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RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 PDF Download

Q.1. Compare the fractions:
(i) 5/8 and 7/12
(ii) 5/9 and 11/15
(iii) 11/12 and 15/16
Ans. We have the following:
(i) 5/8 and 7/12
By cross multiplication, we get:
5 × 12 = 60 and 7 × 8 = 56
However, 60 > 56
∴ 5/8 > 7/12
(ii) 5/9 and 11/15
By cross multiplication, we get:
5 × 15 = 75 and 9 × 11 = 99
However, 75 < 99
∴ 5/9 < 11/15
(iii) 11/12 and 15/16
By cross multiplication, we get:
11 × 16 = 176 and 12 × 15 = 180
However, 176 < 180
∴ 11/12 < 15/16

Q.2. Arrange the following fractions in ascending order:
(i) 3/4, 5/6, 7/9, 11/12
(ii) 4/5, 7/10, 11/15, 17/20
Ans. (i) The given fractions are 3/4, 5/6, 7/9 and 11/12.
LCM of 4, 6, 9 and 12 = 36
Now, let us change each of the given fractions into an equivalent fraction with 72 as its denominator.
3/4 = (3×9) / (4×9) = 27/36
5/6 = (5×6) / (6×6) = 30/36
7/9 = (7×4) / (9×4) = 28/36
11/12 = (11×3) / (12×3) = 33/36
Clearly, 27/36 < 28/36 < 30/36 < 33/36
Hence, 3/4 < 7/9 < 5/6 < 11/12
∴ The given fractions in ascending order are 3/4, 7/9, 5/6 and 11/12
(ii) The given fractions are: 4/5, 7/10, 11/15 and 17/20.
LCM of 5, 10, 15 and 20 = 60
Now, let us change each of the given fractions into an equivalent fraction with 60 as its denominator.
4/5 = (4×12) / (5×12) = 48/60
7/10 = (7×6) / (10×6) = 42/60
11/15 = (11×4) / (15×4) = 44/60
17/20 = (17×3) / (20×3) = 51/60
Clearly, 42/60 < 44/60 < 48/60 < 51/60
Hence, 7/10 < 11/15 < 4/5 < 17/20
∴ The given fractions in ascending order are
7/10, 11/15, 4/5 and 17/20

Q.3. Arrange the following fractions in descending order:
(i) 3/4, 7/8, 7/12, 17/24
(ii) 2/3, 3/5, 7/10, 8/15
Ans. We have the following:
(i) The given fractions are 3/4, 7/8, 7/12 and 17/24.
LCM of 4,8,12 and 24 = 24
Now, let us change each of the given fractions into an equivalent fraction with 24 as its denominator.
3/4 = (3×6) / (4×6) = 18/24
7/8 = (7×3) / (8×3) = 21/24
7/12 = (7×2 / 12×2) = 14/24
17/24 = (17×1) / (24×1) = 17/24
Clearly, 21/24 > 18/24 > 17/24 > 14/24
Hence, 7/8 > 3/4 > 17/24 > 7/12
∴ The given fractions in descending order are 7/8, 3/4, 17/24 and 7/12.
(ii) The given fractions are 2/3, 3/5, 7/10 and 8/15.
LCM of 3,5,10 and 15 = 30
Now, let us change each of the given fractions into an equivalent fraction with 30 as its denominator.
2/3 = (2×10) / (3×10) = 20/30
3/5 = (3×6) / (5×6) = 18/30
7/10 = (7×3) / (10×3) = 21/30
8/15 = (8×2) / (15×2) = 16/30
Clearly, 21/30 > 20/30 > 18/30 > 16/30
Hence, 7/10 > 2/3 > 3/5 > 8/15
∴ The given fractions in descending order are 7/10, 2/3, 3/5 and 8/15.

