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

**1. 30 , 62 , 189 , ? , 3805 , 22836**

**2. 1.5 , 2.5 , 6 , 19 , 77 , ?**

**3. 140, 133, 119, 98, ?, 35**

**4. On 1st January, 2000 the average age of a family of 6 people was ‘A’ years. After 5 years a child was born in the family and one year after that the average age was again found to be ‘A’ years. What is the value of ‘A’? (Assume that there are no other deaths and births.)**

**5. Two trains for Mumbai leave Delhi at 6 am and 6:45 am and travel at 100 kmph and 136 kmph respectively. Approx. how many km from Delhi will the two trains be together?**

**Answers and Explanations**

**1. Answer –** Option B

**Explanation –**

30 × 2 + 2 = 62

62 × 3 + 3 = 189

189 × 4 + 4 = 760

760 × 5 + 5 = 3805

3805 × 6 + 6 = 22836

30 , 62 , 189 , 760 , 3805 , 22836

Hence, option B is correct.

**2. Answer –** Option D

**Explanation –**

**3. Answer –** Option A

**Explanation –**

The Pattern is:-

140

140 – 7 = 133

133 – 14 = 119

119 – 21 = 98

98 – 28 = 70

70 – 35 = 35

Thus, the missing number is 70

So option (A) is the correct answer

**4. Answer –** Option E

**Explanation –**

Initially, the total age of the family is 6A (as average age of family is A and number of members in the family is 6)

Five years down the line the total age will increase by : 6*5

Then a child is born in the family and the total number of members now are 7, after a year total age will increase by 7*1

Given that, the average is still then A, so we can write the equation as:

6A + 6*5 + 7*1 = 7A ⇒ A = 37

**5. Answer –** Option B

**Explanation –**

In 45 min, 1st train will cover = \( 100 \times [\frac {45}{60}]\) = 75 km distance.

Relative speed of trains = 136 – 100 = 36 kmph.

2nd train will catch up the 1st train in = \(\frac {75}{36} = \frac {25}{12} \)hours.

Distance from Delhi = \([\frac {25}{12}] \times 136 \) = 283 km approx.

**2. K and L are two alloys of bronze and iron prepared by mixing the respective metals in the ratio of 5:3 and 5:11 respectively. If the alloys K and L are mixed to form a third alloy M with an equal ratio of bronze and iron then what is the ratio of alloys K and L in the new alloy M?**

**3. Siddhartha wants to buy a 11000 mah powerbank on a condition that he will pay 500 rs at the time of buying, 425 rs after 1 year and 289 rs after 2 years. If the compounded interest rate is 6 1/4 %, then what is the present value of powerbank?**

**Direction[4-5]: In the following number series, a wrong number is given. Find out that wrong number. **

**4. 7, 16, 24, 88, 113, 329**

**5. 75, 85, 110, 135, 165, 200**

**Answers and Explanations**

**1. Answer –** Option B

**Explanation –**

Let the diameter of each ball be 2r.

Length of the box = 3 * 2r=6r

Breadth=2 * 2r = 4r

Height = 2r

Volume=6r * 4r * 2r = 48\({r}^{3}\)

Volume of 6 balls = \( 6 \times (\frac {4}{3}) \times (\frac {22}{7}) \times {r}^{3} = 176 \frac {{r}^{3}}{7}\)

The area of empty space = \(48 {r}^{3} -176 \frac {{r}^{3}}{7}\)

= \(160 \frac {{r}^{3}}{7}\)

The required fraction = \(\frac {160 \frac {{r}^{3}}{7}}{48} = {r}^{3} = \frac {10}{21}\)

**2. Answer –** Option C

**3. Answer –** Option C

**Explanation –**

Let the present value of powerbank = x rs.

After paying 500 rs. Rest cost = (x-500) rs.

∴ compounded amount of this cost for 1 year = (x – 500)\( (1 + \frac {\frac {25}{4}}{100})\)

= \(\frac {17}{16}(x – 500)\)

Now, rest cost after paying 425 rs. = \(\frac {17}{16}(x – 500)\) – 425 rs

∴ compounded amount for next year =

\([\frac {17}{16}(x – 500) – 425 ] \times (1 + \frac {\frac {25}{4}}{100})\)

= \(\frac {17}{16} [\frac {17}{16}(x – 500) – 425]\)rs.

Now, after paying 289 rs. All instalments are completed,

So

\(\frac {17}{16} [\frac {17}{16}(x – 500) – 425]\)rs. = 289

⇒17(x – 500) – 425 * 16 = 272 * 16

⇒17x – 8500 – 6800 = 4352

⇒17x = 19652

⇒x = 1156

⇒Present value of powerbank = 1156 rs.

**4. Answer –** Option B

**Explanation –**

Pattern is \( + {2}^{3}, + {3}^{2}, + {4}^{3}, + {5}^{2}, + {6}^{3}\)

Hence wrong no. is 16

**5. Answer –** Option B

**Explanation –**

The series is:

75 + 3 * 5= 90

90 + 4 * 5= 110

110 + 5 * 5 = 135

135 + 6 * 5 = 165

165 + 7 * 5 = 200

Therefore 85 is wrong

**1. What is the difference between the total sale of Bartaman newspaper and the total sale of ABP newspaper in all the localities together?**

**2. The sale of Bartaman newspaper in Bidhan Nagar is approximately what percent of the total sale of Bartaman newspaper in all the localities together?**

**3. What is the ratio of the sale of Bartaman newspaper in locality Dhiman Nagar to the sale of ABP newspaper in locality Chandan Nagar?**

**4. The sale of Bartaman and ABP newspaper in locality Ananda Nagar is what percent of the sale of the same newspaper in locality English bazar?**

**5. What is the average sale of Bartaman and ABP newspaper in locality Bidhan Nagar and Dhiman Nagar?**

**Answers and Explanations**

**1. Answer –** Option D

**Explanation –**

Bartaman = 1500 + 2000 + 1800 + 3500 + 2500 = 11300

ABP = 2600 + 1000 + 1500 + 3000 + 3500 = 11600

Difference = 11600 – 11300 = 300.

**2. Answer –** Option A

**Explanation –**

Total Bartaman newspapers= 11300

Locality B’s percentage of Bartaman newspaper = \(\frac {2000}{11300} \times 100 \)% = 17.6% ≈ 18%

**3. Answer –** Option D

**Explanation –**

Amount of Bartaman newspaper in Dhiman Nagar and Chandan Nagar are 3500 and 1500.

3500 : 1500

=35 : 15 = 7 : 3.

**4. Answer –** Option B

**Explanation –**

Both newspapers in English bazaar = 3500 + 2500 = 6000 ——100

Both newspaper in Ananda Nagar = 2600 + 1500 = 4100

The required answer = \(\frac {4100}{6000} \times 100 \)% = 68.33%

**5. Answer –** Option A

**Explanation –**

Bidhan Nagar = (2000 + 1000) = 3000

Dhiman Nagar = (3000 + 3500) = 6500

The required answer, Average = \(\frac {(6500 + 3000 )}{2}\)

= \(\frac {9500}{2}\)

= 4750.