Class 7

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 12 Earth, Moon, and the Sun Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 12 Question Answer

Science Curiosity Class 7 Chapter 12 Question Answer

Earth, Moon, and the Sun Class 7 Questions and Answers

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Question 1
In the shown figure, how many hours of sunlight do the North Pole and the South Pole receive during one rotation of the Earth?

Answer:
In the given figure, the North Pole is in darkness, indicating winter in the Northern Hemisphere. During this period, the North Pole receives 0 hours of sunlight, while the South Pole, experiencing summer, receives 24 hours of sunlight in one complete rotation of the Earth.

Question 2.
Fill in the blanks

(i) Stars rise in the ………… and set in the ………….
Answer:
east, west

(ii) Day and night are caused by the Earth’s ………….
Answer:
rotation

(iii) When the Moon fully covers the Sun from our view, it is called a ………… solar eclipse.
Answer:
total

Question 3.
State whether True or False

(i) Lunar eclipse occurs when the Sun comes between the Earth and the Moon.
Answer:
False, A lunar eclipse occurs when the Earth comes in between the Sun and the Moon.

(ii) Sunrise happens earlier in Gujarat than in Jharkhand.
Answer:
False. Sunrise happens earlier in places located more towards the east. Jharkhand lies to the east of Gujarat, so sunrise occurs earlier in Jharkhand.

(iii) In Chennai, the longest day occurs on the summer solstice.
Answer:
True

(iv) We should watch the solar eclipse directly with our naked eye.
Answer:
False. We should not watch the solar eclipse directly with our naked eyes.

(v) Seasons occur due to the tilt of Earth’s axis of rotation and its spherical shape.
Answer:
(v) True

(vi) The Earth’s revolution around the Sun causes day and night.
Answer:
False The Earth’s rotation on its own axis causes day and night.

Question 4.
Padmashree saw the Orion constellation nearly overhead at 8 pm yesterday. When will she see Orion overhead today?
Answer:
A few minutes earlier than 8:00 p.m. This is because as the Earth rotates on its axis, it also moves ahead in its orbit. The Earth’s revolution around the Sun causes stars to appear slightly earlier each night.
[Note: The stars seems to rise 4 min earlier each day due to the Earth’s revolution. So, the Orion will be seen overhead around 7:56 p.m. today.]

Question 5.
Nandhini saw a group of stars rising at midnight on 21 June. When will she see the same group of stars rising at midnight next year?
Answer:
At midnight on 21 June next year. This is because the Earth takes about 365 days to complete one revolution around the Sun, so the same night sky pattern repeats annually.

Question 6.
Abhay noticed that when it was daytime in India, his uncle who was in the USA was generally sleeping as it was night-time there. What is the reason behind this difference?
Answer:
This difference occurs because the Earth rotates on its axis, and at any given time only one half of the Earth faces the Sun (which experience day), while the other half remains in darkness (which experiences night). When it is daytime in India, that part of the Earth is facing the Sun, whereas the United States lies on the side that is turned away from the Sun, where it is night-time.

Question 7.
Four friends used the following ways to see the solar eclipse. Who among them was being careless?
(a) Ravikiran used a solar eclipse goggle.
(b) Jyothi used a mirror to project the Sun’s image.
(c) Adithya saw the Sun directly with his eyes.
(d) Aruna attended a programme arranged by a planetarium.
Answer:
(c) Adithya was being careless because he looked at the Sun directly with his eyes, which can damage eyesight during a solar eclipse.

Question 8.
Fill in the circles in the given figure appropriately with one of the following: Sun, Moon, Earth.

Answer:

Question 9.
The Moon is much smaller than the Sun, yet it can block the Sun completely from our view during a total solar eclipse. Why is it possible?
Answer:
Although the Moon is much smaller than the Sun, it is much closer to the Earth. Thus, both appear nearly the same size as seen from the Earth. So, when the Moon comes exactly between the Sun and the Earth, it covers the Sun completely and blocks its light from reaching the Earth, causing a total solar eclipse.

Question 10.
The Indian cricket team’s matches in Australia are often held in December. Should they pack winter or summer clothes for their trip?
Answer:
Summer clothes. This is because when it is winter in India (in the Northern Hemisphere), it is summer in Australia (in the Southern Hemisphere).

Question 11.
Why do you think lunar eclipses can be seen from a large part of the Earth when they happen, but total solar eclipse can be seen by only a small part of the Earth?
Answer:
A lunar eclipse can be seen from a large part of the Earth because the Earth’s shadow is much larger than the Moon. When the Earth comes between the Sun and the Moon, its shadow falls on the Moon, and all the places on the night side of the Earth can observe the eclipse.

On the other hand, during a solar eclipse, the Moon comes between the Sun and the Earth, and since the Moon’s shadow is small, it falls only on a limited region of the Earth’s surface. Therefore, a total solar eclipse is visible from only a small part of the Earth.

Question 12.
If the Earth’s axis were not tilted with respect to the axis of revolution, explain what would be the effect on seasons?
Answer:
In that case, there would be no change of seasons on the Earth. Every place would receive the same amount of sunlight throughout the year, and the intensity of sunlight would not vary much between different months. As a result, the duration of day and night would be almost equal everywhere on the Earth, and we would not experience distinct seasons.

Class 7 Curiosity Chapter 12 Question Answer

InText Questions

Question 1.
Why does the Sun appear to rise in the East and set in the West? [Page 170]
Answer:
The Earth rotates from West to East; therefore, to an observer on Earth, the Sun appears to rise in the East and set in the West.

Question 2.
While you are sitting on a moving merry-go-round turning anticlockwise, do the objects around you appear to move? Justify your answer. [Page 170]
Answer:
Yes, the objects around us appear to move. They seem to go around us in the clockwise direction, i.e., opposite to the direction of the anticlockwise rotation of the merry-go-round.

Question 3.
If we stand on the Equator facing north, and watch the sky during one full rotation of the Earth from west to east, what will we observe? [Page 173]
Answer:
We would see the Sun appear in the East on our right in the morning, move across the sky to be overhead around noon, and then move towards the west on our left and set, after which stars become visible.

Question 4.
Since the Earth is rotating, shouldn’t the stars also appear to move in the sky like the Sun? [Page 174]
Answer:
Yes. As the Earth rotates on its axis, the stars also

Students can use Class 7 Math Solution Odia Medium and Class 7 Maths Chapter 2 Question Answer Odia Medium ପାଟୀଗାଣିତିକ ପରିପ୍ରକାଶ to check their answers after solving exercises.

7th Class Maths Chapter 2 Question Answer Odia Medium

Class 7 Maths Chapter 2 Odia Medium

2.1 ସରଳ ପରିପ୍ରକାଶ :

Page No. (24 to 26)

ଉଦାହରଣ – 1 

Question 1.
ମଲ୍ଲିକା ପ୍ରତ୍ୟେକ ଦିନ ଜଳଖୁଆରେ 25 ଟଙ୍କା ଖର୍ଚ୍ଚ କରେ । ସେ ସପ୍ତାହର ସୋମବାରଠାରୁ ଶୁକ୍ରବାର ପର୍ଯ୍ୟନ୍ତ ଜଳଖୁଆ ବାବଦରେ ଖର୍ଚ୍ଚ କରୁଥିବା ମୋଟ ରାଶିକୁ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ପ୍ରକାଶ କର ।
Solution:
ସପ୍ତାହର ସୋମବାରଠାରୁ ଶୁକ୍ରବାର ପର୍ଯ୍ୟନ୍ତ ଦିନ ସଂଖ୍ୟା = 5 ଦିନ ।
ମୋଟ ରାଶିର ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ = 5 × 25
5 × 25 ହେଉଛି 25 ର 5 ଗୁଣ କିମ୍ବା 5 ଓ 25 ର ଗୁଣନ ।

ଅନେକ କ୍ଷେତ୍ରରେ ଭିନ୍ନ ଭିନ୍ନ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶର ମାନ ସମାନ ହୋଇଥାଏ । ଏଠାରେ 12 କୁ ବିଭିନ୍ନ ପ୍ରକାରର ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ପ୍ରକାଶ କରିହେବ । (ଦୁଇଟି ସଂଖ୍ୟା ଓ ଚାରି ଗାଣିତିକ ପ୍ରକ୍ରିୟା ‘+’, ‘–’, ‘×’ ଓ ‘÷’ କୁ ବ୍ୟବହାର କରି ।)
10 + 2, 15 – 3, 3 × 4, 24 ÷ 2

Question 2.
ନିଜ ପସନ୍ଦର ସଂଖ୍ୟା ନେଇ ଏହାକୁ ବିଭିନ୍ନ ପ୍ରକାରର ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ପ୍ରକାଶ କର ।
Solution:
ପରିପ୍ରକାଶର ତୁଳନା : ଯେପରି ସଂଖ୍ୟାଦ୍ଵୟ ମଧ୍ଯରେ ତୁଳନା କରିବା ବେଳେ ଆମେ ‘=’, ‘<‘ ଏବଂ ‘>’ ଚିହ୍ନ ବ୍ୟବହାର କରାଯାଇଥାଏ ।
ଉଦାହରଣ : 10 + 2 > 7 + 1 କାରଣ 10 + 2 = 12 ଯାହାକି 7 + 1 = 8 ଠାରୁ ବୃହତ୍ତର ।
ସେହିପରି 13 – 2 < 4 × 3

ନିଜେ କରି ଦେଖ :

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ଯେପରିକି ପରିପ୍ରକାଶ ଗୁଡ଼ିକରେ ”=” ଚିହ୍ନର ଉଭୟ ପାର୍ଶ୍ଵର ମୂଲ୍ୟ ସମାନ ହେବ ।
(a) 13 + 4 = _______ + 6
(b) 22 + = _______ = 6 × 5
(c) 8 × _______ 64 ÷ 2
(d) 34 – _______ = 25
Solution:
(a) 13 + 4 = _______ + 6
LHS = 13 + 4 = 17
RHS = ? + 6 = 17
∴ ? = 17 – 6 = 11

(b) 22 + = _______ = 6 × 5
RHS = 6 × 5 = 30
LHS = ? + 22 = 30
∴ ? 30 – 22 = 8

(c) 8 × _______ 64 ÷ 2
RHS = 64 ÷ 2 = 32
LHS = 8 × ? = 32
∴ ? = 32 + 8 = 4

(d) 34 – _______ = 25
34 – ? = 25
∴ ? = 34 – 25 = 9

Question 2.
ନିମ୍ନଲିଖୂତ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶଗୁଡ଼ିକର ମାନ ଅନୁଯାୟୀ ଊର୍ଦ୍ଧ୍ବକ୍ରମରେ ସଜାଅ ।
(a) 67 – 19
(b) 67 – 20
(c) 35 + 25
(d) 5 × 11
(e) 120 ÷ 3
Solution:
(a) 67 – 19 = 48
(b) 67 – 20 = 47
(c) 35 + 25 = 60
(d) 5 × 11 = 55
(e) 120 ÷ 3 = 40
ସଂଖ୍ୟାଗୁଡ଼ିକ ଊକ୍ରମରେ ସଜାଇଲେ 40, 47, 48, 55, 60
ଅର୍ଥାତ୍ (e), (b), (a), (d), (c)

ଉଦାହରଣ – 2

Question 1.
ଉଦାହରଣ 2 : 1023 + 125 କିମ୍ବା 1022 + 128 ମଧ୍ୟରେ କେଉଁଟି ବଡ଼ ?

Solution:
1023 + 125 = 1148, 1022 + 128 = 1150
ଏଠାରେ 1150 ବଡ଼ ।
ତେଣୁ 1150 > 1148 ବା 1022 + 128 > 1023 + 125

Question 3.
ରାଜା ପାଖରେ 1023 ଟି ମାର୍ବଲ ଥିଲା ଏବଂ ସେ ଆଜି ଆଉ 125 ଟି ମାର୍ବଲ ପାଇଲା, ବର୍ତ୍ତମାନ ତା’ ପାଖରେ 1023 + 125 ମାର୍ବଲ ହେଲା । ଜୟ ପାଖରେ 1022ଟି ମାର୍ବଲ ଥିଲା ଏବଂ ସେ ଆଜି 128ଟି ମାର୍ବଲ ପାଇଲା, ବର୍ତ୍ତମାନ ତା’ ପାଖରେ 1022 + 128 ମାର୍ବଲ ହେଲା । ତେବେ କାହା ପାଖରେ ଅଧୁକ ମାର୍ବଲ ଅଛି ?

Solution:
ରାଜା ପାଖରେ 1023 ଟି ମାର୍ବଲ ଥିଲା ।
ସେ ଆଜି ଆଉ 125 ଟି ମାର୍ବଲ ପାଇଲା ।
ବର୍ତ୍ତମାନ ତା’ ପାଖରେ ମାର୍ବଲ ସଂଖ୍ୟା = 1023 + 125 = 1148ଟି ।
ଜୟ ପାଖରେ 1022ଟି ମାର୍ବଲ ଥିଲା ।
ସେ ଆଜି 128ଟି ମାର୍ବଲ ପାଇଲା ।
ବର୍ତ୍ତମାନ ତା’ ପାଖରେ ହେଲା = 1022 + 128 = 1150 ମାର୍ବଲ ।
∴ ରାଜା ପାଖରେ ଅଧିକ ମାର୍ବଲ ଅଛି ।
ରାଜା ପାଖରେ 1145ଟି ମାର୍ବଲ ଓ ଜୟ ପାଖରେ 1150ଟି ମାର୍ବଲ ଅଛି ।

ଉଦାହରଣ – 3

Question 1.
ଉଦାହରଣ – 3 : କେଉଁଟି ବୃହତ୍ତର ? 113 – 25 କିମ୍ବା 112 – 24 ?

Solution:
ରାଜା ପାଖରେ 113 ଟି ମାର୍ବଲ ଥିଲା, ସେଥୁରୁ 25ଟି ହଳିଗଲା ।
ତା’ ପାଖରେ 113 – 25 = 88ଟି ମାର୍ବଲ ରହିଲା ।
ଜୟ ପାଖରେ 112ଟି ମାର୍ବଲ ଥିଲା ଓ ଆଜି ସେଥୁରୁ 24 ମାର୍ବଲ ହଜିଗଲା ।
ତା’ ପାଖରେ ଆଉ 112 – 24 = 88ଟି ମାର୍ବଲ ରହିଲା ।
ବର୍ତ୍ତମାନ ସେ ଦୁହିଁଙ୍କ ପାଖରେ ସମାନ ସଂଖ୍ୟକ ମାର୍ବଲ ରହିଲା ।
ତେଣୁ 113 – 25 = 112 – 24

Question 2.
ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ‘>’ କିମ୍ବା ‘<‘ କିମ୍ବା ‘=’ ଚିହ୍ନ ବ୍ୟବହାର କରି ତୁଳନା ବିନା ତୁମେ ଏହା କରିପାରିବ କି ? ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ତୁମର ଚିନ୍ତାଧାରା ବ୍ୟାଖ୍ୟା ଉତ୍ତର)
କର । ଜଟିଳ ଗଣନା କର । (ପ୍ରଶ୍ନ ସହିତ
(a) 245 + 289 246 + 285
(b) 273 + 145 272 – 144
(c) 364 + 587 363 + 589
(d) 124 + 245 129 + 245
(e) 213 – 77 214 – 76
Solution:
(a) 245 + 289 246 + 285
ଉଭୟ ପାର୍ଶ୍ଵରେ 1 ଯୋଗକଲେ,
245 + 1 + 289 _____ 246 + 285 + 1
246 + 289 _____ 246 + 286
ବର୍ତ୍ତମାନ 289 > 286 (∵ 246 ସାଧାରଣ)
ତେଣୁ 245 + 289 246 + 285

(b) 273 – 145 272 – 144
ଉଭୟ ପଟେ 1 ଯୋଗକଲେ,
273 – 145 + 1 _____ 272 – 144 + 1
273 – 144 _____ 273 – 144
ବର୍ତ୍ତମାନ 273 – 144 = 273 – 144
ତେଣୁ 273 – 145 272 – 144

(c) 364 + 587 363 + 589
ଉଭୟ ପଟେ 1 ଯୋଗକଲେ,
364 + 587 + 1 _____ 363 + 589 + 1
364 + 588 _____ 364 + 589
ବର୍ତ୍ତମାନ 588 < 589
ତେଣୁ 364 + 587 363 + 589

(d) 124 + 245 129 + 245
ବର୍ତ୍ତମାନ 124 < 129
ତେଣୁ 124 + 245 129 + 245

(e) 213 – 77 214 – 76
ଉଭୟପଟେ 1 ଯୋଗକଲେ
213 – 77 + 1 _____ 214 – 76 + 1
214 – 77 _____ 214 – 75
ବର୍ତ୍ତମାନ – 77 < -75)
ତେଣୁ 213 – 77 214 – 76

2.2 ଜଟିଳ ପରିପ୍ରକାଶର ପଠନ ଓ ମୂଲ୍ୟାୟନ

Page No. (26 to 35)

ଉଦାହରଣ – 4

Question 1.
ମହେଶ ଖେଳ ପଡ଼ିଆକୁ 30ଟି ମାର୍ବଲ ଆଣିଥିଲା । ଅରୁଣ ପ୍ରତ୍ୟେକ ବ୍ୟାଗରେ 4ଟି କରି 5ଟି ବ୍ୟାଗ୍ ମାର୍ବଲ ଆଣିଥିଲା । ତେବେ ମହେଶ ଓ ଅରୁଣ କେତୋଟି ମାର୍ବଲ ଖେଳ ପଡ଼ିଆକୁ ଆଣିଥିଲେ ।
Solution:
ମହେଶ ମାର୍ବଲ ଆଣିଥିଲା = 30ଟି ।
ଅରୁଣ ପ୍ରତ୍ୟେକ ବ୍ୟାଗରେ 4ଟି ଲେଖାଏଁ 5ଟି ବ୍ୟାଗରେ ମାର୍ବଲ ଆଣିଥିଲା ।
ମହେଶ ଏହାକୁ ଗାଣିତିକ ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ଲେଖୁଲା 30 + (5 × 4)
ପରିପ୍ରକାଶରେ ବନ୍ଧନୀର ପ୍ରୟୋଗ : 30 + (5 × 4) ପରିପ୍ରକାଶଟିରେ ମାର୍ବଲ ସଂଖ୍ୟା ପାଇବା ପାଇଁ ଆମେ ପ୍ରଥମେ 5 ଓ 4ର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କରି 30) ସହ ଯୋଗ କରିବା । ଏହି କ୍ରମକୁ ନିମ୍ନଲିଖିତ ଭାବେ ବନ୍ଧନୀର ବ୍ୟବହାରଦ୍ଵାରା ସ୍ପଷ୍ଟ କରାଯାଇପାରିବ ।
30 + (5 × 4) 30 + 20 = 50

ଉଦାହରଣ – 5

Question 1.
ଇରଫାନ୍ 15 ଟଙ୍କା ଦେଇ ଗୋଟିଏ ବିସ୍କୁଟ୍ ପ୍ୟାକେଟ୍ ଓ 56 ଟଙ୍କା ଦେଇ ଗୋଟିଏ ବୁଟ ଡାଲି ପ୍ୟାକେଟ୍ କିଣିଲା, ସେ ଦୋକାନୀକୁ 100 ଟଙ୍କା ଦେଲା । ଦେକାନୀଠାରୁ ଇର୍‌ଫାନ୍ ଫେରି ପାଇଥିବା ଟଙ୍କାର ପରିମାଣ ଜାଣିବା ପାଇଁ ଆବଶ୍ୟକୀୟ ପରିପ୍ରକାଶଟି ଲେଖ ।
Solution:
ଇର୍‌ଫାନ୍ ବିସ୍କୁଟ ପ୍ୟାକେଟ ପାଇଁ 15 ଟଙ୍କା ଓ ବୁଟ ଡାଲି ପ୍ୟାକେଟ୍ ପାଇଁ 56 ଟଙ୍କା ଖର୍ଚ୍ଚ କଲା ।
ମୋଟ ଖର୍ଚ୍ଚ ଟଙ୍କା = 15 + 56
ସେ ଦୋକାନୀକୁ 100 ଟଙ୍କା ଦେଲା ।
ତେଣୁ ସେ 100 ଟଙ୍କାରୁ ମୋଟ୍ ଦାମ୍ ବିୟୋଗ କଲେ ବଳକା ଟଙ୍କା ଫେରି ପାଇବ ।
ଏହାକୁ ନିମ୍ନମତେ ପ୍ରକାଶ କରାଯାଇପାରିବ 100 – (15 + 56) = 100 – 71 = 29

