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

**2. Twenty men were employed to do same work in a certain time. But when one third of the scheduled time was over, it was found that only one – quarter of the total work was completed. How many more men should now be employed to complete the work in three – fourths of the originally scheduled time?**

**3. Amit goes to mall to buy a TV and a washing machine. He bargains for a 10% discount on the washing machine and 25% discount on TV. However, the shopkeeper, by mistake, interchanged the discount percentage figures while making the bill and Amit paid accordingly. When compared to what he should pay for his purchases, what percentage did Amit pay extra given that the washing machine costs 60% as much as the TV. **

**4. P and R entered into partnership business with the capital of Rs.X and Rs. (x+12000), after one year Q joined them with capital of Rs.(x+8000) at the end of second year P and Q with draw their capital and R invest for one more year, if P, Q and R gets profit in the ratio of 8:6:21 respectively. Find sum of capital invested by all three?**

**5. A book-shelf contains 2 English, 3 Hindi and 4 Sanskrit books. If two books are picked at random, then what is the probability that either all are Sanskrit or all are English books?**

**Answers and Explanations**

**1. Answer –** Option E

**Explanation –**

First part amount = (P + 1000)

Second part amount = (P + 1200)

Third part amount = (P – 800)

ATQ-

\(\frac {(P + 1200) \times 10 \times 2}{100} – \frac {(P – 800) \times 8 \times 2}{100} = 500\)

\(\frac {(P + 1200)}{5} – \frac {(4P-3200)}{25} = 500\)

Solving for P from the equation above:

P = 12500 – 9200 = 3300

Total saving of abhishek = (3300 + 1000) + (3300 + 1200)+ (3300 – 800)

= 11300

Total C.I on his saving at 10% = \( 10 + 10 + \frac {10 \times 10}100 \)

= 21%

= \(\frac {11300 \times 21}{100} = 2373 \)

**2. Answer –** Option D

**Explanation –**

20 men were employed to complete the work in say n days. Therefore, the estimated work 20n man days

Work completed in \(\frac {n}{3}\) days is 5n man days.

Remaining time according to revised schedule

= \(\frac {3N}{4} – \frac {n}{3} = \frac {9n – 4n}{12} = \frac {5n}{12}\)

Remaining work = 15n man days

20 men in \(\frac {n}{3}\) days do 5n man days of work

So, number of men needed in \(\frac {5n}{12}\) days to do

15n man days of work = (20) (\(\frac {n}{3}\) ) (\(\frac {12}{5n}\)) (\(\frac {15}{5}\)) = 48

Hence, 28 additional men are needed.

**3. Answer –** Option A

**Explanation –**

Let the cost price (CP) of TV = 100

CP of washing machine = 60

What Amit has paid,

Cost of washing machine= 60 × \(\frac {75}{100}\) = 45

Cost of TV = 100 × \(\frac {90}{100}\) = 90

Total= 90 + 45 = 135

What amit is supoosed to pay

Cost of washing machine= 60 × \(\frac {90}{100}\) = 54

Cost of TV= 100 × \(\frac {75}{100}\) = 75

Total = 54 + 75 = 129

⸫ required % = \( [\frac {(135-129)}{129}]\times 100 \) = 4.65 %

**4. Answer –** Option C

**Explanation –**

P : Q : R

\(X \times 2 : (x + 8000) \times 1 : (x + 12000) \times 3\)

8 : 6 : 21

ATQ-

= \(\frac {2x}{x + 8000} = \frac {8}{6}\)

6x – 4x = 32000

X = 16000

Required sum of capital ( P + Q + R)-

= 16000 + (16000 + 8000) + (16000 + 12000)

=68000

**5. Answer –** Option B

**Explanation –**

The number of books = 2 + 3 + 4 = 9

n(S) = \(^{9}{C}_{2} = \frac {9 \times 8}{2} = 36\)

n(E) = \(^{4}{C}_{2} + ^{2}{C}_{2} = 6 + 1 = 7\)

∴ P(E) = \( \frac {n(E)}{n(S)} = \frac {7}{36} \)

**Direction[1-5]:** **Study the line graph carefully and answer the given questions.**

**Strength of Seven Technology Institutes with Specializations in IT, Electronics and Mechanical in 2012.**

**1. If the number of students with Mechanical specialization in each institute increased by 20% and the number of students with IT specialization in each institute decreased by 10% from 2012-13, what is the difference between total mechanical student to total IT student in 2013.**

**2. What is the ratio between total number of students in institutes R and V respectively?**

**3. What is the difference between total number of students with Electronics specialization from all the institutes together and the total number of students with Mechanical specialization from all the institutes together?**

