ଟି । ଘନ ଏକକ ବିଶିଷ୍ଟ ସମଘନ ଅଛି। ତୁମେ (ଏକକ ସମଘନ) ଭଲଭାବେ ଲକ୍ଷ୍ୟ କଲେ ଜାଣିପାରିବେ, ଏହାର ପ୍ରତ୍ୟେକ ସ୍ତରରେ (4 × 4)ଟି ଏକକ ସମଘନ ଅଛି । ଅର୍ଥାତ୍ ପ୍ରତ୍ୟେକ ସ୍ତରରେ (4 × 4) ବା 16 ଟି ଏକକ ସମଘନ ଅଛି । ଏହିପରି 4ଟି ସ୍ତର ଅଛି, ଅର୍ଥାତ୍ ମୋଟ ଏକକ ସମଘନ ସଂଖ୍ୟା 4 × 4 × 4 = 64 ହେବ।

Question 4.
ନିମ୍ନ ସାରଣୀଟିକୁ ପୂରଣ କର । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

13 = 1	113 = 1331
23 = 8	123 = 1728
33 = 27	133 = 2197
43 = 64	143 = 2744
53 = 125	153 = 3375
63 = 216	163 = 4096
73 = 343	173 = 4913
83 = 512	183 = 5832
93 = 729	193 = 6859
103 = 1000	203 = 8000

Page No. 13

Question 1.
ବର୍ଗସଂଖ୍ୟା ପରି, 1 ଅଙ୍କ, 2 ଅଙ୍କ ଏବଂ 3 ଅଙ୍କ ବିଶିଷ୍ଟ କେତୋଟି ଘନ ସଂଖ୍ୟା ଅଛି କହିପାରିବ କି?
Solution:
ଏକ ଅଙ୍କ ବିଶିଷ୍ଟ ଘନସଂଖ୍ୟା, 1³ = 1, 2³ = 8
ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ଘନସଂଖା, 3³ = 27, 4³ = 64
ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ ଘନସଂଖ୍ୟା, 5³ = 125, 6³ = 216, 7³ = 343, 8³ = 512, 9³ = 729
∴ ଏକ ଅଙ୍କବିଶିଷ୍ଟ 2ଟି, ଦୁଇ ଅଙ୍କବିଶିଷ୍ଟ 2ଟି ଓ ତିନି ଅଙ୍କବିଶିଷ୍ଟ 5ଟି ଘନସଂଖ୍ୟା ଅଛି।

Question 2.
ଗୋଟିଏ ଘନସଂଖ୍ୟାର ଶେଷ ଅଙ୍କ ଦୁଇଟି ଶୂନ (0 0) ହୋଇପାରିବ କି? ବୁଝାଅ।
Solution:
ନା। ଗୋଟିଏ ଘନସଂଖ୍ୟାର ଶେଷରେ ଯେତେ ଶୂନ ଥାଉ ନା କାହିଁକି,
ସେହି ଶୂନ ସଂଖ୍ୟା ସର୍ବଦା 3ର ଗୁଣିତକ ହେବା ଆବଶ୍ୟକ।
ଉଦାହରଣ : 10ର ଘନ = 10 × 10 × 10 = 1000

Question 3.
ସଂଖ୍ୟାଗୁଡ଼ିକର ଘନସଂଖ୍ୟା ନିର୍ଣ୍ଣୟ କର : \(\left(\frac{4}{6}\right)^3\), (13.08)³ ଓ (-6)³
Solution:
\(\left(\frac{4}{6}\right)^3=\left(\frac{4}{6}\right) \times\left(\frac{4}{6}\right) \times\left(\frac{4}{6}\right)=\left(\frac{64}{216}\right)\)
(13.08)³ = 13.08 × 13.08 × 13.08 = 2237.810112
(-6)³ = -6 × -6 × -6 = -216

Question 4.
1729ର ଦୁଇଟି ପରବର୍ତୀ ଟାକ୍ସିକ୍ୟାବ୍ ସଂଖ୍ୟା ହେଲେ 4104 ଓ 13832। ଏହି ଦୁଇଟି ସଂଖ୍ୟାକୁ କେଉଁ ଦୁଇ ଉପାୟରେ ଦୁଇଟି ଘନର ଯୋଗଫଳ ଭାବେ ପ୍ରକାଶ କରାଯାଇପାରିବ, ଚେଷ୍ଟାକର।
Solution:
ଦିଆଯାଇଥବା ଟାକ୍ସିକ୍ୟାବ୍ ସଂଖ୍ୟା ହେଲେ 4104 ଓ 13832
∴ 4104 = 2³ + 16³ = 9³ + 15³
ଏବଂ 13832 = 2³ + 24³ = 18³ + 20³

Page No. 14

Question 1.
ଯୋଗ ନକରି ତୁମେ ଏହି ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ କେତେ କହିପାରିବ କି?
Solution:
ହଁ, ଏହା 10³ ହେବ।

Question 2.
ଗୋଟିଏ ସଂଖ୍ୟା ଘନ କି ନୁହେଁ, ଆମେ କିପରି ଜାଣିବା?
Solution:
ବର୍ଗସଂଖ୍ୟା ଭଳି ମୌଳିକ ଗୁଣନୀୟକର ଉତ୍ପାଦକୀକରଣକୁ ବ୍ୟବହାର କରି ସଂଖ୍ୟାଟି ଘନ କି ନୁହେଁ ଆମେ ଜାଣିପାରିବା।

Question 3.
3375 ଏକ ପୂର୍ଣ ଘନ ସଂଖ୍ୟା କି ନୁହେଁ ପରୀକ୍ଷା କରି ଦେଖୁବା।
Solution:
ଆମେ 3375ରେ ମୌଳିକ ଗୁଣନୀୟକ ନିର୍ଣ୍ଣୟ କରିବା।
3375 = 3 × 3 × 3 × 5 × 5 × 5
ଏବେ ଆମେ ଏହି ଗୁଣନୀୟକଗୁଡ଼ିକୁ ତିନୋଟି ଲେଖାଏଁ ଗ୍ରୁପ୍‌ରେ ସମାନ ଭାଗ କରିବା।
ତେଣୁ, 3375 = (3 × 5) × (3 × 5) × (3 × 5) = (3 × 5)³ = 15³
ଯେହେତୁ ସମସ୍ତ ମୌଳିକ ଗୁଣନୀୟକଗୁଡ଼ିକ ତିନୋଟି ଲେଖାଏଁ ଗ୍ରୁପ୍‌ରେ ରହିବେ,
3375 ଏକ ପୂର୍ଣ୍ଣଘନ ସଂଖ୍ୟା ହେବ।
ଅନ୍ୟ ଏକ ଉପାୟରେ ମଧ୍ୟ ଏହାକୁ ଲେଖୁପାରିବା
3375 = (3 × 3 × 3) × (5 × 5 × 5)³ = 3³ × 5³
ଅର୍ଥାତ୍ \(\sqrt[3]{3375}\) = 15

Question 4.
500 ଏକ ପୂର୍ଣ ଘନ ସଂଖ୍ୟା କି?
Solution:
500 = 2 × 2 × 5 × 5 × 5
ଏଠାରେ ଗୁଣନୀୟକଗୁଡ଼ିକୁ ତିନୋଟି ଲେଖାଏଁ ସମାନ ଭାଗ କରାଯାଇପାରିବ ନାହିଁ ।
ତେଣୁ 500 ଗୋଟିଏ ଗୂର୍ଣ୍ଣ ଘନ ସଂଖ୍ୟା ନୁହେଁ ।

Page No. 15

Question 1.
ଏହି ସଂଖ୍ୟାଗୁଡ଼ିକର ଘନମୂଳ ନିର୍ଣ୍ଣୟ କର।
(i) \(\sqrt[3]{64}\)
(ii) \(\sqrt[3]{512}\)
(iii) \(\sqrt[3]{729}\)
Solution:
(i) 64 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 2 × 2 × 2
64 = (2 × 2 × 2) × (2 × 2 × 2) = 2³ × 2³
ଅର୍ଥାତ୍ \(\sqrt[3]{64}\) = 2 × 2 = 4

(ii) 512 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 2³ × 2³ × 2³
ଅର୍ଥାତ୍ \(\sqrt[3]{512}\) = 2 × 2 × 2 = 8

(iii) 729 ର ମୌଳିକ ଗୁଣନୀୟକ = 3 × 3 × 3 × 3 × 3 × 3
729 = (3 × 3 × 3) × (3 × 3 × 3) = 3³ × 3³
ଅର୍ଥାତ୍ \(\sqrt[3]{729}\) = 3 × 3 = 9

Question 2.
ତୁମେ କ’ଣ ଲକ୍ଷ୍ୟ କରୁଛ?

Solution:

1.3 ଇତିହାସ ପୃଷ୍ଠାରୁ

ନିଜେ କରି ଦେଖ (Page No. 16-17)

Question 1.
27000 ଓ 10678 ର ଘନମୂଳ ନିର୍ଣ୍ଣୟ କର।
Solution:
27000 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
27000 = (2 × 2 × 2) × (3 × 3 × 3) × (5 × 5 × 5) = 2³ × 3³ × 5³
∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5 = 30
10678 ର ମୌଳିକ ଗୁଣନୀୟକ = 2 × 2 × 2 × 11 × 11 × 11
10678 = (2 × 2 × 2) × (11 × 11 × 11) = 2³ × 11³
∴ \(\sqrt[3]{10678}\) = 2 × 11 = 22

Question 2.
କେଉଁ ସଂଖ୍ୟାଦ୍ଵାରା 1323କୁ ଗୁଣନ କଲେ, ଗୁଣଫଳ ଗୋଟିଏ ଘନସଂଖ୍ୟା ହେବ।
Solution:

1323 = 3 × 3 × 3 × 7 × 7
ମୌଳିକ ଗୁଣନୀୟକୁ ତିନୋଟି ଲେଖାଏଁ ଶ୍ରେଣୀରେ ସମାନ କଲେ, ଆଉ ଗୋଟିଏ? ଦରକାର ହୋଇଥାଏ।
ତେଣୁ 1323 କୁ 7 ଦ୍ଵାରା ଗୁଣନ କଲେ ଏହା ଗୋଟିଏ ଘନସଂଖ୍ୟା ହେବ।
ଘନସଂଖ୍ୟା = 1323 × 7 = 9261
∴ 1323କୁ 7 ସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନ କଲେ, ଗୁଣଫଳ ଗୋଟିଏ ଘନସଂଖ୍ୟା ହେବ।

Question 3.
ଭୁଲ୍ (✗) କି ଠିକ୍ (✓), କାରଣ ସହ ଦର୍ଶାଅ।
(i) ଯେକୌଣସି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଘନ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା।
(ii) ଏପରି କୌଣସି ପୂର୍ଣ୍ଣ ଘନ ସଂଖ୍ୟା ନାହିଁ, ଯାହାର ଶେଷ ଅଙ୍କ 8 ହେବ।
(iii) ଗୋଟିଏ ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଘନ, ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହୋଇପାରେ।
(iv) ଗୋଟିଏ ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଘନ ସାତ କିମ୍ବା ଅଧିକ ଅଙ୍କ ବିଶିଷ୍ଟ ହୋଇପାରେ।
(v) ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକର ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଥାଏ।
Solution:
(i) ଯେକୌଣସି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଘନ ଏକ ଯୁଗ୍ମ ସଂଖ୍ୟା। (✗)
କାରଣ – ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଘନ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା।
ଅର୍ଥାତ ଗୋଟିଏ ଅଯୁଗ୍ମ ସଂଖ୍ୟାକୁ ତିନିଥର ଗୁଣନ କଲେ, ଗୁଣଫଳ ଏକ ଅଯୁଗ୍ମ ସଂଖ୍ୟା ହୋଇଥାଏ।
ଉଦାହରଣ – 3³ = 27, 5³ = 125, 7³ = 343

(ii) ଏପରି କୌଣସି ପୂର୍ଣ୍ଣ ଘନ ସଂଖ୍ୟା ନାହିଁ, ଯାହା ଶେଷ ଅଙ୍କ 8 ହେବ। (✗)
କାରଣ – ଗୋଟିଏ ସଂଖ୍ୟାର ଏକକ ଘରେ 2 ଥିଲେ ତାହାର ଘନ ସଂଖ୍ୟାର ଏକକ ଘରେ 8 ର େବ।
ଉଦାହରଣ : 2³ = 8, 12³ = 1728, 22³ = 10648

(iii) ଗୋଟିଏ ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଘନ, ତିନି ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହୋଇପାରେ। (✗)
କାରଣ – ଗୋଟିଏ ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଘନ, ଚାରି କିମ୍ବା ଛଅ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହୋଇପାରେ।
ଉଦାହରଣ – ମନେକର ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ସାନ ସଂଖ୍ୟା = 10
10 ସଂଖ୍ୟାର ଘନ = 10³ = 1000 (ଚାରି ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା)

(iv) ଗୋଟିଏ ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଘନ ସାତ କିମ୍ବା ଅଧ‌ିକ ଅଙ୍କ ବିଶିଷ୍ଟ ହୋଇପାରେ । (✗)
କାରଣ – ଗୋଟିଏ ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଘନ ଚାରି କିମ୍ବା ଛଅ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା ହୋଇପାରେ।
ଉଦାହରଣ – ମନେକର ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ବଡ଼ ସଂଖ୍ୟା = 99
99 ସଂଖ୍ୟାର ଘନ = 99³ = 970299 (ଛଅ ଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟା)

(v) ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକର ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଥାଏ । (✗)
କାରଣ – ଘନ ସଂଖ୍ୟାଗୁଡ଼ିକର ଅଯୁଗ୍ମ କିମ୍ବା ଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ ଥାଏ।
ଉଦାହରଣ – 27 = 3 × 3 × 3 (ଅଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ)
64 = 2 × 2 × 2 × 2 × 2 × 2 (ଯୁଗ୍ମ ସଂଖ୍ୟକ ଗୁଣନୀୟକ)

Question 4.
ମନେକର 1331 ଗୋଟିଏ ପୂର୍ବଘନ ସଂଖ୍ୟା। ଉତ୍ପାଦକ ନିଶ୍ଚୟ ନକରି ତୁମେ ଏହାର ଘନମୂଳ ଅନୁମାନ କରିପାରିବ କି? ସେହିପରି 4913, 12167 ଓ 32768ର ଘନମୂଳ ନିର୍ଣ୍ଣୟ କର।
Solution:
(i) 1331
ସଂଖ୍ୟାକୁ ବାମରୁ ଆରମ୍ଭ କରି ଦୁଇଭାଗ କଲେ ପ୍ରଥମ ଭାଗରେ ତିନୋଟି ଅଙ୍କ ଓ ଦ୍ଵିତୀୟ ଭାଗରେ ଗୋଟିଏ ଅଙ୍କ ରହିବ।
∴ 1331 ସଂଖ୍ୟାର ଶେଷ ଅଙ୍କ = 1, ତେଣୁ ଏହାର ଘନମୂଳର ଶେଷ ଅଙ୍କ! ହେବ।
ପ୍ରଥମ ତିନୋଟି ଅଙ୍କ 331 ପ୍ରଥମ ଭାଗ ଓ 1 ହେଉଛି ଦ୍ଵିତୀୟ ଭାଗ।
∴ 331 ର ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଟି = 1 ଓ ଦ୍ଵିତୀୟ ଭାଗ ‘1’ ଯାହା 1³ = 1, ଏହାର ଘନମୂଳ 1 ହେବ।
∴ √1331 = 11
∴ 1331ର ଘନମୂଳ = 11

(ii) 4913
ଏଠାରେ ପ୍ରଥମ ଭାଗ 913 ଓ ଦ୍ଵିତୀୟଭାଗ 4।
913 ସଂଖ୍ୟାର ଶେଷ ଅଙ୍କ = 3, ତେଣୁ ଏହାର ଘନମୂଳର ଶେଷ ଅଙ୍କ 7 ହେବ। (7³ = 243)
ଶେଷ ତିନୋଟି ଅଙ୍କ (913)କୁ ଛାଡ଼ି ଦେଲେ ଆଉ ବାକି ରହିଲା 4
1³ = 1 ଏବଂ 2³ = 8
4 ହେଉଛି 1 ଓ 8 ମଧ୍ୟରେ, ତେଣୁ ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ 1 ହେବ ।
∴ 4913ର ଘନମୂଳ = 17

(iii) 12167
12167 ସଂଖ୍ୟାର ଶେଷ ଅଙ୍କ = 7, ତେଣୁ ଏହାର ଘନମୂଳର ଶେଷ ଅଙ୍କ 3 ହେବ । (3³ = 27)
ଶେଷ ତିନୋଟି ଅଙ୍କ (167) କୁ ଛାଡ଼ି ଦେଲେ ଆଉ ବାକି ରହିଲା 12
2³ = 8 ଏବଂ 3³ = 27 ତେଣୁ, 2³ < 12 < 3³
ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ 2 ହେବ।
∴ 12167ର ଘନମୂଳ = 23

(iv) 32768
32678 ସଂଖ୍ୟାର ଶେଷ ଅଙ୍କ = 8, ତେଣୁ ଏହାର ଘନମୂଳର ଶେଷ ଅଙ୍କ 2 ହେବ।
ଶେଷ ତିନୋଟି ଅଙ୍କ (768) କୁ ଛାଡ଼ି ଦେଲେ ଆଉ ବାକି ରହିଲା 32
3³ = 27 ଏବଂ 4³ = 64 ତେଣୁ, 27 < 32 < 64
ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ 3 ହେବ।
∴ 32678ର ଘନମୂଳ = 32

