# IBPS RRB PO Quantitative Aptitude Quiz 14

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# IBPS RRB PO Quantitative Aptitude Quiz 14

### Introduction

What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.

A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.

The article IBPS RRB PO Quantitative Aptitude Quiz 14 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. IBPS RRB has released IBPS RRB Officer 2019 Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article IBPS RRB PO Quantitative Aptitude Quiz 14 will assist the students to know the expected questions from Quantitative Aptitude.

### Quiz

1. Fresh fruit contains 68 % water and dry fruit contains 20 % water. How much dry fruit can be obtained from 100 kg of fresh fruits?

A. 32 kg
B. 40 kg
C. 52 kg
D. 80 kg

Explanation –

Quantity of water in 100kg of fresh fruits = ($$\frac {68} {100}$$ x 100) )kg

Quantity of pulp in it = (100 – 68)kg = 32 kg

Let the dry fruit be x kg
Water in it = ($$\frac {20} {100}$$ x 100) )kg = $$\frac {x} {5}$$ kg

Quantity of pulp in it = (x – $$\frac {x} {5}$$)kg = $$\frac {4x} {5}$$ kg
Therefore, $$\frac {4x} {5}$$ kg = 32 => x = $$\frac {160} {4}$$ = 40 kg

2. Fresh grapes contain 80 % water dry grapes contain 10 % water. If the weight of dry grapes is 250 kg. What was its total weight when it was fresh?

A. 1000 kg
B. 1100 kg
C. 1125 kg
D. 1125 kg

Explanation –

Let the weight of fresh grapes be x kg

Quantity of water in it = ($$\frac {80} {100}$$ × x)kg = $$\frac {4x} {5}$$ kg

Quantity of pulp in it = (x – $$\frac {4x} {5}$$ )kg = $$\frac {x} {5}$$ kg

Quantity of water in 250 kg dry grapes

= ($$\frac {10} {100}$$ × 250)kg = 25kg

Quantity of pulp in it = (250 – 25)kg = 225 kg

Therefore, $$\frac {x} {5}$$ = 225

=> x = 1125

3. The quality of water that should be added to reduce 9 ml. Lotion containing 50 % alcohol to a lotion containing 30 % alcohol is

A. 3 ml
B. 4 ml
C. 5 ml
D. 6 ml

Explanation –

Alcohol in 9 ml lotion = ($$\frac {50} {100}$$ × 9)ml = 4.5 ml

Water in it = (9 – 4.5)ml = 4.5 ml

Let x ml of water be added to it, then $$\frac {4.5} {9+x}$$ × 100 = 30

=> $$\frac {4.5} {9+x}$$ = $$\frac {30} {100}$$ = $$\frac {3} {10}$$

=> 3(9+x) = 45 => 27 + 3x = 45

=> 3x = 18

=> x = 6

Water to be added = 6 ml

4. In a school, 40 % of the students play football and 50 % play cricket. If 18 % of the students play neither football nor cricket, the percentage of students playing both is

A. 40 %
B. 32 %
C. 22 %
D. 8 %

Explanation –

Let A = set of students who play football and

B = set of students play cricket.

Then n(A) = 40, n (B) = 50 and

n(A U B) = (100 – 18) = 82

n(A U B) = n(A) + n(B) – n(A ∩ B)

n(A∩B) = n(A) + n(B) – n(AUB) = (40 + 50 -82) = 8

Percentage of the students who play both = 8%

5. If the price of the eraser is reduced by 25% a person buy 2 more erasers for a rupee. How many erasers available for a rupee?

A. 8
B. 6
C. 4
D. 2

Explanation –
Let n erasers be available for a rupee

Reduced Price = ($$\frac {75} {100}$$ × 1) = $$\frac {3} {4}$$

$$\frac {3} {4}$$ rupee fetch n erasers = 1 Rupee will fetch (n × $$\frac {4} {3}$$) erasers

Therefore, $$\frac {4n} {3}$$ = n +2 => 4n = 3n +6 => n = 6

1. The sum of three consecutive integers is 102. Find the lowest of the three?

A. 40
B. 53
C. 29
D. 33

Explanation –

Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33.

2. The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?

A. 34
B. 28
C. 12
D. 17

Explanation –

Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28

3. The sum of three consecutive even numbers is 42. Find the middle number of the three?

A. 14
B. 16
C. 18
D. 24

Explanation –

Three consecutive even numbers (2P – 2), 2P, (2P + 2).
(2P – 2) + 2P + (2P + 2) = 42
6P = 42 => P = 7.
The middle number is: 2P = 14.

4. The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?

A. 50
B. 60
C. 70
D. 80

Explanation –

A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.

5. The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find the their present ages?

A. 30, 40
B. 25, 30
C. 50, 60
D. 20, 40

Explanation –

Their present ages be 3X and 4X.
5 years age, the ratio of their ages was 5:7, then (3X – 5):(4X – 5) = 5:7
X = 35 – 25 => X = 10.
Their present ages are: 30, 40.

1. A man said to his son, “I was two-third of your present age when you were born”. If the present age of the man is 48 years, find the present age of the son?

A. 25.7 years
B. 28 years
C. 29.3 years
D. 28.8 years

Explanation –

Present age of the son be P, he was born P years ago.
The age of the man was: (48 – P).
His age when the son was born should be equal to $$\frac {2} {3}$$ of P.
(48 – P) = $$\frac {2} {3}$$ P
5P = 144 => P = 28.8

2. The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years?

A. 3:4
B. 3:5
C. 4:3
D. 7:8

Explanation –

Let P’s age and Q’s age be 6x and 7x years respectively.

Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8

3. The ratio between the present ages of P and Q is 5:7 respectively. If the difference between Q’s present age and P’s age after 6 years is 2. What is the total of P’s and Q’s present ages?

A. 48 years
B. 52 years
C. 56 years
D. Cannot be determined

Explanation –

Let the present ages of P and Q be 5x and 7x years respectively.

Then, 7x – (5x + 6) = 2
2x = 8 => x = 4
Required sum = 5x + 7x = 12x = 48 years.

4. Present ages of X and Y are in the ratio 5:6 respectively. Seven years hence this ratio will become 6:7 respectively. What is X’s present age in years?

A. 35
B. 42
C. 49
D. Cannot be determined

Explanation –

Let the present ages of X and Y be 5x and 6x years respectively.

Then, $$\frac {(5x + 7)} {(6x + 7)}$$= $$\frac {6} {7}$$
7(5x + 7) = 6(6x + 7) => x = 7
X’s present age = 5x = 35 years.

5. The ratio of the present age of two brothers is 1:2 and 5 years back, the ratio was 1:3. What will be the ratio of their ages after 5 years?

A. 1:4
B. 2:3
C. 3:5
D. 5:6

Explanation –

Their present ages be 3X and 4X.
Let the present ages of the two brothers be x and 2x years respectively.

Then, $$\frac {(x – 5)} {(2x – 5)}$$ = $$\frac {1} {3}$$
3(x – 5) = (2x – 5) => x = 10
Required ratio = (x + 5) : (2x + 5) = 15 : 25 = 3:5