# SSC Ratio and Proportion Quiz Day 1

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# SSC Ratio and Proportion Quiz Day 1

### Introduction

Ratios and Proportions hold a significant place in several competitive exams including recruitment exams and college entrance exams. Ratios are a means to compare quantities and proportions are a means to understand if ratios are equivalent. SSC Ratio and Proportion Quiz Day 1 will provide examples and solved questions to understand the significance of the topic. The Ratios and Proportions topic is crucial for all competitive exams including the Quantitative Aptitude section. SSC Ratio and Proportion Quiz Day 1 is very useful to crack the quantitative aptitude sections of several competitive exams

### Quiz

1. Anmol had 10 paise, 25 paise and 50 paise coins in the ratio of 10 : 8 : 9 respectively. After giving Rs. 20 his mother he has Rs. 40. How many 50 paise coins did he have?

A. 72
B. 60
C. 54
D. 35
E. 21

Explanation:
Total money = 20 + 40 = 60

In this question the total money is given so we need to multiply the value of the coins in the ratio.

10 Ã— 0.10 x + 8 Ã— 0.25x + 9 Ã— 0.50x = 60

x + 2x + 4.5x = 60

7.5 x = 60

x = 8

Number of 50 paise coins = 9 Ã— 8 = 72

2. Three numbers A, B and C are in the ratio of 12 : 15 : 25. If sum of these numbers is 312, ratio between the difference of B and A and the difference of C and B is â€“

A. 3 : 7
B. 10 : 3
C. 3 : 10
D. 5 : 7
E. None of these

Explanation:
Let the number A, B and C be 12a , 15a and 25a respectively

Then from the question 12a + 15a + 25a = 312

â‡’ a = $$\frac{312}{52}$$ = 6

â‡’ required ratio = $$\frac{15 Ã— 6 â€“ 12 Ã— 6}{25 Ã— 6 â€“ 15 Ã— 6}$$

= $$\frac{3}{10}$$ = 3 : 10

3. The ratio of Nicotine to Heroine in four types of drugs of equal quantity is 2 : 3, 3 : 7, 4: 11 and 11 :9 respectively. The four drugs are mixed together. What is the ratio of Nicotine to Heroine after mixing?

A. 213 : 91
B. 418 : 189
C. 91 : 149
D. 149 : 81
E. None of these

Explanation:
Let the quantity of each drugs be â€˜aâ€™

Then the quantity of nicotine after mixing is

($$\frac{2}{5}$$ + $$\frac{3}{10}$$ + $$\frac{4}{15}$$ + $$\frac{11}{20}$$) a

= ($$\frac{2 Ã— 12 + 3 Ã— 6 + 4 Ã— 4 + 11 Ã— 3}{60}$$) a

= ($$\frac{91}{60}$$) a

and quantity of heroine after mixing is

($$\frac{3}{5}$$ + $$\frac{7}{10}$$ + $$\frac{11}{15}$$ + $$\frac{9}{20}$$) a

= ($$\frac{3 Ã— 12 + 7 Ã— 6 +11Ã— 4 + 9 Ã— 3}{60}$$) a

= ($$\frac{149}{60}$$) a

â‡’ the required ratio of nicotine to heroin after mixing

= $$\frac{\frac{91}{60}a}{\frac{149}{60}a}$$
= 91 : 149

4. Out of three positive numbers, the ratio of the first and the second numbers is 3 : 4 that of the second and the third numbers is 5 : 6 if the product of the second and the third numbers is 4320. What is the sum of three numbers?

A. 177
B. 165
C. 185
D. 160
E. None of these

Explanation:
Ratio of first and second numbers = 3 : 4

The ratio of second and third numbers = 5: 6

Let the second number = 5x, third number = 6x

Product of second and third numbers = 4320

5x Ã— 6x = 4320

$${x}^{2}$$ = 144

x = 12

2nd number = 60, 3rd number = 72,

1st number = 60 Ã— $$\frac{3}{4}$$ = 45

Sum of three numbers = 60 + 72 + 45 = 177

5. The cost price of 2 shirts and 3 jeans is Rs. 2200 and the cost price of 2 jeans and 4 shirts is Rs. 2400. Find the ratio between the cost price of the jeans and the shirt.

A. 8 : 9
B. 10 : 7
C. 6 : 5
D. 11 : 10
E. None of these

Explanation:
Let the cost price of one shirt = x Rs., the cost price of one jeans = y Rs.

According to the question,

2x + 3y = 2200 ….1

2y + 4x = 2400

2x + y = 1200 ….2

After solving these 2 equation,

x = 350 Rs. , y = 500 Rs.

Ratio = 500 : 350 = 10 : 7

1. Three numbers are in the ratio 5 : 6 : 7. If the sum of their squares is 990, then the numbers are

A. 10, 12 and 21
B. 25, 30 and 35
C. 15, 18 and 21
D. 20, 24 and 28
E. None of these

Explanation:
Let the Numbers be 5x, 6x and 7x

âˆ´ $${5x}^{2}$$ + $${6x}^{2}$$ + $${7x}^{2}$$ = 990

â‡’ 25$${x}^{2}$$ + 36$${x}^{2}$$ + 49$${x}^{2}$$ = 990

â‡’ 110$${x}^{2}$$ = 990

â‡’ $${x}^{2}$$ = 9

â‡’ x = 3

âˆ´ Numbers = 15, 18 and 21.

2. The ratio of Sitaâ€™s, Riyaâ€™s and Kunalâ€™s monthly income is 84 : 76 : 89. If Riyaâ€™s annual income is Rs. 4,56,000, what is the sum of Sitaâ€™s and Kunalâ€™s annual incomes? (In some cases monthly income is used while in other annual income is used.)