Question 5.
ନିମ୍ନଲିଖୁତ ମଧ୍ୟରୁ କେଉଁଟି ସବୁଠାରୁ ବଡ଼? କାରଣ ଉପସ୍ଥାପନ କର।
(i) 67³ – 66³
(ii) 43³ – 42³
(iii) 67² – 66²
(iv) 43² – 42²
Solution:
(i) 67³ – 66³ = 1 + 67 × 66 × 3
(ii) 43³ – 42³ = 1 + 43 × 42 × 3
(iii) 67² – 66² = 67 + 66 = 133
(iv) 43² – 42² = 43 + 42 = 85
ଆମେ ଜାଣିଲେ ଯେ 67³ – 66³ ସବୁଠାରୁ ବଡ଼।
କାରଣ (n + 1)³ – n³ = 1 + (n + 1) × 3n
(n + 1)² – n² = n + n + 1 = 2n + 1

Class 8 Maths Chapter 1 MCQ Odia Medium

ସମ୍ଭାବ୍ୟ ଚାରୋଟି ଉତ୍ତର ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛି ଲେଖ ।

Question 1.
25 ର ଗୁଣନୀୟକ କେଉଁଟି ?
(a) 15
(b) 10
(c) 5
(d) 20
Answer:
(c) 5

Question 2.
କେଉଁଟି ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ ?
(a) 121
(b) 169
(c) 146
(d) 225
Answer:
(c) 146

Question 3.
ଏକ ସଂଖ୍ୟାର ଶେଷ ଅଙ୍କ ୬ ହେଲେ ଏହାର ବର୍ଗର ଏକକ ଘର ଅଙ୍କଟି କେତେ?
(a) 3
(b) 1
(c) 0
(d) 2
Answer:
(b) 1

Question 4.
କେଉଁଟି ଏକ ଯୁଗ୍ମ ସଂଖାର ବର୍ଗ ?
(a) 441
(b) 529
(c) 484
(d) 225
Answer:
(c) 484

Question 5.
13 ଓ 14 ର ବର୍ଗ ମଧ୍ଯରେ କେତୋଟି ସଂଖ୍ୟା ଅଛି?
(a) 27
(b) 28
(c) 29
(d) 26
Answer:
(d) 26

Question 6.
ଯଦି 45² = 2025 ତେବେ 46² କେତେ?
(a) 2025 + 46
(b) 2025 + 46²
(c) 2025 + 23²
(d) 2025 + 91
Answer:
(d) 2025 + 91

Question 7.
1 + 3 + 5 + 7 + 9 + 11 + 13 ର ମୂଲ୍ୟ କେତେ?
(a) 49
(b) 28
(c) 56
(d) 55
Answer:
(a) 49

Question 8.
198² – 197² ର ମୂଲ୍ୟ କେତେ?
(a) 393
(b) 395
(c) 295
(d) 293
Answer:
(b) 395

Question 9.
196 ର ବର୍ଗମୂଳ କେତେ ?
(a) 16
(b) 18
(c) 14
(d) 24
Answer:
(c) 14

Question 10.
24 ର ଘନ କେତେ?
(a) 17,576
(b) 2,744
(c) 13,824
(d) 13,844
Answer:
(c) 13,824

Question 11.
ନିମ୍ନୋକ୍ତ କେଉଁଟି ଏକ ପୂର୍ଣ ଘନ ସଂଖ୍ୟା?
(a) 1727
(b) 1728
(c) 2164
(d) 3475
Answer:
(b) 1728

Question 12.
ଯୁଗ୍ମ ଗଣନ ସଂଖ୍ୟାମାନଙ୍କର ଘନ କେଉଁ ସଂଖ୍ୟା ହେବ ?
(a) ଅଯୁଗ୍ମ ସଂଖ୍ୟା
(b) ଯୁଗ୍ମ ସଂଖ୍ୟା
(c) ଯୁଗ୍ମ ବା ଅଯୁଗ୍ମ ସଂଖ୍ୟା
(d) କହି ହେବ ନାହିଁ
Answer:
(b) ଯୁଗ୍ମ ସଂଖ୍ୟା

Question 13.
23 ର ଘନର ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଟି କେତେ?
(a) 9
(b) 8
(c) 7
(d) 3
Answer:
(c) 7

Question 14.
21 + 23 + 25 + 27 + 29 ର ମୂଲ୍ୟ କେତେ?
(a) 225
(b) 841
(c) 125
(d) 135
Answer:
(c) 125

Question 15.
23³ – 22³ ର ମୂଲ୍ୟ କେତେ ?
(a) 1519
(b) 1529
(c) 259
(d) 1421
Answer:
(a) 1519

Question 16.
6859 ର ଘନମୂଳ କେତେ?
(a) 29
(b) 17
(c) 19
(d) 13
Answer:
(c) 19

Question 17.
\(\frac{\sqrt[3]{64}+\sqrt[3]{125}}{\sqrt[3]{27}}\) ର ମୂଲ୍ୟ କେତେ?
(a) 3
(b) 7
(c) 8
(d) 9
Answer:
(a) 3

ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

Question 1.
1456 ର ବର୍ଗର ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଟି __________
Answer:
6

Question 2.
26 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟାଟି __________
Answer:
51

Question 3.
24025 ର ବର୍ଗମୂଳ __________ ଅଙ୍କ ବିଶିଷ୍ଟ ।
Answer:
3

Question 4.
125 କୁ __________ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟାଦ୍ଵାରା ଗୁଣନ କଲେ ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ହେବ?
Answer:
5

Question 5.
7³ – 6³ ର ମୂଲ୍ୟ __________
Answer:
127

Question 6.
72 କୁ __________ କ୍ଷୁଦ୍ରତମ ସଂଖ୍ୟା ଦ୍ଵାରା ଭାଗକଲେ ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ହେବ?
Answer:
2

Question 7.
90 ର ଘନରେ __________ ଟି ଶୂନ ରହିବ ।
Answer:
3

Question 8.
25 ର ଗୁଣନୀୟକ __________
Answer:
5

ସଂକ୍ଷେପରେ ଉତ୍ତର ଲେଖ।

Question 1.
1 ଓ 50 ମଧ୍ୟରେ କେତୋଟି ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ଅଛି?
Answer:
1 ରୁ 50 ମଧ୍ୟରେ 6 ଟି ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ଅଛି।
ସେଗୁଡ଼ିକ ହେଲା – 4, 9, 16, 25, 36 ଓ 49 ।

Question 2.
38 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା କେତେ ?
Answer:
n ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ସାଧାରଣ ରୂପ = 2n – 1
∴ 38 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟାଟି = 2 × 38 – 1
= 76 – 1
= 75

Question 3.
6³ କୁ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ରୂପେ ପ୍ରକାଶ କର।
Solution:
ଦତ୍ତ ଅଛି 6³ = 6 × 6 × 6 = 216
∴ n = 6 ଓ n – 1 = 6 – 1 = 5
∴ (6 × 5) + 1 = 30 + 1 = 31
∴ 31 ରୁ ଆରମ୍ଭ କରି ଅଯୁଗ୍ମ ସଂଖ୍ୟାଗୁଡ଼ିକ ହେଲେ
∴ 31 + 33 + 35 + 37 + 39 + 41 = 216

Question 4.
16 ର ଘନ କେତେ?
Solution:
16 ର ଘନ = 16 × 16 × 16 = 4096

Question 5.
17³ – 16³ ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
Solution:
17³ – 16³
ଆମେ ଜାଣୁ b³ – a³ = 1 + 3ab, b > 0
ଯେତେବେଳେ a ଓ b ଦୁଇଟି ନିକଟତମ ସଂଖ୍ୟା
a = 16 ଓ b = 17 ର ପ୍ରୟୋଗ କଲେ
17³ – 16³ = 1 + 3 × 16 × 17 = 817

Question 6.
ଦର୍ଶାଅ ଯେ 29 ଏକ ପୂର୍ଣ୍ଣବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ ।
Solution:
ଆମେ ଜାଣୁ 29 = (1 + 3 + 5 + 7 + 9) + 4
ଏଠାରେ 29 କ୍ରମିକ ଅଯୁଗ୍ମ ସଂଖ୍ୟାମାନଙ୍କର ଯୋଗଫଳ ହେବ ନାହିଁ । ତେଣୁ 29 ଏକ ପୂର୍ବବର୍ଗ ସଂଖ୍ୟା ନୁହେଁ ।

Question 7.
9 + 11 + 13 + 15 + 17 + 19 ର ଯୋଗଫଳ ନିର୍ଣ୍ଣୟ କର।
Solution:
ଆମେ ଜାଣୁ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10²
କିନ୍ତୁ 1 + 3 + 5 + 7 = 4²
∴ 9 + 11 + 13 + 15 + 17 + 19 = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) – (1 + 3 + 5 + 7)
= 10² – 4²
= 100 – 16
= 84

Question 8.
ଯଦି 45² = 2025 ତେବେ 462 ନିର୍ଣ୍ଣୟ କର ।
Solution:
ଆମେ ଜାଣୁ 2025 ସଂଖ୍ୟାଟି ପ୍ରଥମ 45 ଟି ଅଯୁଗ୍ମ ସଂଖ୍ୟାର ଯୋଗଫଳ
∴ n ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2n – 1
46 ତମ ଅଯୁଗ୍ମ ସଂଖ୍ୟା = 2 × 46 – 1 = 92 – 1 = 91
45² ରେ 91 ଯୋଗକଲେ ଏହା 46² ହେବ
∴ 46² = 2025 + 91 = 2116

MCQ Questions for Class 6 Maths with Answers

Class 6 Maths MCQ Chapter Wise

Ganita Prakash Class 6 MCQ Questions

• Chapter 1 Patterns in Mathematics Class 6 MCQ
• Chapter 2 Lines and Angles Class 6 MCQ
• Chapter 3 Number Play Class 6 MCQ
• Chapter 4 Data Handling and Presentation Class 6 MCQ
• Chapter 5 Prime Time Class 6 MCQ
• Chapter 6 Perimeter and Area Class 6 MCQ
• Chapter 7 Fractions Class 6 MCQ
• Chapter 8 Playing with Constructions Class 6 MCQ
• Chapter 9 Symmetry Class 6 MCQ
• Chapter 10 The Other Side of Zero Class 6 MCQ

Also Read Ganita Prakash Class 6 Solutions

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 1 Geometric Twins MCQ improves accuracy in objective exams.

MCQ on Geometric Twins Class 7

Geometric Twins MCQ Class 7

Class 7 Maths Geometric Twins MCQ

Question 1.
In the given figures, select the option in which figures I₁ and I₂ do not appear congruent.

Solution:
(b)
The figures given in options (a), (c) and (d) have same shape and size. So, the figures are congruent.
In figure given in option (b), the shape is same, but sizes are different. So, they are not congruent.

Question 2.
If two squares are congruent, then which of the following is true?
(a) They have same length of sides.
(b) They have same length of diagonals.
(c) both (a) and (b)
(d) None of these
Solution:
(c) both (a) and (b)
If two squares are congruent, they have equal corresponding sides. Since the diagonals depend on the side length, their diagonals are also equal.

Question 3.
When checking congruence, which of the following movement is allowed?
(a) Moving (Sliding)
(b) Rotating
(c) Flipping (reflection)
(d) All of the above
Solution:
(d) All of the above
Moving, rotating or flipping the figure changes only its position or orientation. It does not change the shape or size of the figure. So, all the three movements are allowed.

Question 4.
For two line segments to be congruent, which of the following is correct?
(a) They have the same orientation (direction).
(b) They have at least one common point.
(c) They have the same length.
(d) They have the different orientation (direction).
Solution:
(c) They have the same length.
For two line segments to be congruent, only their lengths need to be equal.

Question 5.
∆ABC and ∆XYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 5 cm and ∠A = ∠X = 30°. ∆ABC and ∆XYZ are congruent by:
(a) SSS congruence rule
(b) SAS congruence rule
(c) ASA congruence rule
(d) The given information is not sufficient.
Solution:
(b) SAS congruence rule

In ∆ABC and ∆XYZ, we have
AB = XY = 6 cm [Given]
AC = XZ = 5 cm [Given]
∠A = ∠X = 30° [Given]
∴ ∆ABC ≅ ∆XYZ [By SAS congruence rule]

Question 6.
If ∆ABC ≅ ∆FDE such that AB = 5 cm, ∠B = 40° and ∠A = 80°, then which of the following is true?
(a) DF = 5 cm, ∠F = 80°
(b) DF = 5 cm, ∠E = 80°
(c) DE = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°
Solution:
(a) DF = 5 cm, ∠F = 80°
Given, ∆ABC ≅ ∆FDE,
AB = 5 cm, ∠B = 40°, ∠d = 80°

The corresponding sides are: AB and FD, BC and DE, AC and FE.
The corresponding angles are: ∠A and ∠F, ∠B and ∠D, ∠C and ∠E.
∴ ∠F = ∠A = 80° and DF = BA = 5 cm

Question 7.
In ∆PQR and ∆XYZ, PQ = XY, ∠P = ∠X and PR = XZ. ∆PQR and ∆XYZ are congruent by:
(a) SSS
(b) SAS
(c) ASA
(d) RHS
Solution:
(b) SAS
In ∆PQR and ∆XYZ, two sides and included angle are the same.

∴ ∆PQR ≅ ∆XYZ [By SAS congruence rule]

Question 8.
In the given figure, if AO = OD and BO = OC then ∆AOB ≅ ∆DOC by:

(a) SSS
(b) SAS
(c) ASA
(d) AAS
Solution:
(b) SAS
In ∆AOB and ∆DOC, we have
AO = DO [Given]
BO = CO [Given]
∠AOB = ∠DOC [Vertically opposite angles]
∴ ∆AOB ≅ ∆DOC [By SAS congruence rule]

Question 9.
Out of four given circles, identify the circles that are congruent?
(i) Circle with radius, r = 10 cm
(ii) Circle with area, A = 100π cm²
(iii) Circle with circumference, S = 30π cm
(iv) Circle with area, A = 25π cm²
Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (ii)
(d) (iii) and (iv)
Solution:
(c) (i) and (ii)
For two circles to he congruent, their radii must be the saune, and hence Perimeter (2πR) and area (πR²) will also be the same.
For (i): r = 10 cm
For (ii): A = 100π cm²
⇒ πr² = 100π cm² ⇒ r² = 100 cm²
⇒ r = 10 cm
For (iii): S = 30π cm
⇒ 2πr = 30π cm ⇒ r = \(\frac{30 \pi}{2 \pi} \mathrm{~cm}\) = 150 cm
For (iv): A = 25π cm²
⇒ πr² = 25π cm² ⇒ r² = 25 cm²
⇒ r = 5 cm
As circles given (i) and (ii) have same radii, they are congruent.

Question 10.
In the given figure, BD = CE and BD and CE are altitudes of ∆ABC.
Which of the following are correct?

(i) ∆BCD £ ∆CBE
(ii) ∠DBE = ∠DCE
(iii)) ∠DCB = ∠ECB
(iv) ∠DPE + ∠DAE = 180°
Choose the correct option from the following:
(a) (i) and (ii) only
(b) (ii), (iii) and (iv)
(c) (i), (ii) and (iv)
(d) (i) and (iv) only
Solution:
(c) (i), (ii) and (iv)
In ∆BDC and ∆CEB, we have

BC = CB [Common side (Hypotenuse)]
BD = CE [Given]
∠CDB = ∠BEC [Each 90°]
∴ ∆BDC ≅ ∆CEB [By RHS congruence rule]
⇒ ∆BCD ≅ ∆CBE
⇒ ∠DCB = ∠EBC and ∠DBC = ∠ECB
[By CPCT]
⇒ ∠DCB – ∠ECB = ∠EBC – ∠DBC
⇒ ∠DCE = ∠EBD
In quadrilateral AEPD,
∠PEA + ∠PDA + ∠DPE + ∠DAE = 360°
[Sum of all interior angles is 360°]
⇒ 90° + 90° + ∠DPE + ∠DAE = 360°
⇒ ∠DPE + ∠DAE = 360° – 180° = 180°
So, (i), (ii) and (iv) are correct.

Geometric Twins 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): If ∆ABC ≅ ∆RPQ, then BC = QR.
(R): Corresponding parts of two congruent triangles are equal.
Solution:
(d) A is false but R is true.
We have, ∆ABC ≅ ∆RPQ
Here, the corresponding vertices are: A and R, B and P, C and Question
⇒ BC – PQ [∵ Corresponding parts of two congruent triangles are equal.]
∴ Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): In ∆PQR, if PQ = PR, then ∠P = ∠R.
(R): Angles opposite to equal sides of a triangle are equal.
Solution:

(d) A is false but R is true.
In ∆PQR, we have
PQ = PR
⇒ ∠PRQ = ∠PQR
⇒ ∠R = ∠Q [Angles opposite to equal sides are equal.]
∴ Assertion (A) is false, but Reason (R) is true.

Question 3.
(A): Two congruent triangles have all corresponding angles equal.
(R): AAA criterion is sufficient to prove congruency of two triangles.
Solution:
(c) A is true but R is false.
Two triangles are said to be congruent, if they can be superimposed exactly, one over the other.
∴ Two congruent triangles have all the corresponding sides and corresponding angles equal.
But AAA criterion is not enough to prove congruency of two triangles, because size of triangles might be different.
∴ Assertion (A) is true, but Reason (R) is false.

