Read RD Sharma Solutions Class 7 Chapter 9 Ratio and Proportion below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations
Exercise 9.1
Question 1: If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y
Solution 1:
x: y = 3: 5
the equation is written as:
=x/y = 3/5
=5x = 3y
=x = 3y/5
By replacing the value of x in given equation
=3x + 4y: 8x + 5y
= 3 (3y/5) + 4y: 8 (3y/5) + 5y
= ((9y + 20y))/5: ((24y + 25y))/5
= 29y/5: 49y/5
= 29y: 49y
= 29: 49
Question 2: If x: y = 8: 9, find the ratio (7x – 4y): 3x + 2y.
Solution 2:
x: y = 8: 9
the equation is written as:
=x/y = 8/9
=9x = 8y
=x = 8y/9
By replacing the value of x in the given equation
=(7x – 4y): 3x + 2y
= 7 (8y/9) – 4y: 3 (8y/9) + 2y
= ((56y – 36y))/9: 42y/9
= 20y/9: 42y/9
= 20y: 42y
= 20: 42
= 10: 21
Question 3: If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers.
Solution 3:
Two numbers are in the ratio 6: 13
Let 6x and 13x be the numbers
Take LCM of 6x and 13x is 78x
= 78x = 312
=x = 312/78
=x = 4
By putting 4 at the place of x
So, the numbers are 6x = 6 (4) = 24
13x = 13 (4) = 52
Question 4: Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers.
Solution 4:
Let 3x and 5x be the numbers
if 8 is added to each other than ratio becomes 2: 3
That the equation is 3x + 8: 5x + 8 = 2: 3
=((3x + 8))/((5x + 8)) = (2/3)
=3 (3x + 8) = 2 (5x + 8)
=9x + 24 = 10x + 16
By transposing
=24 – 16 = 10x – 9x
=x = 8
So, the numbers are
=3x = 3 (8) = 24
=5x = 5 (8) = 40
Question 5: What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3
Solution 5:
Let x be the number to be added
Then ((7 + x))/((13 + x)) = (2/3)
=(7 + x) 3 = 2 (13 + x)
=21 + 3x = 26 + 2x
=3x – 2x = 26 – 21
=x = 5
Hence the required number is 5
Question 6: Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers
Solution 6:
The three numbers are in the ratio 2: 3: 5 and the sum of them is 800
So, sum of the values of the ratio = 2 + 3 + 5 = 10
First number is (2/10) × 800
= 2 × 80
= 160
Second number is (3/10) × 800
= 3 × 80
= 240
Third number is (5/10) × 800
= 5 × 80
= 400
The three numbers are 160, 240 and 400
Question 7: The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages.
Solution 7:
Let 5x and 7x be the present ages of two persons
According to the question ages of two persons are in the ratio 5: 7
Also given that 18 years ago their ages were in the ratio 8: 13
Therefore ((5x – 18))/((7x – 18)) = (8/13)
=13 (5x – 18) = 8 (7x – 18)
=65x – 234 = 56x – 144
=65x – 56x = 234 – 144
=9x = 90
=x = 90/9 = 10
So, the ages are 5x = 5 × 10 = 50 years
And 7x = 7 × 10 = 70 years
Question 8: Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers.
Solution 8:
Let 7x and 11x be the numbers
If 7 is added to each of them then
=((7x + 7))/((11x + 7)) = (2/3)
=3 (7x + 7) = 2 (11x + 7)
=21x + 21 = 22x + 14
=22x – 21x = 21 – 14
=x = 21 – 14 = 7
So, the numbers are 7x = 7 × 7 =49
And 11x = 11 × 7 = 77
Question 9: Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers.
Solution 9:
According to the question the two numbers are in the ratio 2: 7
And their sum = 810
So, the sum of terms in the ratio = 2 + 7 = 9
First number = 2/9 × 810
= 2 × 90
= 180
Second number = 7/9 × 810
= 7 × 90
= 630
Question 10: Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.
Solution 10:
According to the question total amount to be divided = 1350
And the sum of the terms of the ratio = 2 + 3 = 5
Share of Ravish’s money
= 2/5 × 1350
= 2 × 270
= Rs. 540
And Share of Shikha’s money
= 3/5 × 1350
= 3 × 270
= Rs. 810
Question 11: Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5.
Solution 11:
According to the question total amount to be divided = 2000
Sum of the terms of the ratio = 2 + 3 + 5 = 10
Share of P’s money = 2/10 × 2000
= 2 × 200
= Rs. 400
And Share of Q’s money = 3/10 × 2000
= 3 × 200
= Rs. 600
And Share of R’s money = 5/10 × 2000
= 5 × 200
= Rs. 1000
Question 12: The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls.