A. Rs. 11,95,000
B. Rs. 9,83,500
C. Rs. 1,13,000
D. Rs. 10,38,000
E. None of these

Explanation:
Ratio of monthly income and ratio of annual income will be the same,

ie. 84 : 76 : 89

Applying the rule of proportion, we get

Riya’s income in the ratio: Riya’s total income:: Sum of Sita’s and Kunal’s income in ratio: Sum of Sita’s and Kunal’s income in value

76 : 456000 : : (84 + 89) : x

âˆ´ Sum of Sita’s and Kunal’s annual income in value

= $$\frac{456000}{76}$$ Ã— (84 + 89) = Rs. 1038000

3. Two numbers are in ratio of 21 : 26. If 8 is added in each, the new numbers are in ratio of 5 : 6. Find the ratio of numbers, if 6 is subtracted from each number?

A. 18 : 23
B. 19 : 25
C. 6 : 7
D. 9 : 16
E. None of these

Explanation:
Let the numbers be 21x and 26x.

($$\frac{21x + 8}{26x + 8}$$) = $$\frac{5}{6}$$

6(21x + 8) = 5(26x + 8)

126x + 48 = 130x + 40

x = 2

So, numbers will be 42 and 52.

If 6 is subtracted, then numbers will be 36 and 46.

Required ratio = 36 : 46. i.e. 18 : 23

4. If A varies directly as B and inversely as C and A = 6, when B = 2 and C= 3, what is the value of A when B = 8 and C = 6?

A. 12
B. 6
C. 18
D. 24
E. None of these

Explanation:
Let the constant be x,

so putting the first scenario in equation

A = x Ã— $$\frac{B}{C}$$ , we get:

6 = x Ã— $$\frac{2}{3}$$ or x = 9

We can find out A in the second scenario by putting the value of x as 9

A = 9 Ã— $$\frac{8}{6}$$ or A = 12

5. The income of Asaram, Satpal and Rahim in the ratio of 12 : 9 : 7 and their spendings are in the ratio 15 : 9 : 8. If Asaram saves 25% of his income. What is the ratio of the savings of Asaram, Satpal and Rahim?

A. 15 : 18 : 11
B. 5 : 8 : 7
C. 23 : 18 : 11
D. 25 : 16 : 13
E. None of these

Explanation:
Income = Expenditure + Saving

Asaram: 12x = 15y + 3x (3x = 25% of 12x)

Satpal: 9x = 9y + (9x â€“ 9y)

Rahim: 7x = 8y + (7x â€“ 8y)

Therefore, 12x â€“ 3x = 15y

â‡’ $$\frac{x}{y}$$ = $$\frac{5}{3}$$

â‡’ y = $$\frac{3x}{5}$$

Therefore, savings = (income-expenditure)

Asaram = 12x â€“ 9x = 3x
Satpal = 9x â€“ 9y = 9x â€“ $$\frac{27x}{5}$$ = $$\frac{18x}{5}$$

Rahim = 7x â€“ 8y = 7x â€“ $$\frac{24x}{5}$$ = $$\frac{11x}{5}$$

i.e., the ratio of savings of Asaram : Satpal : Rahim
= 3x : $$\frac{18}{5}$$x : $$\frac{11}{5}$$x

= 15 : 18 : 11

1. The ratio of zinc and copper in a brass pieces is 13 : 7. How much zinc will be there in 100 kg of such a piece?

A. 20 kg
B. 35kg
C. 55kg
D. 65kg

Explanation:

Amount of zinc = ($$\frac{100 Ã— 13}{20}$$) kg = 65kg.

2. The ratio of number of men and women in a factory of 720 workers is 7 : 5. How many more women should be joined to make the ratio 1 : 1?

A. 80
B. 100
C. 120
D. 150

Explanation:
Number of men = ($$\frac{720 Ã— 7}{12}$$) = 420.

Number of women = (720 – 420) = 300.

âˆ´ Number of women to be joined = (420 – 300) = 120.

3. What is the ratio whose terms differ by 40 and the measure of which is $$\frac{2}{7}$$ ?

A. 12 : 56
B. 16 : 56
C. 23 : 58
D. None of these

Explanation:
Let the ratio be x : (x + 40).

Then, $$\frac{x}{(x + 40)}$$ = $$\frac{2}{7}$$ â‡” 7x = 2x + 80 â‡” x =16.

âˆ´ Required ratio is 16 : 56.

4. Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B is

A. 4 : 3
B. 3 : 4
C. 1 : 1
D. 2 : 3

Explanation:
From the equation, we get

5% of A + 4% of B = $$\frac{2}{3}$$ (6% of A + 8% of B)

â‡’ 15% of A + 12% of B = 12% of A + 16% of B

â‡’ [15 â€“ 12] % of A = [16 â€“ 12] % of B

â‡’ 3 % of A = 4% of B

â‡’ $$\frac{A}{B}$$ = $$\frac{4}{3}$$

Therefore, A : B = 4 : 3.

5. In a certain school, the ratio of boys to girls is 7 : 5. If there are 2400 students in the school, then how many girls are there?

A. 500
B. 700
C. 800
D. 1000

Explanation:
Let the number of boys and girls are 7x and 5x, respectively.

Given, the total number of students = 2400

â‡’ 7x + 5x = 2400 â‡’ 12x = 2400

âˆ´ x = 200

âˆ´ Required number of girls = 5x = 5 Ã— 200 = 1000.