Question 4.
(A): In ∆ABC, if AB = AC and ∠B = 50°, then ∠C = 50°.
(R): In a triangle, angles opposite to equal sides are equal.
Solution:

(a) Both A and R are true and R is the correct explanation of A.
Given, AB = AC and ∠B = 50°
We know that angles opposite to equal sides of a triangle are equal.
∴ ∠C = ∠B
⇒ ∠C = 50°
∴ Both .Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Geometric Twins Class 7 Fill in the Blanks

Question 1.
A rectangle of length 10 cm and breadth 5 cm is congruent to another rectangle with length _______ and breadth ___________ .
Solution: 10 cm, 5 cm.
For two rectangles to be congruent, their length and breadth must be same.
So, a rectangle of length 10 cm and breadth 5 cm is congruent to another rectangle with length 10 cm and breadth 5 cm.

Question 2.
If two sides and a non-including angle are given, the triangle may or may not uniquely determined. This ambiguous situation is called _______ condition, which can produce ________ different triangles.
Solution: SSA, two
If two sides and a non-including angle are given, the triangle may or may not uniquely determined. This ambiguous situation is called SSA condition, which can produce two different triangles.

Question 3.
For congruent triangles, the perimeters of both triangles are __________ .
Solution: equal
If two triangles are congruent, their corresponding sides and angles are equal.
For congruent triangles, the perimeters of both triangles are equal.

Question 4.
If ∆ABC ≅ ∆XYZ, then AB = ___________, ∠B = ______ and BC = __________ .
Solution: XY, ∠Y, YZ
If two triangles are congruent, their corresponding sides and angles are equal.
Given, ∆ABC ≅ ∆XYZ, then AB = XY, ∠B = ∠Yand BC = YZ.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 1 Large Numbers Around Us MCQ improves accuracy in objective exams.

MCQ on Large Numbers Around Us Class 7

Large Numbers Around Us MCQ Class 7

Class 7 Maths Large Numbers Around Us MCQ

Question 1.
Which of the following numbers has 7 at the lakhs place?
(a) 7,50,812
(b) 70,15,902
(c) 5,74,210
(d) 52,08,701
Solution:
(a) 7,50,812
In 7,50,812, the digit 7 is at lakhs place.
In 70,15,902, the digit 7 is at ten lakhs place.
In 5,74,210, the digit 7 is at ten thousands place.
In 52,08,701, the digit 7 is at hundreds place.

Question 2.
(20 × 10,000) + (53 × 1,000) + (1 × 100) + (2 × 1) represents the number:
(a) 53,20,102
(b) 20,53,102
(c) 2,53,102
(d) 5,20,301
Solution:
(c) 2,53,102
(20 × 10,000) + (53 × 1,000) + (1 × 100) + (2 × 1) = 2,00,000 + 53,000 + 100 + 2
= 2,53,102

Question 3.
The correct way of writing ‘sixty five lakh nine thousand four hundred twenty two’ in numbers is:
(a) 65,90,422
(b) 65,09,422
(c) 6,59,422
(d) 65,40,922
Solution:
(b) 65,09,422
Sixty five lakh nine thousand four hundred twenty two = 65,09,422

Question 4.
10 million is equal to:
(a) 100 crore
(b) 1 lakh
(c) 100 lakh
(d) 10,000 lakh
Solution:
(c) 100 lakh
We know,
1 million = 10 lakh
Thus, 10 million =10 × 10 lakh = 100 lakh

Question 5.
The estimated difference of 1,76,346 and 3,865 rounded off to the nearest thousand is:
(a) 1,70,000
(b) 1,72,000
(c) 1,90,000
(d) 1,54,000
Solution:
(b) 1,72,000
1,76,346 is rounded off to nearest thousands as
1,76,000. [∵ At hundreds place, 3 < 5] 3,865 is rounded off to nearest thousands as 4.000. [∵ At hundreds place, 8 > 5]
Thus, estimated difference = 1,76,000 – 4,000
= 1,72,000

Question 6.
The product 974 × 95 will likely have:
(a) 3 digits
(b) 4 digits
(c) 6 digits
(d) 5 digits
Solution:
(d) 5 digits
Rounding 974 to nearest hundreds we get
1,0 and rounding 95 to nearest tens we get 100.
Thus, estimated product = 1,000 × 100
= 1,00,000.
So, the product will likely have five digits and it will be less than 1,00,000 as the numbers are rounded up and 1,00,000 is the smallest 6-digit number.
Verification: 974 × 95 = 92,530, which has five digits.

Question 7.
The estimated sum of 5,47,86,326 and 45,56,767 rounded off to the nearest lakhs is:
(a) 5,94,00,000
(b) 59,40,000
(c) 54,90,000
(d) 9,47,000
Solution:
(a) 5,94,00,000
5,47,86,326 is rounded off to nearest lakhs as
5,48,0,000. [∵ At ten thousands place, 8 > 5]
45,56,767 is rounded off to nearest lakhs as
46,0,000. [∵ At ten thousands place, 5 = 5]
Thus, estimated sum
= 5,48,00,000 + 46,00,000 = 5,94,00,000

Question 8.
The product 105 × 93 will likely have:
(a) 3 digits
(b) 4 digits
(c) 6 digits
(d) 5 digits
Solution:
(b) 4 digits
Rounding off 105 to nearest hundreds, we get 100 and rounding off 93 to nearest tens we get 90,
Thus, estimated product = 100 × 90 = 9,000
So, the product will likely have four digits and it will be close to 9,000.
Verification: 105 × 93 = 9,765, which has four digits.

Question 9.
1 billion is equal to:
(i) 100 crore
(ii) 100 million
(iii) 100 lakh
(iv) 10,000 lakh
Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Solution:
(c) (i) and (iv)
We know, 1 million = 10 lakh
And, 1 billion = 1,000 million
Thus, 1 billion = 1,000 × 10 lakh
= 10,000 lakh =100 crore [∵ 1 crore = 100 lakh]

Question 10.
In which of the following numbers the comma is placed incorrectly according to the Indian Number System?
(i) 23,56,76,145
(ii) 123,98,34,67
(iii) 780,444,112
(iv) 56,12,34,889
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Solution:
(d) (ii) and (iii)
We know that while writing a number in Indian Number System, the commas are placed in 3-2-2-2… pattern from right to left.

In 23,56,76,145 and 56,12,34,889, the commas are placed in 3-2-2-2… pattern from right to left. They are written correctly according to Indian Number System.

However, in 123,98,34,67 the commas are not placed in 3-2-2-2… pattern from right to left.
Also, in 780,444,112 the commas are placed according to the International Number System,
i. e. in 3-3-3… pattern from right to left.

Thus, the commas are not placed correctly according to the Indian Number System in 123,98,34,67 and 780,444,112.

Large Numbers Around Us 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): 1,000 notes of ₹100 are needed to make ₹ 1,00,000.
(R): Place value helps us break large numbers into smaller, meaningful parts.
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
As ₹1 ,00,000 ÷ 100 = 1,000,
1,000 notes of ₹100 are needed to make ₹1,00,000.
Thus, Assertion (A) is true.
And we know that place value helps us to express larger numbers into smaller, meaningful parts.
Thus, Reason (R) is true, but Reason (R) is not the correct explanation of Assertion (A).

Question 2.
(A): The digit 6 in the number 9,645,123 has a place value of 600,000 in the International Number System.
(R): The place value of a digit is the product of the digit and its position value.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that the place value of a digit in the number system is found by multiplying the
digit with the value of the position it occupies.

Thus, in 9,645, 123, the place value of 6 is 6 × 100,000 = 600,000.
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is tile correct explanation of Assertion (A).

Question 3.
(A): When rounding 3,42,765 to the nearest lakhs, we get 4,00,000.
(R): Rounding to the nearest lakhs depends on the digit in the ten thousands place.
Solution:
(d) A is false but R is true.
When rounding 3,42,765 to nearest lakhs, we check the digit on the just right of lakhs place, i.e. the digit at the ten thousands place.
In 3,42,765, 4 is at ten thousands place and 4 < 5. So, the digit at the lakhs place, i.e. 3 remains unchanged.
Hence, if 3,42,765 is rounded off to nearest lakhs, we get 3,00,000.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 4.
(A): If 98 × 102 is calculated, then the product will be very close to 10,000.
(R): Both numbers 98 and 102 are close to 100.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
98 is rounded off to nearest tens as 100 and 102 is rounded off to nearest hundreds as 100. Thus, the product is estimated as 100 × 100 = 10,000.
This means that the actual product will be close to 10,000.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Large Numbers Around Us Class 7 Fill in the Blanks

Question 1.
In one lakh, 1 is follwed by _____________________ zeros.
Solution: five
We know, 1 lakh = 1,00,000, where the digit 1 is followed by five zeros.
∴ In one lakh, 1 is followed by five zeros.

Question 2.
In the Indian Number Svswm, the correct way to write the number 9625084 is __________.
Solution: 96,25,084
In the Indian Number System, commas are placed in a 3-2-2-2… pattern from the right to left. This helps to separate the digits into hundreds, thousands, lakhs, crores, and so on.
Therefore, in the Indian Number System, the correct way to write the number 9625084 is 96,25,084.

Question 3.
The smallest 7—digit number is _______ lakh.
Solution: ten
The smallest 7-digit number is 10,00,000 i.e. ten lakh.

Question 4.
The largest 8—digit number is _________
Solution: 9,99,99,999
The largest 8-digit number is 9,99,99,999.

Question 5.
The place value of 5 in 7,005,380 is _________.
Solution: 5000
Given number is 7,005,380, which is represented in International Number System.

Periods	Millions			Thousands			Ones
Place Name	HM	TM	M	HTh	TTh	Th	H	T	O
7,005,380			7	0	0	5	3	8	0

Thus, the place value of 5 in 7,005,380 is 5 thousand = 5 × 1000 = 5000.

Question 6.
The expanded form of the number 76,70,905 is _________.
Solution: (7 × 10,00,000) + (6 × 1,00,000) + (7 × 10,000) + (9 × 100) + (5 × 1)
The expanded form of the number 76,70,905 is
(7 × 10,00,000) + (6 × 1,00,000) + (7 × 10,000) + (9 × 100) + (5 × 1).

Question 7.
In a crore, 1 is followed by ________ zeros.
Solution: seven
We know that 1 crore = 1,00,00,000, where 1 is followed by seven zeros.
∴ In a crore, 1 is followed by seven zeros.

Question 8.
If 728 is rounded off to the nearest hundreds, we get ________.
Solution: 700
In 728, the tens digit is 2 and 2 < 5. So, 7 at the hundreds place remains unchanged while rounded off to the nearest hundreds.
Thus, if 728 is rounded off to the nearest hundreds, we get 700.

MCQ Questions for Class 7 Maths with Answers

Class 7 Maths MCQ Chapter Wise

Class 7 Maths MCQ Questions Part 1

• Chapter 1 Large Numbers Around Us Class 7 MCQ
• Chapter 2 Arithmetic Expressions Class 7 MCQ
• Chapter 3 A Peek Beyond the Point Class 7 MCQ
• Chapter 4 Expressions using Letter Numbers Class 7 MCQ
• Chapter 5 Parallel and Intersecting Lines Class 7 MCQ
• Chapter 6 Number Play Class 7 MCQ
• Chapter 7 A Tale of Three Intersecting Lines Class 7 MCQ
• Chapter 8 Working with Fractions Class 7 MCQ

Ganita Prakash Class 7 Part 2 MCQ

• Chapter 1 Geometric Twins Class 7 MCQ
• Chapter 2 Operations with Integers Class 7 MCQ
• Chapter 3 Finding Common Ground Class 7 MCQ
• Chapter 4 Another Peek Beyond the Point Class 7 MCQ
• Chapter 5 Connecting the Dots Class 7 MCQ
• Chapter 6 Constructions and Tilings Class 7 MCQ
• Chapter 7 Finding the Unknown Class 7 MCQ

Also Read Ganita Prakash Class 7 Solutions

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 1 Large Numbers Around Us Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 1 Large Numbers Around Us Solutions

Ganita Prakash Class 7 Chapter 1 Solutions

Class 7 Maths Ganita Prakash Chapter 1 Solutions Large Numbers Around Us

Question 1.
Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems:
(i) 4050678
(ii) 48121620
(iii) 20022002
(iv) 246813579
Solution:
(i) 4050678
Indian System: 40,50,678
Number name: Forty lakh fifty thousand six hundred seventy eight American system: 4,050,678
Number name: Four million fifty thousand six hundred seventy eight

(ii) 48121620
Indian System: 4,81,21,620
Number name: Four crore eighty one lakh twenty one thousand six hundred twenty American System: 48,121,620
Number name: Forty eight million one hundred twenty one thousand six hundred twenty

(iii) 20022002
Indian System: 2,00,22,002
Number name: Two crore twenty two thousand two American System: 20,022,002
Number name: Twenty million twenty two thousand two

(iv) 246813579
Indian System: 24,68,13,579
Number name: Twenty four crore sixty eight lakh thirteen thousand five hundred seventy nine American System: 246,813,579
Number name: Two hundred forty six million eight hundred thirteen thousand five hundred seventy nine

Question 2.
Write the following numbers in Indian place value notation:
(i) One crore one lakh one thousand ten
(ii) One billion one million one thousand one
(iii) Ten crore twenty lakh thirty thousand forty
(iv) Nine billion eighty million seven hundred thousand six hundred
Solution:
(i) 1,01,01,010
(ii) 1,001,001,001
In Indian place value notation: 1,00,10,01,001
(iii) 10,20,30,040
(iv) 9,080,700,600
In Indian place value notation: 9,08,07,00,600

Question 3.
Compare and write ‘<‘ ‘>’ or ‘ = ’:
(i) 30 thousand _________ 3 lakh
(ii) 500 lakh _______ 5 million
(iii) 800 thousand ______ 8 million
(iv) 640 crore _______ 60 billion
Solution:
(i) As 30,000 < 3,00,000
⇒ 30 thousand < 3 lakh (ii) Since 1 million = 10 lakh, 5 million = 50 lakh Clearly, 500 lakh > 50 lakh ⇒ 500 lakh > 5 million

(iii) 800 thousand = 800 × 1000 = 800,000
8 million = 8,000,000
As 800,000 < 8,000,000
⇒ 800 thousand < 8 million

(iv) Since 1 billion = 100 crore, 60 billion = 60 × 100 crores = 6,000 crores
Clearly, 640 crore < 6,000 crore
⇒ 640 crore < 60 billion

Question 4.
Find quick ways to calculate these products:
(i) 2 × 1768 × 50
(ii) 72 × 125
(iii) 125 × 40 × 8 × 25
Solution:
(i) 2 × 1768 × 50 = 2 × 1768 × \(\frac{100}{2}\) = 1768 × 100 = 1,76,800
(ii) 72 × 125 = 72 × \(\frac{1000}{8}\) = 9 × 1000 = 9,000
(iii) 125 × 40 × 8 × 25 = \(\frac{1000}{8}\) × 40 × 8 × \(\frac{100}{4}\) = 1000 × 5 × 2 × 100 = 10,00,000

Question 5.
Using all digits from 0-9 exactly once (the first cannot be 0) to create a 10-digit number, write the —
(i) Largest multiple of 5
(ii) Smallest even number
Solution:
(i) Arranging digits in descending order, we get 9, 8, 7, 6, 5, 4, 3, 2, 1, 0.
All multiple of 5 can end only in 5 or 0.
Hence, the largest multiple of 5 is 9876543210.

(ii) Arranging digits in ascending order, we get 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Since the number cannot start with 0, the smallest number = 1023456789
We know that an even number has either 2, 4, 6 or 8 at its ones place.
Thus, swapping the last two digits, we get the smallest even number = 1023456798

Question 6.
The number 10,30,285 in words is “Ten lakh thirty thousand two hundred eighty five”, which has 41 letters. Give a 7-digit number which has the maximum number of letters.
Solution:
We use 7 (seven) and 8 (eight) to make such a number since both numbers contain five letters when written in words.

One such 7-digit number is 77,77,777 (Seventy seven lakh seventy seven thousand seven hundred seventy seven).

This has 60 letters, making it one of the 7-digit numbers having maximum number of letters.

Question 7.
Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Solution:
To ensure that on exchanging any two digits increases the value of a 9-digit number, the digits must increase from left to right. So, the arrangement would be: 123456789.
There is only 1 number that satisfies the given condition.

Question 8.
Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Solution:
Given, 12345123451234512345
We have 20 digits and need to remove 10 digits, leaving us with 10 digits.
To maximise the resulting number we want the leftmost digit to be as large as possible. Looking at the original number 12345123451234512345, the first few digits are small. We can strike out the initial T234’ to get the larger digit in the first position, i.e. 5.

Now, our numbers starts with 5 as we have struck out 4 digits so far and we need to strike out 6 more.

The remaining number is 5123451234512345. We want the next digit to be as large as possible. So, we strike out ‘1234’ following the 5, leaving us with 551234512345.
We have now struck out 4 + 4 = 8 digits.
We need to strike out 2 metre digits from 551234512345.
To keep the number as large as possible we should strike out ‘ I ’ and ‘2’.

Therefore, by striking out the digits 1234, then 1234, then 1 and 2, the number is 5534512345, which is the largest possible number.

Question 9.
The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’. The words ‘one’ and ‘two’ share a letter ‘o’, and the words ‘two’ and ‘three’ also share a letter ‘t’. How far do you have to count to find two consecutive numbers which do not share an English letter in common?
Solution:
The problem involves finding two consecutive numbers whose English names share no common letters.
Here, words zero and one share e and o. Words one (1) and two (2) share o.
Words two (2) and three (3) share t. Words three (3) and four (4) share r.
Words four (4) and five (5) share/. Words five (5) and six (6) share i.
Words six (6) and seven (7) share s. Words seven (7) and eight (8) share e.
Words eight (8) and nine (9) share i and e.
Words nine (9) and ten (10) share n and e.
.
.
.
.
Words nineteen (19) and twenty (20) share t, e, n and so on.
It shows that all consecutive numbers have atleast one common letter. Hence, their is no such pair of consecutive numbers that do not share an English letter in common.

