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IBPS SO Quantitative Aptitude Quiz 2

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IBPS SO Quantitative Aptitude Quiz 2

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What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.


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.


shape Quiz

Directions(1 – 5): Study the following information carefully and answer the given questions:



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

    A. 197
    B. 243
    C. 312
    D. 206


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?

    A. 2014, 2015, 2016
    B. 2016, 2014, 2015
    C. 2016, 2015, 2014
    D. 2014, 2016, 2015


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?

    A. 126%
    B. 156%
    C. 176%
    D. 167%


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?

    A. 22
    B. 18
    C. 20
    D. 32


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 ?

    A. 35
    B. 29
    C. 33
    D. 27


Answer – Option C

Explanation –
New Delhi 2016 = 50

Maharashtra 2014 = 17

Diff = 50 – 17 = 33

1. The ratio of Ekta’s and Reema’s income last year was 10: 3. The ratio of Ekta’s this year income and last year income is 6: 5 and the ratio of Reema’s this year income and last year income is 2: 3. If the sum of Ekta’s and Reema’s present incomes is Rs. 5124, what was Reema’s income last year?

    A. 1830
    B. 1370
    C. 1372
    D. 1098


Answer – Option D

Explanation –
Let Ekta’s and Reema’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 Ekta’s and Reema’s present incomes is Rs. 5, 124


6x + 2y = 5124


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


Therefore, Reema’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 “x” month and A and B invested for one year. At the end of the year C’s profit is Rs. 3200 and difference between the profit of A and B is Rs. 600 then what is the value of “x”?

    A. 4 months
    B. 3 months
    C. 6 months
    D. 8 months


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.

    A. 80
    B. 20
    C. 70
    D. 50


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 ‘A’ by ram which offers 15% p.a. at simple interest and Rs 15000 in scheme ‘B’ which offers 18% p.a at compound interest. Find the difference between the interests earn from these two schemes after two years?

    A. 396
    B. 456
    C. 486
    D. 482


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 – 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?

    A. 56236
    B. 96336
    C. 85236
    D. 76236


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, \(216k{x}^{3} – 36k{x}^{3} = 80280\)


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


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

Direction(1-3): What approximate value should come in place of the question mark (?) in the following question (Note: You are not expected to calculate the exact value)?


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

    A. 250
    B. 229
    C. 270
    D. 255


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 – 6 +10 = 229


2. 31.85 ÷ 3.90 × 15 = ?

    A. 120
    B. 90
    C. 80
    D. 140


Answer – Option A

Explanation –
31.85 ÷ 3.90 × 15 = ?

Approximates value can be calculated as

≈ 32 ÷ 4 × 15 = 8 × 15

= 120


3. 19750.015 ÷979.82 × 201.04 =?

    A. 45
    B. 200
    C. 4050
    D. 10500


Answer – Option C


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


    A. 58
    B. 33
    C. 107
    D. 186


Answer – Option D

Explanation –
The given number series is based on the following pattern:

33 – 24 = 9 = \({3}^{2}\)

58 – 33 = 25 = \({5}^{2}\)

107 – 58 = 49 = \({7}^{2}\)

188 – 107 = 81 = \({9}^{2}\)

309 – 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


    A. 26
    B. 56
    C. 276
    D. 1008


Answer – Option C

Explanation –
The followed pattern is :

12 × 2 + 2 = 26

26 × 2 + 4 = 56

56 × 2 + 6 = 118

118 × 2 + 8 = 244

244 × 2 + 10 = 498

498 × 2 + 12 = 1008

Hence, the wrong number is 276.


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