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IBPS RRB Officer Scale II Quantitative ...

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IBPS RRB Officer Scale II Quantitative Aptitude Practice Set 1

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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 Officer Scale II Quantitative Aptitude Practice Set 1 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 Officer Scale II Quantitative Aptitude Practice Set 1 will assist the students understanding of the type of questions expected from the topic Quantitative Aptitude.

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Directions(1-5): Permutation and Combinations

1. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

    A. 63
    B. 90
    C. 126
    D. 45


Answer : Option A

Explanation: \({7}_{{C}_{2}}\) x \({3}_{{C}_{1}}\)
= \(\frac{7 × 6}{2 × 1}\)
= 63


2. How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?

    A. 40
    B. 400
    C. 5040
    D. 2520


Answer : Option C

Explanation: ‘LOGARITHMS’ contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= \({10}_{{P}_{2}}\)
= (10 x 9 x 8 x 7)
= 5040.


3. In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?

    A. 10080
    B. 4989600
    C. 4989600
    D. None of these


Answer : Option C

Explanation: In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = \(\frac{8!}{(2!)(2!)}\) = 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters = \(\frac{4!}{2!}\) = 12.
Required number of words = (10080 x 12) = 120960.


4. In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?

    A. 120
    B. 720
    C. 4320
    D. 2160


Answer : Option B

Explanation: The word ‘OPTICAL’ contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.


5. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

    A. 159
    B. 194
    C. 205
    D. 209


Answer : Option C

Explanation: We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (\({6}_{{C}_{1}}\) x \({4}_{{C}_{3}}\)) + (\({6}_{{C}_{2}}\) x \({4}_{{C}_{1}}\)) + (\({6}_{{C}_{3}}\) x \({4}_{{C}_{1}}\) ) + ( \({6}_{{C}_{4}}\) )
= (\({6}_{{C}_{1}}\) x \({4}_{{C}_{1}}\)) + (\({6}_{{C}_{2}}\) x \({4}_{{C}_{2}}\)) + (\({6}_{{C}_{3}}\) x \({4}_{{C}_{1}}\)) + (\({6}_{{C}_{4}}\) )
= (6 x 4) + \(\frac{6 × 5}{2 × 1}\) x \(\frac{4 × 3}{2 × 1}\) + \(\frac{6 × 5 × 4 }{3 × 2 × 1}\) x 4 + \(\frac{6 × 5}{2 × 1}\)
= (24 + 90 + 80 + 15)
= 209.

Directions(1-5): Averages
1. A pupil’s marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:

    A. 10
    B. 20
    C. 40
    D. 73


Answer : Option C

Explanation: Let there be x pupils in the class.
Total increase in marks = [x × \(\frac{1}{2 }\)] = \(\frac{× }{2 }\)
∴ \(\frac{× }{2 }\) = (83 – 63) ⇒ \(\frac{× }{2 }\) = 20
⇒x = 40


2. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

    A. 23 years
    B. 24 years
    C. 25 years
    D. None of these


Answer : Option A

Explanation: Let the average age of the whole team by x years.
∴ 11x – (26 + 29) = 9(x -1)
⇒11x – 9x = 46

⇒ 2x = 46

⇒ x = 23.

So, the average age of the team is 23 years.


3. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:

    A. 3500
    B. 4000
    C. 4050
    D. 5000


Answer : Option B

Explanation: Let P, Q and R represent their respective monthly incomes. Then, we have:

P + Q = (5050 x 2) = 10100 …. (i)

Q + R = (6250 x 2) = 12500 …. (ii)

P + R = (5200 x 2) = 10400 …. (iii)

Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 …. (iv)

Subtracting (ii) from (iv), we get P = 4000.

∴ P’s monthly income = Rs. 4000.


4. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

    A. 35 years
    B. 40 years
    C. 50 years
    D. None of these


Answer : Option B

Explanation: Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.

Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.

∴ Husband’s present age = (90 – 50) years = 40 years.

5. In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?

    A. 67 kg.

    B. 68 kg.
    C. 69 kg.
    D. Data inadequate


Answer : Option A

Explanation: Let Arun’s weight by X kg.

According to Arun, 65 < X < 72

According to Arun's brother, 60 < X < 70.

According to Arun's mother, X <= 68

The values satisfying all the above conditions are 66, 67 and 68.
Required average = \(\frac{66 + 67 + 68}{3}\)
= \(\frac{201}{3}\)
= 67 kg



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