Question 10.
A calculator has only ‘+ 10,000’ and ‘+ 100’ buttons. Write an expression describing the number of button clicks to be made for the following numbers:
(i) 20,800
(ii) 92,100
(iii) 1,20,500
(iv) 65,30,000
(v) 70,25,700
Solution:
(i) 20,800 = (2 × 10,000) + (8 × 100)
Number of clicks = 2 + 8 = 10 clicks

(ii) 92,100 = (9 × 10,000) + (21 × 100)
Number of clicks = 9 + 21 =30 clicks

(iii) 1,20,500 = (12 × 10,000) + (5 × 100)
Number of clicks = 12 + 5 = 17 clicks

(iv) 65,30,000 = (653 × 10,000) + (0 × 100)
Number of clicks = 653 + 0 = 653 clicks

(v) 70,25,700 = (702 × 10,000) + (57 × 100)
Number of clicks = 702 + 57 = 759 clicks

Question 11.
You are given two sets of number cards numbered from 1-9. Place a number card in each box below to get the (i) largest possible sum (ii) smallest possible difference of the two resulting numbers.

Solution:
(i) Since each card numbered 1 – 9 is to be placed in the boxes such that no box is empty, the cards cannot be repeated.
To get the largest possible sum, both the 7-digit and 5-digit numbers need to be the largest.
Largest 7-digit number = 98,76,543; Largest 5-digit number = 98,765
Largest possible sum = 98,76,543 + 98,765 = 99,75,308

(ii) To get the smallest possible difference, the 7-digit number needs to be the smallest, and the 5-digit number needs to be the largest.
Smallest 7-digit number = 12,34,567
Largest 5-digit number = 98,765
Smallest possible difference = 12,34,567 – 98,765 = 11,35,802

Question 12.
A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour.
Solution:
Given, total distance = 13,560 km and duration = 11 days
Distance covered everyday = 13,560 ÷ 11 = 1,232.72 km
Thus, the godwit covers approximately 1,233 km per day.
We know that one day has 24 hours.
Thus, distance covered every hour = 1,233 ÷ 24 = 51.375 km
Hence, the godwit covers approximately 51 km per

InText Questions

Question 1.
Observe the pattern and fill in the boxes given below.

Solution:

Question 2.
What if a person ate 3 varieties of rice every day, will he be able to taste all the lakh varieties in a 100 year lifetime? Find out.
Solution:
With 3 varieties of rice every day, he can taste 365 × 3 = 1095 varieties in a year.
To taste 1 lakh varieties, he would need 1,0,000 ÷ 1095 ≈ 91 years.
Hence, he would be able to eat all 1 lakh varieties of rice in 100 years.

Question 3.
Two of the many different ways to get 5072 are shown below:

These two ways can be expressed as:
(a) (50 × 100) + (7 × 10) + (2 × 1) = 5072
(b) (3 × 1000) + (20 × 100) + (72 × 1) = 5072
Find a different way to get 5072 and write an expression for the same.
Solution:

Buttons	5072
+ 10,00,000
+ 1,00,000
+ 10,000
+ 1,000	5
+ 100	0
+ 10	7
+ 1	2

Expression: (5 × 1000) + (0 × 100) + (7 × 10) + (2 × 1) = 5000 + 70 + 2 = 50724.

Question 4.
The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000?
Solution:
Given, the estimated population of Chintamani in the year 2024 is 1,06,000.
∴ Required difference = 1,06,000 – 1,00,000
= 6,000
Thus, the population of Chintamani in 2024 is 6,000 more than one lakh.

Question 5.
The Thoughtful Thousands only has a + 1000 button. How many times should it be pressed to show:

(i) Three thousand? 3 times
(ii) 10,000? ______
(iii) Fifty-three thousand? _________
(iv) 90,000? __________
(v) One Lakh? ________
(vi) ________? 153 times
(vii) How many thousands are required to make one lakh?
Solution:
(i) \(\frac{3000}{1000}\) = 3 ⇒ 3 times
(ii) \(\frac{10000}{1000}\) = 10 ⇒ 10 times
(iii) \(\frac{53000}{1000}\) = 53 ⇒ 53 times
(iv) \(\frac{90000}{1000}\) = 90 ⇒ 90 times
(v) \(\frac{100000}{1000}\) = 100 ⇒ 100 times
(vi) 153 × 1000 = 1,53,000
(vii) 100000 ÷ 1000 = 100. Thus, 100 thousands make 1 lakh.

Question 6.
How many zeros does a hundred thousand have?
Solution:
100 thousand = 100 × 1000 = 1,00,000
Clearly it has 5 zeros.

Question 7.
With large numbers it is useful to know the nearest thousand, lakh or crore. For example, the nearest neighbours of the number 6,72,85,183 are shown in the table below.

Nearest thousand	6,72,85,000
Nearest ten thousand	6,72,90,000
Nearest lakh	6,73,00,000
Nearest ten lakh	6,70,00,000
Nearest crore	7,00,00,000

Write the five nearest neighbours for these numbers:
(i) 3,87,69,957
(ii) 29,05,32,481
Solution:

Nearest Neighbours	For 3,87,69,957	For 29,05,32,481
Nearest thousands	3,87,70,000	29,05,32,000
Nearest ten thousands	3,87,70,000	29,05,30,000
Nearest lakhs	3,88,00,000	29,05,00,000
Nearest ten lakhs	3,90,00,000	29,10,00,000
Nearest crores	4,00,00,000	29,00,00,000

Question 8.
Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10?
Solution:
We know that 5 × 2= 10 (multiplication fact) gives
two division facts: 10 ÷ 2 = 5 and 10 ÷ 5 = 2.
So, we can use \(\frac{10}{2}\) in place of 5 . Either we multiply a number by 5 or by \(\frac{10}{2}\), we will get the same answer.

Question 9.
Can multiplying a 3-digit number with another 3-digit number give a 4-digit number?
Solution:
The product of smallest 3-digit numbers
= 100 × 100 = 10,000 (5-digit number)
And, the product of largest 3-digit numbers
= 999 × 999 = 9,98,001 (6-digit number)
So, the product of two 3-digit numbers will have either 5 or 6 digits.
Hence, a 4-digit number cannot be obtained by multiplying two 3-digit numbers.

Question 10.
Can multiplying a 4-digit number with a 2-digit number give a 5-digit number?
Solution:
The product of smallest 4-digit number and smallest 2-digit number
= 1000 × 10 = 10,000 (5-digit number)
And, the product of largest 4-digit number and largest 2-digit number
= 9999 × 99 = 9,89,901 (6-digit number)
So, the product of 4-digit number and 2-digit number will have either 5 or 6 digits.
Hence, multiplying a 4-digit number with a 2-digit number can give a 5-digit number.

Question 11.
Roxie wondered, “If I could travel 100 kilometers every day, could I reach the Moon in 10 years?” (The distance between the Earth and the Moon is 3,84,400 km.)
(i) How far would she have travelled in a year?
(ii) How far would she have travelled in 10 years?
Solution:
(i) Distance travelled by Roxie in a day = 100 km
Thus, distance travelled by Roxie in a year = 365 × 100 = 36500 km (As 1 year = 365 days)
(ii) Distance travelled by Roxie in 10 years =100 × 365 x 10 = 36500 × 10 = 365000 km
Since 365000 < 384400, Roxie cannot reach the moon in 10 years.

Large Numbers Around Us Class 7 Extra Questions

Large Numbers Around Us Class 7 Very Short Question Answer

Question 1.
How many thousands are there in 1 million?
Solution:
Place value chart in International Number System is given below:

Periods	Millions			Thousands			Ones
Place Name	HM	TM	M	HTh	TTh	Th	H	T	o
1 million			1	0	0	0	0	0	0
1 thousand						1	0	0	0

1 million is three places to the left of 1 thousand.
Thus, 1 million = 1,000 thousand
Hence, 1,000 thousands are there in one million.

Question 2.
How many hundreds are there in 10 lakhs?
Solution:
Place value chart in Indian Number System is given below:

Periods	Crores		Lakhs		Thousands		Ones
Place Name	TC	C	TL	L	TTh	Th	H	T	o
10 lakh			1	0	0	0	0	0	0
1 hundred							1	0	0

10 lakh is four places to the left of 1 hundred.
Thus, 10 lakh = 10,000 hundred
Hence. 10,000 hundreds are there in 10 lakhs.

Question 3.
The annual wheat production in a region is 673400000 kilograms. Place the commas according to the International Number System format.
Solution:
In the International Number System, commas are placed in a 3-3-3-3… pattern, starting from the right. This helps to separate the digits into hundreds, thousands, millions, billions, and so on.
Number: 673,400,000
Number Name: Six hundred seventy three million four hundred thousand

Question 4.
Without multiplying, predict the number of digits in the product 113 × 98.
Solution:
Rounding 113 to nearest tens we get 110 and rounding 98 to nearest tens, we get 100.
Thus, estimated product = 110 × 100 = 11000, which is a 5-digit number.
Hence, the actual product will have 5 digits.
Verification: 113 × 98 = 11,074, which is a 5-digit number

Question 5.
Compare and write ‘>’, ‘<‘ or ‘=’’:
(i) 80 thousand ______ 8 lakh
(ii) 200 lakh ______ 2 million
Solution:
(i) 80 thousand = 80 × 1,000 = 80,000 and 8 lakh = 8 × 1,00,000 = 8,00,000
As 80,000 < 8,00,000
Thus, 80 thousand < 8 lakh

(ii) We know, 1 million = 10 lakh
So, 2 million = 20 lakh < 200 lakh Thus, 200 lakh > 2 million

Question 6.
How many ten thousands are there in the smallest 6-digit number?
Solution:
The smallest 6-digit number is 1,00,000 i.e. 1 lakh.

Place Name	TL	L	TTh	Th	H	T	o
1 Lakh		1	0	0	0	0	0
10 thousand			1	0	0	0	0

1 lakh is one place to the left often thousands.
Thus, 1 lakh =10 ten thousands

Question 7.
How many thousands are there in 1 lakh?
Solution:

Place Name	L	TTh	Th	H	T	b
1 Lakh	1	0	0	0	0	0
1 thousand			1	0	0	0

1 lakh is 2 places to the left of thousand.
Hence, 1 lakh = 100 thousands.

Large Numbers Around Us Class 7 Short Question Answer

Question 1.
Write the number names of the following numerals in the Indian and International Numbers Systems.
(i) 437065
(ii) 42181602
(iii) 636547150
(iv) 4050607080
Solution:

Sr. No.	Indian System	International System
(0	4,37,065: Four lakh thirty seven thousand sixty five	437,065: Four hundred thirty seven thousand sixty five
(ii)	4,21,81,602: Four crore twenty one lakh eighty one thousand six hundred two	42,181,602: Forty two million one hundred eighty one thousand six hundred two
(iii)	63,65,47,150: Sixty three crore sixty five lakh forty seven thousand one hundred fifty	636,547,150: Six hundred thirty six million five hundred forty seven thousand one hundred fifty
(iv)	4,05,06,07,080: Four arab five crore six lakh seven thousand eighty	4,050,607,080: Four billion fifty million six hundred seven thousand eighty

Question 2.
Find the place value of underlined digits in International Number System.
(i) 3241767
(ii) 98443810
Solution:
(i)

Place Name	M	HTh	TTh	Th	H	T	O
3,241,767	3	2	4	I	7	6	7

From the table, the digit 1 is at thousands place.
Thus, the place value of 1 is 1 × 1,000 = 1,000.

(ii)

Place Name	TM	M	HTh	TTh	Th	H	T	O
98,443,810	9	8	4	4	3	8	1	0

From the table, the digit 8 is at thousands place.
Thus, the place value of 8 is 8 × 1,000 = 1,000.
= 8,000,000.

Question 3.
Calculate the following products:
(i) 4 × 1522 × 50
(ii) 48 × 125
(iii) 125 × 20 × 16 × 25
Solution:
(i) 4 × 1522 × 50 = 4 × 1522 × \(\frac{100}{2}\)
= 2 × 1522 × 100 = 3044 × 100 = 304400

(ii) 48 × 125 = 48 × \(\frac{1000}{8}\) = 6 × 1000 = 6000

(iii) 125 × 20 × 16 × 25 = \(\frac{1000}{8}\) × 20 ×16 × \(\frac{100}{4}\)
= 1000 × 5 × 2 × 100 = 1000 × 10 × 100 = 1000000

Question 4.
A city’s population grew from 5,67,000 to 7.42.0 in five years. Estimate the increase using appropriate rounding.
Solution:
Given, in five years the population of a city grew from 5,67,000 to 7,42,000.
Rounding off both the numbers to nearest ten thousands, we get 5,70,000 and 7,40,000 respectively.
Thus, estimated increase in population
= 7,40,000 – 5,70,000 = 1,70,000

Question 5.
A stadium had 4,73,000 visitors in 2022 and 5.85.0 in 2023. Estimate the increase in visitors by rounding to the nearest ten thousands.
Solution:
Number of visitors in 2022 = 4,73,000
Number of visitors in 2023 = 5,85,000
Rounding both the numbers to nearest ten thousands, we get
4.73.0 → 4,70,000 and 5,85,000 → 5,90,000
Thus, estimated increase in visitors
= 5,90,000 – 4,70,000 = 1,20,000.

Large Numbers Around Us Class 7 Long Question Answer

Question 1.
Round off the following to the given nearest place.
(i) 7,065; hundreds
(ii) 55,777; thousands
(iii) 46,439; ten thousands
(iv) 30,89,732; lakhs
(v) 34,75,68,328; ten crores
Solution:
(i) In 7,065, the digit at the hundreds place is 0.
As the digit at the tens place, 6 > 5, we increase 0 by 1 and we replace the remaining digits to the right of the hundreds place by 0.
Thus, on rounding off 7065 to the nearest hundreds, we get 7100.

(ii) In 55,777, the digit at the thousands place is 5.
As the digit at the hundreds place, 7 > 5, we increase 5 by 1 and we replace the remaining digits to the right of the thousands place by 0.
Thus, on rounding off 55,777 to the nearest thousands, we get 56,000.

(iii) In 46,439, the digit at the ten thousands place is 4.
As the digit at the thousands place, 6 > 5, we increase 4 by 1 and we replace the remaining digits to the right of the ten thousands place by 0.
Thus, on rounding off 46,439 to the nearest ten thousands, we get 50,000.

(iv) In 30,89,732, the digit at the lakhs place is 0.
As the digit at the ten thousands place, 8 > 5, we increase 0 by 1 and we replace the remaining digits to the right of the lakhs place by 0.
Thus, on rounding off 30,89,732 to the nearest lakhs, we get 31,00,000.

(v) In 34,75,68,328, the digit at the ten crores place is 3.
As the digit at the crore place, 4 < 5, 3 remains unchanged and we replace the remaining digits to the right of the ten crores place by 0.
Thus, on rounding off 34,75,68,328 to the nearest ten crores, we get 30,00,00,000.

Question 2.
A clothing store earned ₹6,58,900 in January and ₹ 7,21,600 in February. Estimate the total revenue by rounding each figure to the nearest lakhs. Is your estimated total greater or smaller than the exact total?
Solution:
Revenue earned by clothing store in January = ₹6,58,900
Revenue earned by clothing store in February = ₹ 7,21,600
Rounding the numbers to nearest lakhs, we get
6,58,900 → 7,00,000
[∵ At ten thousands place, 5 = 5]
7,21,600 → 7,00,000
[∵ At ten thousands place, 2 < 5]
Now, estimated total revenue over the two months = ₹ 7,00,000 + ₹ 7,00,000 = ₹ 14,00,000
Actual total revenue = ₹6,58,900 + ₹7,21,600
= ₹ 13,80,500
Thus, the estimated total revenue for the given two months is greater than the actual revenue.

Question 3.
Say True or False.
(i) 6,78,456 is rounded off as 6,80,000 to the nearest ten thousands.
(ii) 52,25,067 is rounded off as 52,00,000 to the nearest thousands.
(iii) 2,31,48,765 is rounded off as 2,40,00,000 to the nearest ten lakhs.
Solution:
We know that, while rounding the number nearest to given place,
If the digit on the right of the given place is 5 or greater than 5, we increase the digit at that place by 1.
If the digit on the right of the given place is less than 5, we keep the digit at that place same.

(i) True
In 6,78,456, the digit at the ten thousands place is 7.
The digit to the right of 7, i.e. the digit at the thousands place is 8 and 8 > 5.
Thus, 6,78,456 is rounded off as 6,80,000 to the nearest ten thousands.

(ii) False
In 52,25,067, the digit at the thousands place is 5.
The digit to the right of 5, i.e. the digit at the hundreds place is 0 and 0 < 5.
So, 5 remains unchanged.
Thus, 52,25,067 is rounded off as 52,25,000 to the nearest thousands.

(iii) False
In 2,31,48,765, the digit at the ten lakhs place is 3.
The digit to the right of 3, i.e. the digit at the lakhs place is 1 and 1 < 5. So, 3 remains unchanged.
Thus, 2,31,48,765 is rounded off as 2,30,00,000 to the nearest ten lakhs.

Large Numbers Around Us Class 7 Case Based Questions

Question 1.
The population details of a town were published in a report.