ପଦଗୁଡ଼ିକ ଏକ ପରିପ୍ରକାଶର ଅଂଶ ବିଶେଷ ହେଇଥାଏ ଯାହାକି ‘+’ ଚିହ୍ନଦ୍ଵାରା ଅଲଗା ହୋଇଥାଏ । ଉଦାହରଣ ସ୍ଵରୂପ, 12 + 7 ରେ ପଦଦ୍ଵୟ 12 ଓ 7 ଅଟେ । ଯାହାକି ନିମ୍ନରେ ଦର୍ଶାଯାଇଛି ।
12 + 7 =

Question 3.
ବର୍ତ୍ତମାନ 83 – 14 ରେ ପଦଦ୍ଵୟ କ’ଣ ?
Solution:
ଆମେ ଜାଣୁ ଯେ, ଏକ ସଂଖ୍ୟାକୁ ବିୟୋଗ କରିବା, ସଂଖ୍ୟାର ବିଲୋମୀ ସଂଖ୍ୟା ଯୋଡ଼ିବା ସହିତ ସମାନ । 83 ରୁ 14 କୁ ବିୟୋଗ କରିବା, ବା – 14 କୁ 83 ସହ ଯୋଗ କରିବା ସହିତ ସମାନ ।
ଅର୍ଥାତ୍ 83 – 14 =
ତେଣୁ 83 – 14 ପରିପ୍ରକାଶର ପଦଗୁଡ଼ିକ ହେଉଛି 83 ଏବଂ -14 ।

ଗୋଟିଏ ସଂଖ୍ୟାର ବିଲୋମୀ ସଂଖ୍ୟା ଏହାର ବିପରୀତ ଚିହ୍ନ ସହିତ ସମାନ ।
ଉଦାହରଣ ସ୍ଵରୂପ : 14 ର ବିଲୋମୀ ସଂଖ୍ୟା -14 ଏବଂ −14 ର ବିଲୋମୀ ସଂଖ୍ୟା 14 ।

Question 4.
ବିୟୋଗ ଚିହ୍ନକୁ ଯୋଗ ଚିହ୍ନରେ ବଦଳାଇବାଦ୍ଵାରା ଗାଣିତିକ ପରିପ୍ରକାଶର ମୂଲ୍ୟରେ କୌଣସି ପରିବର୍ତ୍ତନ ହୁଏ ନାହିଁ ।
Solution:
ଛାତ୍ରଛାତ୍ରୀମାନେ ନିଜେ ଅଭ୍ୟାସ କରନ୍ତୁ ।

Question 5.
ତୁମେ ବୁଝାଇ କହିପାରିବ କି ? ଏକ ସଂଖ୍ୟାକୁ ବିୟୋଗ କରିବା ଅର୍ଥ ଏହାର ବିଲୋମୀ ସହ ଯୋଗ କରିବା ସହିତ ସମାନ ହେବାର କାରଣ କ’ଣ ?
Solution:
ଏକ ପରିପ୍ରକାଶର ସମସ୍ତ ପଦଗୁଡ଼ିକୁ ଚିହ୍ନଟ କରିବା ପାଇଁ ସମସ୍ତ ବିୟୋଗ ପ୍ରକ୍ରିୟାକୁ ଯୋଗ ପ୍ରକ୍ରିୟା ମାଧ୍ଯମରେ ପରିପ୍ରକାଶ କରିବାକୁ ହୁଏ । ଏଠାରେ କେତେକ ପରିପ୍ରକାଶ ଓ ସେମାନଙ୍କର ପଦଗୁଡ଼ିକର ଉଦାହରଣ ଦିଆଗଲା ।

Question 6.
6 × 5, 4 × 6 ପଦଗୁଡ଼ିକ ଏକକ ପଦ ଅଟନ୍ତି କାରଣ ସେହି ସଂଖ୍ୟାଦ୍ଵୟ ମଧ୍ଯରେ କୌଣସି ‘+’ ଚିହ୍ନ ନାହିଁ । ନିମ୍ନ ସାରଣୀରେ କେତେକ ପରିପ୍ରକାଶ ପ୍ରଦତ୍ତ ହୋଇଛି, ସାରଣୀଟିକୁ ସଂପୂର୍ଣ୍ଣ କର । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

ଉଦାହରଣ – 6

Question 1.
ମଧୁ ଏକ ପଡ଼ିଆରେ ଡ୍ରୋନ୍ ଉଡ଼ାଉଛି । ଡ୍ରୋନ୍‌ 6 ମିଟର ଉପରକୁ ଯାଇ 4 ମିଟର ତଳକୁ ଆସିଲା । ପଡ଼ିଆରେ ଡ୍ରୋନ୍‌ ଅନ୍ତିମ ଅବସ୍ଥାନ କେତେ ଉଚ୍ଚରେ ଦର୍ଶାଇବା ପାଇଁ ଏକ ପରିପ୍ରକାଶ ଲେଖ ।
Solution:
ଡ୍ରୋନ୍‌ ପଡ଼ିଆଠାରୁ 6 – 4 = 2 ମିଟର ଉଚ୍ଚରେ ଅବସ୍ଥିତ । ଏହାକୁ ପଦର ସମଷ୍ଟି ଭାବରେ ପ୍ରକାଶ କଲେ-

ଯଦି ଆମେ ପଦଗୁଡ଼ିକର କ୍ରମର ପରିବର୍ତ୍ତନ କରିବା ତେବେ ମୂଲ୍ୟର ପରିବର୍ତ୍ତନ ହେବ ନାହିଁ ।
ଯଥା –
ଦୁଇଟି ଧନାତ୍ମକ ପଦର କ୍ରମର ପରିବର୍ତ୍ତନ କଲେ ଯୋଗଫଳର କୌଣସି ପରିବର୍ତ୍ତନ ହୁଏ ନାହିଁ ।

Question 2.
ଏହି ଧାରଣା ରଣାତ୍ମକ ସଂଖ୍ୟା ଥିବା ପଦଗୁଡ଼ିକ ପାଇଁ ପ୍ରଯୁଜ୍ୟ କି ? ଆଉ କିଛି ପରିପ୍ରକାଶ ନିଅ ଏବଂ ପରୀକ୍ଷା କର ।
Solution:
ଏହି ଧାରଣା ଋଣାତ୍ମକ ସଂଖ୍ୟାଥ‌ିବା ପଦଗୁଡ଼ିକ ପାଇଁ ପ୍ରଯୁଜ୍ୟ ନୁହେଁ ।
କାରଣ,
ପଦ ପରିବର୍ତ୍ତନ କଲେ,
• ଦୁଇଟି ପଦଥ‌ିବା ପରିପ୍ରକାଶରେ ପଦଗୁଡ଼ିକୁ ଅଦଳବଦଳ (କ୍ରମ ପରିବର୍ତ୍ତନ) କରି ଲେଖୁଲେ ଏହାର ମୂଲ୍ୟର କୌଣସି ପରିବର୍ତ୍ତନ ଘଟେ ନାହିଁ ।

• ତିନୋଟି ପଦ ଥ‌ିବା ଏକ ପରିପ୍ରକାଶ : (-7) + 10 + (-11) ଏହି ସଂଖ୍ୟାଗୁଡ଼ିକୁ ନିମ୍ନରେ ଦର୍ଶାଯାଇଥ‌ିବା ଦୁଇଟି ଭିନ୍ନ ଭିନ୍ନ କ୍ରମରେ ଯୋଗ କରିବା । ଯଥା –

(ପ୍ରଥମ ଦୁଇଟି ପଦକୁ ଯୋଗ କରିବା ଏବଂ ଏହି ଯୋଗଫଳକୁ ତୃତୀୟ ପଦ ସହ ଯୋଗ କରିବା)

(ଶେଷ ଦୁଇଟି ପଦକୁ ଯୋଗ କରି ଯୋଗଫଳକୁ ପ୍ରଥମ ପଦ ସହ ଯୋଗ କରିବା)

Question 3.
ଆମେ ଜାଣିଲେ ଯେ, ଧନାତ୍ମକ ସଂଖ୍ୟାଗୁଡ଼ିକୁ ଯୋଗ କରିବାବେଳେ ସେମାନଙ୍କୁ ଏକତ୍ରୀକରଣ କରି ଉପରୋକ୍ତ ଯେକୌଣସି ପଦ୍ଧତିରେ ଯୋଗ କଲେ ଯୋଗଫଳ ସମାନ ହୁଏ ।
ଏହି ଧାରଣା ଋଣାତ୍ମକ ସଂଖ୍ୟାଥବା ପଦଗୁଡ଼ିକ ପାଇଁ ପ୍ରଯୁଜ୍ୟ କି ? ଆଉ କିଛି ପରିପ୍ରକାଶ ନିଅ ଏବଂ ପରୀକ୍ଷା କର ।
Solution:
ଏହି ଧାରଣା ଋଣାତ୍ମକ ପଦଗୁଡ଼ିକ ପାଇଁ ପ୍ରଯୁଜ୍ୟ ନୁହେଁ ।

ଏକ ପରିପ୍ରକାଶରେ ଥିବା ପଦଗୁଡ଼ିକୁ ଏକତ୍ରୀକରଣ କରି ନିମ୍ନଲିଖୁ ଯେକୌଣସି ପଦ୍ଧତିରେ ସମାଧାନ କଲେ, ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ମାନ ଅପରିବର୍ତିତ ରହିବ ।
୧ମ ପଦ + ୨ୟ ପଦ + ୩ୟ ପଦ = ୩ୟ ପଦ + ୨ୟ ପଦ + ୧ ମ ପଦ

Question 4.
(-7) + 10 + (-11) ପରିପ୍ରକାଶକୁ ପରିବର୍ତ୍ତନ କରି -7 ଏବଂ -11 କୁ ପ୍ରଥମେ ଯୋଗ କରିବା ଏବଂ ତା’ପରେ ଯୋଗଫଳକୁ 10 ସହ ଯୋଗ କରିବା ଉତ୍ତର କ’ଣ ହେବ ?
Solution:
(-7) + 10 + (-11)
= (-7) + (-11) + 10
=(-18) + 10 = (-8)
ଆମେ ଦେଖୁଲୁ ଯେ ଯେକୌଣସି କ୍ରମରେ (-7) + 10 + (-11) ପରିପ୍ରକାଶର ପଦଗୁଡ଼ିକ ଯୋଗ କଲେ ଯୋଗପଳ (-8) ହେବ ।

Question 5.
ଯେକୌଣସି କ୍ରମରେ ଏକ ପରିପ୍ରକାଶର ପଦଗୁଡ଼ିକୁ ଯୋଗ କଲେ ସମାନ ମୂଲ୍ୟ ମିଳିଥାଏ କି ? ଆଉ କିଛି ପରିପ୍ରକାଶ ନିଅ ଏବଂ ପରୀକ୍ଷା କର । 3 ରୁ ଅଧିକ ପଦ ବିଶିଷ୍ଟ ପରିପ୍ରକାଶ ନେଇ ପରୀକ୍ଷା କର ।
Solution:
ଯେକୌଣସି କ୍ରମରେ ଏକ ପରିପ୍ରକାଶର ପଦଗୁଡ଼ିକୁ ଯୋଗକଲେ ସମାନ ମୂଲ୍ୟ ମିଳିଥାଏ ।

Question 6.
ସଂଖ୍ୟାକାର୍ଡ଼ର ବ୍ୟବହାର କରି ଏପରି କାହିଁକି ଘଟୁଛି ତୁମେ ବୁଝାଇ ପାରିବ କି ?
Solution:
ଛାତ୍ରଛାତ୍ରୀମାନେ ନିଜେ ଅଭ୍ୟାସ କରନ୍ତୁ ।

• ଯେକୌଣସି କ୍ରମରେ ପଦମାନଙ୍କର ଯୋଗ ସମାନ ମୂଲ୍ୟ ଦେଇଥାଏ । ତେଣୁ କେବଳ ଯୋଗଥିବା ଏକ ପରିପ୍ରକାଶରେ, ପଦଗୁଡ଼ିକ ଯେଉଁ କ୍ରମରେ ଯୋଗ କଲେ ମଧ୍ୟ ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ମୂଲ୍ୟ ସମାନ ହୋଇଥାଏ ।

• ବିନା ବନ୍ଧନୀରେ ନଥୁବା ଏକ ପରିପ୍ରକାଶ ଯେଉଁଥରେ ଗୁଣନ ଓ ଭାଗକ୍ରିୟା ଧ୍ରୁବ । ପ୍ରଥମେ ପଦଗୁଡ଼ିକର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କରି ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ । ପ୍ରତ୍ୟେକ ପଦର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କରାଯିବା ପରେ ସେଗୁଡ଼ିକୁ ଯୋଗ କରାଯାଏ ।

Question 7.
ଉଦାହରଣ ସ୍ଵରୂପ : 30 + 5 × 4 ପରିପ୍ରକାଶଟିକୁ ନିମ୍ନମତେ ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
Solution:
• 30 + 5 × 4 =
• 5 × (3 + 2) + 7 × 8+ 3 ପରିପ୍ରକାଶକୁ ନିମ୍ନମତେ ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
5 × (3 + 2) + 7 × 8 + 3 = = 25 + 56 + 3 = 84
ଯେଉଁଠାରେ ପ୍ରଥମେ (3 + 2) ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କରି ଯୋଗଫଳକୁ 5 ଦ୍ଵାରା ଗୁଣନ କରାଯାଏ
(=25), 7 × 8 ପରିପ୍ରକାଶର ଗୁଣଫଳ (-56)
25 + 56 + 3 = 84 କୁ ଏହିପରି ସରଳୀକରଣ କରାଯାଏ ।
• ପଦମାନଙ୍କର କ୍ରମର ପରିବର୍ତ୍ତନ (ଅଦଳବଦଳ) କରି ଯୋଗ କଲେ ଯୋଗଫଳ ଅପରିବର୍ତିତ ରହେ । ଏହାକୁ ଯୋଗ ପ୍ରକ୍ରିୟାର କ୍ରମବିନିମୟୀ ନିୟମ କୁହାଯାଏ । ସେହିପରି ପଦମାନଙ୍କର ଏକତ୍ରୀକରଣ କରି ଯୋଗ | କଲେ ଯୋଗଫଳ ଅପରିବର୍ତିତ ରହେ । ଏହାକୁ ଯୋଗ ପ୍ରକ୍ରିୟାର ସହଯୋଗୀ ନିୟମ କୁହାଯାଏ ।

ଅଧୁକ ଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ସେମାନଙ୍କର ପଦ :
ଉଦାହରଣ – 7

Question 1.
ଅମୁ, ଚରଣ, ମଧୁ ଏବଂ ସାଧୁ ଏକ ହୋଟେଲକୁ ଯାଇ ଚାରୋଟି ଦୋସା ବରାଦ କଲେ । ପ୍ରତ୍ୟେକ ଦୋସାର ମୂଲ୍ୟ 23 ଟଙ୍କା, ସେମାନେ ଖାଦ୍ୟ ପରଷୁଥିବା ବ୍ୟକ୍ତିକୁ 5 ଟଙ୍କା ବକ୍‌ସିସ୍ ଦେଲେ । ସମୁଦାୟ ଖର୍ଚ୍ଚକୁ ଏକ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ଲେଖ ।
Solution:
1 ଟି ଦୋସାର ମୂଲ୍ୟ = 23 ଟଙ୍କା ।
4 ଟି ଦୋସାର ମୂଲ୍ୟ = 4 × 23 ଟଙ୍କା ।
ଖାଦ୍ୟ ପରଷୁଥ‌ିବା ବ୍ୟକ୍ତି ପାଇଥିବା ଟଙ୍କା ସହ ମୋଟ ଟଙ୍କା = 4 × 23 + 5 =

ଉଦାହରଣ – 9

Question 1.
ରଘୁ ହୋଲ୍‌ସେଲ୍‌ ବଜାରରୁ 100 କି.ଗ୍ରା. ଚାଉଳ କିଣି ଏହାକୁ 2 କି.ଗ୍ରା. ଓଜନ ବିଶିଷ୍ଟ ପ୍ୟାକେଟ୍‌ରେ ଭରିଲା । ପୂର୍ବରୁ ତା’ ପାଖରେ 4 ଟି 2 କି.ଗ୍ରା. ଓଜନ ବିଶିଷ୍ଟ ପ୍ୟାକେଟ୍ ଥିଲା । ବର୍ତ୍ତମାନ ତା’ ପାଖରେ କେତୋଟି 2 କି.ଗ୍ରା. ଓଜନ ବିଶିଷ୍ଟ ପ୍ୟାକେଟ୍ ରହିଲା ? ଏହାକୁ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ମାଧ୍ଯମରେ ଲେଖୁ ପଦଗୁଡ଼ିକୁ ଚିହ୍ନାଅ ।
Solution:
ରଘୁ ପାଖରେ = 4ଟି ପ୍ୟାକେଟ୍ ଥିଲା ।
ନୂଆ ଭାବରେ 2 କି.ଗ୍ରା. ବିଶିଷ୍ଟ ପ୍ୟାକେଟ୍ ହେବ = 100 ÷ 2 ବା \(\frac{100}{2}\)
ବର୍ତ୍ତମାନ ତା’ ପାଖରେ 2 କି.ଗ୍ରା. ଓଜନ ପ୍ୟାକେଟ୍ ସଂଖ୍ୟା 4 + \(\frac{100}{2}\),
ଏହାର ପଦଗୁଡ଼ିକ 4 ଓ \(\frac{100}{2}\) ∴

ଉଦାହରଣ – 10

Question 1.
କାନନ ଜଣେ ଦୋକାନୀକୁ 1 ଟଙ୍କିଆ ଓ 5 ଟଙ୍କିଆ ମୁଦ୍ରା ଏବଂ 10 ଟଙ୍କିଆ, 20 ଟଙ୍କିଆ, 50 ଟଙ୍କିଆ ଓ 100 ଟଙ୍କିଆ ନୋଟ୍‌କୁ ବ୍ୟବହାର କରି 432 ଟଙ୍କା ହେବ । ଏହାକୁ ସେ କିପରି ଦେବ ?
Solution:
ଏଠାରେ ଏକାଧିକ ସମ୍ଭାବନା ଅଛି । ଉଦାହରଣ ସ୍ଵରୂପ, 432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 ଅର୍ଥାତ୍‌,4ଟି 100 ଟଙ୍କିଆ ନୋଟ୍, 1ଟି 20 ଟଙ୍କିଆ ନୋଟ୍, 1ଟି 10 ଟଙ୍କିଆ ନୋଟ୍ ଏବଂ 2ଟି 1ଟଙ୍କିଆ ମୁଦ୍ରା ।
432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1
ଅର୍ଥାତ୍ 8 ଟି 50 ଟଙ୍କିଆ ନୋଟ୍, 1 ଟି 10 ଟଙ୍କିଆ ନୋଟ୍, 4 ଟି 5 ଟଙ୍କିଆ ମୁଦ୍ରା ଏବଂ 2 ଟି 1 ଟଙ୍କିଆ ମୁଦ୍ରା ।

