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 Quantitative Aptitude Quiz Day 3** 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 Quantitative Aptitude Quiz Day 3** will assist the students to know the expected questions from **Quantitative Aptitude**.

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**Answer** : Option B

**Explanation**: Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127.

**2. Which of the following has the most number of divisors?**

**Answer** : Option C

**Explanation**: 99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

**3. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:**

**Answer** : Option C

**Explanation**: Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

The numbers are 16 and 24.

âˆ´ Hence, required sum = (16 + 24) = 40.

**4. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:**

**Answer** : Option C

**Explanation**: Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

âˆ´ The required sum = \(\frac{1}{a}\) + \(\frac{1}{b}\)

= \(\frac{a + b}{ab}\)

= \(\frac{55}{600}\)

= \(\frac{11}{120}\)

**5. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:**

**Answer** : Option C

**Explanation**: Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number = \(\frac{551}{29}\) = 19, Third number = \(\frac{1073}{29}\) = 37

âˆ´ Required sum = (19 + 29 + 37) = 85.

**Answer** : Option C

**Explanation**: S.I. for 3 years = Rs. (12005 – 9800) = Rs. 2205.

S.I. for 5 years = Rs. [\(\frac{2205}{3}\) x 5]= Rs. 3675

âˆ´ Principal = Rs. (9800 – 3675) = Rs. 6125.

Hence, rate = \(\frac{100 Ã— 3675}{6125 Ã— 5}\) = 12%

**2. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6% p.a for 2 years. Find his gain in the transaction per year.**

**Answer** : Option A

**Explanation**: Gain in 2 years = [ [5000 x \(\frac{25}{4}\) x \(\frac{2}{100}\)] – [\(\frac{5000 Ã— 4 Ã— 2}{100}\) ]]

= Rs. (625 – 400)

= Rs. 225.

âˆ´ Gain in 1 year = Rs. \(\frac{225}{2}\) = Rs. 112.50

**3. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?**

**Answer** : Option D

**Explanation**: TS.I. = Rs. (15500 – 12500) = Rs. 3000.

Rate = \(\frac{100 Ã— 3000 }{12500 Ã— 4}\) = 60

**4. A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:**

**Answer** : Option C

**Explanation**: Principal = Rs. \(\frac{100 Ã— 5400 }{12 Ã— 3}\)

= Rs. 15000.

**5. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?**

**Answer** : Option D

**Explanation**: Principal

= Rs.\(\frac{100 Ã— 4016.25 }{9 Ã— 5}\)

= Rs.\(\frac{401625 }{45}\)

= = Rs. 8925

**I.** The average mark obtained by Tarun in four subjects including English is 240.

**II.** The total marks obtained by him in English and Mathematics together are 170.

**III.** The total marks obtained by him in Mathematics and Science together are 180.

**Answer** : Option D

**Explanation**: I gives, total marks in 4 subjects = (60 x 4) = 240.

II gives, E + M = 170

III gives, M + S = 180.

**2. In a cricket team, the average age of eleven players in 28 years. What is the age of the captain?**

**I.** The captain is eleven years older than the youngest player.

**II.** The average age of 10 players, other than the captain is 27.3 years.

**III.** Leaving aside the captain and the youngest player, the average ages of three groups of three players each are 25 years, 28 years and 30 years respectively.

**Answer** : Option C

**Explanation**: Total age of 11 players = (28 x 11) years = 308 years.

I. C = Y + 11 C – Y = 11 …. (i)

II. Total age of 10 players (excluding captain) = (27.3 x 10) years = 273 years.

Age of captain = (308 – 273) years = 35 years.

Thus, C = 35. …. (ii)

From (i) and (ii), we get Y = 24

III. Total age of 9 players = [ (25 x 3) + (28 x 3) + (30 x 3)] years = 249 years.

C + Y = (308 – 249) = 59 …. (iii)

From (i) and (iii), we get C = 35.

Thus, II alone gives the answer.

Also, I and III together give the answer.

**3. The average age of P, Q, R, and S is 30 years. How old is R?**

**I.** The sum of ages of P and R is 60 years.

**II.** S is 10 years younger than R.

**Answer** : Option D

**Explanation**: P + Q + R + S = (30 x 4) P + Q + R + S = 120 …. (i)

I. P + R = 60 …. (ii)

II. S = (R – 10) …. (iii)

From (i), (ii) and (iii), we cannot find R.

**4. How many candidates were interviewed every day by the panel A out of the three panels A, B and C?**

**I.** The three panels on average interview 15 candidates every day.

**II.** Out of a total of 45 candidates interviewed everyday by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B.

**Answer** : Option B

**Explanation**: I. Total candidates interviewed by 3 panels = (15 x 3) = 45.

II. Let x candidates be interviewed by C.

Number of candidates interviewed by A = (x + 2).

Number of candidates interviewed by B = (x + 1).

x + (x + 2) + (x + 1) = 45

3x = 42

x = 14

**5. What is the average age of children in the class?**

**I.** The age of the teacher is as many years as the number of children.

**II.** The average age is increased by 1 year if the teacher’s age is also included.

**Answer**: Option D

**Explanation**: Let there be x children.

I gives, age of teacher = x years.

II gives, average age of (x + 1) persons = (x + 1) years.

Teacher’s age = (x + 1) (x + 1) – \({x}^{2}\) = (\({x}^{2}\) + 1 + 2x) – \({x}^{2}\) = (1 + 2x)

Thus, teacher’s age cannot be obtained.

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