Population of children	12,35,678
Population of adults	34,78,915
Population of senior citizens	5,60,328

The report is also translated for international agencies.
Based on the above information, answer the
following questions:
(i) Create the report for the international agencies using the International Number System.
(ii) What is the face value and place value of digit 3 in the population of children?
(iii) Write the number name for the population of senior citizens in the International Number System.
Solution:
i) The report in International Number System will be:

Population of children	1,235,678
Population of adults	3,478,915
Population of senior citizens	560,328

(ii) Given, the population of children = 12,35,678, which is represented in Indian Number System.

Place Name	12,35,678
C
TL	1
L	2
TTh	3
Th	5
H	6
T	7
O	8

The face value of 3 is 3.
Clearly, digit 3 is at the ten-thousands place.
∴ The place value of 3 is 3 × 10,000 = 30,000.

(iii) In International Number System, population of senior citizens = 560,328
Number name: Five hundred sixty thousand three hundred twenty eight

Question 2.
The principal of Sunrise Public School is preparing a budget for renovating the school. She has received the cost from various departments:
Painting classrooms: ₹4,83,760
Replacing furniture: ₹3,27,450
Electrical work: ₹ 1,65,890
Bathroom renovations: ₹2,49,300
To present a simplified version in a meeting, she rounds off all values to the nearest lakhs.
Based on the above information, answer the following questions:
(i) What is the rounded cost of each item to the nearest lakhs?
(ii) What is the total estimated cost using the rounded values?
(iii) How much is the difference between the estimated and the total actual cost?
Solution:
(i) Rounding off all the costs to the nearest lakhs, we get

Items	Actual Cost	Estimated Cost (to nearest lakh)
Painting classrooms	₹4,83,760	₹5,00,000
Replacing furniture	₹3,27,450	₹ 3,00,000
Electrical work	₹ 1,65,890	₹ 2,00,000
Bathroom renovations	₹2,49,300	₹2,00,000

(ii) The total estimated cost is the sum of the estimated cost of each item.
Thus, total estimated cost = ₹5,00,000 + ₹ 3,00,000 + ₹ 2,00,000 + ₹ 2,00,000
= ₹ 12,00,000

(iii) The total estimated cost = ₹ 12,00,000
Now, total actual cost
= ₹4,83,760 + ₹3,27,450 + ₹ 1,65,890 + ₹2,49,300
= ₹ 12,26,400
Thus, difference between total actual cost and total estimated cost
= ₹ 12,26,400 – 12,00,000 = ₹26,400

Students can refer to BSE Odisha Class 8 Math Solution and Ganita Prakash Chapter 1 A Square and A Cube Class 8 Question Answer to understand textbook questions step by step.

Class 8 Maths Chapter 1 A Square and A Cube Solutions

Ganita Prakash Class 8 Chapter 1 Solutions

Class 8 Maths Ganita Prakash Chapter 1 Solutions A Square and A Cube

Page : 1

Question 1.
Before the process begins, Khoisnam realises that he already knows which lock¬ers will be open at the end. How did he figure out the answer?

Solution:
Khoisan had figured out on the basis of the following:
(i) If a locker is toggled an odd number of times, it will be open.
(ii) If a locker is toggled an even number of times, it will be closed.
(iii) The number of times a locker is toggled is the same as the number of factors of that locker number.
For example for locker #10, person 1 opens it, person 2 closes it, person 5 opens it and person 10 closes it.

The numbers 1, 2, 5, 10 are factors of 10. If the number of factors is even, the locker will be toggled by an even number of people and it will eventually be closed.

In the same manner if we consider locker #4, it will be closed at the end as 4 has 1, 2 and 4 as its factors, which are odd in number.

Page : 2

Question 1.
Does every number have an even number of factors?

We see in some cases, like 2 × 2, that the numbers in the pair are the same.
Solution:
No! Many numbers do not have an even number of factors, for example – 1 has only 1 factor, 4 has 3 factors: 1, 2 and 4, 9 has 3 factors: 1, 3 and 9, 25 has 3 factors: 1, 5 and 25.

Question 2.
Can you use this insight to find more numbers with an odd number of factors?
For instance, 36 has a factor pair 6 × 6 where both numbers are 6. Does this number have an odd number of factors? If every factor of 36 other than 6 has a different factor as its partner, then we can be sure that 36 has an odd number of factors. Check if this is true.

Hence all the following numbers have an odd number of factors –
1 × 1, 2 × 2, 3 × 3, 4 × 4, …

A number that can be expressed as the product of a number with itself is called a square number, or simpiy a square. The only numbers that have an odd number of factors are the squares, because they each have one factor which, when multiplied by itself, equals the number. Therefore, every locker whose number is a square will remain open.
Solution:
Continuing the given insight we can find more numbers with an odd number of factors:


All of the above numbers have odd number of factors.
Note – In the above: 16 : 1 × 16, 2 × 8 and 4 × 4 are called ‘Partner Factors’ and for other numbers as well.

Page : 3

Question 1.
Write the locker numbers that remain open.
Khoisnam immediately collects word clues from these 10 lockers and reads, “The passcode consists of the first five locker numbers that were touched exactly twice.”

Which are these five lockers?
The lockers that are toggled twice are the prime numbers, since each prime number has 1 and the number itself as factors. So, the code is 2 – 3 – 5 – 7 – 11.
Solution:
As we know that each of the square numbers has an odd number of factors. Therefore, the lockers with following numbers will remain open: #1, #4, #9, #16, #25, #36, #49, #64, #81, #100.

Page : 4
1.1 Square Numbers

Question 1.
What patterns do you notice? Share your observations and make conjectures.
Study the squares in the table above. What are the digits in the units places of these numbers? All these numbers end with 0, 1, 4, 5, 6 or 9. None of them end with 2, 3, 7 or 8.

12 = 1	112 = 121	212 = 441
22 = 4	122 =	222 =
32 = 9	132 =
42 = 16	142 =
52 = 25	152 =
62 =	162 =
72 =	172 =
82 =	182 =
92 =	192 =
102 =	202 =

Solution:

12 = 1	112 = 121	212 = 441
22 = 4	122 = 144	222 = 484
32 = 9	132 = 169	232 = 529
42 = 16	142 = 196	242 = 576
52 = 25	152 = 225	252 = 625
62 = 36	162 = 256	262 = 676
72 = 49	172 = 289	272 = 729
82 = 64	182 = 324	282 = 784
92 = 81	192 = 361	292 = 841
102 = 100	202 = 400	302 = 900

We observe that numbers whose unit’s place digit is 1 or 9, squares of these numbers have unit’s place digit 1. The numbers whose unit’s place digit is 2, 3, 7, or 8, squares of these numbers have 4, 9, 9, 4 as their unit’s place digit, respectively The numbers whose unit’s digit have 3 or 7 have their squares having unit’s place digits 9 while the numbers having unit’s place digits 5, 6, or 0 have 5, 6,0 (even number of zeroes) at their unit’s place digit and ten’s place digit for the case of zeroes.

So, we can conclude that squares of numbers end with 0, 1, 4, 5, 6, or 9. None of them end with 2, 3, 7, or 8.

Question 2.
If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
The numbers 16 and 36 are both squares with 6 in the units place. However, 26, whose units digit is also 6, is not a square. Therefore, we cannot determine if a number is a square just by looking at the digit in the units place. But, the units digit can tell us when a number is not a square. If a number ends with 2, 3, 7, or 8, then we can definitely say that it is not a square.
Solution:
No.
If a number ends in ‘0’, then it will not always be a square as numbers 10, 20, 30, 40, 50, 60, etc. are not squares of any number.
If a number ends in ‘0’, it will not always be a square as numbers 11, 21, 31, 41, etc. are not squares.
If a number ends in ‘4’ will not always be a square as numbers 14, 24, 34, 44, etc. are not squares.
If a number ends in ‘5’ will not always be a square as numbers 15, 35, 45, 55, etc. are not squares.
If a number ends in ‘6’ will not always be a square as numbers 26, 46, 56, 66, etc. are not squares.
If a number ends in 9 will not always be a square as numbers 19, 29, 39, 59, 69, etc. are not squares.
So, we can conclude that a number ending at 0, 1, 4, 5, 6, or 9 is not always a square.

Question 3.
Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
The squares, 1², 9², 11², 19², 21², and 29², all have 1 in their units place. Write the next two squares. Notice that if a number has 1 or 9 in the units place, then its square ends in 1.
Solution:
We know that none of square numbers end with 2, 3, 7 or 8.
Based on the above, we can write any number of numbers which can be determined by looking at their units digit that they are not squares.
So, the 5 numbers can be 12, 13, 17, 18, and 28.
Note : We can write any number of numbers which can be determined by looking at their unit’s digits that they are not squares.

Page : 4 – 5

Question 1.
Let us consider square numbers ending in 6 : 16 = 4², 36 = 6², 196 = 14², 256 = 16², 576 = 24², and 676 = 26².
Which of the following numbers have the digit 6 in the units place?
(i) 38²
(ii) 34²
(iii) 46²
(iv) 56²
(v) 74²
(vi) 82²
Solution:
(i) 38² = 1444, No digit 6 in the unit’s place.
(ii) 34² = 1456, Digit 6 in the unit’s place.
(iii) 46² = 2116, Digit 6 in the unit’s place.
(iv) 56² = 3136, Digit 6 in the unit’s place.
(v) 74² = 5476, Digit 6 in the unit’s place.
(vi) 82² = 6724, No, digit 6 in the unit’s place.
Note : The numbers which end at 4 or 6 will have digit 6 in the unit’s place and no other number will have digit 6 in the unit’s place in their squares.

Page : 5

Question 1.
Find more such patterns by observing the numbers and their squares from the table you filled earlier.
Consider the following numbers and their squares.

Solution:
Some more patterns which we find in the table are :
(i) Numbers ending at 1 or 9 have digit 1 at unit’s place in their squares.
(ii) Numbers ending at 2 or 8 have digit 4 at unit’s place in their squares.
(iii) Numbers ending at 3 or 7 have digit 9 at unit’s place in their squares.
(iv) Numbers ending at 5 will have digit 5 at unit’s digit in their squares.
(v) There will be an even number of zeroes in the squares of a number having a zero at their unit’s place.
(vi) No square number will end at 2, 3, 7, or 8.

Question 2.
If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Solution:
If a number contains 3 zeroes at the end, then there will be 6 zeroes at the end of its square.

Question 3.
What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
Solution:
The number of zeroes at the end of square of a number is twice the number of zeroes at the end of the number.

For example, if a number has 2 zeroes at the end then its square will have 4 zeroes and if a number has 5 zeroes at the end then its square will have 10 zeroes at its end. This will always happen. Yes, a square number will always have an even number of zeroes at its end.

Question 4.
What can you say about the parity of a number and its square?
Solution:
The square of an even number has an even parity and the square of an odd number has an odd parity.

Page : 6

Question 1.
Using the pattern above, find 36², given that 35² = 1225.
From the question we know that 1225 is the sum of the first 35 odd numbers. To find 36², we need to add the 36th odd number to 1225.
Solution:
It is given that 35² = 1225
∴, to obtain the 36² we add the 36th odd number to 1225.
We add 36th odd number which is 2 × 36 – 1
= 72 – 1 = 71 to 1225 to get
= 1225 + 71 = 1296
So, we have obtained 36² = 1296

Question 2.
How do we find the 36th odd number?
The 1st odd number is 1, 2nd odd number is 3, 3rd number is 5, … , 6th odd number is 11 and so on.
Solution:
We have obtained 36th odd number by 2 × 36 – 1 = 72 – 1 = 71.

Question 3.
What is the n^th odd number?
The n^th odd number is 2n – 1.
Therefore, the 36th odd number is 71.
By adding 71 to 1225, we get 1296, which is 36².
Consider a number such as 38 that is not a square and subtract consecutive odd numbers starting from 1.

This shows that 38 cannot be expressed as a sum of consecutive odd numbers starting with 1.

Thus, we can say that a natural number is not a perfect square if it cannot be expressed as a sum of successive odd natural numbers starting from 1. We can use this result to find out whether a natural number is a perfect square.
Solution:
n^th odd number is given by 2n – 1.

Page : 7

Question 1.
Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Solution:
Let us consider the following :
2² – 1² = 4 – 1 = 3, there are 3 – 1 = 2 numbers in between.
3² – 2² = 9 – 4 = 5, there are 5 – 1 = 4 numbers in between.
4² – 3² = 16 – 9 = 7, there are 7 – 1 = 6 numbers in between.
5² – 4² = 25 – 16 = 9, there are 9 – 1 = 8 numbers in between.
6² – 5² = 36 – 25 = 11, there are 11 – 1 = 10 numbers in between.
So, we notice that in between (n + 1)² and n², there are 2n numbers.

Question 2.
How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?

Solution:

The largest square less than 1000 is 961, which is the square of 31.

Question 3.
Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.

Solution:

Question 4.
The area of a square is 49 sq. cm. What is the length of its side?
We know that 7 × 7 = 49, or 7² = 49
So, the length of the side of a square with an area of 49 sq. cm is 7 cm.
We call 7 the square root of 49.
In general, if y = x² then x is the square root of y.
Solution:
Area of a square = (Side)²
⇒ (Side)² = 49 sq. cm
⇒ (Side)² = 7 × 7 ⇒ (Side)² = (7)²
⇒ Side = 7
∴, the length of its side = 7 cm.

Page : 8

Question 1.
What is the square root of 64?
We know that 8 × 8 is 64. So, 8 is the square root of 64. What about -8 × -8? That is 64 too!
8² = 64, and (-8)² = 64.
So, the square roots of 64 are +8 and -8.
Every perfect square has two integer square roots. One is positive and the other is negative. The square root of a number is denoted by √
Thus, \(\sqrt{64}\) = ±8 and \(\sqrt{100}\) = ±10.
Note that \(\sqrt{8^2}\) = ±8 and \(\sqrt{10^2}\) = ±10. In general, \(\sqrt{n^2}\) = ± n.
In this chapter, we shall only consider the positive square root.
Solution:
To know the square roots of 64, we should try to get all those numbers whose square is 64. We know that 8 × 8 = 64 and also -8 × (-8) – 64
∴, the square roots of 64 are +8 and -8.

Question 2.
Given a number, such as 576 or 327, how do we find out if it is a perfect square? If it is a perfect square, how can we find its square root?
We know that perfect squares end in 1, 4, 9, 6, 5, or an even number of zeros. But, it is not certain that a number that satisfies this condition is a square.
We can clearly say that 327 is not a perfect square. However, we cannot be sure that 576 is a perfect square.
1. We can list all the square numbers in sequence and find out whether 576 occurs among them. We know that 202 = 400, we can find squares of 21, 22, 23, … and so on until we get 576 or a number greater than 576.
20² = 400 21² = 441 22² = 484 23² = 529 24² = 576
However, this process becomes inefficient for larger numbers.
2. Recall that every square can be expressed as a sum of consecutive odd numbers starting from 1.
Consider \(\sqrt{81}\).

From 81, we successively subtracted consecutive odd numbers starting from 1 until we obtained O at the 9th step. Therefore \(\sqrt{81}\) = 9.
Can we find the square root of 729 using this method? Yes, but it will be time-consuming.
3. We know that a perfect square is obtained by multiplying an integer by itself. Will looking at a number’s prime factorisation help in determining whether it is a perfect square?
Yes, if we can divide the prime factors of a number into two equal groups, then the product of the prime factors in either group combine to form the square root.
Solution:
Given a number, we can decide whether the given number is a perfect square or not by using the following methods:
1. If the given number ends with 2, 3, 7, 8, or an odd number of zeroes, then it can not be a perfect square.

2. We know that every square number can be expressed as a sum of consecutive odd numbers starting from 1.
So, by subtracting consecutive odd numbers, we can check whether the given number is a square or not.

3. We can check whether a given number is a perfect square by prime factorisation. We do the prime factorisation and then group them in pairs. If all prime factors occur in pairs, then it is a perfect square. Otherwise, it is not.
Now we take the given numbers:
Given numbers are 576 and 327.
Clearly 327 cannot be a perfect square as it ends at 7.
For 576, we can do repeated subtraction of consecutive odd numbers starting from 1, or we can find prime factors of 576.
Let us find out prime factors of 576:

∴, 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Grouping prime factors in pairs, we get:
\(\sqrt{576}\) = 2 × 2 × 2 × 3
= 24
So, 576 is a perfect square and its square roots are 24 and -24.

Page : 9

Question 1.
Is 324 a perfect square?
324 = 2 × 2 × 3 × 3 × 3 × 3.
These can be grouped as
324 = (2 × 3 × 3) × (2 × 3 × 3).
= (2 × 3 × 3)² = 18².
We can also write the prime factors in pairs. That is,
324 = (2 × 2) × (3 × 3) × (3 × 3),
which shows that 324 is a perfect square. Thus,
324 = (2 × 3 × 3)² = 18².
Therefore, \(\sqrt{324}\) = 18.
Solution:
Let us find prime factors of 324:

∴, 234 = 2 × 2 × 3 × 3 × 3 × 3
Since, all prime factors of 324 can be grouped in pairs, so 324 is a perfect square.
Hence,
\(\sqrt{324}\) = 2 × 3 × 3 = 18.