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକର ପ୍ରତ୍ୟେକ ପଦଗୁଡ଼ିକ ଲେଖୁ ମାନ ନିର୍ଣ୍ଣୟ କର ।
(a) 28 – 7 + 8
(b) 39 – 2 × 6 + 11
(c) 40 – 10 + 10 + 10
(d) 48 – 10 × 2 + 16 ÷ 2
(e) 6 × 3 – 4 × 8 × 5
Solution:
(a) 28 – 7 + 8 (28 – 7) + 8 = 21 + 8 = 29
(b) 39 – 2 × 6 + 11 = 39 – 12 + 11 = (39 – 12) + 11 = 27 + 11 = 38
(c) 40 – 10 + 10 + 10 = (40 – 10) + (10 + 10) 30 + 20= 50
(d) 48 – 10 × 2 + 16 ÷ 2 = 48 (10 × 2) + (16 ÷ 2)= 48 – 20 + 8 = 28 + 8 = 36
(e) 6 × 3 – 4 × 8 × 5 = (6 × 3) – (4 × 8) × 5 = 18 – 32 × 5 = 18 – 160 = -142

Question 2.
ନିମ୍ନଲିଖୂତ ପ୍ରତ୍ୟେକ ପରିପ୍ରକାଶ ପାଇଁ ଏକ କାହାଣୀ / ପରିସ୍ଥିତି ବର୍ଣ୍ଣନା କର ଏବଂ ମାନ ନିରୂପଣ କର ।
(a) 89 + 21 – 10
(b) 5 × 12 – 6
(c) 4 × 9 + 2 × 6
Solution:
(a) ଗୋଟିଏ ଜାର୍‌ରେ ୫୨ ଟି ବିସ୍କୁଟ ଥିଲା । ସେଥ‌ିରେ ଆଉ 21 ଟି ବିସ୍କୁଟ୍ ରଖାଗଲା । ଏବେ ଜାର୍‌ରୁ 10 ଟି ବିସ୍କୁଟ୍ ବାହାର କରାଯାଇଥିଲା । ଆଉ କେତୋଟି ବିସ୍କୁଟ ରହିଲା ।
= (89 + 21) – 10 = 110 – 10 = 100 ଟି ବିସ୍ଫୁଟ ରହିଲା ।

(b) ଗୋଟିଏ ସ୍କୁଲରେ 5 ଟି ଧାଡ଼ିରେ 12 ଜଣ ଲେଖାଏଁ ଛାତ୍ର ଠିଆ ହୋଇଥିଲେ । ସେଥୁରୁ 6 ଜଣ ଛାତ୍ର ଖେଳକୁଦ ପାଇଁ ଚାଲିଗଲେ । ଆଉ କେତେଜଣ ଛାତ୍ର ଧାଡ଼ିରେ ଠିଆ ହୋଇଥିଲେ ?
= (5 × 12) – 6 = 60 – 6 = 54 ଜଣ ଛାତ୍ର ଧାଡ଼ିରେ ଠିଆ ହୋଇଥିଲେ ।

(c) 4 ଟି ବାସ୍କେଟ୍‌ରେ 9 ଟି ଲେଖାଏଁ ସେଓ ଓ 2 ଟି ବାସ୍କେଟ୍‌ରେ 6 ଟି ଲେଖାଏଁ ସେଓ ଥିଲେ ସମୁଦାୟ ବାସ୍କେଟ୍‌ରେ କେତୋଟି ସେଓ ଅଛି ?
= (4 × 9) + (2 × 6) = 36 + 12 = 48 ଟି ସେଓ ବାସ୍କେଟ୍‌ରେ ଅଛି ।

Question 3.
ନିମ୍ନଲିଖୂତ ପ୍ରତ୍ୟେକ ପରିସ୍ଥିତି ପାଇଁ, ପରିସ୍ଥିତିକୁ ବର୍ଣ୍ଣନା କରୁଥିବା ସାଂଖ୍ୟକ ପରିପ୍ରକାଶଗୁଡ଼ିକ ଲେଖ । ଏହାର ପଦଗୁଡ଼ିକୁ ଚିହ୍ନଟ କର ଏବଂ ମାନ ନିର୍ଣ୍ଣୟ କର ।
(a) ଜଣେ ରାଜା ତାଙ୍କର ଦୁଇ ରାଜକୁମାର କୃଷ୍ଣଚନ୍ଦ୍ର ଓ ରାମଚନ୍ଦ୍ରଙ୍କୁ 100 ଟି ଲେଖାଏଁ ସ୍ୱର୍ଣ୍ଣମୁଦ୍ରା ପ୍ରଦାନ କଲେ । କୃଷ୍ଣଚନ୍ଦ୍ର ଏକ ବ୍ୟବସାୟ ଆରମ୍ଭ କରି ତାଙ୍କ ମୁଦ୍ରାକୁ ଦ୍ୱିଗୁଣିତ କଲେ । ରାମଚନ୍ଦ୍ର କିଛି ଅଳଙ୍କାର କିଣିଛନ୍ତି ଏବଂ ଅଧା ମୁଦ୍ରା ତାଙ୍କ ପାଖରେ ଅଛି । ଏବେ ରାଜକୁମାର କୃଷ୍ଣଚନ୍ଦ୍ର ଏବଂ ରାମଚନ୍ଦ୍ରଙ୍କ ପାଖରେ କେତେ ସ୍ଵର୍ଣ୍ଣମୁଦ୍ରା ଅଛି ତାହାକୁ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ଲେଖ ।
Solution:
ସାଂଖ୍ୟକ ପରିପ୍ରକାଶରେ ପ୍ରକାଶ କଲେ, (100 × 2) + 50 = 200 + 50 = 250 ଟି ସ୍ଵର୍ଣ୍ଣମୁଦ୍ରା ।

(b) ଦୁଇଟି ରେଳ ଷ୍ଟେସନ୍ ମଧ୍ଯରେ ଗୋଟିଏ ମେଟ୍ରୋ ଟ୍ରେନ୍‌ର ଟିକେଟ୍‌ର ମୂଲ୍ୟ ବୟସ୍କଙ୍କ ପାଇଁ 40 ଟଙ୍କା ଏବଂ ଶିଶୁମାନଙ୍କ ପାଇଁ 20 ଟଙ୍କା । ସମୁଦାୟ ଟିକେଟର ମୂଲ୍ୟ କେତେ ହେବ ?
(i) 4 ଜଣ ବୟସ୍କ ଓ 3 ଜଣ ଶିଶୁଙ୍କ ପାଇଁ
(ii) 3 ଜଣ ଲେଖାଏଁ ବୟସ୍କ ଥବା 2 ଟି ଦଳ ପାଇଁ
Solution:
(i) 4 ଜଣ ଓ 3 ଜଣ ଶିଶୁଙ୍କ ପାଇଁ ଟିକେଟ୍‌ର ମୂଲ୍ୟ
= (4 × 40) + (3 × 20) = 160 + 60 = 220 ଟଙ୍କା ।

(ii) 3 ଜଣ ଲେଖାଏଁ ବୟସ୍କ ଥ‌ିବା 2 ଟି ଦଳ ପାଇଁ ଟିକେଟ୍‌ର ମୂଲ୍ୟ
= (3 × 40) + (3 × 40) = 120 + 120 = 240 ଟଙ୍କା ।

(c) ପାର୍ଶ୍ୱସ୍ଥ ଚିତ୍ରରେ ଦିଆଯାଇଥିବା ପରିମାଣକୁ ବ୍ୟବହାର କରି ଝରକାର ଉଚ୍ଚତା ନିର୍ଣ୍ଣୟ ପାଇଁ ଆବଶ୍ୟକ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ଲେଖ ଓ ମାନ ନିର୍ଣ୍ଣୟ କର ।

ଝରକାର ଉଚ୍ଚତା
= 2 ବନ୍ଧର ଓସାର + 6 ଗ୍ରୀଲ୍‌ର ଉଚ୍ଚତା + 7 ଟି ଫାଙ୍କର ଓସାର
= (2 × 3 + 6 × 2 + 7 × 5) ସେ.ମି.
= 6 + 12 + 35 = 53 ସେ.ମି. ।

2.3 ବନ୍ଧନୀ ଅପସାରଣ – 1 :

Page No. (35 to 43)

ଉଦାହରଣ – 12

Question 1.
ଆମେ ପୂର୍ବରୁ ଦେଖୁଥିଲେ ଯେ, ଇରଫାନ୍ ଏକ ପ୍ୟାକେଟ୍ ବିସ୍କୁଟ ଟ. 15 ଏବଂ ବୁଟ ଡାଲି ପ୍ୟାକେଟ୍ (ଟ. 56)ରେ କିଣି ଟ. 100 ଦୋକାନୀକୁ ଦେଇଥିଲେ ସେ ଫେରିପାଇଥବା ଟଙ୍କା 100 – (15 + 56) = 29 ଟଙ୍କା ।
Solution:
ଏହାକୁ ନିମ୍ନମତେ ନିର୍ଣ୍ଣୟ କରାଯାଇପାରିବ ।

(a) ପ୍ରଥମେ ବିସ୍କୁଟ୍ ପ୍ୟାକେଟ୍‌ର ଦାମ୍ 15 ଟଙ୍କାକୁ 100 ଟଙ୍କାରୁ ବିୟୋଗ କରିବା = 100 – 15 = 85 ଟଙ୍କା ।
ଯଦି ସେ କେବଳ ବିସ୍କୁଟ କିଣିବ, ତେବେ ଏହି ପରିମାଣର ଟଙ୍କା ଦୋକାନୀ ଇରଫାନ୍‌କୁ ଫେରସ୍ତ କରିବେ, ମାତ୍ର ଇରଫାନ୍ ବୁଟ ଡାଲି କିଣିଥିବାରୁ ଏହାର ମୂଲ୍ୟ 56 କୁ 85 ଟଙ୍କାରୁ ବାଦ ଦେବେ ।

(b) ତେଣୁ ଫେରସ୍ତ ଟଙ୍କା ପାଇଁ 85 ଟଙ୍କାରୁ ଡାଲିର ମୂଲ୍ୟ ବିୟୋଗ କରିବା = 85 – 56 = 29 ଟଙ୍କା ।
ଆମେ ଏଠାରେ କ’ଣ କଲେ, 100 – 15 – 56 = 29 ଟଙ୍କା ।
ତେଣୁ 100 – (15 + 56) = 100 – 15 – 56 = 29 ଟଙ୍କା ।
> ‘-‘ ଚିହ୍ନ ଥିଲେ ବନ୍ଧନୀ ଅପସାରଣ କରିବା ପରେ ବନ୍ଧନୀ ମଧ୍ୟରେ ଥିବା ପଦଗୁଡ଼ିକର ଚିହ୍ନର ପରିବର୍ତ୍ତନ ହେବ, ପ୍ରଥମ ଉଦାହରଣରେ 40 ଓ 3 ର ଚିହ୍ନ ଏବଂ ଦ୍ଵିତୀୟ ଉଦାହରଣରେ 15 ଓ 56 ଚିହ୍ନର ପରିବର୍ତ୍ତନ ହେଉଛି ।

ଉଦାହରଣ – 13 

Question 1.
500 – (250 – 100) ପରିପ୍ରକାଶଟିକୁ ବିଚାରକୁ ନିଆଯାଉ । ଏହି ପରିପ୍ରକାଶକୁ ବନ୍ଧନୀ ଅପସାରଣ କରି ଲେଖୁହେବ କି ?
Solution:
ଏହାକୁ ସରଳୀକରଣ କରିବା ପାଇଁ ପ୍ରଥମେ 250 – 100 = 150 ନିର୍ଣ୍ଣୟ କରି 500 ରୁ ବିୟୋଗ କରିବା ।
500 – (250 – 100) = 500 – 150 = 350

ଉଦାହରଣ – 14

Question 1.
ହୀରାଙ୍କର ମୁଦ୍ରା ସଂଗ୍ରହରେ ଆଗ୍ରହ ଅଛି, ସେ ଗୋଟିଏ ବ୍ୟାଗରେ 28ଟି ମୁଦ୍ରା ଓ ଅନ୍ୟ ଏକ ବ୍ୟାଗରେ 35 ଟି ମୁଦ୍ରା ସଂଗ୍ରହ କଲେ । ସେ ତାଙ୍କ ସାଙ୍ଗକୁ ଦ୍ୱିତୀୟ ବ୍ୟାଗରୁ 10ଟି ମୁଦ୍ରା ଉପହାର ଦେଲେ । ତେବେ ହୀରାଙ୍କ ପାଖରେ ଥିବା ମୁଦ୍ରାକୁ ନେଇ ଏକ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ଲେଖ ।
Solution:
ଏହା 28 + (35 – 10) ରୂପେ ପ୍ରକାଶ କରାଯାଇପାରେ ।
ଆମେ ଜାଣିଛୁ ଯେ ଏହା 28 + 35 + (−10) କିମ୍ବା 28 + 35 – 10 ରୂପେ ଲେଖୁପାରିବା ।
ତେଣୁ 28 + (35 + 10) = 28 + 35 – 10 = 53

ବନ୍ଧନୀ ପୂର୍ବରୁ ‘-‘ ଚିହ୍ନ ନଥାଇ ‘+’ ଚିହ୍ନ ଥିଲେ ଅପସାରଣ ସମୟରେ ବନ୍ଧନୀ ମଧ୍ୟରେ ଥ‌ିବା ପଦଗୁଡ଼ିକର ଚିହ୍ନର ପରିବର୍ତ୍ତନ ହେବ ନାହିଁ ।
ଏଠାରେ 35 ଓ −10 ଚିହ୍ନର କୌଣସି ପରିବର୍ତ୍ତନ ହୋଇନାହିଁ ।

ସାଂଖ୍ୟକ ପଦ ନିରୀକ୍ଷଣ – 1

Question 1.
ନିମ୍ନରେ କେତେକ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ତିନୋଟି ସ୍ତମ୍ଭରେ ପ୍ରଦତ୍ତ ହୋଇଛି, ପ୍ରତ୍ୟେକ ସ୍ତମ୍ଭରେ ଗୋଟିଏ ବା ଅଧୁକ ପଦର ପରିବର୍ତ୍ତନ କରାଯାଇଛି, ଯଥା ସମ୍ଭବ କମ୍ ହିସାବ କରି ପରିପ୍ରକାଶର ପ୍ରଥମ ସ୍ତମ୍ଭ ଅନୁଯାୟୀ ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର । (ପ୍ରଶ୍ନ ସହିତ ଉତ୍ତର)

Solution:

ନିଜେ କରି ଦେଖ :

Question 1.
ଶୂନ୍ୟସ୍ଥାନରେ ସଂଖ୍ୟା ଏବଂ ବାକ୍ସରେ ଚିହ୍ନ ବ୍ୟବହାର କରି ପୂରଣ କର ଯେପରିକି ପରିପ୍ରକାଶର ଉଭୟ ପାର୍ଶ୍ଵର ମାନ ସମାନ ହେବ ।
(a) 24 + (6 – 4) = 24 + 6 _______
(b) 38 + (_____ ______) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 6 – 4
(d) 24 – 6 – 4 = 24 – 6 _______
(e) 27 – (8 + 3) = 27 8 3
(f) 27 – (____ ____ = 27 – 8 + 3
Solution:
(a) 24 + (6 – 4)= 24 + 6 4
(b) 38 + (9 4) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 6 – 4
(d) 24 – 6 – 4 = 24 – 6 4
(e) 27 – (8 + 3) = 27 8 3
(f) 27 – (8 3) = 27 – 8 + 3

Question 2.
ବନ୍ଧନୀ ଅପସାରଣ କରି ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକ ଲେଖ ଯେପରିକି ପରିପ୍ରକାଶ ମାନ ଅପରିବର୍ତ୍ତିତ ରହିବ ।
(a) 14 + (12 + 10)
(b) 14 – (12 + 10)
(c) 14 + (12 – 10)
(d) 14 – (12 – 10)
(e) – 14 + 12 – 10
(f) 14 – (- 12 – 10)
Solution:
(a) 14 + (12 + 10) = 14 + 12 + 10
(b) 14 – (12 + 10) = 14 – 12 – 10
(c) 14 + (12 – 10) = 14 + 12 – 10
(d) 14 – (12 – 10) 14 – 12 + 10
(e) – 14 + (12 – 10) = – 14 + 12 – 10
(f) 14 – (- 12 – 10) = 14 + 12 + 10

Question 3.
ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକର ମାନ ନିର୍ଣ୍ଣୟ କର । ପ୍ରତ୍ୟେକ ଯୋଡ଼ାର ମାନ ସମାନ କି ନୁହେଁ ପ୍ରଥମେ ଅନୁମାନ କର । କେତେବେଳେ ଦୁଇଟି ପରିପ୍ରକାଶର ମାନ ସମାନ ହେବ କି ?
(a) (6 + 10) – 2 ଏବଂ 6 + (10 – 2)
(b) 16 – (8 – 3) ଏବଂ (16 – 8) – 3
(c) 27 – (18 + 4) 27 + (- 18 – 4)
Solution:
(a) LHS = 6 + 10 – 2 = 16 – 2 = 14,
RHS = 6 + (10 – 2) = 6 + 8 = 14
∴ LHS = RHS

(b) 16 – (8 – 3) ଏବଂ (16 – 8) – 3
LHS = 16 – (8 – 3) = 16 – 5 = 11, RHS = (16 – 8) – 3 = 8 – 3 = 5
∴ LHS > RHS

(c) 27 – (18 + 4) 27 + (- 18 – 4)
LHS = 27 – (18 + 4) = 27 – 22 = 5,
RHS = 27 + (- 18 – 4) = 27 + (-22) = 27 – 22 = 5
∴ LHS = RHS

Question 4.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ପରିପ୍ରକାଶଗୁଡ଼ିକ ମଧ୍ୟରୁ କେଉଁ ଯୋଡ଼ା ପରିପ୍ରକାଶର ମାନ ସମାନ ଚିହ୍ନାଅ । ସେଗୁଡ଼ିକର ହିସାବ ନକରି ପଦଗୁଡ଼ିକୁ ବୁଝି ଲେଖ ।
(a) 319 + 537, 319 – 537, -537 +319, 537 – 319
(b) 87 + 46 – 109, 87 + (46 – 109), (87 + 46) – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109
Solution:
(a) 319 – 537ଓ – 537 + 319 ପରିପ୍ରକାଶର ମାନ ସମାନ ।
(b) 87 – 46 + 109 ଓ (87 – 46) + 109 ପରିପ୍ରକାଶର ମାନ ସମାନ ।

Question 5.
ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶଗୁଡ଼ିକର ସୂଚିତ ମୂଲ୍ୟ ଅପରିବର୍ତିତ ରଖୁ ଉପଯୁକ୍ତ ସ୍ଥାନରେ ବନ୍ଧନୀ ବ୍ୟବହାର କର ।
(a) 34 – 9 + 12 = 13
(b) 56 – 14 – 8 = 34
(c) – 22 – 12 + 10 + 22 = -22
Solution:
(a) 34 – (9 + 12) = 13
(b) (56 – 14) – 8 = 34
(c) – 22 – (12 + 10) + 22 = -22

Question 6.
ଶୂନ୍ୟସ୍ଥାନରେ ଏପରି ସଂଖ୍ୟା ପୂରଣ କର ଯେପରିକି ଉଭୟ ପାର୍ଶ୍ଵର ମାନ ସମାନ ହେବୁ ।
(a) 423 + _______ = 419 + _______
(b) 207 – 68 = 210 – _______
Solution:
(a) 423 + 419 = 419 + 423
(b) 207 – 68 = 210 – 71

Question 7.
2, 3 ଓ 5 ଏବଂ ‘+’, ‘-‘ ଓ ବନ୍ଧନୀ ବ୍ୟବହାର କରି ଯେତେ ସମ୍ଭବ ଭିନ୍ନଭିନ୍ନ ମାନ ଥୁବା ସାଂଖ୍ୟକ ପରିପ୍ରକାଶଗୁଡ଼ିକ ଗଠନ କର । ଯେପରି : 2 – 3 + 5 = 4 ଏବଂ 3 – (5 – 2) = 0
Solution:
(a) 2 + (3 + 5) = 10
(b) 2 – (3 + 5) = -6
(c) 3 – (2 + 5) = -4
(d) 5 – (2 + 3) = 0
(e) 5 + 3 – 2 = 6
(f) 2 + 3 – 5 = 0
(g) 2 – 3 + 5 = 4