Solution 12:
According to the question that boys and the girls in a school are in the ratio 7:4
Sum of the ratio is 7 + 4 = 11
Total strength = 550
Boys strength = 7/11 × 550
= 7 × 50
= 350
Girls strength = 4/11 × 550
= 4 × 50
= 200
Question 13: The ratio of monthly income to the savings of a family is 7: 2. If the savings be of Rs. 500, find the income and expenditure.
Solution 13:
According to the question that the ratio of income and savings is 7: 2
Let be 2x is the savings
2x = 500
So, x = 250
Income = 7x
Income = 7 × 250 = 1750
Expenditure = Income – savings
= 1750 – 500
= Rs.1250
Question 14: The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides.
Solution 14:
According to the question sides of a triangle are in the ratio 1: 2: 3
And the Perimeter = 36cm
Sum of the ratio = 1 + 2 + 3 = 6
First side = 1/6 × 36
= 6cm
Second side = 2/6 × 36
= 2 × 6
= 12cm
Third side = 3/6 × 36
= 6 × 3
= 18cm
Question 15: A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get?
Solution 15:
According to the question total amount to be divided = 5500
Sum of the ratio = 2 + 3 = 5
Share of Raman’s money = 2/5 × 5500
= 2 × 1100
= Rs. 2200
And share of Aman’s money = 3/5 × 5500
= 3 × 1100
= Rs. 3300
Question 16: The ratio of zinc and copper in an alloy is 7: 9. It the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.
Solution 16:
According to the question that ratio of zinc and copper in an alloy is 7: 9
Let their ratio = 7x: 9x
Weight of copper = 11.7kg
=9x = 11.7 = x = 11.7/9
= x = 1.3
Weight of the zinc in the alloy = 1.3 × 7
= 9.10kg
Question 17: In the ratio 7: 8. If the consequent is 40, what a the antecedent
Solution 17:
According to the question the ratio is 7: 8
Let the ratio of consequent and antecedent 7x: 8x
Consequent = 40
=8x = 40
=x = 40/8
=x = 5
Antecedent = 7x
= 7 × 5
= 35
Question 18: Divide Rs 351 into two parts such that one may be to the other as 2: 7.
Solution 18:
According to the question total amount is to be divided = 351
Ratio is 2: 7
The sum of ratio = 2 + 7= 9
First ratio of amount = 2/9 × 351
= 2 × 39
= Rs. 78
Second ratio of amount = 7/9 × 351
= 7 × 39
= Rs. 273
Question 19: Find the ratio of the price of pencil to that of ball pen, if pencil cost Rs.16 per score and ball pen cost Rs.8.40 per dozen.
Solution 19:
One score contains 20 pencils and cost per score = 16
So, the cost of pencil = 16/20
= Rs. 0.80
Cost of one dozen ball pen = 8.40 [1 dozen = 12]
Thus, cost of pen = 8.40/12
= Rs 0.70
Ratio of the price of pencil to that of ball pen = 0.80/0.70
= 8/7
= 8: 7
Question 20: In a class, one out of every six students fails. If there are 42 students in the class, how many pass?
Solution 20:
According to the question total number of students = 42 one out of 6 student fails
x out of 42 students
1/6 = x/42
x = 42/6
x = 7
Number of students who fail = 7 students
No of students who pass =Total students – Number of students who fail
= 42 – 7
= 35 students.
Exercise 9.2
Question 1: Which ratio is larger in the following pairs?
(i) 3: 4 or 9: 16
(ii) 15: 16 or 24: 25
(iii) 4: 7 or 5: 8
(iv) 9: 20 or 8: 13
(v) 1: 2 or 13: 27
Solution 1:
(i) 3: 4 or 9: 16
Take LCM of 4 and 16 is 16
3 : 4 we can written as = 3/4
3/4 × 4/4 = 12/16
And we have 9/16
Clearly 12/16 > 9/16
Therefore 3: 4 > 9: 16
(ii) 15: 16 or 24: 25
Take LCM of 16 and 25 is 400
15: 16 we written as = 15/16
15/16 × 25/25 = 375/400
And we have 24/25
24/25 × 16/16 = 384/400
Clearly 384/400 > 375/400
Therefore 15: 16 < 24: 25
(iii) 4: 7 or 5: 8
Take the LCM of 7 and 8 is 56
4: 7 we written as = 4/7
4/7 × 8/8 = 32/56
And we have 5/8
5/8 × 7/7 = 35/56
Clearly 35/56 > 32/56
Therefore 4: 7 < 5: 8
(iv) 9: 20 or 8: 13
Take the LCM of 20 and 13 is 260
9: 20 we written as = 9/20
9/20 × 13/13 = 117/260
And we have 8/13
8/13 × 20/20 = 160/260
Clearly 160/260 > 117/260
Therefore 9: 20 < 8: 13
(v) 1: 2 or 13: 27
Take the LCM of 2 and 27 is 54
1: 2 we written as = 1/2
1/2 × 27/27 = 27/54
And we have 13/27
13/23 × 2/2 = 26/54
Clearly 27/54 > 26/54
Therefore 1: 2 > 13: 27
Question 2: Give the equivalent ratios of 6: 8.
