Class 7 Maths RS Aggarwal Solutions

RS Aggarwal Class 7 Solutions Chapter 3 Decimals Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3B
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3C
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3D
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3E
• RS Aggarwal Solutions Class 7 Chapter 3 Decimals CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
(i) 6.5, 16.03, 0.274, 119.4
In these decimals, the greatest places of decimal is 3
6.5 = 6.500
16.03 = 16.030
0. 274 = 0.274
119.4 = 119.400 are like decimals.
(ii) 3.5, 0.67, 15.6, 4
In these decimal, the greatest place of decimal is 2
3.5 = 3.50
0.67 = 0.67
15.6 = 15.60
4 = 4.00 are the like decimals

Question 5.
Solution:
(i) Among 78.23 and 69.85,
78.23 is greater than 69.85 (78 > 69)
78.23 > 69.85
(ii) Among 3.406 and 3.46,
3.406 is less than 3.46 (40 < 46)
3.406 < 3.46
(iii) Among 5.68 and 5.86,
5.68 is less than 5.86 (68 < 86)
5.68 < 5.86
(iv) Among 14.05 and 14.005
14.5 is greater than 14.005 (05 > 00)
14.5 >14.005
(v) Among 1.85 and 1.805,
1.85 is greater than 1.805 (85 > 80)
1.85 > 1.805
(vi) Among 0.98 and 1.07,
0.98 is less than 1.07 (0 < 1)
0.98 < 1.07

Question 6.
Solution:
(i) 4.6, 7.4, 4.58, 7.32, 4.06
Converting the given decimals into like decimals, we get:
4.60, 7.40, 4.58, 7.32, 4.06.
We see that 4.06 < 4.58 < 4.60 < 7.32 < 7.40.
Writing in ascending order, 4.06, 4.58, 4.6, 7.32, 7.4
(ii) 0.5, 5.5, 5.05, 0.05, 5.55
Converting the given decimals into like decimals, we get:
0. 50, 5.50, 5.05, 0.05, 5.55
We see that 0.05 < 0.50 < 5.05 < 5.50 < 5.55.
Writing in ascending order, 0.05, 0.50, 5.05, 5.5, 5.55
(iii) 6.84, 6.84, 6.8, 6.4, 6.08
Converting the given decimals into like decimals
6.84, 6.48, 6.80, 6.40, 6.08
We see that 6.08 < 6.40 < 6.48 < 6.80 < 6.84
Writing in ascending order,
6.08, 6.4, 6.48, 6.8, 6.84
(iv) 2.2, 2.202, 2.02, 22.2, 2.002
Converting them into like decimals
2.200, 2.202, 2.020, 22.200, 2.002 we see that
2.002 < 2.020 < 2.200 < 2.202 < 22.200
Now writing in ascending order,
2.002, 2.020, 2.2, 2.202, 22.2

Question 7.
Solution:
(i) 7.4, 8.34, 74.4, 7.44, 0.74
Converting them into like decimals,
7.40, 8.34, 74.40, 7.44, 0.74
we see that
74.40 > 8.34 > 7.44 > 7.40 > 0.74
Writing in descending order,
74.4, 8.34, 7.44, 7.4, 0.74
(ii) 2.6, 2.26, 2.06, 2.007, 2.3
Converting them into like decimals,
2.600, 2.260, 2.060, 2.007, 2.300
We see that
2.600 > 2.300 > 2.260 > 2.060 > 2.007
Writing in descending order,
2.6, 2.3, 2.26, 2.06, 2.007

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 3 Decimals Ex 3A are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:
(i) A number of the form \(\frac { a }{ b }\), where a and b are rational numbers, is called a natural number.
Here, a is the numerator and b is the denominator.
(a) \(\frac { 2 }{ 3 }\) is a fraction with 2 as the numerator and 3 as the denominator.
(b) \(\frac { 12 }{ 5 }\) is a fraction with 12 as the numerator and 5 as the denominator.
(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples : \(\frac { 2 }{ 5 }\) and \(\frac { 4 }{ 15 }\).
(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Example : \(\frac { 11 }{ 3 }\) and \(\frac { 41 }{ 35 }\)

Question 2.
Solution:
Required number to be added

Hence, the required number is 8\(\frac { 2 }{ 5 }\)

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Mark (√) against the correct answer in each of the following:

Question 10.