Question 8.
ଯେତେବେଳେ କୌଣସି ସଂଖ୍ୟାରୁ 9 ବିୟୋଗ କରିବାକୁ ଥିଲେ ଯଶୋଦା ସଂଖ୍ୟାରୁ 10 ବିୟୋଗ କରି ସେଥ‌ିରେ 1 ଯୋଗ କରିଥାଏ । ଉଦାହରଣ ସ୍ଵରୂପ 36 – 9 = 26 + 1)
(a) ସେ ସର୍ବଦା ଠିକ୍ ଉତ୍ତର ପାଇବ କି ? କାହିଁକି ?
(b) ତୁମେ ସେହିଭଳି ଅନ୍ୟ କୌଶଳ ବିଷୟରେ ଚିନ୍ତା କରିପାରିବ କି ? କେତେକ ଉଦାହରଣ ଦିଅ ।
Solution:
(a) ହଁ । – 9 = – 10 + 1 ଉଦାହରଣ : 25 – 9 = 25 – 10 + 1 = 16
(b) ଯଦି ଗୋଟିଏ ସଂଖ୍ୟାରେ 99 ଯୋଗ କରାଯିବ, 100 ଯୋଗକରି 1 ବିୟୋଗ କରିବା ସହିତ ସମାନ ।
ଉଦାହରଣ : 176 + 100 – 1 = 276 – 1 = 275

Question 9.
(i) 73 – 14 + 1
(ii) 73 – 14 – 1 ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ଦୁଇଟିକୁ ଲକ୍ଷ୍ୟକର । ପ୍ରତ୍ୟେକର ମାନ ସମାନ ଥ‌ିବା ପରିପ୍ରକାଶଗୁଡ଼ିକୁ, ନିମ୍ନଲିଖତ ପରିପ୍ରକାଶ ମଧ୍ଯରୁ ଚିହ୍ନଟ କର ।
(a) 73 – (14 + 1)
(b) 73 – (14 – 1)
(c) 73 + (- 14 + 1)
(d) 73 + (- 14 – 1)
Solution:
(i) 73 – 14 + 1 = 59 + 1 = 60
(ii) 73 – 14 – 1 = 59 – 1 = 58
(a) 73 – (14 + 1) = 73 – 15 = 58
(b) 73 – (14 – 1) = 73 – 13 = 60
(c) 73 + (14 + 1) = 73 – 13 = 60
(d) 73 + (- 14 – 1) = 73 – 15 = 58
∴ (b) ଓ (c), (i) ସହିତ ସମାନ ଏବଂ (a) ଓ (d), (ii) ସହିତ ସମାନ ।

2.4 ବନ୍ଧନୀ ଅପସାରଣ – II :

ଉଦାହରଣ – 15

Question 1.
ସରୋଜ ଓ ମନୋଜ ଦୁହେଁ ଜଳଖୁଆ ଦୋକାନକୁ ଯାଇ ପ୍ରତ୍ୟେକ ଗୋଟିଏ ଲେଖାଏଁ କଟଲେଟ୍ ଓ ରସଗୋଲା ଖାଇଲେ । ଗୋଟିଏ କଟ୍‌ଲେଟ୍‌ର ଦାମ୍ 43 ଟଙ୍କା ଓ ଗୋଟିଏ ରସଗୋଲାର ଦାମ୍ 24 ଟଙ୍କା । ସେମାନେ ଦୋକାନୀକୁ ଦେଇଥିବା ମୋଟ ଟଙ୍କାର ପରିମାଣକୁ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ପ୍ରକାଶ କର ।
Solution:
ଗୋଟିଏ କଟ୍‌ଲେଟ୍‌ର ଦାମ୍ = 43 ଟଙ୍କା ଓ ଗୋଟିଏ ରସଗୋଲାର ଦାମ୍ = 24 ଟଙ୍କା ।
ପ୍ରତ୍ୟେକ ଗୋଟିଏ ଲେଖାଏଁ କଟ୍‌ଲେଟ୍‌ ଓ ରସଗୋଲା ଖାଇଛନ୍ତି ।
ତେଣୁ ପ୍ରତ୍ୟେକ = 43 + 24 ଟଙ୍କା ଦେବେ ।

Question 2.
ତେବେ ସେମାନେ ମୋଟ କେତେ ଟଙ୍କା ଦେବେ ?

Solution:
ସେମାନେ ଟଙ୍କା ଦେବେ = 2 × (43 + 24) ଟଙ୍କା ।
ଏହାକୁ ପଦ ଅନୁଯାୟୀ ଲେଖୁଲେ =
ତେଣୁ 2 × 43 + 2 × 24 = 86 + 48 = 134 ଟଙ୍କା ।
∴ ସେମାନେ 134 ଟଙ୍କା ଦେବେ ।

Question 3.
ଯଦି ଅନ୍ୟ ଜଣେ ସାଙ୍ଗ ଆରିଫ୍ ସେମାନଙ୍କ ସହ ଯୋଗ ଦିଏ ତେବେ ସେମାନଙ୍କୁ ପୈଠ କରିବାକୁ ଥିବା ମୋଟ ଟଙ୍କାର ପରିମାଣକୁ ସାଂଖ୍ୟକ ପରିପ୍ରକାଶରେ ଲେଖ ।
Solution:
ସେମାନେ ଟଙ୍କା ଦେବେ = 3 × (43 + 24) ଟଙ୍କା ।
ଏହାକୁ ପଦ ଅନୁଯାୟୀ ଲେଖୁଲେ =
ତେଣୁ 3 × 43 + 3 × 24 = 129 + 72 = 201 ଟଙ୍କା ।
∴ ସେମାନେ 201 ଟଙ୍କା ଦେବେ ।

ଉଦାହରଣ – 16 

Question 1.
ଜନରାଜ୍ୟ ଦିବସ ଉପଲକ୍ଷେ ବାଳକ ସ୍କାଉଟ୍ ଓ ବାଳିକାଗାଇମାନଙ୍କର ପ୍ୟାରେଡ଼୍ ଅନୁଷ୍ଠିତ ହେଉଥିଲା । ସ୍କାଉଟ୍‌ରେ 5 ଜଣ କରି 4ଟି ଧାଡ଼ି ଏବଂ ଗାଇଡ଼ରେ 5 ଜଣ କରି 3 ଟି ଧାଡ଼ି ହୋଇଥିଲା (ନିମ୍ନ ଚିତ୍ର ଅନୁଯାୟୀ) । ତେବେ ଏହି ପ୍ୟାରେଡ୍‌ରେ ମୋଟ କେତେ ଜଣ ଗାଇଡ୍ ଓ ସ୍କାଉଟ୍ ଛାତ୍ରଛାତ୍ରୀ ଅଂଶଗ୍ରହଣ କରିଥିଲେ ?

Solution:
ପ୍ୟାରେଡ୍‌ରେ ଅଂଶଗ୍ରହଣ କରିଥିବା ବାଳକ ସ୍କାଉଟ୍ ସଂଖ୍ୟା = 4 × 5 ।
ଅଂଶଗ୍ରହଣ କରିଥିବା ବାଳିକା ଗାଇଡ଼୍ ସଂଖ୍ୟା = 3 × 5 ।
ମୋଟ ଅଂଶଗ୍ରହଣ କରିଥିବା ସ୍କାଉଟ୍ ଓ ଗାଇଡ଼୍ ସଂଖ୍ୟା = 4 × 5 + 3 × 5 ।
ଏହାକୁ ଆମେ ପ୍ରଥମେ ମୋଟ ଧାଡ଼ି ସଂଖ୍ୟା 4 + 3 ନିର୍ଣ୍ଣୟ କରି ସେଥ‌ିରେ ପ୍ରତ୍ୟେକ ଧାଡ଼ିର ପିଲା ସଂଖ୍ୟା 5 ଗୁଣନ କରିପାରିବା ।
ଏହା (4 + 3) × 5 । ତେଣୁ 4 × 5 + 3 × 5 = (4 + 3) × 5
ଏହାକୁ ଗଣନା କରି ଆମେ ପାଇବା

(4 + 3) × 5 = 7 × 5 = 35

Question 5.
5 × 4 + 3 ≠ 5 × (4 + 3) କାହିଁକି, ବର୍ଣ୍ଣନା କରିପାରିବ କି ?
ବାମପଟ : 5 × 4 + 3 = 20 + 3 = 23,
ଦକ୍ଷିଣପଟ : 5 × (4 + 3) = 5 × 7 = 35
ବାମପଟ ଦକ୍ଷିଣପଟ ସହିତ ସମାନ ନୁହେଁ ।
∴ ତେଣୁ 5 × 4 + 3 ≠ 5 × (4 + 3)

Question 6.
5 × (4 + 3) = 5 × (3 + 4) × 5 ସମାନ କି ?
Solution:
ବାମପଟ : 5 × (4 + 3) = 5 × 7 = 35,
ଦକ୍ଷିଣପଟ : 5 × (3 + 4) × 5 = 5 × 7 × 5 = 175
∴ 5 × (4 + 3) = 5 × (3 + 4) × 5 ସମାନ ନୁହେଁ ।

• ପୂର୍ବ ଦୁଇ ଉଦାହରଣ ଅନୁଯାୟୀ ଦେଖୁବା 10 × 98 + 3 × 98 ଅର୍ଥ 98 ର 10 ଗୁଣ ଓ 98 ର 3 ଗୁଣର ସମଷ୍ଟି ।
ସ୍ପଷ୍ଟଭାବେ ଏହା 98 ର 10 + 3 = 13 ଗୁଣ । ତେଣୁ, 10 × 98 + 3 × 98 = (10 + 3) × 98
ଅନ୍ୟ ପ୍ରକାରରେ (10 + 3) × 98 = 10 × 98 + 3 × 98
ସଂଖ୍ୟାର କ୍ରମ ପରିବର୍ତ୍ତନ କରି ଲେଖୁଲେ ଆମେ ପାଇବା
98 × 10 + 98 × 3 = 98 × (10 + 3) କିମ୍ବା 98 × (10 + 3) = 98 × 10 + 98 × 3

• ସେହିପରି 14 × 10 – 6 × 10 ପରିପ୍ରକାଶକୁ ବିଚାର କରାଯାଉ ।
14 × 10 – 6 × 10 ର ଅର୍ଥ 10 ର 14 ଗୁଣରୁ 10 ରୁ 6 ଗୁଣ ବିୟୋଗ ।
ଏଥ‌ିରୁ ସ୍ପଷ୍ଟ ଯେ ଏହା 14 – 6 = 8 ର 10 ଗୁଣ ।
ତେଣୁ 14 × 10 – 6 × 10 = (14 – 6) × 10 କିମ୍ବା (14 – 6) × 10 = 14 × 10 – 6 × 10
ଏହି ଧର୍ମକୁ ସଂକ୍ଷେପରେ ଯୋଗ (ବିୟୋଗ)ର ଗୁଣନ, ଗୁଣନର ଯୋଗ (ବିୟୋଗ) ସହ ସମାନ ।

ସାଂଖ୍ୟକ ପଦ ନିରୀକ୍ଷଣ – II

ଉଦାହରଣ – 17

Question 1.
ଆସ ଦେଖୁବା, ଗୋଟିଏ ଗୁଣନ ପ୍ରକ୍ରିୟାର ସଂଖ୍ୟାର ପରିବର୍ତ୍ତନ ହେଲେ କ’ଣ ହେବ ?
53 × 18 = 954 ହେଲେ, 63 × 18 ର ମାନ ନିର୍ଣ୍ଣୟ କର ।
Solution:
63 × 18 ର ଅର୍ଥ 63 ର 18 ଗୁଣ
63 × 18 = (53 + 10) × 18 = 53 × 18 + 10 × 18 = 954 + 180 = 1134

ଉଦାହରଣ – 18 

Question 1.
97 × 25ରେ ମାନ ପାଇଁ ଏକ ଫଳପ୍ରଦ ଉପାୟ ନିର୍ଣ୍ଣୟ କର ।
Solution:
97 × 25 ର ଅର୍ଥ 97 ର 25 ଗୁଣ ।
ଏହାକୁ ଆମେ ଲେଖୁ ପାରିବା (100 – 3) × 25
ଏହାର ଅର୍ଥ 100 ର 25 ଗୁଣରୁ 3 ର 25 ଗୁଣ ବିୟୋଗ
97 × 25 = 100 × 25 – 3 × 25 = 2500 – 75 = 2425

Question 3.
ଏହି ପଦ୍ଧତି ଅବଲମ୍ବନ କରି ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର ।
(a) 95 × 8
(b) 104 × 15
(c) 49 × 50
Solution:
(a) 95 × 8 = (100 – 5) × 8 = 100 × 8 – 5 × 8 = 800 – 40 = 760
(b) 104 × 15 = (100 + 4) × 15 = 100 × 15 + 4 × 15 = 1500 + 60 = 1560
(c) 49 × 50 = (50 – 1) × 50 = 50 × 50 – 1 × 50 = 2500 – 50 = 2450

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନ ପରିପ୍ରକାଶଗୁଡ଼ିକର ଉଭୟ ପାର୍ଶ୍ଵରମାନ ପରିବର୍ତ୍ତନ ନକରି ଶୂନ୍ୟସ୍ଥାନରେ ସଂଖ୍ୟା ଓ ବାକ୍ସ ମଧ୍ୟରେ ଚିହ୍ନ ବ୍ୟବହାର କର ।
(a) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 3 × _______
(d) (9 + 2) × 4 = 9 × 4 2 × _______
(e) 3 × ( _______ + 4) = 3 _______ + _______
(f) ( _______+ 6) × 4 = 13 × 4 + _______
(g) 3 × ( _______ + _______) = 3 × 5 + 3 × 2
(h) (______ + ______) × ______ = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 ×_____
(j) (5 – 2) × 7 = 5 × 7 – 2 × ______
(k) 5 × (8 – 3) = 5 × 8 5 × ______
(l) (8 – 3) × 7 = 8 × 7 3 × 7
(m) 5 × (12 – ____) = 5 × ______
(n) (15 – ______) × 7 = ______ 6 × 7
(0) 5 × (5 _____ – _____= 5 × 9 – 5 × 4
(p) (_____ – _____) × ______ = 17 × 7 – 9 × 7
Solution:
(a) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 3 × 8
(d) (9 + 2) × 4 = 9 × 4 2 × 4
(e) 3 × (1 + 4) = 3 × 1 + 3 × 4
(f) (13 + 6) × 4 = 13 × 4 + 6 × 4
(g) 3 × (5 + 2) = 3 × 5 + 3 × 2
(h) (2 + 3) × 4 = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × 2
(j) (5 – 2) × 7 = 5 × 7 – 2 × 7
(k) 5 × (8 – 3)= 5 × 8 5 × 3
(l) (8 – 3) × 7 = 8 × 7 3 × 7
(m) 5 × (12 – 4) = 5 × 12 5 × 4
(n) (15 – 6) × 7 = 15 × 7 6 × 7
(0) 5 × (9 – 4) = 5 × 9 – 5 × 4
(p) (17 – 9) × 7 = 17 × 7 – 9 × 7

Question 2.
ନିମ୍ନ ଗାଣିତିକ ପରିପ୍ରକାଶ ଗୁଡ଼ିକର ଉଭୟ ପାର୍ଶ୍ଵକୁ ଦେଖୁ ଶୂନ୍ୟ ସ୍ଥାନରେ ‘<‘, ‘>’ ବା ‘=’ ଚିହ୍ନ ବ୍ୟବହାର କର ।
(a) (8 – 3) × 29 (3 – 8) × 29
(b) 15 + 9 × 18 (15 + 9) × 18
(c) 23 × (17 – 9) 23 × 17 + 23 × 9
(d) (34 – 28) × 42 34 × 42 – 28 × 42
Solution:
(a) (8 – 3) × 29 (3 – 8) × 29
ବାମପାର୍ଶ୍ଵ : (8 – 3) × 29 = 5 × 29 = 145,
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ : (3 – 8) × 29 = -5 × 29 = -145
ବାମପାର୍ଶ୍ଵ > ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ (8 – 3) × 29 (3 – 8) × 29

(b) 15 + 9 × 18 (15 + 9) × 18
ବାମପାର୍ଶ୍ଵ : 15 + 9 × 18 = 15 + 162 = 177,
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ : (15 + 9) × 18 = 24 × 18 = 432
ବାମପାର୍ଶ୍ଵ < ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴15 + 9 × 18  (15 + 9) × 18

(c) 23 × (17 – 9) 23 × 17 + 23 × 9
ବାମପାର୍ଶ୍ଵ : 23 × (17 – 9) = 23 × 8 = 184
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ : 23 × 17 + 23 × 9 = 391 + 207 = 598
ବାମପାର୍ଶ୍ଵ < ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 23 × (17 – 9) 23 × 17 + 23 × 9

(d) (34 – 28) × 42 34 × 42 – 28 × 42
ବାମପାର୍ଶ୍ଵ : (34 – 28) × 42 = 6 × 42 = 252
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ : 34 × 42 – 28 × 42 = 1428 – 1176 = 252
ବାମପାର୍ଶ୍ଵ = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ ।
∴ (34 – 28) × 42 34 × 42 – 28 × 42

Question 3.
2 × (1 + 6) = 14 ଏହିପରି 14 ପାଇବା ପାଇଁ ଅଧୁକ ସଂରଚନା ଅଛି କି ? ନିମ୍ନରେ ଥିବା ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(a) ______ × (______ + ______) = 14
(b) ______ × (_______ + ______) = 14
(c) ______ × (______ + ______) = 14
(d) ______ × (______ + ______) = 14
Solution:
(a) 2 × (2 + 5) = 14
(b) 2 × (3 + 4) = 14
(c) 7 × (1 + 1) = 14
(d) 7 × (2 + 0) = 14

Question 4.
ନିମ୍ନରେ ଥ‌ିବା ପ୍ରତ୍ୟେକ ଚିତ୍ରରେ ଦିଆଯାଇଥିବା ସଂଖ୍ୟାଗୁଡ଼ିକର ସମଷ୍ଟି ଅତି କମ୍‌ରେ ଦୁଇଟି ଭିନ୍ନ ଉପାୟରେ ନିର୍ଣ୍ଣୟ କର । ଗାଣିତିକ ପରିପ୍ରକାଶ ମାଧ୍ୟମରେ ଏହାକୁ କିପରି ସମାଧାନ କଲ ଲେଖ ।
(i)
Solution:
(i) ଉପାୟ – 1 : ଧାଡ଼ିଗୁଡ଼ିକୁ ଯୋଗକଲେ, ଧାଡ଼ି – 1 + ଧାଡ଼ି – 2 + ଧାଡ଼ି – 3
= (4 + 8 + 4) + (8 + 4 + 8) + (4 + 8 + 4) = (16 + 20 + 16) = 52
ଉପାୟ – 2 : 4 × 5 + 8 × 4 = 20 + 32 = 52

(ii) ଉପାୟ 1:
ପ୍ରଥମ ଧାଡ଼ି = 5 + 6 + 6 + 5 = 22 ତୃତୀୟ ଧାଡ଼ି = 6 + 5 + 5 + 6 = 22
ଦ୍ଵିତୀୟ ଧାଡ଼ି = 6 + 5 + 5 + 6 = 22 ଚତୁର୍ଥ ଧାଡ଼ି = 5 + 6 + 6 + 5 = 22
ପ୍ରତ୍ୟେକ ଧାଡ଼ିର ଯୋଗ ସଂଖ୍ୟା = 22
ଏଠାରେ ଚାରୋଟି ଧାଡ଼ି ଅଛି । ∴ ସମୁଦାୟ = 4 × 22 = 88
ସାଂଖ୍ୟକ ପରିପ୍ରକାଶ କଲେ = 4 × (5 + 6 + 6 + 5))
ଉପାୟ – 2 : 5 × 8 + 6 × 8 = 40 + 48 = 88