Solution 2:
6: 8
To make equivalent ratio we multiplying numerator and denominator by 2
=6/8 × 2/2 = 12/16
And also to make equivalent ratio we dividing numerator and denominator by 2
=((6/2))/((8/2) ) = 3/4
Two equivalent ratios are 3: 4 = 12: 16
Question 3: Fill in the following blanks:
12/20 = …. /5 = 9/….
Solution 3:
12/20 = 3/5 = 9/5
Description:
Consider 12/20 = …. /5
Let x be the unknown value
Therefore 12/20 = x/5
On cross multiplying
= 12 × 5 = 60
=x = 60/20
=x = 3
Consider 12/20 = 9/….
Let y be the unknown value
Therefore 12/20 = 9/y
On cross multiplying
= 20 × 9 = 180
y = 180/12
y = 15
Exercise 9.3
Question 1: Find which of the following are in proportion?
(i) 33, 44, 66, 88
(ii) 46, 69, 69, 46
(iii) 72, 84, 186, 217
Solution 1:
(i) 33, 44, 66, 88
Product of extremes = 33 × 88 = 2904
Product of means = 44 × 66 = 2904
product of extremes = product of means
Hence given numbers are in proportion.
(ii) 46, 69, 69, 46
Product of extremes = 46 × 46 = 2116
Product of means = 69 × 69 = 4761
product of extremes is not equal to product of means
Hence given numbers are not in proportion.
(iii) 72, 84, 186, 217
Product of extremes = 72 × 217 = 15624
Product of means = 84 × 186 = 15624
product of extremes = product of means
Hence given numbers are in proportion.
Question 2: Find x in the following proportions:
(i) 16: 18 = x: 96
(ii) x: 92 = 87: 116
Solution 2:
(i) 16: 18 = x: 96
In proportion we know that product of extremes = product of means
16/18 = x/96
On cross multiplying
x = ((16 × 96))/18
x = 1536/18
Dividing both numerator and denominator by 6
x = 256/3
(ii) x: 92 = 87: 116
In proportion we know that product of extremes = product of means
x/92 = 87/116
On cross multiplying
x = ((87 × 92))/116
x = 69
Question 3: The ratio of income to the expenditure of a family is 7: 6. Find the savings if the income is Rs.1400.
Solution 3:
Income of a family 1400
The ratio of income and expenditure = 7: 6
7x = 1400
Therefore x = 200
Expenditure = 6x
= 6 × 200 = Rs.1200
Savings = Income – Expenditure
= 1400 -1200
= Rs.200
Question 4: The scale of a map is 1: 4000000. What is the actual distance between the two towns if they are 5cm apart on the map?
Solution 4:
The scale of map = 1: 4000000
Let x cm assume the actual distance between towns
1: 4000000 =5: x
x = 5 × 4000000
x = 20000000 cm [1km = 1000 m; 1m = 100 cm]
Therefore
x = 200 km
Question 5: The ratio of income of a person to his savings is 10: 1. If his savings for one year is Rs.6000, what is his income per month?
Solution 5:
The ratio of income of a person to his savings is 10: 1
Savings per year = 6000
Savings per month = 6000/12
= Rs.500
Then let income per month be x
x: 500 = 10:1
x = 500 × 10
x = 5000
Income per month is Rs. 5000
Question 6: An electric pole casts a shadow of length 20 meters at a time when a tree 6 meters high casts a shadow of length 8 meters. Find the height of the pole.
Solution 6:
The length electric pole shadow is 20m
Height of the tree: Length of the shadow of tree
Height of the pole: Length of the shadow of pole
=x: 20 = 6: 8
=x = 120/8
=x = 15
Therefore height of the pole is 15 meters