Q.4. Reenu got 2/7 part of an apple while Sonal got 4/5 part of it. Who got the larger part and by how much?
Ans. We will compare the given fractions 2/7 and 4/5 in order to know who got the larger part of the apple.
We have,
By cross multiplication, we get: 2 × 5 = 10 and 4 × 7 = 28
However, 10 < 28
∴ 2/7 < 4/5
Thus, Sonal got the larger part of the apple.
Now, 4/5 - 2/7 = (28-10) / 35 = 18/35
∴ Sonal got 18/35 part of the apple more than Reenu.

Q.5. Find the sum
(i) 5/9 + 3/9
(ii) 8/9 + 7/12
(iii) 5/6 + 7/8
(iv) 7/12 + 11/16 + 9/24

Ans. (i) 5/9 + 3/9 = 8/9
(ii) 8/9 + 7/12
= 32/36 + 21/36 [∵ LCM of 9 and 12 = 36]
= (32+21) / 36
= 53/36
= RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
(iii) 5/6 + 7/8 = 20/24 + 21/24 [∵ LCM of 6 and 8 = 24]
= (20+21) / 24
= 41/24
=
(iv) 7/12 + 11/16 + 9/24
28/48 + 33/48 + 18/48 [∵ LCM of 12, 16 and 24 = 48]
= (28+33+18) / 48
= 79/48
=
(v)= 19/5 + 23/10 + 16/15
= 114/30 + 69/30 + 32/30 [∵ LCM of 5, 10 and 15 = 30]
= (114+69+32) / 30
= 215/30
= RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
(vi)
= 35/4 + 52/5
= 175/20 + 208/20 [∵ LCM of 4 and 5 = 20]
= (175+208) / 20
= 383/20
=

Q.6. Find the difference:
(i) 5/7 - 2/7
(ii) 5/6 - 3/4
(iii)
(iv) 7 -
(v)
(vi)
Ans. (i) 5/7 - 2/7 = (5-2) / 7 = 3/7
(ii) 5/6 - 3/4 = 10/12 - 9/12 [∵ LCM of 6 and 4 = 12]
= (10-9) / 12
= 1/12
(iii) RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 - 7/10 = 16/5 - 7/10
= 32/10 - 7/10 [∵ LCM of 5 and 10 = 10]
= (32-7) / 10
= 25/10
= 5/2
=
(iv) 7 -
= 7/1 - 14/3
= (21-14) / 3 [∵ LCM of 1 and 3 = 3]
= 7/3
=
(v)
= 33/10 - 22/15
= (99-44) / 30 [∵ LCM of 10 and 15 = 30]
= 55/30
= 11/6
= RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
(vi)
= 23/9 - 22/15
= (115-66) / 45 [∵ LCM of 9 and 15 = 45]
= 49/45
=

Q.7. Simplify:
(i) 2/3 + 5/6 - 1/9
(ii) 8 -
(iii)
Ans. (i) 2/3 + 5/6 - 1/9
= (12+15-2) / 18 [∵ LCM of 3, 6 and 9 = 18]
= (27-2) / 18
= 25/18
=
(ii) 8 - RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
= 8/1 - 9/2 - 9/4
= (32-18-9) / 4 [∵ LCM of 1, 2 and 4 = 4]
= (32-27) / 4
= 5/4
=
(iii)
= 53/6 - 27/8 + 19/12
= (212-81+38) / 24 [∵ LCM of 6, 8 and 12 = 24]
= (250-81) / 24
= 169/24
=

Q.8. Aneeta boughtkg apples andkg guava. What is the total weight of fruits purchased by her?
Ans. Total weight of fruits bought by Aneeta = RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 kg
Now, we have:
= 15/4 + 9/2
= (15 + 18) / 4 [∵ LCM of 2 and 4 = 4]
= (15 + 18) / 4
= 33/4
=
Hence, the total weight of the fruits purchased by Aneeta iskg.