ନିଜେ କରି ଦେଖ :

Question 1.
ନିମ୍ନରେ ଦିଆଯାଇଥିବା ପରିସ୍ଥିତିଗୁଡ଼ିକୁ ପଢ଼ । ପ୍ରତ୍ୟେକ ପରିସ୍ଥିତି ପାଇଁ ଉପଯୁକ୍ତ ଗାଣିତିକ ପରିପ୍ରକାଶଗୁଡ଼ିକ ଲେଖୁ ସମାଧାନ କର ।
(a) ନବରଙ୍ଗପୁର ଜିଲ୍ଲା ହାଟ ସପ୍ତାହର ସାତ ଦିନ ବସେ । ରହିମ୍ ନିଜ ବାଡ଼ିରେ ଫଳୁଥୁବା ଆମ୍ବ ପ୍ରତ୍ୟେକ ଦିନ ୨ କି.ଗ୍ରା. କରି ଏବଂ ରାମ ନିଜ ବାଡ଼ିରେ ଫଳୁଥିବା ଆମ୍ବ ପ୍ରତ୍ୟେକ ଦିନ 11 କି.ଗ୍ରା. କରି ଏହି ହାଟକୁ ଯୋଗାଇ ଥାଆନ୍ତି । ସେମାନଙ୍କଦ୍ଵାରା ହାଟକୁ ସପ୍ତାହଯାକ ଯୋଗାଇଥୁବା ସମୁଦାୟ ଆମ୍ବର ପରିମାଣ କେତେ ?
Solution:
ରହିମ୍ ହାଟକୁ ଆମ୍ବ ନିଏ = 9 କି.ଗ୍ରା. ।
ରାମ ହାଟକୁ ଆମ୍ବ ନିଏ = 11 କି.ଗ୍ରା. ।
ଦୁଇଜଣ ମିଶିକରି ହାଟକୁ ଆମ୍ବ ନିଅନ୍ତି = 9 + 11 = 20 କି.ଗ୍ରା. ।
ଗୋଟିଏ ସପ୍ତାହରେ ରହିମ୍ ଓ ରାମ ଆମ୍ବ ନିଅନ୍ତି = 7 × 20 କି.ଗ୍ରା. = 140 କି.ଗ୍ରା. ।
∴ ସେମାନଙ୍କଦ୍ଵାରା ହାଟକୁ ସପ୍ତାହଯାକ ଯୋଗାଇଥିବା ସମୁଦାୟ ଆମ୍ବର ପରିମାଣ 140 କି.ଗ୍ରା. ।

(b) ବିନୁର ମାସିକ ଆୟ 20,000 ଟଙ୍କା । ସେ ମାସକୁ ଘରଭଡ଼ା ବାବଦକୁ 5,000 ଟଙ୍କା, ଖାଦ୍ୟ ବାବଦକୁ 5,000 ଟଙ୍କା ଏବଂ ଅନ୍ୟାନ୍ୟ ଖର୍ଚ୍ଚ ବାବଦକୁ 2,000 ଟଙ୍କା ଖର୍ଚ୍ଚ କରେ । ତେବେ ସେ ବର୍ଷକୁ କେତେ ଟଙ୍କା ସଞ୍ଚୟ କରିବ ?
Solution:
ବିନୁ ଗୋଟିଏ ମାସରେ ଘରଭଡ଼ା ବାବଦକୁ ଖର୍ଚ୍ଚ କରେ = 5,000 ଟଙ୍କା ।
ଖାଦ୍ୟ ବାବଦକୁ ଖର୍ଚ୍ଚ କରେ = 5,000 ଟଙ୍କା ।
ଅନ୍ୟାନ୍ୟ ଖର୍ଚ୍ଚ = 2,000 ଟଙ୍କା ।
ସମୁଦାୟ ଖର୍ଚ୍ଚ = 5,000 + ‘5,000 + 2,000 = 12,000 ଟଙ୍କା ।
ବିଜୁର ମାସିକ ଆୟ = 20,000 ଟଙ୍କା ।
ମାସକୁ ସଞ୍ଚୟ କରେ = 20,000 – 12,000 = 8,000 ଟଙ୍କା ।
1 ବର୍ଷ = 12 ମାସ ।
ଗୋଟିଏ ବର୍ଷରେ ସଞ୍ଚୟ କରିବ = 8,000 × 12 = 96,000 ଟଙ୍କା ।
∴ ତେବେ ସେ ବର୍ଷକୁ ୨୦,୦୦୦ ଟଙ୍କା ସଞ୍ଚୟ କରିବ ?

(c) ଗୋଟିଏ ଗେଣ୍ଡା ଦିନବେଳେ ଏକ ଖୁଣ୍ଟ ଉପରକୁ 3 ସେ.ମି. ଚଢ଼େ ଏବଂ ରାତିରେ ସେ 2 ସେ.ମି. ଓହ୍ଲାଏ । ଖୁଣ୍ଟଟି 10) ସେ.ମି. ଉଚ୍ଚ ଏବଂ ଏହାର ଉପରେ ଏକ ସ୍ଵାଦିଷ୍ଟ ଖାଦ୍ୟ ଅଛି । ଖୁଣ୍ଟର ଅଗ୍ରଭାଗରେ ଏହି ବସ୍ତୁକୁ ପାଇବା ପାଇଁ ତାକୁ କେତେ ଦିନ ଲାଗିବ ?
Solution:
ପ୍ରତ୍ୟେକ ଦିନ ଗେଣ୍ଡାଟି ଦିନବେଳେ ଖୁଣ୍ଟ ଉପରେ ଚଢ଼େ = 3 ସେ.ମି. ।
ରାତିରେ ଓହ୍ଲାଏ = 2 ସେ.ମି. ।
ପ୍ରତ୍ୟେକ ଦିନ ଖୁଣ୍ଟରେ ରହେ = (3 – 2) ସେ.ମି. = 1 ସେ.ମି. ।
ସପ୍ତମ ଦିନ = (7 × 1) ସେ.ମି. = 7 ସେ.ମି. ।
ଅଷ୍ଟମ ଦିନ = (7 + 3) ସେ.ମି. = 10 ସେ.ମି. ।
∴ ଖୁଣ୍ଟର ଅଗ୍ରଭାଗକୁ ପହଞ୍ଚିବା ପାଇଁ ତାକୁ ଆଠ ଦିନ ଲାଗିବ ।

Question 2.
ମହେନ୍ଦ୍ର ମଙ୍ଗଳବାର ଓ ଶନିବାର ବ୍ୟତୀତ ଅନ୍ୟ ଦିନମାନଙ୍କରେ ପ୍ରତିଦିନ ଦୁଇ ପୃଷ୍ଠାର ଏକ ଗପ ପଢ଼େ । ସେ 8 ସପ୍ତାହରେ କେତୋଟି ଗପ ପଢ଼ିବ ? ନିମ୍ନରେ ଦିଆଯାଇଥବା କେଉଁ ଗାଣିତିକ ପରିପ୍ରକାଶଟି ଏହାକୁ ସୂଚିତ
କରେ ?
(a) 5 × 2 × 8
(b) (7 – 2) × 8
(c) 8 × 7
(d) 7 × 2 × 8
(e) 7 × 5 – 2
(f) (7 + 2) × 8
(g) 7 × 8 – 2 × 8
(h) (7 – 5) × 8
Solution:
ମହେନ୍ଦ୍ର ମଙ୍ଗଳବାର ଓ ଶନିବାର ବ୍ୟତୀତ ଅନ୍ୟ ଦିନମାନଙ୍କରେ ପ୍ରତିଦିନ ଦୁଇ ପୃଷ୍ଠାର ଏକ ଗପ ପଢ଼େ ।
ଗୋଟିଏ ସପ୍ତାହରେ ପଢ଼େ = (7 – 2) ଦିନ ।
ପ୍ରତିଦିନ ଦୁଇ ପୃଷ୍ଠାର ଗପ ପଢ଼ିଲେ, ଗୋଟିଏ ସପ୍ତାହରେ ପଢ଼ିବେ = (7 – 2) × 2
ଆଠ ସପ୍ତାହରେ ପଢ଼ିବେ = (7 – 2) × 2 × 8 = 5 × 2 × 8
∴ ଉତ୍ତର (a) ଟି ଠିକ୍ ।

Question 3.
ନିମ୍ନଲିଖୁତ ଗାଣିତିକ ପରିପ୍ରକାଶଗୁଡ଼ିକର ମାନ ନିର୍ଣ୍ଣୟର ବିଭିନ୍ନ ପଦ୍ଧତି ଲେଖ ।
(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Solution:
(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)
= (- 1) + (- 1) + (- 1) + (- 1) + (- 1) = (- 1) × 5 = -5
କିମ୍ବା, 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= (1 + 3 + 5 + 7 + 9) – (2 + 4 + 6 + 8 + 10) = 25 – 30 = -5

(b) 1 – 1 + 1 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) = 0 + 0 + 0 + 0 + 0 = 0
କିମ୍ବା, 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 + 1 + 1 + 1 + 1) – (1 + 1 + 1 + 1 + 1) = 5 – 5 = 0

Question 4.
‘<‘, ‘>’ କିମ୍ବା ‘=’ ବ୍ୟବହାର କରି ନିମ୍ନଯୋନା ଗାଣିତିକି ପରିପ୍ରକାଶଗୁଡ଼ିକର ତୁଳନା କର ।
(a) 49 – 7 + 8 49 – 7 + 8
(b) 83 × 42 – 18 83 × 40 – 18
(c) 145 – 17 × 8 145 – 17 × 6
(d) 23 × 48 – 35 23 × (48 – 35)
(e) (16 – 11) × 12 -11 × 12 + 16 × 12
(f) (76 – 53) × 88 88 × (53 – 76)
(g) 25 × (42 + 16) 25 × (43 + 15)
(h) 36 × (28 – 16) 35 × (27 – 15)
Solution:
(a) ବାମ ପାର୍ଶ୍ଵ = 49 – 7 + 8 = (49 – 7) + 8 = 42 + 8 = 50
ଦକ୍ଷିଣ ପାର୍ଶ୍ୱ = 49 – 7 + 8 = 49 – (7 – 8) = 49 (-1) = 49 + 1 = 50
ବାମ ପାର୍ଶ୍ଵ = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 49 – 7 + 8 49 – 7 + 8

(b) ବାମ ପାର୍ଶ୍ଵ = 83 × 42 – 18 (83 × 42) – 18 = 3486 – 18 = 3468
ଦକ୍ଷିଣ ପାର୍ଶ୍ୱ = 83 × 40 – 18 = (83 × 40) – 18 = 3320 – 18 = 3302
ବାମ ପାର୍ଶ୍ଵ > ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 83 × 42 – 18 83 × 40 – 18

(c) ବାମ ପାର୍ଶ୍ଵ = 145 – 17 × 8 = 145 – (17 × 8) = 145 – 136 = 9
ଦକ୍ଷିଣ ପାର୍ଶ୍ୱ = 145 – 17 × 6 = 145 – (17 × 6) = 145 – 102 = 43
ବାମ ପାର୍ଶ୍ଵ < ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 145 – 17 × 8 145 – 17 × 6

(d) ବାମ ପାର୍ଶ୍ଵ = 23 × 48 – 35 (23 × 48) – 35 = 1104 – 35 = 1069
ଦକ୍ଷିଣ ପାର୍ଶ୍ୱ = 23 × (48 – 35) = 23 × 13 = 299
ବାମ ପାର୍ଶ୍ଵ > ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 23 × 48 – 35 23 × (48 – 35)

(e) ବାମ ପାର୍ଶ୍ଵ = (16 – 11) × 12 = 5 × 12 = 60
ଦକ୍ଷିଣ ପାର୍ଶ୍ୱ = -11 × 12 + 16 × 12 = (-11 × 12) + (16 × 12) = -132 + 192 = 60
ବାମ ପାର୍ଶ୍ଵ – ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ (16 – 11) × 12 – 11 × 12 + 16 × 12

(f) ବାମ ପାର୍ଶ୍ଵ = (76 – 53) × 88 = 23 × 88 = 2024
ଦକ୍ଷିଣ ପାର୍ଶ୍ୱ = 88 × (53 – 76) = 88 × -23 = -2024
ବାମ ପାର୍ଶ୍ଵ > ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ (76 – 53) × 88 = 88 × (53 – 76)

(g) ବାମ ପାର୍ଶ୍ଵ = 25 × (42 + 16) = 25 × 58 = 1450
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ = 25 × (43 + 15) = 25 × 58 = 1450
ବାମ ପାର୍ଶ୍ଵ = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 25 × (42 + 16) 25 × (43 + 15)

(h) ବାମ ପାର୍ଶ୍ଵ = 36 × (28 – 16) = 36 × 12 = 432
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ = 35 × (27 – 15) = 35 × 12 = 420
ବାମ ପାର୍ଶ୍ଵ > ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ । ∴ 36 × (28 – 16)  35 × (27 – 15)

Question 5.
ହିସାବ ନକରି ନିମ୍ନରେ ଦିଆଯାଇଥିବା ଗାଣିତିକ ପରିପ୍ରକାଶ ସହ ସମାନ ପରିପ୍ରକାଶକୁ ଚିହ୍ନଟ କର । ତୁମେ ପରିପ୍ରକାଶଗୁଡ଼ିକୁ ପଦ ବ୍ୟବହାର କରି କିମ୍ବା ବନ୍ଧନୀ ଅପସାରଣ କରି ଲେଖୁର । ଏଠାରେ ଏକାଧ‌ିକ ପରିପ୍ରକାଶ, ଦତ୍ତ ପରିପ୍ରକାଶ ସହ ସମାନ ହୋଇପାରେ ।
(a) 83 – 37 – 12
(i) 84 – 38 – 12
(ii) 84 – (37 + 12)
(iii) 83 – 38 – 13
(iv) – 37 + 83 – 12

(b) 93 + 37 × 44 + 76
(i) 37 + 93 × 44 + 76
(ii) 93 + 37 × 76 + 44
(iii) (93 + 37) × (44 + 76)
(iv) 37 × 44 + 93 + 76
Solution:
(a) 83 – 37 – 12 = (83 – 37) – 12 = 46 – 12 = 34
(i) 84 – 38 – 12 = (84 – 38) – 12 = 46 – 12 = 34
(ii) 84 – (37 + 12) = 84 – 49 = 35
(iii) 83 – 38 – 13 = (83 – 38) – 13 = 45 – 13 = 32
(iv) – 37 + 83 – 12 (37 + 83) – 12 = 46 – 12 = 34
∴ (i) ଓ (iv) ର ମୂଲ୍ୟ ସମାନ ।

(b) 93 + 37 × 44 + 76 = 93 + (37 × 44) + 76 = 93 + 1628 + 76 = 1797
(i) 37 + 93 × 44 + 76 = 37 + (93 × 44) + 76 = 37 + 4092 + 76 = 4205
(ii) 93 + 37 × 76 + 44 = 93 + (37 × 76) + 44 = 93 + 2812 + 44 = 2949
(iii) (93 + 37) × (44 + 76) = 130 × 120 = 15600
(iv) 37 × 44 + 93 + 76 = (37 × 44) + 93 + 76 = 1628 + 93 + 76 = 1797
∴ (iv) ଠିକ୍ ମୂଲ୍ୟ ପ୍ରକାଶ କରୁଛି ।

Question 6.
ଏକ ସଂଖ୍ୟା ନିଅ ଏବଂ ଦଶଟି ଭିନ୍ନଭିନ୍ନ ଗାଣିତିକ ପରିପ୍ରକାଶ ଲେଖ ଯାହାର ମାନ ଉକ୍ତ ସଂଖ୍ୟା ସହ ସମାନ ହେବ ।
Solution:
ମନେକର ସଂଖ୍ୟାଟି = ‘0’
(a) 4 – 4 = 0
(b) (8 – 3) × 0 = 0
(c) 3 × 6 – 18 = 0
(d) (7 – 2) × 5 – 25 = 0
(e) 7 + 4 – 11 = 0
(f) – 4 + 8 × 2 – 12 = 0
(g) 2 × (6 – 6) = 0
(h) (9 + 4) (10 + 3) = 0
(i) 7 × (6 + 2) – 56 = 0
(j) (7 × 3) + 4 – 25 = 0

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 11 Light: Shadows and Reflections Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 11 Question Answer

Science Curiosity Class 7 Chapter 11 Question Answer

Light: Shadows and Reflections Class 7 Questions and Answers

Let Us Enhance Our Learning

Question 1.
Which of the following are luminous objects?
Mars, Moon, Pole Star, Sun, Venus, Mirror
Answer:
The Sun and the Pole Star are luminous objects because they emit their own light.

Question 2.
Match the items in Column I with those in Column II.

Answer:

Column I	Column II
(i) Pinhole camera	C. Forms an inverted image
(ii) Opaque object	A. Blocks light completely
(iii) Transparent object	D. Light passes almost completely through it
(iv) Shadow	B. The dark region formed behind the object

Question 3.
Sahil, Rekha, Patrick, and Qasima are trying to observe the candle flame through the pipe as shown in the figure alongside. Who can see the flame?

Answer:
Rekha can see the candle flame. Light travels in a straight line. Only Rekha’s line of sight through the straight pipe is directly aligned with the candle flame, allowing light to reach her eyes. Sahil, Patrick, and Qasima cannot see the flame because the light cannot bend around corners.

Question 4.
Look at the images shown in the following figure, and select the correct image showing the shadow formation of the boy.

Answer:
(d) The boy’s shadow forms on the side opposite to the Sun, as light travels in a straight line and cannot pass through opaque objects. Hence, figure (d) correctly shows the shadow formation.

Question 5.
The shadow of a ball is formed on a wall by placing the ball in front of a fixed torch as shown in the figure given. In scenario (i) the ball is closer to the torch, while in scenario (ii) the ball is closer to the wall. Choose the most accurate representation of the shadows formed in both scenarios from the options provided (A and B).

Answer:
(i)-(B). The ball is placed closer to the torch and farther from the wall, so the shadow formed on the wall appears larger and blurrier.
(ii)-(A). The ball is closer to the wall and farther from the torch, so the shadow appears smaller and sharper.

Question 6.
Based on above figures in question, match the position of the torch in Column I with the characteristics of the ball’s shadow in Column II.
Answer:

Column A	Column B
(i) If the torch is close to the ball	B. The shadow would be larger
(ii) If the torch is far away	A. The shadow would be smaller
(iii) If the ball is removed from the set-up	D. A bright spot would appear on the screen
(iv) If two torches are present in the set-up on the left side of the ball	C. Two shadows would appear on the screen

Question 7.
Suppose you view the tree shown in the alongside figure, through a pinhole camera. Sketch the outline of the image of the tree formed in the pinhole camera.

Answer:
The image of the tree formed in the pinhole camera will be inverted (upside down).

Question 8.
What difference do you notice? Explain the reason for the difference.
Answer:

When the paper with the name written on it is held in front of a plane mirror, the image of the name, along with the letters, appears reversed from left to right. This happens because a plane mirror produces a laterally inverted image. In lateral inversion, the left side of the object appears on the right side of the image, and the right side of the object appears on the left side of the image.

Question 9.
Measure the length of your shadow at 9 AM, 12 PM, and 4 PM with the help of your friend. Write down your observations:
(i) At which of the given times is your shadow the shortest?
(ii) Why do you think this happens?
Answer:
(i) The shadow is the shortest at 12 PM (noon).
(ii)At noon, the Sun is almost directly overhead. When the Sun is high in the sky, the light rays fall almost vertically on us, so the shadow formed is short and close to our feet.

Question 10.
On the basis of following statements, choose the correct option.
Statement A: Image formed by a plane mirror is laterally inverted.
Statement B: Images of alphabets T and O appear identical to themselves in a plane mirror.
(a) Both statements are true
(b) Both statements are false
(c) Statement A is true, but statement B is false
(d) Statement A is false, but statement B is true
Answer:
(a) A plane mirror produces an image that is laterally inverted, which means the left side of the object appears on the right side in the image and vice versa. However, certain letters, such as T and O, appear identical to themselves even after lateral inversion, because their shapes are symmetrical about the vertical axis.

