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 SO Quantitative Aptitude Quiz 2** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. The article **IBPS SO Quantitative Aptitude Quiz 2** will assist the students to know the expected questions from **Quantitative Aptitude**.

**1. How many conferences totally organised by five states in the year 2014 and 2015?**

**Answer –** Option D

**Explanation –**

Total = (21 + 18) + (12 + 6) + (20 + 27) +(34 + 41) + (17 + 10 )

= 39 + 18 + 47 + 75 + 27

= 206

**2. If the years are arranges in descending order of the total no of conferences organised by given states, which of the following would be the correct order?**

** Answer –** Option B

**Explanation –**

2014 = 104

2015 = 102

2016 = 138

**3. Approximately how much % increase in the no of conferences organised from 2014 to 2016 by Goa?**

**Answer –** Option D

**Explanation –**

% = \( \frac {32-12}{12} \times 100 = 167 \) %

**4. What is the average no of conferences organised by the state Andhra Pradesh in all the three years?**

**Answer –** Option A

**Explanation –**

AP Avg = \(\frac {20 + 27 + 19}{3}\) = 22

**5. What is the difference between the no of conferences organised by New Delhi in 2016 to no of conferences organised by Maharashtra in 2014 ?**

**Answer –** Option C

**Explanation –**

New Delhi 2016 = 50

Maharashtra 2014 = 17

Diff = 50 â€“ 17 = 33

**Answer –** Option D

**Explanation –**

Let Ektaâ€™s and Reemaâ€™s current income be 6x and 2y respectively

Hence, their last year income was 5x and 3y respectively.

Since the ratio of their income last year was 10 : 3

5x : 3y = 10: 3

x = 2y

The sum of Ektaâ€™s and Reemaâ€™s present incomes is Rs. 5, 124

6x + 2y = 5124

7x = 5124; x = 732 and y = 366

Therefore, Reemaâ€™s income last year = 3y = Rs. 1098

So option (D) is the correct answer.

**2. A, B and C invested in the ratio 5:6:8 respectively. If C invested for â€œxâ€ month and A and B invested for one year. At the end of the year Câ€™s profit is Rs. 3200 and difference between the profit of A and B is Rs. 600 then what is the value of â€œxâ€?**

** Answer –** Option D

**Explanation –**

Let A, B & C invested in the ratio of 5a,6a & 8a respectively and the total profit be P.

Their investment period is 12, 12 & x months respectively.

Hence, the total investment is \((5a \times 12) + (6a \times 12) + (8a \times x)\)

The difference in the profits of A & B is Rs 600.

\(\frac {(6a \times 12) – (5a \times12)}{(6a \times 12) + (5a \times 12) + (8a \times 12)}\)

\(\frac {12a}{132a + 8ax}P\)= 600

Profit earned by C is Rs. 3200

\(\frac {(8a \times x)}{(6a \times 12) + (5a \times 12) + (8a \times 12)}P\) = 3200

\(\frac {(8ax)}{(132a \times 8ax)}\) = 3200

By dividing the two equations obtained, we will get:

\((\frac {8x}{12}) = \frac {3200}{600}\)

x = 8 months

**3. If the ratio between the cost price and marked price of an article is 2:3. If the selling price of that article is Rs 1920 and the shopkeeper gave two successive discounts of 20% on the marked price. Find the loss during this transaction.**

**Answer –** Option A

**Explanation –**

The ratio between the cost price and marked price of an article is 2:3

Let the cost price = 2x & marked price = 3x

Two consecutive discounts of 20% are given.

So, one cummulative discount for two discounts of 20%

20% + 20% – \(\frac {20 \times 20}{100}\)% = 36%

Selling price = 1920 Rs

Hence,

\( 3x \times (\frac {64}{100}) = 1920\)

x = 1000

So, Cost price = 2000 Rs & marked price = 3000 Rs

Loss = (Selling price – Cost price) = 1920 – 2000 = 80 Rs

**4. Rs 18000 is invested in scheme â€˜Aâ€™ by ram which offers 15% p.a. at simple interest and Rs 15000 in scheme â€˜Bâ€™ which offers 18% p.a at compound interest. Find the difference between the interests earn from these two schemes after two years?**

**Answer –** Option D

**Explanation –**

SI in scheme A = \(\frac {18000 \times 2 \times 15}{100} = 5400\)

CI in scheme B = 15000 \(((\frac {118}{100} – \frac {118}{100}) – 1)\) = 5886

Required difference = 5886 â€“ 5400 = 486

So, option (D) is the correct answer.

**5. The cost of a precious stone varies as the cube of its weight. A certain precious stone broke into three pieces whose weighs are in the ratio 1 : 2 : 3, as a result of which its cost reduces by 80280. What was the cost of the stone before breaking?**

**Answer –** Option B

**Explanation –**

let the weights be x 2x and 3x grams

Previously, the weight of the stone was x + 2x +3x = 6x gms

Cost of the stone before breaking = \(k{(6x)}^{3} = 216k{x}^{3}\)

Cost the stone after breaking = \(k [{x}^{3} + {(2x)}^{3} + {(3x)}^{3}] = 36k{x}^{3}\)

Now, \(216k{x}^{3} â€“ 36k{x}^{3} = 80280\)

K\({x}^{3} = \frac {80280}{180}\)

216k\({x}^{3} = \frac {80280}{180} \times \) 216 = 96336

**1. \({(14.999)}^{2} – \sqrt[4] {1300} + \sqrt {99} = \)?**

**Answer –** Option B

**Explanation –**

Take a nearest values

\({(14.999)}^{2} – \sqrt [4]{1300} + \sqrt {99} = \)?

Since \(6 \times 6 \times 6 \times 6 = 1296\) very near to 1300

therefore,

? =225 â€“ 6 +10 = 229

**2. 31.85 Ã· 3.90 Ã— 15 = ?**

** Answer –** Option A

**Explanation –**

31.85 Ã· 3.90 Ã— 15 = ?

Approximates value can be calculated as

â‰ˆ 32 Ã· 4 Ã— 15 = 8 Ã— 15

= 120

**3. 19750.015 Ã·979.82 Ã— 201.04 =?**

**Answer –** Option C

**4. Find out the wrong number in the following given series.
24 33 58 107 186 309**

**Answer –** Option D

**Explanation –**

The given number series is based on the following pattern:

33 â€“ 24 = 9 = \({3}^{2}\)

58 â€“ 33 = 25 = \({5}^{2}\)

107 â€“ 58 = 49 = \({7}^{2}\)

188 â€“ 107 = 81 = \({9}^{2}\)

309 â€“ 188 = 121 = \({11}^{2}\)

Obviously, 186 is the wrong number.

**5. In the following number series, a wrong number is given. Find out that wrong number.
12, 26, 56, 118, 276, 498, 1008**

**Answer –** Option C

**Explanation –**

The followed pattern is :

12 Ã— 2 + 2 = 26

26 Ã— 2 + 4 = 56

56 Ã— 2 + 6 = 118

118 Ã— 2 + 8 = 244

244 Ã— 2 + 10 = 498

498 Ã— 2 + 12 = 1008

Hence, the wrong number is 276.