RD Sharma Solutions Class 7 Ratio And Proportion Exercise 9.1

Exercise 9.1

Q1. If x: y=3:5, find the ratio 3x+4y:8x+5y.

x:y=3:5

\(\frac{x}{y}=\frac{3}{5}\)

5x=3y

\(x=\frac{3y}{5}\)

\(3x+4y:8x+5y=\frac{3\times3y}{5}+4y:\frac{8\times3y}{5}+5y\\ =\frac{9y+20y}{5}:\frac{24y+25y}{5}\\ =\frac{29y}{5}:\frac{49y}{5}\\ =29y:49y =29:49\)

Q2. If x: y=8:9, find the ratio (7x-4y):3x+2y.

x:y=8:9

\(\frac{x}{y}=\frac{8}{9}\)

9x=8y

\(x=\frac{8y}{9}\)

\(7x-4y:3x+2y=\frac{7\times8y}{9}-4y:\frac{3\times8y}{9}+2y\\ =\frac{56y-36y}{9}:\frac{42y}{9}\\ =20:42\\ =10:21\)

Q3. If two numbers are in the ratio 6:13 and their L.C.M is 312, find the numbers.

Let the required numbers be 6x and 13x

Then their L.C.M is 78x

78x=312

\(x=\frac{312}{78}\)

x=4

Thus,

The numbers are

\(6x=6\times4=24\)

\(13x=13\times4=52\)

Q4. Two numbers are in the ratio 3:5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers

Let the required numbers be 3x and 5x

If 8 is added to each other

3x+8:5x+8=2:3

\(\frac{3x+8}{5x+8}=\frac{2}{3}\)

3(3x+8)=2(5x+8)

9x+24=10x+16

10x-9x=24-16

x=8

Thus the numbers are 3x=3(8)=24

5x=5(8)=40

Q5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3

Let the number to be added be x

Then

\(\frac{7+x}{13+x}=\frac{2}{3}\)

(7+x)3=2(13+x)

3x-2x=26-21

x=5

Hence the required number is 5

Q6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers

Given that

Three numbers are in ratio 2:3:5

Sum of these numbers=800

Sum of the terms of the ratio=2+3+5=10

\(First\;number=\frac{2}{10}\times800\\ =160 Second\;number=\frac{3}{10}\times800\\ =240 Third\;number=\frac{5}{10}\times800\\ =400\)

Q7. 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.

Let the required ages be 5x and 7x

18 years ago their age ratios

\(\frac{5x-18}{7x-18}=\frac{8}{13}\)

\(65x-13\times18=8\times7x-8\times18\)

65x-234=56x-144

65x-56x=234-144

9x=90

x=10

Thus the ages are 5x=\(5\times10\)=50years

7x=\(7\times10\)=70 years

Q8. 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.

Let the required numbers be 7x and 11x

If 7 is added to each of the numbers it becomes

\(\frac{7x+7}{11x+7}=\frac{2}{3}\)

21x+21=22x+14

X=21-14=7

Thus

The numbers are 7x=\(7\times7\)=49

11x=\(11\times7\)=77

Q9. Two numbers are in the ratio 2: 7.11the sum of the numbers is 810. Find the numbers.

Two numbers are in theratio=2:7

Sum of the numbers=810

We have,

Sum of the terms in the ratio=2+7=9

First number=\(\frac{2}{9}\times810\)

=\(2\times90\)

=180

Second number=\(\frac{7}{9}\times810\)

=\(7\times90\)

=630

Q10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.

We have

Sum of the terms of the ratio=2+3=5

Ravish money=\(\frac{2}{5}\times1350\)

=\(2\times270\)

=Rs.540

Shikha money=\(\frac{3}{5}\times1350\)

=\(3\times270\)

=Rs.810

Q11. Divide Rs 2000 among P, Q, R in the ratio 2:3:5.

We have

Sum of the terms of the ratio=2+3+5=10

P-share= \(\frac{2}{10}\times total\;money\\ =\frac{2}{10}\times2000\\ =2\times200\\ =Rs.400\)

Q-share= \(\frac{3}{10}\times total\;money\\ =\frac{3}{10}\times2000\\ =3\times200\\ =Rs.600\)

R-share= \(\frac{5}{10}\times total\;money\\ =\frac{5}{10}\times2000\\ =5\times200\\ =Rs.1000\)

Q12. 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.

We have,

The boys and girls is in the ratio 7:4

Sum of the terms in the ratio=7+4=11

Total strength=550

Boys strength= \(\frac{7}{11}\times550\\ =7\times50\\ =350\;boys\)

Girls strength= \(\frac{4}{11}\times550\\ =4\times50\\ =200\;girls\)

Q13. 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.

It is given that

The ratio of income and savings is 7:2

Savings

2x=500

So, x=250

Therefore,

\(Income=7\times250\\ =1750\)

Expenditure=Income-savings

=1750-500

=Rs.1250

Q14. The sides of a triangle are in the ratio 1 : 2 : 3. If the perimeter is 36cm, find its sides.

The sides of the triangle are in the ratio 1:2:3

Sum of the terms in the ratio=1+2+3=6

Perimeter=36cm

First side= \(\frac{1}{6}\times36\\ =6cm\)

Second side= \(\frac{2}{6}\times36\\ =12cm\)

Third side= \(\frac{3}{6}\times36\\ =18cm\)

Q15. A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get ?

We have,

Sum of the terms in the ratio=2+3=5

Raman share=\(\frac{2}{5}\times5500\\ =2\times1100\\=Rs.2200\)

Aman share=\(\frac{3}{5}\times5500\\ =3\times1100\\=Rs.3300\)

Q16. 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.

 We have,

The ratio of zinc and copper in an alloy=7:9

Weight of copper in the alloy=11.7kg

9x=11.7kg

\(x=\frac{11.7}{9}\)

Weight of zinc in the alloy=\(1.3\times7\)

=9.10kg

Therefore weight of zinc =9.10kg

Q17. In the ratio 7: 8. if the consequent is 40,  what a the antecedent

Given ratio=7:8

Consequent

8x=40

\(x=\frac{40}{8}\)

x=5

antecedent=7x=\(7\times5\)=35

Q18. Divide Rs 351 into two parts such that one may be to the other as 2 : 7.

Ratio=2:7

The sum of the terms in the ratio=2+7=9

First ratio=\(\frac{2}{9}\times351\\=2\times39\\=Rs.78\)

Second ratio=\(\frac{7}{9}\times351\\=7\times39\\=Rs.273\)

Q19. 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.

One score=20

It is Rs.16 per score for pencil

Pencil cost=\(\frac{16}{20}\)

=Rs.0.80

Cost of one dozen ball pen=Rs. 8.40

Cost of one ball pen=\(\frac{8.40}{12}\)

=Rs.0.70

Ratio of price of pencil to that of ball pen=\(\frac{0.80}{0.70}\)

=\(\frac{8}{7}\)

Q20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass?

Given,

One out of 6 student fails

X out of 42 students

\(\frac{1}{6}=\frac{x}{42}\)

\(x=\frac{42}{6}\)

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.