Question 11.
Suppose you are given a tube of the shape shown in the figure alongside, and two plane mirrors smaller than the diameter of the tube. Can this tube be used to make a periscope? If yes, mark where you will fix the plane mirrors.

Answer:
Yes, the given tube can be used to make a periscope. To do so, two plane mirrors should be fixed inside the tube at 45° angles to its length, facing each other, one at the top bend and the other at the bottom bend of the tube, as shown in the figure alongside.

Question 12.
We do not see the shadow on the ground of a bird flying high in the sky. However, the shadow is seen on the ground when the bird swoops near the ground. Think and explain why it is so.
Answer:
When a bird flies high in the sky, it is very far from the ground, and the light rays from the Sun falling on it are almost parallel. The shadow formed at such a large distance becomes extremely faint and spreads over a very large area, making it invisible to our eyes. However, when the bird flies close to the ground, the distance between the bird and the ground is small. The bird blocks the same amount of light, but now in a concentrated area, so a clear and distinct shadow is formed on the ground.

Class 7 Curiosity Chapter 11 Question Answer

InText Questions

Question 1.
When a shiny steel plate is used in a shadow experiment, why does it form a shadow on the screen and a bright spot on the opposite wall? [Page 159]
Answer:
The steel plate is opaque, so it forms a shadow. However, its smooth, shiny surface also reflects the incident light, creating a bright spot on the opposite wall.

Question 2.
When a person looks into a mirror, they can see their face. Name and explain the responsible phenomenon. [Page 160]
Answer:
A person can see the image of their face in a mirror due to reflection of light. When light from a source falls on the face, it is reflected and scattered in many directions. Light rays from the face strike the smooth, reflective surface of the mirror and bounce back to the eyes, allowing one to see an image of the face.

Question 3.
Can we see an image of an object only in a mirror? [Page 162]
Answer:
No. Images can be seen in any reflecting surface like polished metal or water, and also in a pinhole camera.

Question 4.
Why is “ЯОИАЛШЯМА” written on an ambulance? [Page 162]
Answer:
It appears as “AMBULANCE” when viewed in the rear-view mirrors of the vehicles ahead of the ambulance. This helps the drivers see it properly and give way to the ambulance.

BSE Odisha Guru Class 7 Science Notes

BSE Odisha Class 7 Science Notes

BSE Odisha 7th Class Science Notes

• Class 7 Science Chapter 1 Notes Odia Medium – ଚିର ବିକାଶଶୀଳ ବିଜ୍ଞାନ ଜଗତ
• Class 7 Science Chapter 2 Notes Odia Medium – ଅମ୍ଳୀୟ, କ୍ଷାରୀୟ ଓ ନିରପେକ୍ଷ ପଦାର୍ଥର ଅନୁସନ୍ଧାନ
• Class 7 Science Chapter 3 Notes Odia Medium – ବିଦ୍ୟୁତ୍ : ଏହାର ପରିପଥ ଓ ଉପାଦାନ
• Class 7 Science Chapter 4 Notes Odia Medium – ଧାତୁ ଓ ଅଧାତୁର ଦୁନିଆ
• Class 7 Science Chapter 5 Notes Odia Medium – ଆମ ଚତୁଃପାର୍ଶ୍ବରେ ଭୌତିକ ଓ ରାସାୟନିକ ପରିବର୍ତ୍ତନ
• Class 7 Science Chapter 6 Notes Odia Medium – କିଶୋରାବସୁ : ବୃଦ୍ଧି ଓ ପରିବର୍ତ୍ତନ

Class 7 Science Odia Medium Notes

• Class 7 Science Chapter 7 Notes Odia Medium – ପ୍ରକୃତିରେ ତାପ ସ୍ଥାନାନ୍ତରଣ
• Class 7 Science Chapter 8 Notes Odia Medium – ସମୟ ଏବଂ ଗତିର ପରିମାପ
• Class 7 Science Chapter 9 Notes Odia Medium – ପ୍ରାଣୀମାନଙ୍କ ଜୀବନ ପ୍ରକ୍ରିୟା
• Class 7 Science Chapter 10 Notes Odia Medium – ଉଦ୍ଭିଦରେ ଜୀବନ ପ୍ରକ୍ରିୟା
• Class 7 Science Chapter 11 Notes Odia Medium – ଆଲୋକ : ଛାୟା ଓ ପ୍ରତିଫଳନ
• Class 7 Science Chapter 12 Notes Odia Medium – ପୃଥ‌ିବୀ, ଚନ୍ଦ୍ର ଓ ସୂର୍ଯ୍ୟ

Class 7 Science Curiosity Notes

Class 7 Curiosity Notes

• Chapter 1 The Ever-Evolving World of Science Class 7 Notes
• Chapter 2 Exploring Substances: Acidic, Basic, and Neutral Class 7 Notes
• Chapter 3 Electricity: Circuits and Their Components Class 7 Notes
• Chapter 4 The World of Metals and Non-metals Class 7 Notes
• Chapter 5 Changes Around Us: Physical and Chemical Class 7 Notes
• Chapter 6 Adolescence: A Stage of Growth and Change Class 7 Notes

Curiosity Class 7 Science Notes

• Chapter 7 Heat Transfer in Nature Class 7 Notes
• Chapter 8 Measurement of Time and Motion Class 7 Notes
• Chapter 9 Life Processes in Animals Class 7 Notes
• Chapter 10 Life Processes in Plants Class 7 Notes
• Chapter 11 Light: Shadows and Reflections Class 7 Notes
• Chapter 12 Earth, Moon, and the Sun Class 7 Notes

Also Read BSE Odisha Class 7 Science Solutions

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 10 Life Processes in Plants Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 10 Question Answer

Science Curiosity Class 7 Chapter 10 Question Answer

Life Processes in Plants Class 7 Questions and Answers

Let Us Enhance Our Learning

Question 1.
Complete the following table.

Answer:

Question 2.
Imagine a situation where all the organisms that carry out photosynthesis on the Earth have disappeared. What would be the impact of this on living organisms?
Answer:
If all photosynthesising organisms disappear, no food will be produced, because they are the only living beings that can make food using sunlight, carbon dioxide, and water. Without photosynthesis, oxygen will not be released into the air. As a result, there would be neither any source of food nor oxygen for other organisms, and life on Earth would not be able to continue.

Question 3.
A potato slice shows the presence of starch with iodine solution. Where does the starch in potatoes come from? Where is the food synthesised in the plant, and how does it reach the potato?
Answer:
The starch in a potato comes from the food made by the leaves of the potato plant through photosynthesis. This food is then transported from the leaves to other parts of the plant through the phloem. The potato, which is an underground storage part (a tuber), receives this transported food and stores it in the form of starch.

Question 4.
Does the broad and flat structure of leaves make plants more efficient for photosynthesis? Justify your answer.
Answer:
Yes, broad leaves make photosynthesis more efficient. In general, leaves are broad and flat, which allows them to capture more sunlight. Since sunlight is essential for photosynthesis, having a larger surface area helps maximise food production.

Question 5.
X is broken down using Y to release carbon dioxide, Z, and energy.
X+Y → Carbon dioxide +Z+ Energy
X, Y, and Z are three different components of the process. What do X, Y, and Z stand for?
Ans.
In the respiration equation,

• X = Glucose
• Z = Water
• Y = Oxygen

During respiration, plants use oxygen to break down glucose and release carbon dioxide, water, and energy.

Question 6.
Krishna set-up an experiment with two potted plants of same size and placed one of them in sunlight and the other in a dark room, as shown in the figure. Answer the following questions:
(i) What idea might she be testing through this experiment?
(ii) What are the visible differences in plants in both the conditions?
(iii) According to you, leaves of which plants confirm the iodine test for the presence of starch?
Answer:
(i) Krishna is testing whether sunlight is necessary for photosynthesis and production of food in plants.
(ii)

• The plant kept in sunlight looks healthy, green, and fresh with lot of green leaves.
• The plant kept in the dark room appears pale, yellowish, weak with wilted leaves because it could not prepare food without sunlight.

(iii) Only the leaves of the plant kept in sunlight will confirm the presence of starch by turning blue-black with iodine solution. This is because photosynthesis takes place only when sunlight is available.

Question 7.
Vani believes that ‘carbon dioxide is essential for photosynthesis’. She puts an experimental setup, as shown in the figure, to collect evidence to support or reject her idea. Answer the following questions:

(i) In which plant(s) in the set-up(s) will starch be formed?
(ii) In which plant(s) in the set-up(s) will starch not be formed?
(iii) In which plant(s) in the set-up(s) will oxygen be generated?
(iv) In which plant(s) in the set-up(s) will oxygen not be generated?
Answer:
(i) Starch will be formed in the plant kept in sunlight with carbon dioxide, i.e., only setup (A) will form starch.
(ii) Plants cannot make starch if either sunlight or carbon dioxide is missing. So, starch will not be formed in:

• Set-up (B): Sunlight is present but no carbon dioxide
• Set-up (C): Carbon dioxide is present but no sunlight
• Set-up (D): Neither sunlight nor carbon dioxide is present

(iii) Oxygen is released only during photosynthesis, and for that, both sunlight and carbon dioxide must be present. Thus, oxygen will be generated only in set-up (A).
(iv) Oxygen will not be generated in the following set-ups as they will fail to perform photosynthesis:

• Set-up (B): No carbon dioxide is present
• Set-up (C): No sunlight is present
• Set-up (D): Neither sunlight nor carbon dioxide are present

Question 8.
Ananya took four test tubes and filled three-fourth of each test tube with water. She labelled them A, B, C, and D. In test tube A, she kept a snail; in test tube B, she kept a water plant; in test tube C, she kept both a snail and a plant. In test tube D, she kept only water. Ananya added a carbon dioxide indicator to all the test tubes. She recorded the initial colour of water and observed if there are any colour changes in the test tubes after 2-3 hours. What do you think she wants to find out? How will she know if she is correct?

Answer:
Ananya wants to find out how living organisms affect the carbon dioxide level in water. She is checking whether plants or animals release or use carbon dioxide.
She will look for colour changes in the carbon dioxide indicator:

• If the indicator changes colour showing more carbon dioxide, it means the organism is releasing carbon dioxide (like the snail in test tube A).
• If the indicator shows less carbon dioxide, it means the organism is using carbon dioxide for photosynthesis (like the plant in test tube B).
• Test tube C (snail + plant) will show a balanced result depending on how much carbon dioxide is used or released.
• Test tube D (only water) will show no change in absence of any living organisms.

Thus, by comparing colour changes, she can confirm if her idea about carbon dioxide usage or release by plants and animals is correct.

Question 9.
Design an experiment to observe if water transportation in plants is quicker in warm or cold conditions.
Answer:
Objective: To demonstrate that temperature affects the speed of water movement (transpiration pull) in plants.
Materials required: Two similar fresh leafy stems, two beakers, water, red or blue ink (food colour can also be used), heat source (to boil water).

Procedure:

• Take two similar leafy stems of the same size and type.
• Fill two beakers with equal amounts of water mixed with coloured ink.
• Boil the water for one beaker and allow it to remain warm. Keep the water in the other beaker at room temperature.
• Place one leafy stem in each beaker.
• Leave the set-up undisturbed for some time.
• Observe the movement of coloured water upward through the stems.

Result: The stem placed in warm (boiled) water shows faster upward movement of coloured water compared to the stem in room-temperature water. This indicates that higher temperature increases the rate of transpiration pull and speeds up water transport in plants.

Question 10.
Photosynthesis and respiration are essential to maintain balance in nature. Discuss.
Answer:
Photosynthesis helps plants absorb carbon dioxide from the air and release oxygen. This oxygen is needed by all living beings for respiration. Respiration in plants and animals uses oxygen and releases carbon dioxide back into the air.
Because of these two processes, the amounts of oxygen and carbon dioxide in the atmosphere remain balanced. Thus, photosynthesis and respiration work together to maintain the natural balance of gases needed to support life on Earth.

Class 7 Curiosity Chapter 10 Question Answer

InText Questions

Question 1.
Observe a plant during its life span. Describe the changes that occur at different stages and explain the factors responsible for these changes. [Page 138]
Answer:
A plant grows from a seed into a seedling, then into a mature plant that produces flowers and fruits. These changes occur due to sunlight, water, air, nutrients, and suitable temperature.

Question 2.
How do plants obtain the food they need to grow? [Page 140]
Answer:
Plants prepare their own food by the process of photosynthesis using sunlight, carbon dioxide, water, and chlorophyll.

Question 3.
Why did we decolourise the leaf at the beginning of the activity to estimate the presence of starch? [Page 141]
Answer:
The leaf was decolourised to remove chlorophyll so that the colour change during the starch test could be clearly observed.

Question 4.
How does sunlight contribute in the production of starch in plants? [Page 141]
Answer:
Sunlight provides the energy required for photosynthesis, during which glucose is formed and later stored as starch in plants.

Question 5.
Apart from sunlight, what other factors are required by plants to prepare their food? [Page 143]
Answer:
Apart from sunlight, plants require water, carbon dioxide, and chlorophyll to prepare their food.

Question 6.
Which gas from the air is essential in the process of food preparation in plants? [Page 143]
Answer:
Carbon dioxide is the essential gas required for the preparation of food in plants.

Question 7.
Which part of the plant is involved in the synthesis of starch? [Page 144]
Answer:
The leaves are mainly involved in the synthesis of starch as they contain chlorophyll.

Question 8.
During the process of photosynthesis, do plants only take in substances from their surroundings, or do they also release any substances? [Page 145]
Answer:
Plants take in carbon dioxide and water from their surroundings, and release oxygen during photosynthesis.

Question 9.
Which part of the plant helps in the exchange of carbon dioxide and oxygen? [Page 146]
Answer:
Stomata present on the leaves help in the exchange of carbon dioxide and oxygen.

Question 10.
How do water and minerals taken up by the roots move to all parts of the plant? [Page 147]
Answer:
Water and minerals are transported to all parts of the plant through xylem.

Question 11.
How does food get transported from leaves to other parts of a plant? [Page 148]
Answer:
Food prepared in the leaves is transported to other parts of the plant through phloem.

Question 12.
Do plants respire as we do? [Page 149]
Answer:
Yes, plants also respire continuously to release energy, but the process is comparatively slower than in animals.

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 9 Life Processes in Animals Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 9 Question Answer

Science Curiosity Class 7 Chapter 9 Question Answer

Life Processes in Animals Class 7 Questions and Answers

Let Us Enhance Our Learning

Question 1.
Complete the journey of food through the alimentary canal by filling up the boxes with appropriate parts:

Answer:
Food → Mouth → Oesophagus → Stomach → Small intestine → Large intestine → Anus

Question 2.
Sahil placed some pieces of chapati in test tube A. Neha placed chewed chapati in test tube B, and Santushti took boiled and mashed potato in test tube C. All of them added a few drops of iodine solution to their test tubes A, B, and C, respectively. What would be their observations? Give reasons.
Answer:

• Sahil will observe that the solution in test tube A turns blue-black because chapati contains starch, which reacts with iodine to give a blue-black colour.
• Neha will observe that the solution in test tube B does not turn blue-black because the saliva in the chewed chapati has broken down the starch into sugar, which does not react with iodine.
• Santushti will observe that the solution in test tube C turns blue-black because boiled and mashed potato contains starch, which reacts with iodine to give a blue-black colour.

Question 3.
What is the role of the diaphragm in breathing?
(a) To filter the air
(b) To produce sound
(c) To help in inhalation and exhalation
(d) To absorb oxygen
Answer:
(c) Diaphragm helps in inhalation and exhalation.

Question 4.
Match the following.

Answer:

Name of the part	Functions
(i) Nostrils	A. Fresh air from outside enters
(ii) Nasal passages	D. Tiny hairs and mucus help to trap dust and dirt from the air we breathe
(iii) Windpipe	E. Air reaches our lungs through this part
(iv) Alveoli	B. exchange of gases occurs
(v) Rib cage	C. protects lungs

Question 5.
Anil claims to his friend Sanvi that respiration and breathing are the same process. What question(s) can Sanvi ask him to make him understand that he is not correct?
Answer:
Sanvi can ask the following questions to Anil to help him understand the difference between breathing and respiration:

• “If we stop breathing for a few minutes, can the body still utilise energy?”
• “Can we say that moving air in and out of the lungs is the same as breaking down food in the presence of oxygen to produce energy?”
• “Is it possible to produce energy without using oxygen?”

Question 6.
Which of the following statements is correct and why?
Anu: We inhale air.
Shanu: We inhale oxygen.
Tanu: We inhale air rich in oxygen.
Answer:

• The correct statement is of Tanu.
• This is because, when we inhale, we take in air that contains 21% oxygen, along with other gases (air is a mixture of gases).

Question 7.
We often sneeze when we inhale a lot of dust-laden air. What can be possible explanations for this?
Answer:
When we inhale air containing dust particles, they get trapped in the tiny hair and mucus lining our nasal passages. This irritates the sensitive inner lining of the nose, and triggers sneezing. Sneezing helps expel the dust particles from our nasal passages, allowing clean, fresh air to enter the body.

Question 8.
Paridhi and Anusha of Grade 7 started running for their morning workout. After they completed their running, they counted their breaths per minute. Anusha was breathing faster than Paridhi. Provide at least two possible explanations for why Anusha was breathing faster than Paridhi.
Answer:

• Anusha might be less physically fit than Paridhi, so her body needs more oxygen to meet the energy demands, resulting in faster breathing.
• Anusha might have run faster or for a longer duration than Paridhi, increasing her breathing rate to supply more oxygen to keep up with the energy required.

Question 9.
Yadu conducted an experiment to test his idea. He took two test tubes, A and B, and added a pinch of rice flour to the test tubes, half-filled with water and stirred them properly. To test tube B, he added a few drops of saliva. He left the two test tubes for 35-45 min. After that, he added iodine solution into both the test tubes. Experimental results are as shown in the figure given below. What do you think he wants to test?

Answer:
Yadu wants to test whether saliva can break down starch into sugar. Test tube A contains rice flour and water only, while test tube B contains rice flour, water, and saliva. By adding iodine to both test tubes, he can check if starch is still present. In the result of experiment, since iodine did not give a blue-black colour in test tube B, it confirms that saliva has broken down starch into sugar.

Question 10.
Rakshita designed an experiment taking two clean test tubes, A and B and filled them with lime water as shown in the figure. In test tube A, the surrounding air that we inhale was passed on by sucking air from the pipe, and in test tube B, the exhaled air was blown through the pipe. What do you think she is trying to investigate? How can she confirm her findings?
Answer:

• Rakshita is trying to investigate whether exhaled air contains carbon dioxide.
• Lime water turns milky in the presence of carbon dioxide.
• The lime water in test tube A will remain clear for a longer time because inhaled air has very little amount of carbon dioxide, whereas in test tube B, it will turn milky immediately because exhaled air contains carbon dioxide in higher amount.

Hence, she can confirm her findings by comparing the appearance of lime water in both the test tubes.

Class 7 Curiosity Chapter 9 Question Answer

InText Questions

Question 1.
How do different animals break down complex food molecules into simpler substances to be utilised by their bodies? Does this process work the same way in all animals? [Page 122]
Answer:
Complex food components like carbohydrates, proteins, and fats are broken down into simpler substances through digestion so the body can absorb and use them for growth and development. While all animals perform digestion, the process varies depending on their body structure and eating habits, for e.g., cows have a rumen, and birds have a gizzard to help in the digestion process.

Question 2.
Think about your favourite food. Does your mouth feel watery? [Page 122]
Answer:
Yes, the mouth feels watery when we think of our favourite food. This happens because more saliva is released when we recall or imagine our favourite food.

Question 3.
What role does the saliva play in the mouth? [Page 123]
Answer:
Saliva moistens the food, helps in chewing and swallowing, and begins the digestion of starch (complex carbohydrate) in the mouth.