**4. If the number of students in institutes P, Q and R with IT specialization increased by 15%, 22% and 10% respectively from 2012 to 13, what was the total number of students with IT specialization in the three institutes together in 2013?**

**5. In institutes P, T and U the percentage of girls out of total number of students with Electronics specialization in respective institute is 50%, 55% and 48% respectively, what is the approx total number of boys in these three institutes with Electronics specialization?**

**Answers and Explanations**

**1. Answer –** Option B

**Explanation –**

Total mechanical student = \([\frac {(280 + 360 + 200 + 260 + 320 + 320 + 380)}{100}] \times 120 = 2544\)

Total IT student = \([\frac {(220 + 240 + 320 + 140 + 280 + 150 + 280)}{100} ]\times 90 = 1467\)

Difference = 2544 – 1467 = 1077

**2. Answer –** Option B

**Explanation –**

Total student in R = 200 + 260 + 320 = 780

Total student in V = 220 + 280 + 380 = 880

Ratio = 780 : 880 = 39 : 44

**3. Answer –** Option D

**Explanation –**

Total mechanical student = (280 + 360 + 200 + 260 + 320 + 320 + 380) = 2120

Total Electronics student = (340 + 300 + 260 + 340 + 190 + 260 + 220) = 1910

Difference = 2120 – 1910 = 210

**4. Answer –** Option C

**Explanation –**

Total number = \( [\frac {(220 \times 115)}{100}]+ [\frac {(240 \times 122)}{100}]+ [\frac {(320 \times 110)}{100}] = 898 \)

**5. Answer –** Option A

**Explanation –**

Total no. of boys (Approx.) in P, T and U = \( [\frac {(340 \times 50)}{100}]+ [\frac {(190 \times 45)}{100}]+ [\frac {(260 \times 52)}{100}] = 391 \)

**1. I. 3\( {X}^{2}\) + 7X = 6
II. 3\( {Y}^{2}\) – 7Y + 2 = 0**

**2. I. 6\( {X}^{2}\) – 7X + 2 = 0
II.2\( {Y}^{2}\) – 15Y + 25 = 0**

**Direction[3-4]: What will come in place of the question marks (?) in the following questions?**

**3. 67.39 -11.78 + 19.63 =? + 22.41**

**4. 125 of 44% + 840 of 75% = ?**

**5. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 2 % higher rate, it would have fetched Rs. 400 more. Find the sum.**

**Answers and Explanations**

**1. Answer –** Option E

**Explanation –**

3\( {X}^{2}\) + 7X = 6

3\( {X}^{2}\) + 7X – 6 = 0

3\( {X}^{2}\) + 9X – 2X – 6 = 0

3X(X + 3) -2(X + 3) = 0

(X + 3)(3X – 2) = 0

X = -3 or X= \( \frac{2}{3}\)

3\( {Y}^{2}\) – 7Y + 2 = 0

3\( {Y}^{2}\) – 6Y – Y + 2 = 0

3Y(Y – 2) -1(Y – 2) = 0

(Y – 2)(3Y – 1) = 0

Y = 2 or Y = \( \frac{1}{3}\)

No relation can be established.

**2. Answer –** Option B

**Explanation –**

6\( {x}^{2}\) – 7x + 2 = 0

6\( {x}^{2}\) – 4x – 3x + 2 = 0

2x(3x – 2)-1(3x – 2) = 0

(3x – 2)(2x – 1) = 0

x = \( \frac{2}{3}\) or x = \( \frac{1}{2}\)

2\( {y}^{2}\) -15y + 25 = 0

2\( {y}^{2}\) – 10y – 5y + 25 = 0

2y(y – 5)- 5(y – 5) = 0

(y – 5)(2y – 5) = 0

y = 5 or y = \( \frac{5}{2}\)

x<y

**3. Answer –** Option B

**Explanation –**

67.39 -11.78 + 19.63 = ? + 22.41

? = 75.24 – 22.41 \(\rightarrow\) 52.83

**4. Answer –** Option C

**Explanation –**

44% of 125 + 75% of 840 = 55 + 630 = 685

**5. Answer –** Option C

**Explanation –**

Let sum be x.

Original rate be R.

1st condition:

Rate of interest = R

Time = 2 years

2nd condition:

Rate of interest increased by 2 = (R + 2) %

Time = 2 years

Therefore,

\(\frac {[P × (R + 2) × 2]}{100} – \frac {(P × R × 2)}{100} = 400\)

2PR + 4P – 2PR – = 40000

4P = 40000

P = Rs.10000

The required sum is Rs. 10,000