RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion CCE Test Paper
These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper.
Other Exercises
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper
Question 1.
Solution:
Question 2.
Solution:
The sum of ratio terms is = 2 + 3 + 5 = 10
Then, we have :
A’s share = ₹ \(\frac { 2 }{ 10 }\) x 1100 = ₹ 220
B’s share = ₹ \(\frac { 3 }{ 10 }\) x 1100 = ₹ 330
C’s share = ₹ \(\frac { 5 }{ 10 }\) x 1100 = ₹ 550
Question 3.
Solution:
Product of the extremes = 25 x 6 = 150
Product of the means = 36 x 5 = 180
The product of the extremes is not equal to that of the means.
Hence, 25, 36, 5 and 6 are not in proportion.
Question 4.
Solution:
x : 18 :: 18 : 108
⇒ x x 108 = 18 x 18
(Product of extremes = Product of means)
⇒ 108x = 324
⇒ x = 3
Hence, the value of x is 3.
Question 5.
Solution:
Suppose that the numbers are 5x and 7x
Then, 5x + 7x = 84
12x = 84
x = 7
Hence, the numbers are (5 x 7) = 35 and (7 x 7) = 49.
Question 6.
Solution:
Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively
Eight years ago, age of A = (4x – 8) yrs
Eight years ago, age of B = (3x – 8) yrs
Then,
(4x – 8) : (3x – 8) = 10 : 7
⇒ \(\frac { 4x – 8 }{ 3x – 8 }\) = \(\frac { 10 }{ 7 }\)
⇒ 28x – 56 = 30x – 80
⇒ 2x = 24
⇒ x = 12
Question 7.
Solution:
Distance covered in 60 min = 54 km
Distance covered in 1 min = \(\frac { 54 }{ 60 }\) km
Distance covered in 40 min = \(\frac { 54 }{ 60 }\) x 40 = 36 km
Question 8.
Solution:
Suppose that the third proportional to 8 and 12 is x
Then, 8 : 12 :: 12 : x
⇒ 8x = 144 (Product of extremes = Product of means)
x = 18
Hence, the third proportional is 18.
Question 9.
Solution:
40 men can finish the work in 60 days
1 man can finish the work in 60 x 40 days [Less men, more days]
75 men will finish the work in = \(\frac { 60 x 40 }{ 75 }\) = 32 days
Hence, 75 men will finish the same work in 32 days.
Mark (✓) against the correct answer in each of the following :
Question 10.
Solution:
(d) 6 : 4 : 3
A = \(\frac { 3 }{ 2 }\) B
C = \(\frac { 3 }{ 4 }\) B
A : B : C = \(\frac { 3 }{ 2 }\) B : B : \(\frac { 3 }{ 4 }\) B
= 6 : 4 : 3
Question 11.
Solution:
(a) 2 : 3 : 4
Question 12.
Solution:
(c) 11 : 3
Question 13.
Solution:
(a) 3
Let us assume that the number to be subtracted is x
Then, (15 – x) : (19 – x) = 15 : 3
⇒ \(\frac { 15 – x }{ 19 – x }\) = \(\frac { 3 }{ 4 }\)
⇒ 60 – 4x = 57 – 3x
⇒ x = 3
Question 14.
Solution:
(b) ₹ 360
Sum of the ratio terms = 4 + 3 = 7
B’s share = ₹ 840 x \(\frac { 3 }{ 7 }\) = ₹ 360
Question 15.
Solution:
(c) 40 years
Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively
After 5 years, the ages of A and B will be (5x + 5) yrs and (2x + 5) yrs, respectively
Then, (5x+ 5) : (2x + 5) = 15 : 7
⇒ \(\frac { 5x+ 5 }{ 2x + 5 }\) = \(\frac { 15 }{ 7 }\)
Cross multiplying; we get:
⇒ 35x + 35 = 30x + 75
⇒ 5x = 40
⇒ x = 8
Then, the present age of A is 5 x 8 = 40 yrs.
Question 16.
Solution:
(a) 896
Suppose that the number of boys in the school is x
Then, x : 320 = 9 : 5
⇒ 5x = 2880
⇒ x = 576
Hence, total strength of the school = 576 + 320 = 896
Question 17.
Solution:
Question 18.
Solution:
(i) True
Suppose that the men proportional is x
Then, 0.4 : x :: x : 0.9
⇒ 0.9 x 0.4 = x x x (Product of extremes = Product of means)
⇒ x² = 0.36
⇒ x = 0.6
(ii) False
Suppose that the third proportional is x.
Then, 9: 12 :: 12 : x
⇒ 9x = 144 (Product of extremes = Product of means)
⇒ x = 16
(iii) True
8 : x :: 48 : 18
⇒ 144 = 48x (Product of extremes = Product of means)
⇒ x = 3
(iv) True
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