Question 4.
Have you ever wondered why the taste of chapati becomes slightly sweet as you chew? [Page 123]
Answer:
Chapati tastes slightly sweet while chewing because saliva in the mouth breaks down the starch in chapati into simple sugars, which tastes sweet.

Question 5.
How does the food move down the food pipe? [Page 124]
Answer:
The walls of the food pipe contract and relax in a wave-like motion to push food down into the stomach. This movement continues throughout the alimentary canal, pushing the food forward.

Question 6.
How does the small intestine absorb nutrients from food? [Page 126]
Answer:
The small intestine absorbs nutrients through tiny finger-like projections which increase the surface area for absorption and transfer the digested nutrients into the blood, which carries them to different body parts.

Question 7.
After most of the nutrients are digested and absorbed in the small intestine, what happens to the undigested food? [Page 126]
Answer:
Undigested food passes into the large intestine, where water and minerals are absorbed and the remaining waste (in the form of stool) is stored in the rectum, and later expelled through anus by a process called egestion.

Question 8.
Why is the large intestine called ‘large’ even though it is shorter than the small intestine? [Page 126]
Answer:
It is called the large intestine because its diameter (width) is larger than that of the small intestine, even though it is shorter in length.

Question 9.
We have seen cows keep chewing the food even when they are not actively grazing or eating anything. Why? [Page 127]
Answer:
Cows are ruminants that chew partially digested food returned to the mouth through a process called rumination. This process helps in the proper digestion of food.

Question 10.
Is the process of respiration the same in all animals? [Page 128]
Answer:
No, respiration is different in different animals. Some use lungs (goat, lizard, etc.), some gills (tadpole and fish), and others breathe through skin (earthworm and adult frog underwater).

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 8 Measurement of Time and Motion Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 8 Question Answer

Science Curiosity Class 7 Chapter 8 Question Answer

Measurement of Time and Motion Class 7 Questions and Answers

Let Us Enhance Our Learning

Question 1.
Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.
Answer:
The speed of a car that travels 150 m in 10 s is
Speed = \(\frac{\text { Distance }}{\text { Time }}\) = \(\frac{150}{10}\) = 15 m/s
Convert metre per second to kilometre per hour, as follows:
= \(\frac{15 \mathrm{~m} \times \frac{\mathrm{km}}{1000 \mathrm{~m}}}{1 \mathrm{~s} \times \frac{\mathrm{h}}{3600 \mathrm{~s}}}\) = 54 km/h
[Trick: To convert m/s to km/h, multiply by \(\frac{18}{5}\)]

Question 2.
A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?
Answer:
Distance covered by each runner =400 m
Time taken by first runner = 50s
Time taken by second runner = 45s
Speed of first runner = \(\frac{400}{50}\) = 8m/s
Speed of second runner = \(\frac{400}{45}\) = 8.89m/s
Difference in speed = 8.89-8=0.89 m/s
Therefore, the second runner is faster, having a greater speed by 0.89 m/s

Question 3.
A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?
Answer:
To convert m/s to km/h, multiply by \(\frac{18}{5}\):
So, speed = 25 × \(\frac{18}{5}\) = 90 km/h
Time taken = \(\frac{\text { Distance }}{\text { Speed }}\)=\(\frac{360}{90}\) = 4 hours
Hence, the train takes 4 hours to cover a distance of 360 km.

Question 4.
A train travels 180 km in 3 h. Find its speed in:
(i) km/h
(ii) m/s
(iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?
Answer:
(i) Speed in km/h:
Speed = \(\frac{\text { Distance }}{\text { Time }}\)= \(\frac{180}{3}\)=60km/h

(ii) Speed in m/s:
To convert km/h to m/s, we multiply by \(\frac{5}{18}\). Thus, 60 × \(\frac{5}{18}\)=16.67 m/s

(iii) Distance travelled in 4 h:
Distance = Speed × Time = 60 × 4 = 240km
Hence, the train’s speed is 60 km/h or 16.7 m/s, and it will travel 240 km in 4 hours.

Question 5.
The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h ?
Answer:
To convert km/h into m/s, multiply by \(\frac{5}{18}\)
Speed of train in m/s=72 km /h × \(\frac{5}{18}\) =20m/ s
Thus, the train moves faster than the horse.
Difference in their speeds =20-18=2m/s.
Hence, the horse’s speed is 2 m/s less than the train’s speed.

Question 6.
Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.
Answer:
A car moving on a straight highway with no traffic covers equal distances in equal intervals of time. Its speed remains constant throughout the journey. Such motion, where the object moves at a constant speed along a straight line, is called uniform linear motion. On the other hand, a car moving in city traffic slows down and speeds up frequently because of traffic signals and other vehicles on the road. So it covers unequal distances in equal intervals of time. Such motion, where the speed keeps changing, is called non-uniform motion.

Question 7.
Data for an object covering distances in different intervals of time are given in the following table. If the object is in uniform motion, fill in the gaps in the table.

Answer:
The motion is uniform, which means the object covers equal distances in equal intervals of time. From the table, in every 10 -second interval, the object covers 8 metres. Hence, the distances increase by 8 m for each 10s.

Time (S)	0	10	20	30	40	50	70
Distance(m)	0	8	16	24	32	40	56

Question 8.
A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.
Answer:
Given: Total time = 3h
The car covers unequal distances (60km, 70km, and 50 km ) in equal time intervals of one hour each. Hence, the motion is non-uniform because equal intervals of time do not correspond to equal distances.
Average speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
= \(\frac{60+70+50}{3}\) = 60km/h

Question 9.
Which type of motion is more common in daily life-uniform or non-uniform? Provide three examples from your experience to support your answer.
Answer:
In daily life, non-uniform motion is more common because most objects do not move at a constant speed for long durations. Their speeds keep changing due to various factors such as road conditions, obstacles, or the need to stop and start. Examples:

• A car moving through traffic, which slows down and speeds up repeatedly.
• A bicycle on a road slows down at turns and speeds up on straight paths.
• A vehicle moving on an uneven or bumpy road experiences irregular changes in speed.

Question 10.
Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed.

Answer:
From the table, the distances covered in equal intervals of 10 seconds are not the same. For example, between 0-10 s, the object covers 6 m , and between 10-20s, it covers 4 m. Thus, the object’s motion is non-uniform.
Average speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
= \(\frac{60}{100}\) =0.6 m/s

Question 11.
A vehicle moves along a straight line and covers a distance of 2 km. In the first 500 m, it moves with a speed of 10m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?
Answer:
Given: Total distance = 2km=2000m
Total time for completing the journey =200 s
Time taken to cover the first part:
t₁= \(\frac{\text { Distance of first part }}{\text { Speed in first part }}\)=\(\frac{500}{10}\)=50s

Time taken to cover the second part:
t₁= \(\frac{\text { Distance of first part }}{\text { Speed in first part }}\)=\(\frac{500}{5}\)=100s
Total time taken so far =50+100=150s
Time left =200-150=50s
Remaining distance =2000-(500+500)=1000m

Speed required for the remaining part:
Speed = \(\frac{\text { Remaining distance }}{\text { Time left }}\)=\(\frac{1000}{50}\) = 20m/s
Now, the average speed of the vehicle for the entire journey is given by:
Average speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }} \)
= \(\frac{2000}{200}\) = 10m/s
Thus, the vehicle must move at 20 m/s for the remaining 1000 m; its average speed for the entire journey is 10 m/s.

Class 7 Curiosity Chapter 8 Question Answer

InText Questions

Question 1.
How was time measured when there were no clocks and watches? [Page 106]
Answer:
People used repeating natural events such as sunrise, sunset, moon phases and seasons. Later, simple devices like sundials, water clocks, hourglasses and candle clocks were used to measure shorter intervals of time.

Question 2.
Why does the Samrat Yantra sundial require correction to match Indian Standard Time? [Page 107]
Answer:
The Samrat Yantra at Jantar Mantar, Jaipur, is a huge stone sundial whose shadow moves at a steady rate and can measure very small time intervals as short as 2 seconds, though it shows local solar time rather than standard clock time. Local solar time depends on the Sun’s actual position in the sky at a specific place, whereas IST is a uniform time standard adopted for the entire country.

Question 3.
Why were early water clocks not very accurate? [Page 108]
Answer:
Early water clocks were not very accurate because their time measurement depended on the rate of water flow, which was difficult to keep constant. The flow changed as the water level in the container fell. Since there was no method to keep the flow of water constant, the clocks could not measure equal intervals of time accurately.

Question 4.
Does the mass of the bob affect the time period of a pendulum? [Page 110]
Answer:
No, the time period is independent of the bob’s mass. For small oscillations, it depends only on the length of the pendulum.

Question 5.
How can we compare the speeds of the runners in different races? [Page 112]
Answer:
The speeds of runners in different races can be compared by calculating their average speeds in the respective races.

Question 6.
How do we decide who is running faster in an ongoing race? [Page 113]
Answer:
The runner who has covered more distance in the same time is running faster.

Question 7.
When speed varies during a journey, what type of speed is calculated using total distance and total time? [Page 115]
Answer:
The speed calculated using the total distance covered and the total time taken is called the average speed.

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 7 Heat Transfer in Nature Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 7 Question Answer

Science Curiosity Class 7 Chapter 7 Question Answer

Heat Transfer in Nature Class 7 Questions and Answers

Let Us Enhance Our Learning

Question 1.
Choose the correct option in each case.
(i) Your father bought a saucepan made of two different materials, A and B, as shown in figure. The materials A and B have the following properties

(a) Both A and B are good conductors of heat.
(b) Both A and B are poor conductors of heat.
(c) A is a good conductor and B is a poor conductor of heat.
(d) A is a poor conductor and B is a good conductor of heat.
Answer:
(c) The body of the saucepan (A) is made of metal, which is a good conductor of heat The handle (B) is made of plastic which is a poor conductor of heat.

(ii) Pins are stuck to a metal strip with wax and a burning candle is kept below the rod, as shown in figure. Which of the following will happen?

(a) All the pins will fall almost at the same time
(b) Pins I and II will fall earlier than pins III and IV
(c) Pins I and II will fall later than pins III and IV
(d) Pins II and III will fall almost at the same time
Answer:
(d) The portion of the metal strip just above the candle flame (between Pins II and III) gets heated first. Heat will be conducted equally in both directions, causing Pins II and III to fall almost together.

(iii) A smoke detector is a device that detects smoke and sounds an alarm. Suppose you are fitting a smoke detector in your room. The most suitable place for this device will be:(a) Near the floor
(b) In the middle of a wall
(c) On the ceiling
(d) Anywhere in the room
Answer:
(c) Hot smoke rises due to convection, so it accumulates on the ceiling, allowing the detector to sense it at the earliest.

Question 2.
A shopkeeper serves you cold lassi in a tumbler. By chance, the tumbler had a small leak. You were given another tumbler by the shopkeeper to put the leaky tumbler in it. Will this arrangement help to keep the lassi cold for a longer time? Explain.
Answer:
Yes, this arrangement will help keep the lassi cold for a longer time. When the tumbler with a small leak is placed inside another tumbler, a layer of air gets trapped between the two tumblers. Since air is a poor conductor of heat (insulator), so it reduces the flow of heat from the warm surroundings into the cold lassi. Additionally, the two layers of tumblers further helps in insulation. Thus, the lassi remains cold for a longer time because heat transfer is minimised.

Question 3.
State with reason(s) whether the following statements are True [T] or False [F].
(i) Heat transfer takes place in solids through convection.
Answer:
False In solids, heat cannot be transferred by convection, because the particles cannot move freely from one place to another.

(ii) Heat transfer through convection takes place by the actual movement of particles.
Answer:
True

(iii) Areas with clay materials allow more seepage of water than those with sandy materials.
Answer:
False Clay particles are very fine and tightly packed, with very small pores, so water seeps through them very slowly. Sandy soils have larger pores that allow faster seepage.

(iv) The movement of cooler air from land to sea is called a land breeze.
Answer:
True

Question 4.
Some ice cubes placed in a dish melt into water after some time. Where do the ice cubes get heat for this transformation?
Answer:
The ice cubes get heat from their warmer surroundings. The ice absorbs heat from the air, the dish, and nearby objects at higher temperatures, causing it to melt into water.

Question 5.
A burning incense stick is fixed, pointing downwards. In which direction would the smoke from the incense stick move? Show the movement of smoke with a diagram.
Answer:

The smoke from the burning incense stick fixed pointing downwards will move upward. Smoke is a mixture of hot gases and tiny solid particles that are released during burning. As it is warmer than the surrounding air, it rises up. This upward movement of smoke happens due to convection. The movement of smoke can be represented as shown in the figure above.

Question 6.
Two test tubes with water are heated by a candle flame as shown in figure. Which thermometers (figure (A) or figure (B)) will record a higher temperature? Explain.

Answer:
The thermometer in figure (A) will record a higher temperature. The flame heats the water at the bottom directly, causing convection currents that spread heat evenly through the water. In figure (B), the heated water at the top becomes less dense but cannot rise further as it is already on top, so convection currents are not properly established. As a result, heat does not spread. efficiently through the water, and the thermometer records a lower temperature.

Question 7.
Why are hollow bricks used to construct the outer walls of houses in hot regions?
Answer:
Hollow bricks are used to construct the outer walls of houses in hot regions because the air trapped inside the holes acts as an insulator. Air is a poor conductor of heat, so it reduces the transfer of heat from the outside to the inside of the house. This helps keep the interiors cooler even in hot weather.

Question 8.
Explain how large water bodies prevent extreme temperatures in area around them.
Answer:
Water heats up and cools down more slowly than land. During the day, the land becomes hot faster, while the water remains relatively cooler. The cool air from above the water moves towards the land (sea breeze) and lowers the temperature in surrounding area.

At night, the land cools down faster, while the water remains warmer. The warmer air over the water rises, and the cooler air from the land moves toward the sea (Land Breeze). This circulation helps keep the coastal area from becoming too cold. Thus, due to the different rates of heating/cooling of land and water, areas near large water bodies experience moderate temperatures and do not face extreme temperatures.

Question 9.
Explain how water seeps through the surface of the Earth and gets stored as groundwater.
Answer:
When rain falls on the Earth’s surface, some of the water flows over the ground, while a part of it seeps down through the soil and rocks. The spaces between soil and rock particles allow water to pass through. This downward movement of water into the ground is called infiltration. The water then collects in the pore spaces of rocks and sediments beneath the surface, forming underground water reserves known as aquifers. This stored water is called groundwater, which can be obtained through wells and bore wells.

Question 10.
The water cycle helps in the redistribution and replenishment of water on the Earth. Justify the statement.
Answer:
The water cycle is a continuous process through which water moves between the atmosphere, land, and water bodies, thereby helping in the redistribution and replenishment of water on the Earth. Heat from the Sun causes water in oceans, rivers, and lakes to evaporate, and plants also release water vapour through transpiration.

This water vapour rises, cools, and condenses to form clouds. The clouds bring rain, snow, or hail through precipitation. The rainwater that falls on the Earth’s surface flows into ponds, rivers, and oceans or seeps into the ground to become groundwater.

Class 7 Curiosity Chapter 7 Question Answer

InText Questions

Question 1.
Why we generally use metal utensils for cooking, whereas we prefer clay and porcelain cups for drinking tea or coffee? [Page 91]
Answer:
Generally, metal utensils are used for cooking because metals are good conductors of heat. On the other hand, clay and porcelain are poor conductors of heat. Tea and coffee kept in these cups stays hot for a longer time.

Question 2.
Using scientific reasoning, explain why two thin blankets provide more warmth than one thick blanket. [Page 92]
Answer:
Air trapped between the layers of two thin blankets acts as a poor conductor of heat. It does not allow body heat to escape quickly. This insulation makes two thin blankets warmer than one thick blanket.

Question 3.
Is it possible to construct houses that are not affected much by the outside heat and cold? [Page 92]
Answer:
Yes, it is possible to construct houses that are not affected much by outside heat and cold by utilising the concept of heat transfer. This can be done by using insulating materials, thick walls, air gaps, and reflective surfaces, which reduce heat transfer by conduction, convection, and radiation, keeping the inside temperature more comfortable.

Question 4.
Why does smoke rise up when an incense stick (agarbatti) is burnt? [Page 93]
Answer:
Smoke is a mixture of hot gases and tiny solid particles that are released when something burns. As it is warmer than the surrounding air, it rises up.

Question 5.
How does heat transfer take place in liquids? Do liquids also rise up when heated like air? [Page 94]
Answer:
Heat transfer takes place in liquids by the process of convection. In convection, heat transfer takes place by the actual movement of particles of liquids from one place to another. Yes, liquids also rise up when heated like air. When a liquid is heated, it expands, become lighter, and rises.

Question 6.
Why are windows of the houses in coastal areas placed facing the sea? [Page 95]
Answer:
As the land gets heated faster than water during the day, it causes warm air above the land to rise. In coastal areas, this causes cooler air to move from the sea towards the land. This movement of cooler air from the sea to the land is called sea breeze. Hence, in hot places, sea breeze relieves people from the heat. That is why, windows of the houses in coastal areas are placed facing the sea.

Question 7.
Why do we feel warm sitting around a fire? [Page 96]
Answer:
We feel warm because heat is transferred directly from the fire (hot object) to us by radiation.

Question 8.
Identify the various heat transfer processes when heating water in a pan. [Page 97]
Answer:
Heat is transferred from the flame to the utensil by conduction. Subsequently, water in the utensil gets heated up by convection. The warmth that we feel around the flame and the hot utensil is due to radiation.

Question 9.
What is water cycle? List the processes that make up the water cycle. [Page 98]
Answer:
The continuous movement of water upward as water vapour and downward through precipitation, passing through soil, rocks, and plants, and finally returning to water bodies, is called water cycle. The processes that make up the water cycle are: evaporation, transpiration, condensation, precipitation.

Question 10.
How does water seep through the surface of the Earth? [Page 99]
Answer:
Water seeps through the surface of the Earth by the process of infiltration. When it rains, water falls on the ground and enters the soil through tiny pores and spaces present between soil particles.

Question 11.
Why does water seep fastest through gravel, slower through sand, and slowest through clay? [Page 100]
Answer:
Water seeps fastest through gravel, slower through sand, and slowest through clay due to differences in the size of soil particles and the spaces between them. Gravel has large particles with big gaps, allowing water to pass through quickly. Sand has smaller particles and narrower spaces, so water seeps more slowly. Clay has very fine particles with extremely small pores, which restrict the movement of water and cause it to seep very slowly.

Easy-to-read Ganita Prakash Class 7 Notes and Part 2 Chapter 7 Finding the Unknown Class 7 Notes save valuable study time during exam season.

Class 7 Maths Chapter 7 Finding the Unknown Notes

Class 7 Finding the Unknown Notes

The Weighing Scale and the Idea of Equality
A weighing scale compares weights placed on its two pans.
The number in the middle shows the total weight on both sides.

The scale is balanced only when both sides have equal total weights. The scale tilts in the direction of the heavier weight.

This is similar to the mathematical principle of equality (a = b) which states:
“Lets assume two expressions represent the same value. To maintain this equality, any operation applied to one side must be exactly applied to the other.”

Meaning of an Equation
A statement of equality between two algebraic expressions is called an equation.
We write an equation as two algebraic expressions with an equal sign ‘= ’ between them.
Here are some examples of equations:
2x + 7 = 11, 5 – 3y = 2y + 12, \(\frac{m+5}{3}\) = 3p + 2 etc.
An equation may contain one or more letter numbers (called variables) representing unknown quantities.

Solving Equations Systematically
The process of finding the value(s) of the letter-numbers for which the equality holds, or for which the value of LHS of the equation becomes equal to the value of RHS of the equation, is called solving the equation.

The specific value that satisfies the equation is known as its solution.
For example, x = 5 is the solution of equation 3x – 5 = 10 as x = 5 satisfies the equation.

