# IBPS SO Ratio and Proportion Quiz 1

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# IBPS SO Ratio and Proportion Quiz 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. IBPS SO Ratio and Proportion Quiz 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. IBPS SO Ratio and Proportion Quiz 1 is very useful to crack the quantitative aptitude sections of several competitive exams

### Quiz

1. In an army selection process, the ratio of selected to unselected was 6:1. If 90 less had applied and 30 less selected, the ratio of selected to unselected would have been 8:1. How many candidates had applied for the process?

A. 3150
B. 6300
C. 4725
D. 1575

Explanation –
Let the number of selected and unselected candidates be 6x and x respectively.

Then, number of candidates applied = 6x + x = 7x

According to the question,

If 90 less applied and 30 less selected, the ratio of selected to unselected candidates = $$\frac {8}{1}$$, i.e.

New number of selected candidates = 6x – 30

And new number of applied candidates = 7x – 90

∴new number of candidate unselected = 7x – 90 – (6x – 30)

= x – 60

Thus, $$\frac {6x – 30}{x – 60} = \frac {8}{1}$$

$$\Rightarrow$$ 6x – 30 = 8x – 480

$$\Rightarrow$$ x = 225

Thus, the number of candidates applied for the process = 7x

= 7 × 225

= 1575

2. The price of gold is directly proportional to square of its weight. A person broke down the gold 3:2:1 and face a loss of Rs 4,620. Find the initial price of gold.

A. 7560
B. 6560
C. 7450
D. 7440

Explanation –
3:2:1 $$\rightarrow$$ 6 (According to Question) $$\rightarrow$$ 36 (On squaring)

In this question if we brake it in the ratio and then squaring it then we get 14 (9 + 4 + 1)

So, we loss 22 (36 – 14) if we broke it

Loss = Rs. 4620

Initial price = $$\frac {4620}{22} \times 36$$

= Rs 7560

3. The no. of mangoes on three baskets are in the ratio of 3 : 5 : 7. In which ratio the no. of mangoes in first two baskets must be increase so that the new ratio becomes 5 : 6 : 8.

A. 3 : 13
B. 12 : 7
C. 11 : 2
D. 13 : 3

Explanation –
A.T.Q.

$${B}_{1} : {B}_{2} : {B}_{3} = 3X : 5X : 7X$$

and again,

$${B}_{1} : {B}_{2} : {B}_{3} = 5X : 6X : 8X$$

According to question there is increase in first two baskets only. It means the no. of mangoes remain constant in the third basket.

So, 7x = 8y or x = $$\frac {8}{7} y$$ Hence, 3x : 5x : 7x = $$3 \times \frac {8}{7} y : 5 \times \frac {8}{7} y : 7 \times \frac {8}{7} y$$

= 24y : 40y : 56y and 5y : 6y : 8y $$\Rightarrow$$ 35y : 42y: 56y

i.e, increase in first basket = 11 increases in second basket = 2 Required ratio = 11 : 2

4. In a mixture of 60 liters, the ratio of milk and water 2:1. If this ratio is to be 1:2 , then the quantity of water to be further added is:

A. 20 liters
B. 30 liters
C. 40 liters
D. 60 liters

Explanation –
Quantity of milk =$$60 \times \frac {2}{3}$$ = 40 liters

Quantity of water = 60 – 40 = 20 liters

Let x liters of water to be added so as to get the ratio 1:2. Then,

$$\frac {40}{20 + x} = \frac {1}{2}$$

x = 80 – 20 = 60 liters

5. Two numbers are in the ratio 3: 5. If 9 is subtracted from each number, then they are in the ratio of 12: 23. What is the second number?

A. 44
B. 55
C. 66
D. 77

Explanation –
Let the numbers be 3x & 5x (since they are in the ratio of 3:5)

So, subtraction 9, we get

3x – 9 and 5x – 9

Now, the ratio is 12 :12: (given)

$$\frac {3x – 9}{5x – 9} = \frac {12}{23}$$

(By cross multiplication)

23(3x – 9) = 12(5x -9)

69x – 207 = 60x – 108

9x = 99

x = 11

So, the original numbers are 3x = 3 $$\times$$ 11 = 33 and 5 $$\times$$ 11 = 55 , with the second number being 5.

1. Sachin and Anurag have monthly incomes in the ratio 6 : 7 and Their monthly expenditures in the ratio 5: 4. If they save Rs.700 and Rs.2100 respectively, find the monthly expenditure of Anurag.

