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IBPS RRB Quantitative Aptitude Quiz Day 3

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IBPS RRB Quantitative Aptitude Quiz Day 3

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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 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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Directions(1-5): LCM and HCF Problems

1. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

    A. 123
    B. 127
    C. 235
    D. 305


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?

    A. 99
    B. 101
    C. 176
    D. 182


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:

    A. 28
    B. 32
    C. 40
    D. 64


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.

∴ 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:

    A. \(\frac{55}{601}\)
    B. \(\frac{601}{55}\)
    C. \(\frac{11}{120}\)
    D. \(\frac{120}{11}\)


Answer : Option C

Explanation: Let the numbers be a and b.

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

    A. 75
    B. 81
    C. 85
    D. 89


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
∴ Required sum = (19 + 29 + 37) = 85.

Directions(1-5): Simple Interest
1. A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:

    A. 5%
    B. 8%
    C. 12%
    D. 15%


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
∴ Principal = Rs. (9800 – 3675) = Rs. 6125.
Hence, rate = \(\frac{100 × 3675}{6125 × 5}\) = 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.

    A. Rs. 112.50
    B. Rs. 125
    C. Rs. 225
    D. Rs. 167.50


Answer : Option A

Explanation: Gain in 2 years = [ [5000 x \(\frac{25}{4}\) x \(\frac{2}{100}\)] – [\(\frac{5000 × 4 × 2}{100}\) ]]
= Rs. (625 – 400)
= Rs. 225.
∴ 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?

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


Answer : Option D

Explanation: TS.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = \(\frac{100 × 3000 }{12500 × 4}\) = 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:

    A. Rs. 2000
    B. Rs. 10,000
    C. Rs. 15,000
    D. Rs. 20,000


Answer : Option C

Explanation: Principal = Rs. \(\frac{100 × 5400 }{12 × 3}\)
= 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?

    A. Rs. 4462.50
    B. Rs. 8032.50
    C. Rs. 8900
    D. Rs. 8925


Answer : Option D

Explanation: Principal
= Rs.\(\frac{100 × 4016.25 }{9 × 5}\)
= Rs.\(\frac{401625 }{45}\)
= = Rs. 8925

Directions(1-5): Averages
1. How many marks did Tarun secure in English?

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.


    A. I and II only
    B. II and III only
    C. I and III only
    D. All I, II and III


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.

    A. Any two of the three
    B. All I, II and III
    C. II only or I and III only
    D. II and III only


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.

    A. I alone sufficient while II alone not sufficient to answer
    B. II alone sufficient while I alone not sufficient to answer
    C. Either I or II alone sufficient to answer
    D. Both I and II are not sufficient to answer


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.

    A. I alone sufficient while II alone not sufficient to answer
    B. II alone sufficient while I alone not sufficient to answer
    C. Either I or II alone sufficient to answer
    D. Both I and II are not sufficient to answer


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.


    A. I alone sufficient while II alone not sufficient to answer
    B. II alone sufficient while I alone not sufficient to answer
    C. Either I or II alone sufficient to answer
    D. Both I and II are not sufficient to answer


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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