There are three methods of solving an equation:

1. Trial and error method
2. Balancing / Systematic method
3. Transposition method

Trial and error method
To solve an equation using trial and error method, we follow the given steps:
Step 1: Obtain the LHS and RHS of the given equation.
Step 2: Check the different values of the variable in LHS and RHS one by one until LHS = RHS.
Step 3: The value of the variable for which LHS = RHS is the solution of the given equation.

The trial and error method is considered a crude approach because it involves testing many possible solutions one by one.

Balancing / Systematic method
An equation remains balanced when we perform the same operation (addition/subtraction/ multiplication/division) on both sides.
There are four rules which must be followed while balancing an equation.

1. If a number is added to one side of the equation, it must also be added to the other side.
2. If a number is subtracted from one side of the equation, it must also be subtracted from the other side.
3. If one side of an equation is multiplied by a number, the other side must also be multiplied by the same number.
4. If one side of an equation is divided by a non-zero number, the other side must also be divided by the same non-zero number.

Transposition Method
Transposition method is an efficient version of the balancing method.
If a term is transposed from one side of the equation to the other side, its sign changes.

If a number in multiplication (or division) is transposed from one side of the equation to the other side, its operation changes to division (or multiplication).

Solving Real Life Situations Using Equations
The word problems are first translated into an equation containing variables and constants.
Then, the equation is solved using the methods discussed earlier.

Follow the steps given below to solve a word problem:
Step 1: Obtain the given word problem.
Step 2: Study the problem carefully and identify what information is provided and what is the “unknown” or required to be found.
Step 3: Denote the unknown quantities by variables x,y, z, etc.
Step 4: Convert the language statements (or the practical situation) into a mathematical statement. For example, twice of a number is written as 2x.
Step 5: Accordingly, apply all the conditions which are given in the problem to form the equation.
Step 6: Solve the equation obtained in step 5.

The Weighing Scale and Equality

1. A weighing scale compares weights on two pans.
2. The scale is balanced when both sides have equal weight.
3. If one side is heavier and other side is lighter, then the heavier side goes down and lighter side goes up.
4. Adding (or removing) the same weight to (or from) both sides keeps the scale balanced.

Idea of an Equation
An equation shows equality between two expressions.
Example: 5x = 2x + 21, 3m – 4 = 11, 8y = 4 + 7y, etc.
It uses the equals sign (=).
Value of both sides of an equation are equal.
The value of variable that makes LHS = RHS is called the solution of the equation.

Variables and Expressions
Letters like x, y, n represent unknown numbers.
Expressions are formed using numbers and variables.
The same variable always represents the same value.

Methods of Solving Equations
The process of finding the solution of the equation is called “solving the equation’
Important methods of solving equations are:

Balancing Method
An equation remains balanced when we perform the same operation (addition / subtraction / multiplication / division) on both sides.
Rules:
Anything added to one side of the equation must also be added to the other side.

Anything subtracted from one side of the equation must also be subtracted from the other side.

If one side of an equation is multiplied by a number, the other side must also be multiplied by the same number.

If one side of an equation is divided by a non-zero number, the other side must also be divided by the same number.

Transposition Method
Transposition method is an efficient version of balancing method.

If a term is transposed from one side of the equation to the other side, its sign changes. For example, 5x + 2 = 4 ⇒ 5x = 4 – 2 If a number in multiplication (or division) is transposed from one side of the equation to the other side, its operation changes to division (or multiplication).
For example, 3x = 15 ⇒ x = \(\frac{15}{3}\)

Solving Word Problems
Follow the path given below to solve a word problem:

Easy-to-read Ganita Prakash Class 7 Notes and Part 2 Chapter 6 Constructions and Tilings Class 7 Notes save valuable study time during exam season.

Class 7 Maths Chapter 6 Constructions and Tilings Notes

Class 7 Constructions and Tilings Notes

A geometric construction is a step-by-step method of drawing lines, angles and shapes with accuracy, without guessing or drawing freehand.
Each construction is based on mathematical reasoning.

Perpendicular Bisector
A line that bisects a given line and is perpendicular to it, is called the perpendicular bisector.

Construction of Perpendicular Bisector:

Step 1: Draw the given line segment, say AB.
Step 2: Open the compass such that its radius is greater than half of AB. Place the compass’s pointer on point A and draw two arcs, one above and one below the line segment.
Step 3: Keeping the same radius (do not change the compass setting), place the pointer on B. Draw another two arcs intersecting the previous arcs at points P and Q.
Step 4: Using a ruler, draw a line passing through P and Q. The line PQ is the required perpendicular bisector of AB, intersecting it at point R. Here, R is the midpoint of AB.

Angle Bisector
An angle bisector is a ray drawn from the vertex of an angle that makes the two parts of the angle exactly of the same measure.

For example, in the figure, ray KM divides the ∠LKJ measuring 60° into two equal parts, the measure of each smaller angle is equal to 30°, i.e. ∠LKM = ∠JKM = 30°.

Additionally, ray KM is called angle bisector of ∠LKJ.

Construction of an Angle Bisector:

Step 1: Draw the given angle, say ∠ABC.
Step 2: Place compass pointer on B. Open it to an arbitrary radius and draw an arc intersecting both arms of ∠ABC at points P and Q. This ensures BP = BQ.
Step 3: With P as the centre and a radius more than half of PQ, draw an arc. With the same radius and Q as the centre, draw another arc intersecting the previous arc at point X.
Step 4: Join BX. The ray BX is the required angle bisector of ∠ABC.

Construction of an Angle Equal to Given Angle
Suppose we have to copy an angle ∠BAC i.e. to make an angle equal to ∠BAC. However, we do not know the measure of ∠BAC. For this we follow the given steps:

Step 1: Draw a ray PQ with initial point P.
Step 2: Place the compass pointer at vertex A. Open it to a convenient/arbitrary radius and draw an arc intersecting both arms of ∠BAC at points J and K.
Step 3: Taking caution and making sure that the opening (or the radius) of the compass does not change, place the pointer on P. Draw an arc intersecting ray PQ at point M.

Step 4: Place the compass pointer on K and adjust the width so that pencil lead rests exactly on measuring the distance KJ).
Step 5: Again, taking caution that the radius of the compass has not changed, place the pointer on M and draw a new arc that intersects the previous arc at point L.

Step 6: Draw a ray starting from point P and passing through point L using a ruler.
Step 7: Label a point R on ray PL. ∠QPR is required angle, which is an exact copy of ∠BAC.

Construction of an Angle Measuring 60°

To draw 60° angle, we follow the given steps:
Step 1: Draw a ray OQ with initial point O.
Step 2: Taking 0 as the centre and with the convenient radius, draw an arc intersecting the ray OQ at point A.
Step 3: With A as the centre and using the same radius as step 2, draw another arc to intersect the first arc at point B.
Step 4: Draw ray OB. ∠AOB is required angle of measure 60°.

Construction of a line parallel to a Given Line
If a transversal intersects two lines and the corresponding angles formed are equal, then the two lines are parallel.

To draw a line through B which is parallel to line m, we follow the given steps:
Step 1: Take any point A on line m. Join B and A to form a transversal line l.
Step 2: With A as centre and any convenient radius, draw an arc intersecting line m at point
C and line l at point D.
Step 3: Using the same radius as in step 2, draw another arc with B as centre. Let this arc intersect line l at point E.
Step 4: Place the pointed tip of the compass at D and adjust the opening so that the pencil lead is at C. This measures the length CD.
Step 5: With E as centre and the radius equal to length CD, draw an arc. This arc will cut the previous arc (drawn in step 3) at point F.
Step 6: Join B and F with a ruler to get a new line n.
The line n drawn through point B is parallel to line m, because the corresponding angles formed are equal, i.e. ∠EBF = ∠DAC.

Arch Designs

Arch designs have a symmetrical and geometric structure.

To create these arches, one must first draw accurate supporting lines on a flat surface like paper or stone.

A trefoil arch is a type of arch that has three rounded curves joined together, forming a shape like three small overlapping circles. The word “trefoil” means three leaves, so the arch looks a bit like a three-leaf clover.

Symmetry is essential for a trefoil arch. It must posses line symmetry to ensure the structure is balanced.

A pointed arch is formed by two arcs that meet at a sharp top point.

It uses two equal line segments placed at an angle. By locating their midpoints, arcs are drawn using the wavy wave method

Regular Hexagons

A regular polygon is a shape whose all sides are equal in length and all angles are equal in measure.

A regular hexagon can be visualised as being made up of six identical equilateral triangles.

The six triangles form a closed hexagonal shape.

The outer boundary becomes a regular hexagon.

Each interior angle of the hexagon is 120° (since two 60° angles meet at each vertex of hexagon).

Lines joining opposite vertices (diagonals) of the hexagon pass straight through the centre.

Tiling
Tiling means covering a region completely with given shapes, without any gaps or overlaps. One important type of tiling problem is to check whether the total number of unit squares in a given region can be fully covered by a given type of tile.

In any tiling problem, the first and most important step is a counting check, i.e. the total number of unit squares in the grid must be divisible by the area of one tile.

In general, an m × n rectangular grid can be tiled using 2×1 dominoes if and only if m × n is even, which means at least one of m orn must be even.

Certain regular shapes, like squares, equilateral triangles, and regular hexagons, can tile the entire plane without leaving gaps or overlaps. For example, regular hexagons form a natural tiling pattern, commonly seen in bee hives.

Other regular polygons (like Pentagon) result in unavoidable gaps or overlaps, so they cannot tile a flat surface (plane) perfectly.

Types of Bisector
Perpendicular bisector:
A line that bisects a given line and is perpendicular to it, is called the perpendicular bisector

PQ is perpendicular bisector of AB.

Angle bisector: A ray drawn from the vertex of an angle that divides the angle into two equal parts is called the angle bisector.

Here, BX is angle bisector of ∠ABC.

Copying An Angle
Copying an angle means constructing an angle that is exactly equal in measure to a given angle. We can copy a given angle using geometric instruments (a ruler and a compass), without measuring the angle with a protractor.
Consider the given angle to be ∠BAC.
Steps of construction to copy an angle:

Step 1: Draw a ray PQ with P as the initial point.
Step 2: With the compass at A and any radius, draw an arc cutting the arms of ∠BAC at J and K.

Step 3: Without changing the compass opening, place it at P and draw an arc cutting PQ at M.
Step 4: Place the compass at K and open it to reach J (measure KJ).

Step 5: Keeping the same opening, place the compass at M and draw an arc intersecting the previous arc at L.
Step 6: Using a ruler, draw a ray from P through L.
Step 7: Mark any point R on this ray. ∠RPQ is the required angle, equal to ∠BAC.

Tilings
Tiling (or tessellation) is the process of covering a flat region using a set of shapes so that there are absolutely no gaps or overlaps.

In general, a regular polygon can tile the plane only if the interior angle divides 360° exactly (whole number of times).

The hexagonal tiling is highly efficient; it is famously used by bees in honeycombs because it covers the area without wasting space and uses the minimum amount of wax to store the maximum amount of honey.

Tangram
A tangram is a mathematical puzzle made by cutting a square into seven geometric pieces, called tans.

Go through BSE Odisha Class 7 Science Solutions and Class 7 Science Chapter 6 Adolescence: A Stage of Growth and Change Question Answer to understand textbook questions more clearly.

Class 7 Science Curiosity Chapter 6 Question Answer

Science Curiosity Class 7 Chapter 6 Question Answer

Adolescence: A Stage of Growth and Change Class 7 Questions and Answers

Let Us Enhance Our Learning

Question 1.
Ramesh, an 11-year-old boy, developed a few pimples on his face. His mother told him that this is because of ongoing biological changes in his body.
(i) What could be the possible reasons for the development of these pimples on his face?
(ii) What can he do to get some relief from these pimples?
Answer:
(i) The development of pimples on Ramesh’s face could be due to the hormonal changes and oily secretions from the skin that occur during adolescence.
(ii)

• Avoid squeezing or picking the pimples.
• Maintain good body hygiene.
• Drink plenty of water.
• Use a gentle cleanser for the face.

Question 2.
Which of the following food groups would be a better option for adolescents and why?

Answer:
The food group (B), would be a better choice for adolescents because it contains sources of all essential nutrients, i.e., carbohydrates, proteins, fats, vitamins, and minerals that are necessary for supporting the rapid growth and development occurring during this period. On the other hand, food group (A) contains various junk foods that are high in fats and sugar but low in essential nutrients.

Question 3.
Unscramble the underlined word in the following sentences:
(i) The discharge of blood in adolescent girls which generally occurs every 28-30 days is nstmnoiaretu.
(ii) The hoarseness in the voice of adolescent boys is due to enlarged iceov xob.
(iii) Secondary sexual characteristics are natural signs that the body is preparing for adulthood and mark the onset of urtypeb.
(iv) We should say NO to lahoclo and srugd as they are addictive.
Answer:
(i) nstmnoiaretu – menstruation
(ii) iceov xob – voice box
(iii) urtypeb – puberty
(iv) lahoclo – alcohol, srugd – drugs

Question 4.
Shalu told her friend, “Adolescence brings only physical changes, like growing taller or developing body hair.” Is she correct? What would you change in this description of adolescence?
Answer:
No, Shalu is partially correct because adolescence not only brings physical changes like growing taller or the appearance of body hair, but also includes internal (biological) changes indicating reproductive capability, as well as emotional and behavioural changes.

Question 5.
During a discussion in the class, some of the students raised the following points. What questions would you ask them to check the correctness of these points?
(i) Adolescents do not need to worry about behavioural changes.
(ii) If someone tries a harmful substance once, they can stop anytime they want.
Answer:
The following questions can be asked to check the correctness of points made by the students:
(i)

• Can ignoring certain behavioural changes affect mental health or social relationships later on?
• Why do you think it is important for adolescents to adapt to changes in a positive way rather than worrying about them?
• Are behavioural changes during adolescence linked to hormonal changes, and how can one handle these changes?

(ii)

• Can harmful substances like tobacco and alcohol cause addiction, and is it easy to stop using them once a person becomes dependent?
• How does peer pressure or emotional state influence the ability to stop using a substance?
• Has any research shown that people can easily quit using harmful substances after trying them just once?

Question 6.
Adolescents sometimes experience mood swings. On some days, they feel very energetic and happy, while on other days, they may feel low. What other behavioural changes are associated with this age?
Answer:
Adolescents undergo several behavioural changes, such as:

• A strong desire for independence, developing their own identity, and personal decision making.
• A desire to explore new interests and engage in varied activities like music, dance, or sports.
• Increased influence of peers, attraction to them, and trying to mimic their behaviour.
• Intense emotions, frequent mood swings and increased sensitivity.
• Getting involved in social work and initiatives to support the needy and the disadvantaged.

Question 7.
While using a toilet, Mohini noticed that used sanitary pads were scattered near the bin. She got upset and shared her feelings with her friends. They discussed the importance of menstrual hygiene and healthy sanitary habits. What menstrual hygiene and sanitary habits would you suggest to your friends?
Answer:
The following menstrual hygiene and sanitary habits should be followed:

• Properly disposing used sanitary pads by wrapping them in paper and disposing them in a dustbin.
• Using clean, reusable cloth pads or sanitary napkins.
• Washing hands with soap and water before and after handling sanitary products.
• Cleaning the pubic area regularly to maintain hygiene and prevent bacterial or fungal infections.

Question 8.
Mary and Manoj were classmates and good friends. On turning 11, Mary developed a little bulge on the front of her neck. She visited the doctor who gave her medication and asked to take iodine-rich diet. Similarly, a bump was developed on the front of Manoj’s neck when he turned 12. However, the doctor told him that it was a part of growing up. According to you, what could be the possible reason for advising Mary and Manoj differently?
Answer:
The doctor advised Mary and Manoj differently because the causes of their neck bulges were different.

• In Mary’s case, the bulge on her neck was likely due to goitre, a condition caused by iodine deficiency. Iodine is essential for performing physical and mental activities efficiently. When the body lacks iodine, it leads to a disease called goitre and a prominent bulge or swelling on the front of the neck is a characteristic symptom of this disease. Therefore, the doctor advised her to take an iodine-rich diet and prescribed required medication.
• On the other hand, in Manoj’s case, the bulge on his neck was due to the natural growth of the voice box during puberty. This results in the formation of the Adam’s apple, which is a normal part of male development and also causes the voice to become deeper and hoarse.

Question 9.
During adolescence, the boys and girls undergo certain physical changes, a few of which are given below.
(i) Change in voice
(ii) Development of breasts
(iii) Growth of moustache
(iv) Growth of facial hair
(v) Pimples on the face
(vi) Growth of hair in the pubic region
(vii) Growth of hair in armpits
Categorise these changes in the table given below:

Answer:

Physical changes during adolescence
Observed only in boys	Observed only in girls	Common in boys and girls
Change in voice	Development of breasts	Pimples on the face
Growth of moustache		Growth of hair in the pubic region
Growth of facial hair		Growth of hair in armpits

Question 10.
Prepare a poster mentioning the tips for adolescents to live a healthy lifestyle.
Answer:
The students should prepare the poster themselves.

Class 7 Curiosity Chapter 6 Question Answer

InText Questions

Question 1.
What changes are most commonly seen during adolescence? [Page 75]
Answer:
Commonly observable changes during adolescence are:

• Increase in height and weight
• Appearance of hair in different parts of the body, like the armpits and pubic region
• Change in voice (more prominent in boys)
• Oily secretions from the skin causing pimples on the face

Question 2.
Is adolescence just about physical changes or changes associated with reproductive capability? [Page 78]
Answer:
No, adolescence also brings about significant emotional and behavioural changes along with other changes.

Question 3.
Does an ideal nutritious diet for adolescents mean consuming only protein and carbohydrate-rich foods? [Page 79]
Answer:
For proper growth and to perform well in studies and sports, we not only require proteins and carbohydrates, but also an adequate amount of fats, vitamins, and minerals, along with a proper intake of dietary fibre and water.

Question 4.
In what ways do nutrients support our growth and development? [Page 79]
Answer:
Nutrients such as proteins, fats, carbohydrates, vitamins and minerals play a crucial role in keeping our body healthy. These nutrients are obtained from foods like milk, curd, cheese, spinach, millets, kidney beans, and dried fruits such as raisins and figs.
These nutrients help our body in the following ways:

• Support the growth and development of bones and muscles
• Increase strength and help build a healthy body
• Aid in the formation of blood
• Improve energy levels

Question 5.
Adolescents, especially girls, may suffer from blood-related health problems due to a deficiency of iron or vitamin B12. Name and describe such health problem(s). [Page 80]
Answer:
Anaemia is a blood-related health problem caused by a deficiency of iron or vitamin B12. It leads to low haemoglobin levels, resulting in weakness, fatigue, and pale skin. Adolescent girls are more prone to anaemia due to rapid growth and blood loss during menstruation.

Question 6.
How can we manage iron deficiency in our body? [Page 80]
Answer:
Iron deficiency can be prevented and managed by including iron-rich foods in our diet, such as beans, nuts, dried apricots, spinach, lentils, legumes, broccoli, and fortified cereals. In some cases, doctors may also recommend iron supplements to maintain healthy iron levels in the body.

Question 7.
What are some sources of vitamin B12? [Page 80]
Answer:
Vitamin B12 is mainly obtained from foods such as eggs, meat, liver, fish, cheese, milk, fortified cereals, nutritional yeast, and green leafy vegetables.

Question 8.
List some government schemes that aim to prevent deficiency diseases like anaemia. [Page 80]
Answer:
In India, the Adolescent Anaemia Control Programme focuses on reducing the prevalence and severity of anaemia among adolescents and supports the government’s Anaemia Mukt Bharat initiative under the National Health Mission.
The Weekly Iron and Folic Acid Supplementation (WIFS) programme is a key component of the Anaemia Mukt Bharat strategy. It involves the supervised weekly consumption of Iron and Folic Acid (IFA) tablets by both school-going and outof-school adolescents to prevent iron-deficiency anaemia.

Question 9.
Why do changes occur during the adolescent life stage? [Page 84]
Answer:
Many changes during adolescence, including menstruation and other signs of puberty, are primarily due to the production of certain chemicals in our bodies called hormones.