Question 2.
Is 156 a perfect square?
The prime factorisation of 156 is 2 × 2 × 3 × 13.
We cannot pair up these factors.
Therefore, 156 is not a perfect square.
Solution:
Let us find prime factors of 156:

∴, 156 = 2 × 2 × 3 × 13
Here, we can see that all prime factors of 156 cannot be grouped in pairs.
Hence, 156 is not a perfect

Question 3.
Find whether 1156 and 2800 are perfect squares using prime factorisation.
Solution:
Prime factors of:

∴, 1156 = 2 × 2 × 17 × 17
All prime factors of 1156 can be grouped in pairs.
Hence, 1156 is a perfect square and \(\sqrt{1156}\) = 34.
Also, 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, all prime factors cannot be grouped in pairs.
Hence, 2800 is not a perfect square.

Figure it Out: Page : 10 – 11

Question 1.
Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Solution:
(i) 2032,
(ii) 2048,
(iii) 1027 are not perfect squares as they end with 2, 8 and 7, respectively. 1089 is a perfect square as it is square of 33.

Question 2.
Which one among 64², 108², 292², 36² has last digit 4?
Solution:
Any number whose last digit is 8 will have 4 as last digit in its square.
So, 108² has the last digit 4.
Also, if the last digit is 2, then its square will have 4 at its units place.
So, 292² has 4 at its last digit.

Question 3.
Given 125² = 15625, what is the value of 126²?
(i) 15625 + 126
(ii) 15625 + 262
(iii) 15625 + 253
(iv) 15625 + 251
(v) 15625 + 51²
Solution:
To get the value of 126², we add 126th odd number to 125² = 15625.
126th odd number is 2 × 126 – 1 = 252 – 1 = 251
So, we get 126² = 15625 + 251
i.e., (iv) 15625 + 251, is the correct option.

Question 4.
Find the length of the side of a square whose area is 441 m².
Solution:
Area of the square = 441
So, (side)² = 441 (∴, area of square = (side)²)
⇒ (side)² = (3 × 3 × 7 × 7)
⇒ (side)² = (3² × 7²)
⇒ (side)² = (3 × 7)²
⇒ (side)² = (21)²
⇒ Side = 21 m,
(∴, length cannot be negative)
∴, The length of the side = 21 m.

Question 5.
Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Solution:
To get the smallest number that is divisible by 4, 9 and 10, will be the LCM of these numbers.
Now,
4 = 2 × 2 = 2²
9 = 3 × 3 = 3²
10 = 2 × 5 = 2 × 5
To get the LCM, we collect the factors with highest powers.
So, LCM(4, 9, 10) = 2² × 3² × 5
= 4 × 9 × 5 = 180
Therefore, 180 is the smallest number that is divisible by 4, 9 and 10.

Question 6.
Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Solution:
We first of all get the prime factors of 9408:

∴, 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
Here, we can notice that 3 remains unpaired. So, we must multiply 9408 by 3 to get the product, which is a perfect square.
Square root of the new number is:
2 × 2 × 2 × 3 × 7 = 168

Question 7.
How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100
Solution:
(i) There are 32 numbers between 162 and 17²
(ii) There are 198 numbers between 992 and 100²

Question 8.
In the following pattern, fill in the missing numbers:
1² + 2² + 2² = 3²
2² × 3² × 6² = 7²
3² × 4² + 12² = 13²
4² + 5² + 20² = (_)²
9² + 10² + (_)² = (_)²
Solution:
4² + 5² + 20² = (21)²
9² + 10² + (90)² = (91)²

Question 9.
How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.

Solution:
There are 81 tiny squares.
Prime factorisation of 81 :

∴, 81 = 3 × 3 × 3 × 3

Page : 11
1.2 Cubic Numbers

Question 1.
How many cubes of side 1 cm will make a cube of side 3 cm?
Consider the numbers 1, 8, 27, …
These numbers are called perfect cubes. Can you see why they are named so?
Each of them is obtained by multiplying a number by itself three times. We note that
1 = 1 × 1 × 1
8 = 2 × 2 × 2
27 = 3 × 3 × 3
Solution:
27 cubes of 1 cm will make a cube of side 3 cm.

Page : 12

Question 1.
Is 9 a cube?
We see that 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27. This shows that 9 is not a perfect cube. Nor is any number from 10 to 26.

Solution:
No, 9 is not a cube as 9 = 3 × 3. We cannot group the prime factors of 9 as a group of 3.
So, 9 is not a perfect cube.

Question 2.
Can you estimate the number of unit cubes in a cube with an edge length of 4 units?
It has 64 unit cubes! If you notice carefully, each layer of this cube has 4 × 4 unit cubes. Each square layer has 16 unit cubes (4 × 4), and there are 4 such layers, so the
total number of unit cubes is 4 × 4 × 4 = 64.
Since 5³ = 5 × 5 × 5= 125, 125 is a cube.
In general, for any number n, we write the cube
n × n × n as n³.

Solution:
Total number of unit cubes in a cube with an edge length of 4 units = 4 × 4 × 4 = 64.

Question 3.
Complete the table below.

13 = 1	113 = 1331
23 = 8	123 =
33 = 27	133 = 2197
43 = 64	143 = 2744
53 = 125	153 =
63 =	163 =
73 =	173 = 4913
83 =	183 = 5832
93 =	193 = 6859
103 =	203 =

Solution:

Question 4.
What patterns do you notice in the table above?
Solution:
We have noticed that cubes of numbers ending at 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0 end at 1, 8, 7, 4, 5, 6, 3, 2, 9, and three zeroes, respectively.

Page : 12 – 13

Question 1.
We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?
Solution:
The possible last digit of cubes can be any digit from 0 to 9. There is no exception as we have for square numbers.
There can be even or odd number of zeroes in a cube.

Question 2.
Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Solution:
Cubes with 1-digit are 1 and 8.
Cubes with 2-digit are 27 and 64.
Cubes with 3-digit are 125, 216, 343, 512, and 729.
We observe that all these cubes are the cubes of 1-digit numbers only.
Clearly, 1 is the cube of 1, and 729 is the cube of 9.

Question 3.
Can a cube end with exactly two zeroes (00)? Explain.
Just as we can take squares of fractions/decimals – (\(\frac{4}{6}\))² (13.08)², and
(6)² – we also can compute cubes of such numbers – (\(\frac{4}{6}\))³, (13.08)³, and (-6)³.
(\(\frac{4}{6}\))³ = (\(\frac{4}{6}\)) × (\(\frac{4}{6}\)) × (\(\frac{4}{6}\)) = ((\(\frac{64}{216}\)))
(13.08)³ = 13.08 × 13.08 × 13.08 = 2237.810112
(-6)³ = -6 × -6 × -6 = -216.
Solution:
No cube can end with exactly two zeroes. If a number ends at 0 (single), then its cube will have three zeroes.
So, no cube number can have two zeroes at the end.

Question 4.
The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes.

How did Ramanujan know this? Well, he loved numbers. All through his life, he tinkered with numbers. During Ramanujan’s time in Cambridge, his colleagues often marveled at his ability to see deep patterns in numbers that seemed arbitrary to others. His colleague, John Littlewood, once said,
“Every positive integer was one of his [Ramanujan’s] personal friends”.
Solution:
4104 = 2³ + 16³ and 9³ + 15³ 13832 = 18³ + 20³ and 3³ + 24³
Note : 21³ = 9261, 22³ = 10648, 23³ = 12167, 24³ = 13824.

Page : 14

Question 1.
Can you tell what this sum is without doing the calculation?
91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109
Solution:
We see here that starting from the 10 × 9 + 1 = 91, we have 10 odd numbers added in the sequence.
So, the sum will be 10³ = 1000.

Question 2.
Let us check if 3375 is a perfect cube.
3375 = 3 × 3 × 3 × 5 × 5 × 5.
Can the factors be split into three identical groups? For 3375, we can
form three groups of (3 × 5). So,
3375 = (3 × 5) × (3 × 5) × (3 × 5)
= (3 × 5)³ = 15³.
Another way is to check if the factors can be grouped into triplet(s):
3375 = (3 × 3 × 3) × (5 × 5 × 5) = 3³ × 5³.
This means \($\sqrt[3]{3375}$\) = 15.
Solution:
Let us find prime factors of 3375 :

∴, 3375 = 3 × 3 × 3 × 5 × 5 × 5
Since we can group all prime factors in groups of 3,
So, we conclude that 3375 is a perfect cube.

Question 3.
Is 500 a perfect cube?
500 = 2 × 2 × 5 × 5 × 5. We see that the factors cannot be split into three identical groups. Therefore, 500 is not a perfect cube.

Observe that each prime factor of a number appears three times in the prime factorisation of its cube.
Solution:
Let us find prime factors of 500:

∴, 500 = 2 × 2 × 5 × 5 × 5
Here, we see that all prime factors cannot be grouped in a group of 3.
So, 500 is not a perfect cube.

Page : 15

Question 1.
Find the cube roots of these numbers:
(i) \(\sqrt[3]{64}\) =
(ii) \(\sqrt[3]{512}\) =
(iii) \(\sqrt[3]{729}\) =
Solution:

Question 2.
Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?

Solution:

So, at level – 3, we get the differences same as 6.

Figure it Out : Page : 16 – 17

Question 1.
Find the cube roots of 27000 and 10648.
Solution:
Prime factors of 27000 :

∴, 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
⇒ \(\sqrt{27000}\) = 2 × 3 × 5
= 2 × 3 × 5 = 30

∴, 10648 = 2 × 2 × 2 × 11 × 11 × 11
⇒ \(\sqrt{10648}\) = 2 × 11
= 2 × 11 = 22

Question 2.
What number will you multiply by 1323 to make it a cube number?
Solution:
To get the number by which we should multiply 1323 to make it a cube number.
We first of all find prime factors of 1323.

∴, 1323 = 3 × 3 × 3 × 7 × 7
Here, we can see that 7 remains ungrouped in group of three.
So we multiply 1323 by 7 to get a perfect cube. The new number which is a perfect cube is: 9261. Its cube root is : 21.

Question 3.
State true or false. Explain your reasoning.
1. The cube of any odd number is even.
2. There is no perfect cube that ends with 8.
3. The cube of a 2-digit number may be a 3-digit number.
4. The cube of a 2-digit number may have seven or more digits.
5. Cube numbers have an odd number of factors.
Solution:
1. FALSE: The cube of any odd number is odd as 1³ = 1, 3³ = 27, 5³ = 125, 7³ = 343, …

2. FALSE : There are perfect cubes that end with 8 as 2³ = 8, 12³ = 1728, ³ = 10648, …

3. FALSE : The cube of a 2-digit number can never be a 3-digit number.
For example, 10³ = 1000
Here, we see that the smallest two-digit number 10 has cube which has 4-digits.

4. FALSE : The cube of a 2-digit number can never be of 7-digit or more digit.
As we can see that 99 is the largest 2-digit number and its cube is 9,70,299, which is a 6-digit number.

5. FALSE : Cube numbers have an odd number of factors is a false statement. We can see that 8, which is a perfect cube, has factors 1, 2, 4, and 8, which are 4 in numbers.

Question 4.
You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
Solution:
1331 is a perfect cube of ‘11’,
4913 is a perfect cube of ‘17’,
12167 is a perfect cube of ‘23’, and
32768 is a perfect cube of ‘32’.

The cubes of numbers ending at 1, 2, 3, and 7, end with 1, 8, 7, and 3, respectively.

Question 5.
Which of the following is the greatest? Explain your reasoning.
(i) 67³ – 66³
(ii) 43³ – 42³
(iii) 67² – 66²
(iv) 43² – 42²
Solution:
(i) 63³ – 66³ = 13267
(ii) 43³ – 42³ = 5419
(iii) 67² – 66² = 133
(iv) 43² – 42² = 85
Clearly, 63³ – 66³ is the greatest and 43² – 42² is the least.

A Square and A Cube Class 8 Extra Questions

Multiple Choice Questions

Question 1.
Which of the following is not a perfect square?
(a) 225
(b) 144
(c) 288
(d) 256
Solution:
Here, 225 = 15², 144 = 12², and 256 = 16².
But 288 cannot be written as a square of any natural number.
(c) 288

Question 2.
Which of the following cannot be a digit at the end of a perfect square?
(a) 1
(b) 2
(c) 4
(d) 6
Solution:
The digit at the end of a perfect square can be 1, 4, 5, 6, 9 or an even number of zeroes.
(b) 2

Question 3.
Total number of factors of a perfect square number is:
(a) Even
(b) Odd
(c) Can be even or odd
(d) None of these
Solution:
This is a fact that the number of factors of a perfect square number is always odd in numbers.
This is a fact that the number of factors of a perfect square number is always odd in numbers.
For example:
1² = 1, factors: 1, number of factors = 1
2² = 4, factors: 1, 2, 4, number of factors = 3
3² = 9, factors: 1, 3, 9, number of factors = 3
4² = 16, factors: 1, 2, 4, 8, 16, number of factors = 5, etc.
(b) Odd

Question 4.
Which of the following is not a perfect square?
(a) 1024
(b) 576
(c) 729
(d) 927
Solution:
1024 = 32², 576 = 24², 729 = 27²
But 927 is not a square of any natural number.
(d) 927

Question 5.
Which of the following is a perfect square?
(a) 124632
(b) 207936
(c) 732423
(d) 783228
Solution:
In the given options, we can notice that (a), (c), and (d) end with 2, 3, and 8 respectively. Hence, these numbers cannot be perfect squares.
Answer:
(b) 207936

Assertion and Reasoning

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

Question 1.
Assertion (A) : The number 102325462 cannot be a square number.
Reason (R) : A square number never ends at 2.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Question 2.
Assertion (A) : \(\sqrt{2704}\) = 52
Reason (R) : (52)² = 2704.
Answer:
(a) Assertion (A) and Reason (R) both are true and Reason (R) is the correct explanation for the Assertion (A).

Case Based Questions

Question 1.
During a dance practice in school, 6570 students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that few children are left out.

Answer the following questions with proper steps and reasons:
(i) How many students were left out in the arrangement?
(ii) What is the number of students forming a square arrangement?
(iii) Find the number of children in each row of the square arrangement.
Solution:
(i) Total number of students = 6570
Using long division, let us find the square root of 6570:

Since, remainder = 9
Hence, (i) 9 students were left out in the arrangement.

(ii) Number of students forming a square arrangement = 6570 – 9 = 6561

(iii) The number of children in each row = \(\sqrt{6561}\) = 81
So, there are 81 children in each row.

Question 2.
Aarya visited her home in a village. She went to her orchard in which she counted the trees and found that there were 52 trees in all. She argued that even trees are arranged in any pattern, but they cannot be arranged in a square arrangement.

Based on the above, answer the following:
(i) Is Aarya right in saying that trees in her orchard cannot be arranged in a square?
(ii) How many trees will be left out if they are to be arranged in a square?
(iii) How many more trees will be required if Aarya wants to arrange them as in a square?
Solution:
(i) Yes. If the orchard has 52 trees then they cannot be arranged in a square arrangement. As there is 2 in its unit’s place, it cannot be a square.

(ii) Three trees will be left out if they are to be arranged in a square.
As 52 – 3 = 49 = (7)²

(iii) Above 52,64 is the square number.
So, 12 more trees will be required to arrange the trees in a square arrangement.

Go through BSE Odisha Class 8 Science Solutions Chapter 2 The Invisible Living World: Beyond Our Naked Eye Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 2 Question Answer

Class 8 Science Ch 2 The Invisible Living World: Beyond Our Naked Eye Question Answer

Class 8 Science Chapter 2 The Invisible Living World Beyond: Our Naked Eye Question Answer

Probe and Ponder Questions

Question 1.
Have you ever wondered what you might see if the invisible world around you became visible?
Answer:
Yes, it would be fascinating to see the microorganisms like bacteria, fungi, protozoa, and viruses that are constantly interacting with our environment and even inside our bodies. It would change how we perceive cleanliness, health, and the complexity of ecosystems around us

Question 2.
How do you think your observation of this hidden world might change the way you think about size, complexity, or even what counts as ‘living’?
Answer:
Observing this hidden world would reveal that even microscopic organisms exhibit complex behaviors such as movement, reproduction, response to stimuli, and forming colonies. This would deepen our understanding of what qualifies as a living being and show that size does not limit the complexity of life.

Question 3.
Have you thought how these tiny living beings interact with each other?
Answer:
Yes, microorganisms constantly interact; some live symbiotically, while others compete for resources or prey on one another. For example, bacteria can help in digestion or cause diseases, and fungi can decompose organic matter, enriching the soil. These interactions form the foundation of many ecological processes.

Question 4.
Share your questions……
Answer:
Seeing the invisible world and learning about microorganisms sparks so many questions! Here are some questions you might have, and you can add your own:

• What do microorganisms eat to stay alive?
• Can microorganisms see or feel each other, or do they just bump into things?
• Why are some bacteria helpful (like in curd) and others harmful (like causing diseases)?
• How do microorganisms survive in tough places, like hot springs or salty water?

InText Questions

Question 1.
Why do microorganisms not infect the pickles and murabbas?
Answer:
Preservation with Salt and Sugar: Pickles and murabbas are made with high concentrations of salt or sugar. These act as preservatives and prevent the growth of microbes, so the food does not spoil easily.

The Invisible Living World: Beyond Our Naked Eye Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 25-26)

Question 1.
Various parts of a cell are given below. Write them in the appropriate places in the following diagram.
Nucleus , Cytoplasm
Chloroplast , Cell wall
Cell membrane , Nucleoid

Answer:

Question 2.
Aanandi took two test tubes and marked them A and B. She put two spoonfuls of sugar solution in each of the test tubes. In test tube B, she added a spoonful of yeast. Then she attached two incompletely inflated balloons to the mouth of each test tube. She kept the set-up in a warm place, away from sunlight.
(i) What do you predict will happen after 3-4 days? She observed that the balloon attached to testtube B was inflated. What can be a possible explanation for this?
(a) Water evaporated in test tube B and filled the balloon with the water vapour.
(b) The warm atmosphere expanded the air inside the test tube B, which inflated the balloon.
(c) Yeast produced a gas inside the test tube B which inflated the balloon.
(d) Sugar reacted with warm air, which produced gas, eventually inflating the balloon.

