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 2** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc.

**Total number of students appeared and qualified from various colleges in a scholarship exam.**

**1. The average number of students qualified in the examination from colleges R and S is what per cent of the average number of students appeared for the examination from the same colleges?**

**2. What is the ratio of the total number of students appeared to the total number of students qualified in the scholarship exam from R and T?**

**3. What is the ratio of the number of students qualified in the scholarship examination from college P and the number of students qualified in the examination from college Q?**

**4. The number of students appeared for the scholarship exam from college S is approximately what percent of the total number of students appeared for the exam from all the colleges together?**

**5. What is the difference between the average number of students appeared in the scholarship exam from all the given colleges and the average number of students qualified from all the colleges together?**

**Answers and Explanations**

**1. Answer –** Option A

**Explanation –**

Total number of students qualified in the examination from colleges R and S

= (3250 + 1500) = 4750

Average number of students qualified in the examination from colleges R and S

= \( \frac {4750}{2}\) = 2375

Total number of students appeared in the examination from colleges R and S

= (3750 + 2500) = 6250

Average number of students appeared in the examination from colleges R and S

= \( \frac {6250}{2}\) = 3125

Required percentage = \( \frac { (2375 \times 100)}{3125}\) (2375*100) = 76%

**2. Answer –** Option C

**Explanation –**

Total number of students appeared in the scholarship exam from R and T

= (3750 + 3000) = 6750

Total number of students qualified in the scholarship exam from R and T

= (3250 + 2250) = 5500

Required ratio = \( \frac {6250}{5500}\) = 27 : 22

**3. Answer –** Option D

**Explanation –**

Required ratio = \( \frac {2250}{1500}\) = 3 : 2

**4. Answer –** Option C

**Explanation –**

Total number of students appeared for the scholarship exam from colleges = 2500

Total number of students appeared for the exam from all the colleges

= (3500 + 2750 + 3750 + 2500 + 3000) = 15500

Required percentage = \( \frac { (2500 \times 100)}{15500}\) = 16.12%

**5. Answer –** Option A

**Explanation –**

Total number of students appeared for the exam from all the colleges

= (3500 + 2750 + 3750 + 2500 + 3000) = 15500

Average = \( \frac {15500}{5}\) = 3100

Total number of students qualified for the exam from all the colleges

= (2250 + 1500 + 3250 + 1500 + 2250) = 10750

Average = \( \frac {10750}{5}\) = 2150

Required difference = (3100 – 2150) = 950

**1. \(\sqrt {676} = \sqrt {225} + ?\)**

**2. \(\frac {3}{19} of \frac {6}{15} of 855 = ?\)**

**3. 672Ã·24Ã—18+153-345= ?**

**Direction[4-5]:** In each question numbered I and II have been given. You have to solve both the equations and mark the appropriate option.

**4. I. \({X}^{2} – x – 6 = 0\)**

**II. \({Y}^{2} – 6Y + 8 = 0\)**

**5. I. \({X}^{2} + 11x + 30 = 0\)**

**II. \({Y}^{2} + 7Y + 12 = 0\)**

**Answers and Explanations**

**1. Answer –** Option A

**Explanation –**

\(\sqrt {676} = \sqrt {225} + ?\)

26 = 15 + x

So x = 11

**2. Answer –** Option B

**Explanation –**

? = 855 Ã— (6 Ã· 15) Ã— (3 Ã· 19) = 54

Hence option B is correct

**3. Answer –** Option C

**Explanation –**

672 Ã· 24 Ã— 18 + 153 – 345

504 + 153 -345 = 312

**4. Answer –** Option E

**Explanation –**

I. x=3,âˆ’2

II. Y=2, 4

Hence answer=(E)

**5. Answer –** Option B

**Explanation –**

From I,

\({X}^{2} + 11x + 30 = 0\)

\({X}^{2} + 6X + 5X + 30 = 0\)

X (X + 6) + 5(X + 6) = 0

(X + 6)(X + 5) = 0

X = -5, -6

From II,

\({Y}^{2} + 7Y + 12 = 0\)

\({Y}^{2} + 4Y + 3(y – 4) = 0\)

Y = -3, -4

**2. If the numerator of a fraction is increased by 200% and the denominator is increased by 300%, the resultant fraction is 15/26. What was the original fraction?**

**3. Four men can complete a piece of work in 40 days. Four women can complete the same work in 64 days. Find the time in which 10 men and 16 women working together can complete the same work?**

**4. The respective ratio between two numbers is 4: 7. If 4 is added to each of the two numbers, their respective ratio is 3 : 5. What is the larger number?**

**5. 4% of 599.78 Ã— 19.1919 Ã· (2.987 +1.004) =?**

**Answers and Explanations**

**1. Answer –** Option A

**Explanation –**

Let Adamâ€™s investment be 7x

Shaneâ€™s investment = 2x

Let total revenue for two years be 2Y

Adamâ€™s share in revenue = Y

Shaneâ€™s share in revenue = Y

Now to equalise initial investment \( \rightarrow 2x + \frac {Y}{2} + \frac {Y}{4} = 7x\)

\(\frac {3Y}{4} = 5x \rightarrow \frac {x}{Y} = \frac {3}{20}\)

Ratio of initial investment and total reveue= \(\frac {9x}{2Y} = \frac {27}{40}\)

**2. Answer –** Option D

**Explanation –**

Let the original fraction be \( \frac {x}{y}\)

Then according to questions

\( \frac {300x}{400y} = \frac {15x}{20y} = \frac {3x}{4y} \)

\( \frac {3x}{4y} = \frac {15}{26} \)

\( \frac {x}{y} = \frac {10x}{13y} \)

**3. Answer –** Option A

**Explanation –**

4 men take 40 days to complete the work.

10 men would take = \(\frac {4 \times 40}{10}\) = 16 days

Similarly, 16 women would take 16 days to complete it.

10 men as well as 16 women would take 16 days to complete it.

So, working together they would take \(\frac {16}{2}\) = 8 days to complete it.

**4. Answer –** Option C

**Explanation –**

\(\frac {4x + 4}{7x + 4} = \frac {3}{5}\)

x = 8

larger number = 7x = 56

hence option C is correct

**5. Answer –** Option C

**Explanation –**

4% of 600 = 24

2.987 + 1.004 = 4

\(\frac {24 Ã— 19}{4}\) = 456

6 Ã— 19 = 114