Solution:
(d) \(\frac { 5 }{ 8 }\)
\(\frac { 5 }{ 8 }\) is a vulgar fraction, because its denominator is other than 10,100, 1000, etc.

Question 11.
Solution:
(c) \(\frac { 46 }{ 63 }\)
A fraction \(\frac { a }{ b }\) is said to be irreducible or in its lowest terms if the HCF of a and b is 1
46 = 2 x 23 x 1
63 = 3 x 3 x 21 x 1
Clearly, the HCF of 46 and 63 is 1.
Hence, \(\frac { 46 }{ 63 }\) is an irreducible fraction.

Question 12.
Solution:
(d) None of these
Reciprocal of 1\(\frac { 3 }{ 5 }\) = Reciprocal of \(\frac { 8 }{ 5 }\) = \(\frac { 5 }{ 8 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
(b) 33 km
Distance covered by the car on

Question 17.
Solution:

Question 18.
Solution:
(i) False.
By cross multiplication, we have:
9 x 24 = 216 and 13 x 16 = 208
However, 216 > 208
\(\frac { 9 }{ 16 }\) > \(\frac { 13 }{ 24 }\)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following:
Question 1.
Solution:
(c)
Denominator in (a) and (b) is 10
These are decimal fractions
But denominator of (c) is 3
\(\frac { 10 }{ 3 }\) is a vulgar fraction

Question 2.
Solution:
(c)
\(\frac { 7 }{ 10 }\) and \(\frac { 7 }{ 9 }\) are proper fractions as each of these have numerator less than its denominator
\(\frac { 9 }{ 7 }\) is improper fraction

Question 3.
Solution:
(a)
\(\frac { 105 }{ 112 }\) is reducible fraction because HCF 112 of 105 and 112 is 7

Question 4.
Solution:
(c)

Question 5.
Solution:
(c)

Question 6.
Solution:
(d)

Question 7.
Solution:
(c)

Question 8.
Solution:
(d)

Question 9.
Solution:
(d)

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(c)

Question 13.
Solution:
(d)

Question 14.
Solution:
(d)

Question 15.
Solution:
(b)
The correct statement will be

Question 16.
Solution:
(c)
A car runs in 1 litre of petrol = 16 km

Question 17.
Solution:
(a)

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Total weight of 18 boxes

Question 6.
Solution:
Total amount of oranges=Rs 378

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Total quantity of milk = 24 litres
and quantity of milk got by one student = \(\frac { 2 }{ 5 }\) litres

Question 14.
Solution:
Quantity of water in a bucket

Question 15.
Solution:

Question 16.
Solution:
Product of two numbers = 42

Question 17.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Cost of 1 metre pf cloth

Question 6.
Solution:
Distance covered in 1 hour

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Weight of Amit = 35 kg.

Question 12.
Solution:
Total number of students in a class = 42
Number of boys = \(\frac { 5 }{ 7 }\) of 42 = 5 x 6 = 30
Number of girls = 42 – 30 = 12

Question 13.
Solution:
Sapna total income for one month = Rs 24000
Amount spent = \(\frac { 7 }{ 8 }\) of her income
= \(\frac { 7 }{ 8 }\) x 24000
= Rs (7 x 3000) = Rs 21000
Amount deposited in the bank per month = Rs 24000 – 21000 = Rs 3000

Question 14.
Solution:
Length of each side of a square

Question 15.
Solution:
Length of rectangular field (l)

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