Answer:
(c) Yeast produced a gas inside test tube B which inflated the balloon. The yeast fermented the sugar, releasing carbon dioxide gas (CO₂), which inflated the balloon.

(ii) She took another test tube, 1/4 filled with lime water. She removed the balloon from test tube B in such a manner that the gas inside the balloon did not escape. She attached the balloon to the test tube with lime water and shook it well. What do you think she wants to find out?
Answer:
She wants to test whether the gas produced is carbon dioxide CO₂). If the lime water turns milky, it confirms the presence of CO₂, because CO₂ reacts with lime water to form calcium carbonate.

Question 3.
A farmer was planting wheat crops in his field. He added nitrogen-rich fertiliser to the soil to get a good yield of crops. In the neighbouring field, another farmer was growing bean crops, but she preferred not to add nitrogen fertiliser to get healthy crops. Can you think of the reasons?
Answer:
Beans are leguminous crops that form a symbiotic relationship with Rhizobium bacteria present in their root nodules. These bacteria fix nitrogen from the atmosphere into the soil, providing a natural source of nitrogen. Therefore, the second farmer does not need to add nitrogen-rich fertiliser.

Question 4.
Snehal dug two pits, A and B, in her garden. In pit A, she put fruit and vegetable peels and mixed it with dried leaves. In pit B , she dumped the same kind of waste without mixing it with dried leaves. She covered both the pits with soil and observed after 3 weeks. What is she trying to test?
Answer:
She is trying to test the effectiveness of composting.

• In Pit A, the mixture of green waste (fruit/vegetable peels) and dry leaves provides the correct carbon-nitrogen balance needed for decomposition.
• Pit B lacks this balance and will decompose more slowly and may smell. This experiment shows that decomposing agents like microorganisms work better when both carbonrich and nitrogen-rich materials are present.

Question 5.
Identify the following microorganisms
(i) I live in every kind of environment, and inside your gut.
(ii) I make bread and cakes soft and fluffy.
(iii) I live in the roots of pulse crops and provide nutrients for their growth.
Answer:
(i) Bacteria
(ii) Yeast
(iii) Rhizobium.

Question 6.
Devise an experiment to test that microorganisms need optimal temperature, air, and moisture for their growth.
Answer:
Set up 3 bread slices:
Slice A: Warm, moist environment (near sink). Slice B: Dry environment (sealed container).
Slice C: Cold environment (refrigerator).
After 3 days, observe: Slice A will have maximum microbial (fungal) growth.
Conclusion: Microorganisms grow best when temperature, moisture, and air are optimal.

Question 7.
Take 2 slices of bread. Place one slice in a plate near the sink. Place the other slice in the refrigerator. Compare after three days. Note your observations. Give reasons for your observations.
Answer:

• Take two slices of bread.
• Place one slice of bread in a plate near the sink. Name it as ‘A’.
• Now, take another slice of bread in a plate and place in the refrigerator. (Say B).

Observations :

• Bread in plate ‘A’ shows fungal/mould growth due to air, warmth and moisture.
• Bread in plate ‘ B ‘ which is kept in refrigerator shows less or no growth as cold inhibits the growth of fungi.
  Reason : Warm and moist conditions near the sink promote microbial growth, while cold temperatures in the refrigerator inhibit it.

Question 8.
A student observes that when curd is left out for a day, it becomes more sour. What can be two possible explanations for this observation?
Answer:
Two reasons why curd becomes more sour when left out for a day:

• Microorganisms (lactic acid bacteria) continue to grow and multiply, producing more actic acid, which makes the curd more sour.
• Warmer temperature speeds up bacterial activity, increasing acid production and sourness.

Question 9.
Observe the set-up given in Fig. and answer the following questions.
(i) What happens to the sugar solution in flask A?
(ii) What do you observe in test tube B after four hours? Why do you think this happened?
(iii) What would happen if yeast was not added in flask A?

Answer:
(i) The yeast ferments the sugar in the warm solution, producing carbon dioxide gas and a small amount of alcohol.
(ii) Lime water in test tube B turns milky. This happens because \mathrm{CO}_2 produced in flask A travels through the delivery tube into flask B and reacts with lime water, confirming the presence of carbon dioxide.
(iii) Fermentation would not occur, so no carbon dioxide would be produced. As a result, lime water would remain clear in test tube B.

Class 8 Science Chapter 2 Question Answer

Activity 1

Let us observe
Aim : Let us observe through a round-bottom flask.
Materials Required : A round-bottom flask made of glass, water, a cork, a news paper.

Procedure :

• First of all we will take a round-bottom flask made up of glass as
  The Invisible Living World : Beyond Our Naked Eye (Curiosity) shown in figure. Fill it with water.
• Now, close the mouth of the flask using a cork.
• Place the flask on an open book and look at the letters through it.

Observations :

• We can observe that the letters appear larger when seen through the flask! This happens because the flask filled with water acts like a magnifying glass.
• If we use a real magnifying glass to look at small organisms, like an ant, we are able to see the details of its body more clearly.

Inference:

• People were curious to know the tiny organisms around them, but they could not see them with their naked eyes.
• With the discovery of microscope we are able to explore tiny world first time.

Activity 2

Let us study a cell (Teacher demonstration activity)
Aim : To prepare a temporary mount of onion peel.
Materials Required : Onion bulb, a forceps, petri dish, safranin glycerin, a needle, coverslip, a microscope.

Procedure :
1. Take an onion bulb from your kitchen or garden and wash it thoroughly with water.
2. Now, cut the onion bulb vertically into pieces.
3. Take one piece of onion and pull out the thin, transparent layer from its inner surface with the help of forceps. This layer is called the onion peel.
4. Now, put the peel in a petri dish containing a few drops of safranin (red-coloured stain) for about 30 seconds. The stain will give a pinkish colour to the cells and help us see them clearly.
5. Using a thin brush transfer the onion peel to another petri dish containing water to rinse the peel and remove extra stain.
6. Now, we will place the stained onion peel on the glass slide carefully with the help of a thin brush, ensuring it does not break or fold.
7. Put a drop of glycerin over the onion peel on the slide. The glycerin will enable you to
8. Now, slowly place a coverslip over the peel with the help of a needle, such that no air bubbles get trapped.
9. Use blotting paper to gently wipe off any extra glycerin around the edges of the covers’v.
10. Observe the slide under a microscope or a foldscope. Compare it with fig. (c).

Observations :

• There are many rectangular structures under the microscope. These are the cells of the onion peel, which are closely arranged without any space between them.
• The structure of cells are similar to a wall made of bricks.

Inference:

• All plants are made of cells.
• These small microscopic structures that we see in an onion peel are the basic building units of onion bulb. These structures are called cells.

Activity 3

Let us investigate
Aim : To observe the mount of human cheek cell.
Materials Required : Toothpick, glass slide methylene blue (a blue coloured stain), glycerin coverslip, blotting paper, a microscope.

Procedure:

• Rinse your mouth with clean water.
• Now, with the help of the blunt end of a clean toothpick gently scrape the inside of your cheek.
• Put the scraped material in a drop of water on a clean glass slide and spread it evenly.
• Now, mix a drop of methylene blue (a blue-coloured stain) over the material on slide. When you add stain it improves the visibility of the material under the microscope by increasing contrast.
• Wait for one minute, add a drop of glycerin over the material on the slide to prevent the cells from drying.
• Now, carefully place a clean coverslip on the material, and remove the excess glycerin from the edges of the coverslip using blotting paper.
• Observe the slide under a microscope and draw what you see in your notebook.

Observations :

• A polygonshaped structure as shown in Fig. These are cheek cells, which form the inner lining of your mouth.
• We can observe that the cells have three main parts- a thin outer lining, a central region, and a small round structure inside it. The outer layer is called the cell membrane. The round structure in the middle is the nucleus, which is also covered by a thin membrane. The space between the cell membrane and nucleus is filled with cytoplasm.

Inference : Cell membrane, cytoplasm and nucleus are the basic parts of a cell (Both plants and animal cells).

Activity 4

Let us observe pond water/stagnant water
Aim : To observe pond water/stagnant water.
Materials Required : A dropper, a microscope or foldscope slide, cover slip.

Procedure :

• Take a container and collect pond or stagnant water in it. You can take help of your teacher or elder(s).
• Now, by using a dropper place a drop of pond or stagnant water on a microscope or foldscope slide. Put a coverslip and observe it under the microscope or foldscope.
• Look at the tiny organisms found in the pond or stagna t water and note down your observations.

Observations :

• Microorganisms like bacteria and Amoeba are seen.
• Some multicellular organisms like fungi and algae are also present.

Inference : Microorganisms unicellular and multicellular both are found in stagnant water/ pond water.

Activity 5

Let us observe soil suspension
Aim : To observe soil suspension.
Materials Required : A beaker, moist soil, spoon, gloves, glass rod, a dropper, microscope, coverslip.

Procedure :

• First of all we will take a beaker and collect some moist soil in it from the nearby farm (field) or garden. Never touch the soil with your bare hands- use a spoon or gloves.
• Now, pour some water into the beaker and stir it with a glass rod. Actually the liquid, which you are looking dirty, has very fine particles of soil, and is called soil suspension. Keep it aside for some time and let the mixture settle.
• With the help of a dropper take out a drop of water from the top layer. Place the drop on a microscope slide.
• Cover it gently with a coverslip and observe it under the microscope (Fig.)

Observations : You may observe small moving organisms similar to those saw in activity 4.
Inference:

• This shows that even soil suspension contains a variety of tiny creatures that cannot be seen with the unaided eye.
• These tiny creatures that cannot be seen with the naked eye are said to be microorganisms (micro means very small; organisms means living beings) or microbes.

Activity 6.

Let us study
Aim : Let us study pond water and soil suspension.

Procedure :

• A group of students studying in Grade 8 performed Activities – 4 and 5 and they collected information from the library and internet.
• They recorded the data obtained in the table given below. They identified the microorganisms as protozoa, algae, fungi and bacteria.

Observations :
1. Table -1. Organisms present in pond water.

2. Table -2 : Organisms present in soil suspension

Inference : They are everywhere, and we can only see them with a microscope – a device that magnifies them 100 to 400 times. Though microorganisms are small in size, they play an important role in our lives.

Activity 7.

Let us do
Aim : To make manure from kitchen waste or plant waste.
Materials Required : Fruits and vegetable peels, soil.

Procedure :

• Take an empty container and fill it halfway with farm soil or garden soil.
• Now, add some kitchen or plant waste like fruit and vegetable peels to the container. Thereafter, put a layer of soil on it and leave it aside.
• Do not disturb it and after 2-3 weeks, observe the changes that have taken place.

Observations :

• You may find that peels of fruits and vegetables have turned into a dark-coloured material.
  This is manure, which is rich in nutrients and helps increase the fertility of the soil.

Inference : Soil contains various kinds of microorganisms. Some of these microorganisms, like fungi and bacteria, act on the plant waste and slowly break it down into simpler, nutrient-rich manure.

Activity 8.

Let us perform
Aim : To study the process of fermentation.
Materials Required : Two bowls, 200 g of flour (atta or maida), sugar, yeast powder.

Procedure:

• Take two bowls X and Y.
• Take 200 g of flour (atta or maida) in both the bowls and add a pinch of sugar.
• Now, in bowl X, add a small amount of yeast powder and mix it well with the flour.
• In bowl Y, do not add any yeast, so that we can compare the results of the two bowls.
• Knead the flour of the two bowls by warm water to make soft dough.
• Now, cover the dough with a damp cloth and keep it in a warm place.
• Observe both the bowls after 4-5 hours.

Observations :

• The dough in bowl A, where yeast was added, has risen slightly, become fluffy, and has a different smell compared to the dough kneaded without yeast.
• The volume of dough rises, smell or texture has been changed.

Inference:

• Yeast is a type of microorganism and belongs to fungi.
• Yeast respires and breaksdown food to release energy for their growth.
• During this process carbon dioxide is released which make the dough soft and fluffy.

Activity 9

Let us prepare
Aim : Proper warm condition is necessary for the growth of bacteria.
Materials Required : Two small glass bowls, milk, curd.

Procedure :

• Take two small glass bowls – label them ‘A’ and ‘B’.
• Pour lukewarm milk in bowl A, and cold milk in bowl B.
• Now, add a small spoonful of curd to each bowl and mix well using a spoon.
• Cover both bowls. Keep bowl A in a warm place and bowl B in a cool place (like a refrigerator) for a few hours or overnight.

Observations :

• Table : Testing for curd formation using milk in different conditions
  
• You will observe that in bowl A, the milk has turned inte curd after a few hours and has become little sour.
• Whereas in bowl B, the milk has not curdled, but it might be a little sour.

Inference:

• The curd contains several types of bacteria, one of them is Lactobacillus which ferments the milk to form curd.
• These bacteria grow well in warm conditions. That is why curd is formed in bowl A but not in bowl B.

The Invisible Living World: Beyond Our Naked Eye Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
How do microorganisms act as a cleaning agent of nature ?
Answer:
The organic wastes like vegetable peels and remains of animals are broken down into harmless and usable substances by the action of microorganisms. In agriculture they are used to increase soil fertility by fixing nitrogen and by making manure.

Question 2.
How is bread formed ?
Answer:
When yeast together with some sugar and warm water is mixed with flour (atta or maida) the dough begins to rise. The presence of sugar and the warmth stimulates rapid growth of the yeast cells. During their reproduction, yeast cells produce carbon dioxide. Bubbles of CO₂ fill the dough and make it rise. When baked into a loaf, the bread becomes light and spongy.

Question 3.
Write any three functions of the plant cell wall.
Answer:
The plant cell wall have the following functions :

• It maintains the shape of the cells.
• It protects the cells from mechanical injury and prevents their desiccation.
• It provides mechanical support against gravity. It is due to the rigid cell walls that the aerial parts of the plants are able to keep erect and expose their leaves to sunlight.

Question 4.
Why do plant cells possess large sized vacuole?
Answer:
Plant cells have large distinct vacuoles, They use their vacuoles for transporting and storing nutrients, metabolites and waste products. Plants also accumulate water in their vacuoles as they became larger through turgor driven by cell-wall expansion.

Long Answer Type Questions

Question 1.
Make a sketch of the human nerve cell. What function do nerve cell perform ?
Answer:

Functions :
(a) Nerve cells help in the transfer of messages from various body parts to brain and brain to various parts of the body.
(b) They help also in the coordination of the functions of the organs of the body.

Question 2.
Write short notes on the following :
(a) Cytoplasm
(b) Nucleus of a cell
Answer:
(a) Cytoplasm : Cytoplasm is a thick jelly-like fluid inside the cell membrane. All the life functions take place in the cytoplasm. There are many small cytoplasmic bodies in cytoplasm. These are called cell organelles. All organelles in cytoplasm play an active role in the functioning of the cell.

(b) Nucleus of a Cell : Nucleus is a large, spherical organelle presents in all the cells. Nucleus controls all the activities of cell. It is surrounded by a thin membrane called nuclear membrane. In animal cells, the nucleus is present at the centre of the cell. However, in the case of plant cells, it is present at the periphery of the cells. The nucleus consists of a nuclear membrane, nucleoplasm, nucleolus and chromosomes.

Case-Study Based Questions

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

Every living beings, from tiny bacteria to large animals, is made of cells. These organisms are very much useful in our daily life. During fermentation yeast breakdown sugar and produce carbon dioxide. This gas make idli and dosas soft and spongy. In olden days people knows how to prevent the growth of microbes. They add salt and sugar as a preservatives in pickles and murabbas. Do you know microalgae, plant like microscopic organisms, release oxygen while making food through photosynthesis ? They produce over half of earth’s oxygen. Similarly, ‘spirulina’ a microalga, is labbled as a super food because it is rich in protein and contain vitamin B₁₂.

(i) What property of yeast helps to make idli and dosas soft ?
(a) Releasing oxygen during fermentation.
(b) Releasing carbon dioxide during fermentation.
(c) Formation of Nitrogen gas
(d) None of these
Answer:
(a) Releasing oxygen during fermentation.

(ii) Why does pickles and murabbas not get spoiled easily?
(a) Stored in metal jar
(b) Contains vinegar
(c) High salt and sugar concentration
(d) None of these
Answer:
(c) High salt and sugar concentration

(iii) What do microalgae contribute most to earth?
(a) Oxygen production
(b) Soil Formation
(c) Nitrogen production
(d) Cloud formation
Answer:
(a) Oxygen production

(iv) What makes ‘spirulina’ a super food ?
(a) due to fat content
(b) due to high protein and vitamin B₁₂
(c) due to rich in carbohydrate
(d) due to rich in fibre
Answer:
(b) due to high protein and vitamin B₁₂

Picture Based Questions

I. Study the picture given below carefully and answer the following questions :

(i) Name the microorganism and the group to which it belongs.
(a) bread mould
(b) fungi
(c) bacteria
(d) none of these
Answer:
(a) Bread mould

(ii) Name the food item on which the organism grows.
Answer:
Bread, fruits, starchy food like pasta, potatoes etc.

(iii) Does it grow well in dry or in moist conditions?
Answer:
It grows well in moist condition.

(iv) How can you prevent it from growing?
Answer:
We can prevent it from growing by the following ways:

• store in a cool and dry place.
• keep in a air tight container.

