Rd Sharma Solutions Class 7 Chapter 9 Ratio and Proportion
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 9, Ratio and Proportion. Here students can easily find Exercise wise solution for chapter 9, Ratio and Proportion. Students will find proper solutions for Exercise 9.1, 9.2 and 9.3. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Ratio and Proportion Exercise 9.1 Solution
Question no – (1)
Solution :
In the given question,
x : y = 3 : 5
∴ x/y = 3/5
= x = 3y/5
Now, 3x + 4y : 8x + 5y
= 3 × 3y/5 + 4y : 8 × 3y/5 + 5y
= 9y + 20y/5 : 24y + 25y/5
= 29 : 49
Therefore, the ratio will be 29 : 49
Question no – (2)
Solution :
In the given question,
x : y = 8 : 9
∴ x/y = 8/9
x = 8y/9
So, 7x – 4y : 3x + 2y
= 7 × 8y/9 – 4y : 3 × 8y/9 + 2y
= 56y – 36y/9 : 24y + 18y/9
= 20 : 42
Therefore, the ratio will be 20 : 42
Question no – (3)
Solution :
Let, the number 6x and 13x
So, 78x = 312
x = 312/78
= x = 4
∴ The numbers are = 24
And, 13 × 4 = 52
Therefore, the required numbers are 24, 52.
Question no – (4)
Solution :
Let, the number 3x and 5x
∴ 3x + 8 : 5x + 8 = 2:3
3x + 8/5x + 8 = 2/3
= 9x + 24 = 10x + 16
= 9x – 10x = 16 – 24
= – x = – 8
x = 8
∴ The numbers are,
= 3 × 8 = 24
= 5 × 8 = 40
Therefore, 24, and 40 will be the required numbers.
Question no – (5)
Solution :
Let, x be added
Then, 7 + x/13 + x = 2/3
= 21 + 3x = 26 + 2x
= 3x – 2x = 26 – 21
∴ x = 5
Hence, 5 should be added to each term of the ratio to ratio become 2 : 3.
Question no – (6)
Solution :
Let, the numbers 2x, 3x and 5x
So, 2x + 3x + 5x = 800
= x = 80
∴ The numbers are,
2 × 80 = 160,
3 × 80 = 240
And 5 ×80 = 400
Therefore, the required numbers are 160, 240, and 400.
Question no – (7)
Solution :
Let, ages of the persons = 5x and 7x
5x – 18/7x – 18 = 8/13
= 65x – 234 = 56x – 144
= 65x – 56x = – 144 + 234
= x = 90/9
∴ x = 10
Therefore, their present ages of the persons are 50 and 70.
Question no – (8)
Solution :
Let, the numbers 7x and 11x
= 7x + 7/11x + 7 = 2/3
= 21x + 21 = 22x + 14
= 21x – 22x = 14 – 21
= – x = -7
= x = 7
∴ The numbers are,
= 7 × 7 = 49
And, 11 × 7 = 77
Thus, the required numbers are 49 and 77.
Question no – (9)
Solution :
Let, the numbers 2x and 7x
So, 2x + 7x = 810
= x = 810/9
= x = 90
∴ The numbers are,
= 2 × 90 = 180
And 7 × 90 = 630
Therefore, the required numbers are 180 and 630.
Question no – (10)
Solution :
Let, Ravish and Shikha get 2x and 3x
= 2x + 3x = 1350
= x = 1350/5
= x = 270
∴ Ravish get,
= 2 × 270
= 540
∴ Shikha get,
= 3 × 270
= 810
Therefore, Ravish get Rs. 540 and Shikha gets Rs 810.
Question no – (11)
Solution :
Let, P,Q,R get 2x, 3x,and 5x
2x + 3x + 5x = 2000
= x = 2000/10
= x = 200
∴ P get,
= 2 × 2000
= ₹400
∴ Q get,
= 3 × 2000
= ₹600
∴ R get,
= 5 × 2000
= ₹1000
Question no – (12)
Solution :
Let, Boys and girls in schools = 7x and 4x
∴ 7x + 4x = 550
= x = 550/11
x = 50
∴ Boys in school,
= 7 × 50
= 350
∴ Girls in school,
= 4 × 50
= 200
Therefore, the number boys are 350 and the girls are 200.
Question no – (13)
Solution :
Let, Income and savings = 7x and 2x
So, = x = 500/5
= x = 100
Savings,
= 2x = 500
= x = 250
∴ Income,
= 7 × 250
= 1750
∴ Expenditure,
= 1750 – 500
= 1250
Hence, the income is Rs 1750 and expenditure is Rs 1250.
Question no – (14)
Solution :
Let, sides of the triangle are x, 2x and 3x
x + 2x + 3x = 36
= 6x = 36
= x = 6
∴ Sides are,
= 6, 6 × 2 = 12
And 6 × 3 = 18
Therefore, its sides are 6, 12, and 18 cm.
Question no – (15)
Solution :
Let, Raman and Aman get 2x and 3x
2x + 3x = 5500
= 5x = 5500
= x = 1100
∴ Raman will get,
= 2 × 1100
= 2200
∴ Aman will get,
= 3 × 1100
= 3300
Hence, Raman will get Rs 2200 and Aman will get Rs 3300.
