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 Clerk Numerical Ability Quiz 6** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc.

**2. In how many different ways can the letters of the word “RADIUS” be arranged in such a way that the vowels occupy only the odd positions?**

**3. If 5:7 is the ratio of the present ages of Shahul and Ravi respectively. The difference between their ages is 6 years then what is the age of Shahul?**

**4. The ratio of present ages of Mani and Dilip is 4:3, after 3 years Mani’s age will be 39 years then the present age of Dilip is :**

**5. The ratio of the age of Hari to that of Charan is 6:7. If Hari is 4 years younger than Charan then what will be the ratio of the ages of Hari and Charan after 4 years?**

**Answers and Explanations**

**1. Answer –** Option A

**Explanation –**

There are 9 letters in the given word, out of which 4 are vowels.

In the word “CANDIDATE” we treat the vowels “AIAE” as one letter.

Thus, we have CNDDT(AIAE).

Now, we have to arrange 6 letters, out of which D occurs twice.

Therefore, number of ways of arranging these letters = \(\frac {6!}{2!} = \frac {720}{2} = 360\) ways.

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

Number of ways of arranging these letters = \(\frac {4!}{2!} = \frac {1 x 2 x 3 x 4}{2} = 12\)

Therefore, required number of words = (360 x 12) = 4320.

**2. Answer –** Option D

**Explanation –**

There are 6 different letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

[1] [2] [3] [4] [5] [6]

Now, 3 vowels can be placed at any of the three places out of 3 marked 1, 3 and 5.

Number of ways of arranging the vowels = \(^{3}{P}_{3} = 3! = 6\) = 3! = 6 ways.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = \(^{3}{P}_{3} = 3! = 6\) ways.

Therefore, total number of ways = 6 x 6 = 36.

**3. Answer –** Option A

**Explanation –**

Let the present age of Shahul and Ravi be 5X and 7X respectively.

Given that 7X – 5X = 6

2X = 6 \(\Rightarrow\) X = 3

Then Shahul’s present age is 5X = 5(3) = 15.

Hence, the answer is 15 years.

**4. Answer –** Option B

**Explanation –**

Let the present ages of Mani and Dilip is 4X and 3X respectively.

After 3 years Mani’s age is 4X + 3

4X + 3 = 39

4X = 36

X = 9.

Then the present age of Dilip is 3X = 3(9) = 27

Hence the answer is 27 years.

**5. Answer –** Option A

**Explanation –**

Ratio of the ages of Hari and Charan is 6:7.

Let the ages of Hari and Charan be 6X years and 7X years respectively.

Then 7X – 6X = 4 (since Hari is 4 years younger)

X = 4.

Now, the required ratio is 6X + 4 : 7X + 4

6(4) + 4 : 7(4) + 4 = 28 : 32

7 : 8

Hence the answer is 7 : 8

**2. 10% of ?% of 1000 = 150**

**3. The average of four consecutive odd integers is 24 less than the sum of these integers. Then which will be the largest of them?**

**4. The arithmetic mean of the five consecutive numbers a, a + 1, a+ 2, a + 3 and a + 4 is A, then the value of the median of these numbers is equal to:**

**5. A man lent out Rs.9600 at \(\frac {9}{2}\) % per annum for a year and 9 months. At the end of the duration, the amount he earned as S.I was:**

**Answers and Explanations**

**1. Answer –** Option B

**Explanation –**

Difference between successive numbers divisible by 3 will be 3. Therefore, the series takes the form as below.

x – 6,x – 3, x, x + 3, x +6, x + 9

Sum of all terms in the series = 99 = x – 6 + x – 3 + x + x + 3 + x + 6 + x + 9 = 6x + 9

6x + 9 = 99

Or 6x = 90

Or x = 15

Sum of first two numbers = x – 6 + x – 3 = 2x – 9 = 2(15) – 9 = 21

**2. Answer –** Option A

**Explanation –**

Let the number to be found be N

Then, 10% of N% of 1000 = 150

Or \((\frac {10}{100})x(\frac {N}{100}) x 1000 = 150\)

Or N = \(\frac {(150 x 100 x 100)}{(10 x 1000)} = 150\)

**3. Answer –** Option A

**Explanation –**

Let the x, x + 2, x + 4 and x + 6 be the 4 consecutive odd integers.