The Invisible Living World: Beyond Our Naked Eye Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
What tool first helped humans see tiny living organisms?
(a) Spectrometer
(b) Microscope
(c) Thermometer
(d) Telescope
Answer:
(b) Microscope

Question 2.
What term was introduced by Robert Hooke after observing cork ?
(a) Nucleus
(b) Molecules
(c) Microbe
(d) Cell
Answer:
(d) Cell

Question 3.
Who was known as Father of Microbiology ?
(a) Antonie Van Leeuwenhoek
(b) Isaac Newton
(c) Charls Darwin
(d) Robert Hooke
Answer:
(a) Antonie Van Leeuwenhoek

Question 4.
Chloroplast is found in the :
(a) Plant cell only
(b) Animal cell only
(c) Both a and b
(d) None of these
Answer:
(a) Plant cell only

Question 5.
Yeast is used in the production of :
(a) sugar
(b) alcohol
(c) hydrochloric acid
(d) oxygen
Answer:
(b) alcohol

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): Rhizobium bacteria lives in the root nodules of leguminous plants.
Reason (R): These bacteria trap nitrogen from the air and make it useful for the plants.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

2. Assertion (A): Spirulina, a microalga called superfood.
Reason (R): Because it is a good source of vitamin B₁₂ and protein.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

Fill in the blanks

1. Microorganisms can be seen with the help of a ………….
Answer:
microscope

2. Alcohol is produced with the help of …………
Answer:
yeast

3. A muscle cell is shaped like a ………….
Answer:
spindle

4. A herve cell is very ………… and has ………….
Answer:
long, branches

5. The cell wall in the plant cell provides ………… and ………… to plants.
Answer:
rigidity, strength.

True or False

1. We can see microorganisms with our naked eyes.
Answer:
False

2. Mushroom is a fungi.
Answer:
True

3. The yolk (the yellow part of an egg) of an ostrich egg is a single cell.
Answer:
True

4. A group of similar cells forms a type of organ.
Answer:
False

5. Plant cells do not have a vacuole.
Answer:
False

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(a)

Question 1.
If A = {a,b,c,d} mention the type of relations on A given below, which of them are equivalence relations?
(i) {(a, a), (b, b)}
(ii) {(a, a), (b, b), (c, c), (d, d)}
(iii) {(a, b), (b, a), (b, d), (d, b)}
(iv) {(b, c), (b, d), (c, d)}
(v) {(a, a), (b, b), (c, c), (d, d), (a, d), (a, c), (d, a), (c, a), (c, d), (d, c)}
Solution:
(i) Symmetric and transitive but not reflexive.
(ii) Reflexive, symmetric as well as transitive. Hence it is an equivalence relation.
(iii) Only symmetric
(iv) Only transitive
(v) Reflexive, symmetric and transitive. Hence it is an equivalence relation.

Question 2.
Write the following relations in tabular form and determine their type.
(i) R = {(x, y) : 2x – y = 0] on A = {1,2,3,…, 13}
(ii) R = {(x, y) : x divides y} on A = {1,2,3,4,5,6}
(iii) R = {(x, y) : x divides 2 – y} on A = {1,2,3,4,5}
(iv) R = {(x, y) : y ≤, x ≤, 4} on A = {1,2,3,4,5}.
Solution:
(i) R = {(x, y) : 2x- y = 0} on A
= {(x, y) : y = 2x} on A
= {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12)}
R is neither reflexive nor symmetric nor transitive.

(ii) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1,5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5,5), (6, 6)}
R is reflexive transitive but not symmetric.

(iii) R = {(x, y) : x divides 2 – y} on A
= {1, 2, 3, 4, 5}
= {(x, y) : 2-y is a multiple of x}
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 4), (3, 2), (3, 5), (4, 2), (5, 2)}
R is neither reflexive nor symmetric nor transitive.

(iv) R = {(x, y) : y ≤ x ≤ 4} on A
= {1, 2, 3, 4, 5}
= {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4, 4)}
R is neither reflexive nor symmetric but transitive.

Question 3.
Test whether the following relations are reflexive, symmetric or transitive on the sets specified.
(i) R = {(m,n) : m-n ≥ 7} on Z.
(ii) R = {(m,n) : 2|(m+n)} on Z.
(iii) R = {(m,n) : m+n is not divisible by 3} Z.
(iv) R = {(m,n) : is a power of 5} on Z – {0}.
(v) R = {(m,n) : mn is divisible by 2} on Z.
(vi) R = {(m,n) : 3 divides m-n} on {1,2,3…,10}.
Solution:
(i) R = {{m, n) : m- n ≥ 7} on Z
Reflexive:
∀ m ∈ Z, m – m = 0 < 7
⇒ (m, m) ∉ R
Thus, R is not reflexive.
Symmetry:
Let (m, n) ∈ R
⇒ m – n ≥ 7
⇒ n – m < 7
∴ (n, m) ∉ R
⇒ R is not symmetric.
Transitive:
Let (m, n), (n, p) ∈ R
m – n ≥ 7
and n – p > 7
⇒ m – p ≥ 7
⇒ (m, p) ∈ R
⇒ R is transitive.

(ii) R = {(m, n) : 2 | (m + n)} on Z
Reflexive:
∀ m ∈ Z, m + m = 2m
which is divisible by 2.
⇒ 2 | (m + m)
⇒ (m, m) ∈ R
⇒ R is reflexive.
Symmetry:
Let (m, n) ∈ R
⇒ 2 | (m + n)
⇒ 2 | (n + m)
(n, m) ∈ R
⇒ R is symmetric.
Transitive:
Let (m, n), (n, p), ∈ R
⇒ 2 | (m + n) and 2 | (n + p)
⇒ m + n = 2k₁
⇒ n + p = 2k₂
⇒ m + 2n + p = 2k₁ + 2k₂
⇒ m + p = 2(k₁ + k₂ – 1)
⇒ 2 | (m + p)
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus, R is an equivalence relative.

(iii) R = {(m, n) : m + n is not divisible by 3} on Z
Reflexive:
As 3 + 3 is divisible by 3
we have (3, 3) ∉ R
⇒ R is not reflexive.
Symmetric:
Let (m, n) ∈ R
⇒ m + n is not divisible by 3
⇒ n + m is not divisible by 3
⇒ (n, m) ∈ R
⇒ R is symmetric.
Transitive:
(3, 1), (1, 6) ∈ R
But (3, 6) ∉ R
⇒ R is not transitive.

(iv) R = {(m, n) : \(\frac{m}{n}\) is a power of 5} on Z – {0}
Reflexive:
∀ m ∈ Z – {0}
\(\frac{m}{m}\) = 1 = 5°
⇒ (m, m) ∈ R
⇒ R is reflexive.
Symmetric:
Let (m, n) ∈ R
\(\frac{m}{n}\) = 5^k
\(\frac{n}{m}\) = 5^-k
⇒ (n, m) ∈ Z
⇒ R is symmetric.
Transitive:
Let (m, n), (n, p) ∈ R
⇒ \(\frac{m}{n}\) = 5^k1 , \(\frac{n}{p}\) = 5^k2
⇒ \(\frac{m}{n}\) . \(\frac{n}{p}\) = 5^k1 . 5^k2
⇒ \(\frac{m}{p}\) = 5 ^k1+k2
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.

(v) R = {(m, n) : mn is divisible by 2} on Z
Reflexive:
3 ∈ Z
3 x 3 = 9
which is not divisible by 2.
∴ (3, 3) ∉ R
⇒ R is not reflexive.
Symmetric:
Let (m, n) ∈ R
⇒ mn is divisible by 2
⇒ nm is divisible by 2
⇒ (n, m) ∈ R
⇒ R is symmetric.
Transitive:
⇒ (3, 2), (2, 5) ∈ R
⇒ But 3 x 5 = 15,
⇒ which is not divisible by 2.
⇒ (3, 5) ∉ R
R is not transitive.

(vi) R = {(m, n) : 3 divides m-n} on A = {1, 2, 3……,10}
Reflexive:
Clearly ∀ m ∈ A, m – m = 0
which is divisible by 3
⇒ (m, m) ∈ R
⇒ R is reflexive
Symmetric:
Let (m, n) ∈ R
⇒ m – n is divisible by 3
⇒ n – m is also divisible by 3
⇒ (n, m) ∈ R
⇒ R is symmetric
Transitive:
Let (m, n), (n, p) ∈ R
⇒ m – n and n – p are divisible by 3
⇒ m – n + n – p is also divisible by p.
⇒ m – p is divisible by p.
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.

Question 4.
List the members of the equivalence relation defined by the following partitions on X= {1,2,3,4}. Also find the equivalence classes of 1,2,3 and 4.
(i) {{1},{2},{3, 4}}
(ii) {{1, 2, 3},{4}}
(iii) {{1,2, 3, 4}}
Solution:
(i) The equivalence relation is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3)}
[1] = {1}, [2] = {2}, [3] = {3, 4} and [4] = {3, 4}

(ii) The equivalence relation is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}
[1] = [2] = [3] = {1, 2, 3}
[4] = {4}

(iii) The equivalence relation is
R = A x A, [1] = [2] = [3] = [4] = A

Question 5.
Show that if R is an equivalence relation on X then dom R = rng R = X.
Solution:
Let R is an equivalence relation on X.
⇒ R is reflexive
⇒ (x, x) ∈ R ∀ x ∈ X
⇒ Dom R = Rng R = X

Question 6.
Give an example of a relation which is
(i) reflexive, symmetric but not transitive.
(ii) reflexive, transitive but not symmetric.
(iii) symmetric, transitive but not reflexive.
(iv) reflexive but neither symmetric nor transitive.
(v) transitive but neither reflexive nor symmetric.
(vi) an empty relation.
(vii) a universal relation.
Solution:
(i) The relation R = {(a, b), (b, a), (a, c), (c, a), (a, a), (b, b), (c, c)} defined on the set {a, b, c} is reflexive, symmetric but not transitive.
(ii) “The relation x ≤ y on z” is reflexive, transitive but not symmetric.
(iii) The relation R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)} defined on the set {a, b, c, d} is symmetric, transitive but not reflexive.
(iv) The relation R = {(a, a), (b, b), (c, c), (a, b), (b, c)} defined on the set A = {a, b, c} is reflexive but neither symmetric nor transitive.
(v) R = {(a, b), (b, c), (a, c)} on A = {a, b, c} is transitive but neither reflexive nor symmetric.
(vi) On N the relation R= {(x, y) : x + y = – 5} is an empty relation.
(vii) On N the relation R = {(x, y) : x + y > 0} is an universal relation.

Question 7.
Let R be a relation on X, If R is symmetric then xRy ⇒ yRx. If it is also transitive then xRy and yRx ⇒ xRx. So whenever a relation is symmetric and transitive then it is also reflexive. What is wrong in this argument?
Solution:
Let R is a relation on X.
If R is symmetric then xRy ⇒ yRx
If R is also transitive then xRy and yRx ⇒ xRx
⇒ Whenever a relation is symmetric and transitive, then it is reflexive. This argument is wrong because the symmetry of R does not imply dom R = X and for reflexive xRx ∀ x ∈ X.

Question 8.
Suppose a box contains a set of n balls (n ≥ 4) (denoted by B) of four different colours (may have different sizes), viz. red, blue, green and yellow. Show that a relation R defined on B as R={(b1, b2): balls b1 and b2 have the same colour} is an equivalence relation on B. How many equivalence classes can you find with respect to R?
[Note: On any set X a relation R={(x, y): x and y satisfy the same property P} is an equivalence relation. As far as the property P is concerned, elements x and y are deemed equivalent. For different P we get different equivalence relations on X]
Solution:
On B, R = {(b1, b2) : balls b1 and b2 have the same colour}

Reflexive:
∀ b ∈ B, b and b are of same colour
⇒ (b, b) ∈ R
⇒ R is reflexive.

Symmetric:
Let (b1, b2) ∈ R
⇒ b1 and b2 are of same colour
⇒ b2 and b1 are of same colour
⇒ (b2, b1) ∈ R
⇒ R is symmetric.

Transitive :
Let (b1, b2) and (b2, b3) ∈ R
⇒ b1 and b2 are of same colour
b2 and b3 are of same colour
⇒ b1, b3 are of same colour
⇒ (b1, b3) ∈ R
⇒ R is transitive
∴ R is an equivalence relation.
As there are 4 types of balls there are 4 equivalence relations with respect to R.

Question 9.
Find the number of equivalence relations on X={1,2,3}. [Hints: Each partition of a set gives an equivalence relation.]
Solution:
Method – 1: Number of equivalence relations on a set A with | A | = n.
= The number of distinct partitions of A
= B_n
where B_n+1 = \(\sum_{k=0}^n \frac{n !}{k !(n-k) !} \mathrm{B}_k\)
with B₀ = 1
Here n = 3
B₁ = 1
B_{2 =} \(\frac{1 !}{0 ! 1 !}\) B₀ + \(\frac{1 !}{1 ! 1 !}\) B₁
= 1 + 1 = 2
B₃ = \(\frac{2 !}{0 ! 2 !}\) B₀ + \(\frac{2 !}{1 ! 1 !}\) B₁ + \(\frac{2 !}{2 ! 0 !}\) B₂
= 1 + 2 + 2 = 5
Thus there are 5 equivalence relations.

Method – 2:
X= {1, 2, 3}
Number of equivalence relations = number of distinct partitions.
Different partitions of X are
{{1} {2}, {3}}
{{1}, {2, 3}}, {{2}, {1,3}},
{{3}, {1,2}} and {{1, 2,3}}
Thus number of equivalence relations = 5.

Question 10.
Let R be the relation on the set R of real numbers such that aRb iff a-b is an integer. Test whether R is an equivalence relation. If so find the equivalence class of 1 and ½ w.r.t. this equivalence relation.
Solution:
The relation R on the set of real numbers is defined as
R = {(a, b) : a – b ∈ Z}

Reflexive:
∀ a ∈ R (set of real numbers)
a – a = 0 ∈ Z
⇒ (a, a) ∈ R
⇒ R is reflexive.

Symmetric:
Let (a, b) ∈ R
⇒ a – b ∈ Z
⇒ b – a ∈ Z
⇒ (b, a) ∈ R
⇒ R is symmetric.

Transitive:
Let (a, b), (b, c) ∈ R
⇒ a – b and b – c ∈ Z
⇒ a – b + b – c ∈ Z
⇒ a – c ∈ Z
⇒ (a, c) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.
[1] = {x ∈ R : x -1 ∈ Z} = Z
\(\begin{aligned}
{\left[\frac{1}{2}\right] } &=\left\{x \in \mathrm{R}: x-\frac{1}{2} \in \mathrm{Z}\right\} \\
&=\left\{x \in \mathrm{R}: x=\frac{2 k+1}{2}, k \in \mathrm{Z}\right\}
\end{aligned}\)

Question 11.
Find the least positive integer r such that
(i) 185 ∈ [r]₇
(ii) – 375 ∈ [r]₁₁
(iii) -12 ∈ [r]₁₃
Solution:
(i) 185 ∈ [r]₇
⇒ 185 – r = 7k, k ∈ z and r < 7
⇒ r = 3
(ii) – 375 ∈ [r]₇
⇒ – 375 – r = 11k, k ∈ z and r < 11
⇒ r = 10
(iii) – 12 ∈ [r]₁₃
⇒ – 12 – r = 13k, k ∈ z and r < 13
⇒ r= 1

Question 12.
Find least non negative integer r such that
(i) 7 x 13 x 23 x 413 r (mod 11)
(ii) 6 x 18 x 27 x (- 225) = r (mod 8)
(iii) 1237(mod 4) + 985 (mod 4) = r (mod 4)
(iv) 1936 x 8789 = r (mod 4)
Solution:
(i) 7 x 13 x 23 x 413 ≡ r (mod 11)
Now 7 x 13 ≡ 3 mod 11
23 ≡ 1 mod 11
413 ≡ 6 mod 11
∴ 7 x 13 x 23 x 413 ≡ 3 x 1 x 6 mod 11
≡ 18 mod 11
≡ 7 mod 11
∴ r = 7

(ii) 6 x 18 x 27 x – 225 ≡ r (mod 8)
Now 6 x 18 ≡ 108 = 4 mod 8
27 ≡ 3 mod 8
– 225 ≡ 7 mod 8
⇒ 6 x 18 x 27 x – 225 ≡ 4 x 3 x 7 mod 8
≡ 84 mod 8
≡ 4 mod 8
∴ r = 4

(iii) 1237 (mod 4) + 985 (mod 4) r (mod 4)
Now 1237 ≡ 1 mod 4
985 ≡ 1 mod 4
⇒ 1237 (mod 4) + 985 (mod 4)
≡ (1 + 1) mod 4
≡ 2 mod 4
⇒ r = 2

(iv) 1936 x 8789 ≡ r (mod 4)
1936 x 8789 ≡ 0 mod 4
∴ r = 0

Question 13.
Find least positive integer x satisfying 276x + 128 ≡ (mod 7)
[Hint: 276 ≡ 3, 128 ≡ 2 (mod 7)]
Solution:
Now 128 ≡ 2 mod 7
Now 176 x + 128 ≡ 4 mod 7
⇒ 176 x ≡ (4 – 2) mod 7
⇒ 176 x ≡ 2 mod 7
176 x x ≡ 2 mod 7,
But 276 ≡ 3 mod 7
Thus x = 3.

Question 14.
Find three positive integers x_i, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)
[Hint: If X₁ is a solution then any member of [X₁] is also a solution]
Solution:
3x ≡ 2 mod 7
Least positive value of x ≡ 3
Each member of [3] is a solution
∴ x = 3, 10, 17 …..