Question no – (16)
Solution :
Zinc and copper in alloy = 7x and 9x
So, 9x = 11.7
= x = 11.7/9 = 1.3
∴ Weight of zinc in the alloy,
= 1.3 × 7
= 9.10 kg
∴ Weight of Zinc,
= 9.10 kg
Therefore, the weight of zinc in the alloy is 9.10 kg.
Question no – (17)
Solution :
Let, the number are 7x and 8x
So, 8x = 40
= x = 5
∴ The antecedent,
= 7 × 5
= 35
Hence, the antecedent will be 35.
Question no – (18)
Solution :
The two parts = 2x and 7x
So, 2x + 7x = 357
= x = 357/9
= x = 39
∴ One part will be,
= 2 × 39
= Rs. 78
∴ Second part will be,
= 7 × 39
= Rs. 273
Question no – (20)
Solution :
If one out of six student fail term ratio = 1 : 6
Let, fails students in the class be x
Hence, 1 : 6 = x : 42
= 1/6 = x/42
= 1/x = 6/42
= x = 7
∴ 7 student fails out of 42 students.
∴ Passed Students,
= (42 – 7)
= 35
Therefore, 35 students will pass.
Ratio and Proportion Exercise 9.2 Solution
Question no – (1)
Solution :
(i) 3 : 4 or 9 : 6
= 3 : 4 = 3/4 = 12/10
= 9 : 6 = 9/16
Therefore, 3 : 4 > 9 : 6
(ii) 15 : 16 or 24 : 25
15 : 16 = 15/16 = 15 × 25/16 × 25 = 375/400
24 : 25 = 24/25 = 24 × 16/25 + 16 = 384/400
Hence, 24 : 25 > 15 : 16
(iii) 4 : 7 or 5 : 8
= 4/7 = 4 × 8/7 × 8 = 32/56
= 5/8 = 5 × 7/8 × 7 = 35/56
So, 5/8 > 4/7
Thus, 5 : 8 > 4 : 7
(iv) 9 : 20 or 8 : 13
= 9/20 = 9 × 13/20 × 13 = 117/260
= 8/13 = 8 × 20/13 × 20 = 160/260
= 8/13 > 9/20 8
Hence, 8 : 13 > 9 : 20
(v) 1 : 2 or 13 : 12
= 1/2 = 1 × 27/2 × 27 = 27/54
= 13/27 = 13 × 2/27 × 2 = 26/54
Therefore, 1 : 2 > 13 : 27
Question no – (2)
Solution :
Two equivalent ratios of 6 : 8 are, 3 : 4 and 18 : 24
Explanation :
First, 6 : 8
= 6/2 and 8/2
= 3 : 4
Second, 6 : 8
= 6 × 3 = 18
= 8 × 3 = 24
= 18 : 24
Question no – (3)
Solution :
= 12/20 = 3/5 = 9/15
Ratio and Proportion Exercise 9.3 Solution
Question no – (1)
Solution :
(i) 33, 44, 66, 88,
= 33 : 44 = 66 : 88
= 3 : 4 = 3 : 4
Hence they are proportion
(ii) 46, 69, 69, 46
= 46 : 69 ≠ 69 : 46
So, they are not proportion
(iii) 72, 84, 186, 217
= 72 : 84 and 186 : 217
= 72/84
= 6/7
= 186/217 = 6/7
= 6/7
= 6/7 = 6/7
Hence they are in proportion.
Question no – (2)
Solution :
(i) 16 : 18 = x : 96
= 16 : 18 =x : 96
= 16/18 = x/96
= x = 16 × 96/18
∴ x = 256/3
Therefore, the x will be 256/3
(ii) x : 92 = 87 : 116
= x/92 = 87/116
= x = 87 × 92/116
∴ x = 69
Hence, the x will be 69
Question no – (3)
Solution :
Let, Income = 7x
Expenditure = 6x
Then, 7x = 1400
= x = 200
Then saving = 7x – 6x = x
= Rs 200
Therefore, the savings of the family is Rs 200.
Question no – (4)
Solution :
As per the given question,
Distance ratio = 1 : 4000000
∴ Distance between two towns,
= 5 × 4000000
= 20000000 cm
= 200 km
Therefore, the actual distance between the two towns is 200 km.
Question no – (5)
Solution :
According to the question,
Income = 10x
Savings = x
∴ x = 6000
∴ Income per year,
= 10 × 6000
= ₹60,000
∴ Income per month,
= 60000/12
= 5000
Therefore, his monthly income is Rs. 5000.
Question no – (6)
Solution :
Let, the height of the pole = x
Given, The pole costs a shadow of length 20 metre
and 6m cost a shadow of length 8 metre
∴ x : 20 = 6 : 8
= x/20 = 6/8
= x = 20 × 6/8
= 15 m
Therefore. the height of the electric pole will be 15 meter.
Next chapter solution :
👉 Chapter 10 👈