Q.9. A rectangular sheet of paper iscm long andcm wide. Find its perimeter.
Ans. We have:
RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
Perimeter of the rectangle ABCD = AB + BC + CD +DA
=cm
= (63/4 + 25/2 + 63/4 + 25/2) cm
= (63 + 50 + 63 + 50) / 4 cm [∵ LCM of 2 and 4 = 4]
= (226 / 4) cm
=(113 / 2) cm
=cm
Hence, the perimeter of ABCD iscm

Q.10. A picture iscm wide. How much should it be trimmed to fit in a framecm wide?
Ans. Actual width of the picture = RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 cm = 38/5 cm
Required width of the picture =cm = 73/10 cm
∴ Extra width = (38/5−73/10)cm
= (76−73) / 10 cm [∵ LCM of 5 and 10 is 10]
= 3/10 cm
Hence, the width of the picture should be trimmed by 3/10 cm.

Q.11. What should be added toto get 18?
Ans. Required number to be added = 18− RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
=18/1−38/5
= (90−38) / 5 [∵ LCM of 1 and 5 = 5]
= 52/5 =
Hence, the required number is

Q.12. What should be added toto get?
Ans. Required number to be added =
= 42/5−109/15
= (126−109)/15 [∵ LCM of 5 and 15 = 15]
= 17/15
= RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
Hence, the required number should be 12151215.

Q.13. A piece of wirem long broke into two pieces. One piece ism long. How long is the other piece?
Ans. Required length of other piece of wire = m
=(15/4−3/2) m
= ((15−6)/4) m [∵ LCM of 4 and 2 = 4]
= 9/4 m
= RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 m
Hence, the length of the other piece of wire ism.

Q.14. A film show lasted ofhours. Out of this timehours was spent on advertisements. What was the actual duration of the film?
Ans. Actual duration of the film = ( − RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 )hours
= (11/3−3/2) hours
= (22−9)/ 6 hours [∵ LCM of 3 and 2 = 6]
= 13/6 hours
= hours
Hence, the actual duration of the film was hours

Q.15. Of 2/3 and 5/9, which is greater and by how much?
Ans. First we have to compare the fractions: 2/3 and 5/9.
By cross multiplication, we have:
2 × 9 = 18 and 5 × 3 = 15
However, 18 > 15
∴2/3 > 5/9
So, 2/3 is larger than 5/9.
Now, 2/3−5/9
= (6−5) / 9 [∵ LCM of 3 and 9 = 9]
= 1/9
Hence, 2/3 is 1/9 part more than 5/9.

Q.16. The cost of a pen is Rsand that of a pencil is Rs. Which costs more and by how much?
Ans. First, we have to compare the cost of the pen and the pencil.
Cost of the pen = Rs RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7 = Rs 83/5
Cost of the pencil = Rs = Rs 19/4
Now, we have to compare fractions 83/5 and 19/4.
By cross multiplication, we get: 83 × 4 = 332 and 19 × 5 = 95
However, 332 > 95
∴ 83/5 > 19/4
So, the cost of pen is more than that of the pencil.
Now, Rs (83/5 − 19/4)
= Rs (332 − 95) / 20 [∵ LCM of 4 and 5 = 20]
= Rs 237
= Rs RS Aggarwal Solutions: Integers (Exercise 2A) | Mathematics (Maths) Class 7
∴ The pen costs Rsmore than the pencil.

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FAQs on RS Aggarwal Solutions: Integers (Exercise 2A) - Mathematics (Maths) Class 7

1. What are integers?

Ans. Integers are a set of whole numbers, including both positive and negative numbers, along with zero.

2. How are integers different from natural numbers?

Ans. Integers include all the natural numbers along with their negatives and zero, whereas natural numbers only include positive whole numbers starting from 1.

3. Can integers be fractions or decimals?

Ans. No, integers cannot be fractions or decimals. Integers are whole numbers without any fractional or decimal part.

4. Are all negative numbers considered integers?

Ans. Yes, all negative numbers are considered integers. In the set of integers, negative numbers are included along with positive numbers and zero.

5. How are integers used in real-life situations?

Ans. Integers are used in various real-life situations such as temperature measurements (positive and negative temperatures), money transactions (positive and negative values), and determining the direction (positive and negative coordinates).

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