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 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 Quantitative Aptitude Quiz Day 1** will assist the students to know the expected questions from **Quantitative Aptitude**.

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**Answer** : Option C

**Explanation**: The word ‘APPLE’ contains 5 letters, 1A, 2P, 1L.and 1E.

âˆ´ Required number of ways = \(\frac{5!}{(1!)(2!)(1!)(1!)}\) = 60

**2. The value of \({75}_{{P}_{2}}\) is :**

**Answer** : Option C

**Explanation**: \({75}_{{P}_{2}}\) = \(\frac{75!}{75-2!}\)

= \(\frac{75!}{73!}\)

= \(\frac{75*74*(73!)}{73!}\)

= (75 * 74) = 5550.

**3. In how many ways can the letters of the word ‘LEADER’ be arranged ?**

**Answer** : Option C

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

The word ‘LEADER’ contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

âˆ´ Required number of ways = \(\frac{6!}{(1!)(2!)(1!)(1!)(2!)}\) = 360

**4. How many words with or without meaning, can be formed by using all the letters of the word, ‘DELHI’, using each letter exactly once ?**

**Answer** : Option D

**Explanation**: The word ‘DELHI’ contains 5 different letters.

Required number of words = Number of arrangements of 5 letters, taken all at a time = \({5}_{{P}_{5}}\) = 5 !

= (5 *4 *3 *2 *1) = 120.

**5. In how many different ways can the letters of the word ‘RUMOUR’ be arranged ?**

**Answer** : Option A

**Explanation**: The word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1U.

âˆ´ Required number of ways = \(\frac{6!}{(2!)(2!)(1!)(1!)}\) = 180

**Answer** : Option B

**Explanation**: Total age of 5 members, 3 years ago = (17 x 5) years = 85 years

Total age of 5 members now = (85 + 3 x 5) years = 100 years

Total age of 6 members now = (17 x 6) years = 102 years

Age of the baby = (102 â€“ 100) years = 2 years

**2. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is:**

**Answer** : Option B

**Explanation**: Let the ratio be k:1

Then, k x 16.4 + 1 x 15.4 = (k + 1) x 15.8

(16.4 â€“ 15.8) k = (15.8 â€“ 15.4)

k = \(\frac{2}{3}\)

**3. The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books?**

**Answer** : Option A

**Explanation**: Total price of the two books = Rs. [(12 x 10) â€“ (11.75 x 8)] = Rs. (120 â€“ 94) = Rs. 26

Let the price of one book be Rs.x

Then, the price of other book = Rs. (x + 60% of x ) = x + \(\frac{3}{5}\)x = (\(\frac{8}{5}\))x

so, x + \(\frac{8}{5}\)x = 26 , x = 10

The prices of the two books are Rs. 10 and Rs. 16

**4. The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher?**

**Answer** : Option C

**Explanation**: Total ages of 30 boys = 14 x 30 = 420 years

Total age when class teacher is included = 15 x 31 = 465 years

Age of class teacher = 465 â€“ 420 = 45 years

**5. The Average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of failed candidates is 15, what is the number of candidates who passed the examination?**

**Answer** : Option B

**Explanation**: Let the number of passed candidates be a

Then total marks =>120 x 35 = 39 a + (120 â€“ a) x 15

4200 = 39 a + 1800 â€“ 15 a

a = 100

**Answer** : Option C

**Explanation**: 6*2 + 10 = 22

22*2 â€“ 10 = 34

34*2 + 10 = 78

78*2 â€“ 10 = 146

146*2 + 10 = 302

302*2 â€“ 10 = 594

**2. 4, 4, 15, 74, 524, 4724, 51974**

**Answer** : Option B

**Explanation**: 4*1 + 0 = 4

4*3 + 2 = 14

14*5 + 4 = 74

74*7 + 6 = 524

524*9 + 8 = 4724

4724*11 + 10 = 51974

**3. 3, 5, 13, 39, 177, 891**

**Answer** : Option D

**Explanation**: 3*1 + 2 = 5

5*2 + 3 = 13

13*3 + 4 = 43

43*4 + 5 = 177

177*5 + 6 = 891

**4. 3, 7, 12, 32, 57, 93, 142**

**Answer** : Option B

**Explanation**: 3 + \({2}^{2}\) = 7

7 + \({3}^{2}\) = 16

16 + \({4}^{2}\) = 32

32 + \({5}^{2}\) = 57

16 + \({4}^{2}\) = 32

32 + \({5}^{2}\) = 57

57 + \({6}^{2}\) = 93

93 + \({7}^{2}\) = 142

**5. 2, 13, 38, 130, 522, 2625**

**Answer** : Option D

**Explanation**: 2*1 + 11 = 13

11*2 + 12 = 38

38*3 + 13 = 127

127*4 + 14 = 522

522*5 = 2625

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