Class 7 Exam  >  Class 7 Notes  >  Mathematics (Maths) Class 7 (Old NCERT)  >  RS Aggarwal Solutions: Fractions (Exercise 2A)

RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT) 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
RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
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: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
(iii) 5/6 + 7/8 = 20/24 + 21/24 [∵ LCM of 6 and 8 = 24]
= (20+21) / 24
= 41/24
=RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
(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
=RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
(v)RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)= 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: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
(vi)RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
= 35/4 + 52/5
= 175/20 + 208/20 [∵ LCM of 4 and 5 = 20]
= (175+208) / 20
= 383/20
=RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)

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

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

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

Q.9. A rectangular sheet of paper isRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm long andRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm wide. Find its perimeter.
Ans. We have:
RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
Perimeter of the rectangle ABCD = AB + BC + CD +DA
=RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)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
=RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm
Hence, the perimeter of ABCD isRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm

Q.10. A picture isRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm wide. How much should it be trimmed to fit in a frameRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm wide?
Ans. Actual width of the picture = RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)cm = 38/5 cm
Required width of the picture =RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)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 toRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)to get 18?
Ans. Required number to be added = 18−RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
=18/1−38/5
= (90−38) / 5 [∵ LCM of 1 and 5 = 5]
= 52/5 =RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
Hence, the required number isRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)

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

Q.13. A piece of wireRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)m long broke into two pieces. One piece isRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)m long. How long is the other piece?

Ans. Required length of other piece of wire = RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)m
=(15/4−3/2) m
= ((15−6)/4) m [∵ LCM of 4 and 2 = 4]
= 9/4 m
=RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)m
Hence, the length of the other piece of wire isRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)m.

Q.14. A film show lasted ofRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)hours. Out of this timeRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)hours was spent on advertisements. What was the actual duration of the film?

Ans. Actual duration of the film = (RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT) −RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT))hours
= (11/3−3/2) hours
= (22−9)/ 6 hours [∵ LCM of 3 and 2 = 6]
= 13/6 hours
RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)hours
Hence, the actual duration of the film was RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)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 RsRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)and that of a pencil is RsRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT). 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 = RsRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)= Rs 83/5
Cost of the pencil = Rs RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)= 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
= RsRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)
∴ The pen costs RsRS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT)more than the pencil.

The document RS Aggarwal Solutions: Fractions (Exercise 2A) | Mathematics (Maths) Class 7 (Old NCERT) is a part of the Class 7 Course Mathematics (Maths) Class 7 (Old NCERT).
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FAQs on RS Aggarwal Solutions: Fractions (Exercise 2A) - Mathematics (Maths) Class 7 (Old NCERT)

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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