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

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*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=\(5\times10\)

7x=\(7\times10\)

*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=\(7\times7\)

11x=\(11\times7\)

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

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*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.*

*Â *

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 ?*

*Â *

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.*

** Â **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, Â what a the antecedent *

*Â *

Given ratio=7:8

Consequent

8x=40

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

x=5

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

*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=\(\frac{0.80}{0.70}\)

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

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*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.