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**.

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**Answer –** Option B

**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?**

**Answer –** Option C

**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**

**Answer –** Option D

**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**

**Answer –** Option D

**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?**

**Answer –** Option B

**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

**Answer –** Option D

**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?**

**Answer –** Option B

**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?**

**Answer –** Option A

**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?**

**Answer –** Option C

**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?**

**Answer –** Option A

**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.

**Answer –** Option D

**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?**

**Answer –** Option D

**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?**

**Answer –** Option A

**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?**

**Answer –** Option A

**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?**

**Answer –** Option C

**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