A. 3200
B. 2480
C. 2800
D. 3140

Explanation –
Ratio of income = 6x : 7x

Ratio of expenditure = 5y : 4y

A.T.Q

6x – 5y = 700

7x – 4y = 2100

On Solving we get x = 700, y = 700

Required

Monthly Expenditure of anurag = 4y = 4x 700 = Rs.2800

2. In Shyam’s wallet there are total Rs 36, consisting of 10 paise, 20 paise and Re 1 coins such that there is at least one coin of each denomination. The ratio of coins of 10 paise to 20 paise is 8 : 5. Find the minimum no. of Rs 1 coins is

A. 9
B. 7
C. 8
D. 11

Explanation –
Let the coins of 10 paise, 20 paise and Rs 1 be x, y & z respectively.

A.T.Q.

10x + 20y + 100z = 3600

x : y = 8 : 5

80y + 100y + 100z = 3600

180y + 100z = 3600

9y + 5y = 180

Putting z = 1, 2, 3, 4, 5…., we get z = 9, y = 15 (an integer)

Hence, minimum 9 coins of `1 will be there.

3. In an army camp ration is available for 100 soldiers for 10 days. After 2 days, 60 soldiers joined. Then for how many more days will the remaining ration last?

A. 7
B. 6
C. 5
D. 4

4. If 15A = 20B = 30C, and A is 300, then find the value of B + C – A:

A. 75
B. 80
C. 60
D. 50

Explanation –
A: B: C::20 $$\times 30 : 15\times 30 : 15 \times 20 = 4 : 3 :2$$

B = $$\frac {300}{4} \times 3 = 225$$

c = $$\frac {300}{4} \times 2 = 150$$

B + C – A = 225 + 150 – 300 = 75

5. If x – y : y – z : z – k = 2 : 3 : 5, then what is the value of z -k : (x – k)?

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

Explanation –
Given: x – y : y – z : z – k = 2 : 3 : 5

Let x – y = 2x ———(i)

y – z = 3x ———(ii)

And z – k = 5x ———(iii)

Adding eq. (i), (ii) and (iii), we get

x – k = 10x

z -k : (x – k) = 5x : 10x = 1 : 2

1. Before a battle there were the ratio of captains to soldiers was 2 : 7. During the war 25 captains and 100 soldiers were martyred. The new ratio of captains to soldiers became 3 : 10. What is the number of soldiers after the war?

A. 250
B. 200
C. 150
D. 100

Explanation –
Let the No. of captain and soldiers = 2x & 7x

$$\frac {2x – 25}{7x – 100} = \frac {3}{10}$$

20x – 250 = 21x – 300

x = 50

No. of soldiers after war = 7x – 100

= 7 $$\times$$50 – 100 = 250

2. If 50 less had applied and 25 less selected, the ratio of selected to unselected would have been 9 : 4. So how many candidates had applied if the ratio of selected to unselected was 2 : 1.

A. 125
B. 250
C. 375
D. 500

Explanation –

Selected Unselected Total Results
2 1 3 Initial
9 4 13 Final

Initially there were total 3x candidates and 2x were selected.

According to question

Ratio of selected to applied:

$$\frac {2x – 25}{3x – 50} = \frac {9}{13}$$

26x – 25 $$\times$$ 13 = 27x – 9 $$\times$$ 50

27x – 26 x = 450 – 325

x = 125

Total candidates = 3x = 3 $$\times$$ 125 = 375

3. If 3A = 7B = 13C, then A : B : C is equal to

A. 21:39:91
B. 39:91:21
C. 13:7:3
D. 91:39:21

Explanation –
Given: 3A = 7B = 13C

$$\frac {A}{B} = \frac {7}{3}$$

And $$\frac {B}{C} = \frac {13}{7}$$

To get the relation between A, B and C,

Taking LCM of 3, 7 and 13, i.e. 273

Now, divide the given as

A : B : C = $$\frac {273}{3} : \frac {273}{7} : \frac {273}{13} = 91 : 39 : 21$$

4. In an MBA selection process, the ratio of selected to unselected was 11:2. If 40 less had applied and 20 less selected, the ratio of selected to unselected would have been 10:1. How many candidates had applied for the process?

A. 220
B. 260
C. 300
D. 340

Explanation –
Let the selected candidates = 11x

Unselected candidates = 2x

Total candidates = 11x + 2x = 13x

New applied candidates = 13x – 40

New selected candidates = 11x – 20

Ratio between the selected and unselected candidates = 10:1

ratio of total candidates to selected candidates = 11:10

$$\frac {13x – 40}{11x – 20} = \frac {11}{10}$$

130x – 400 = 121x – 220

9x = 180

X = 20

Total number of applied candidates = 13x = 13 × 20 = 260

5. If P : Q : R = 2 : 3 : 5, then what is the value of (P + Q) : (Q + R) : (R + P)?

A. 5:8:7
B. 2:3:5
C. 5:8:10
D. 4:9:25