We have to find the largest integer x + 6.

It is given that the average of these numbers is 24.

i,e., \(\frac {(x + x + 2 + x + 4 + x + 6)}{4} \) = 24

4x + 12 = 24 x 4

4x = 84

x = \(\frac {84} {4} \) = 21

Then x + 6 = 21 + 6 = 27

Hence, the answer is 27.

**4. Answer –** Option A

**Explanation –**

Let a, a + 1, a + 2, a + 3 and a + 4 be the given 5 consecutive numbers.

We have to find value of their median.

i.e, middle number of them = a + 2

It is given that, their average = arithmetic mean = A = \(\frac {(a + a + 1 + a + 2 + a + 3 + a + 4)}{5}\)

A = \(\frac {(5a + 10)}{5} = \frac {5(a+2)}{5} = a + 2 \)

a + 2 = A.

**5. Answer –** Option B

**Explanation –**

Given that, principal = P = Rs.9600, R = \(\frac {9}{2}\) % and T = 1 year and 9 months = 1 + \(\frac {9}{12}\) year = \(\frac {7}{4}\) years.

Now, we have to find the S.I for \(\frac {7}{4}\) years.

S.I = \(\frac {PRT}{100}\) = Rs. \(9600 \times \frac {9}{2} \times \frac {7}{4} \times \frac {1}{100}\) = 12 x 9 x 7 = 756

Hence, the required S.I amount is Rs.756

**1. The number of men visiting Temples S are approximately what percent of the total number of person visiting all the temples together?**

**2. The number of children visiting Temple R is what per cent of the number of children visiting Temple T and U together?**

**3. What is the total number of visitor children and men together visiting Temple Q and Temple R together?**

**4. What is the approximate average number of women visiting all the Temples together?**

**5. What is the ratio of the number of women visiting Temple P to that of those visiting Temple R?**

**Answers and Explanations**

**1. Answer –** Option A

**Explanation –**

Total number of men visiting Temples S

= \(\frac {(56800 \times 45)}{100}\) = 25560

Total number of person visiting all the temples

= (34500 + 72500 + 45600 + 56800 + 42500 + 64600) = 316500

Required percentage = \(\frac {(25560 \times 100)}{316500}\) = 8%

**2. Answer –** Option C

**Explanation –**

Total number of children visiting Temple R

= \(\frac {(45600 \times 23)}{100} = 10488\)

Total number of children visiting Temple T and U

= \(\frac {(42500 \times 20)}{100} + \frac {(64600 \times 12)}{100}\)

= 8500 + 7752 = 16252

Required percentage = \(\frac {(10488 \times 100)}{16252} = 64.5 \)%

**3. Answer –** Option D

**Explanation –**

Total number of visitors children and men together visiting Temple Q

= \(\frac {(72500 \times 55)}{100}\) = 39875

Total number of visitors children and men together visiting Temple R

= \(\frac{(45600 \times 58)}{100}\) = 26448

Required number = (39875 + 26448) = 66323

**4. Answer –** Option C

**Explanation –**

Total number of women visiting all the Temples

= \(\frac {(34500 \times 55)}{100} + \frac {(72500 \times 45)}{100} + \frac {(45600 \times 42)}{100} + \frac {(56800 \times 28)}{100} + \frac {(42500 \times 65)}{100} + \frac {(64600 \times 58)}{100}\)

= (18975 + 32625 + 19152 + 15904 + 27625 + 37468) = 151749

Required average = \(\frac {151749}{6}\) = 25291.5 = 25292

**5. Answer –** Option A

**Explanation –**

Total number of women visiting Temple P

= \(\frac {(34500 \times 55)}{100}\) = 18975

Total number of women visiting Temple R

= \(\frac {(45600 \times 42)}{100} = 19152\)

Required ratio = \(\frac {18975}{19152} = 6325 : 6384 \)