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TS CAB Recruitment Quantitative Aptitude Quiz 1

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TS CAB Recruitment Quantitative Aptitude Quiz 1

shape Introduction

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 TS CAB Recruitment Quantitative Aptitude Quiz 1 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. The article TS CAB Recruitment Quantitative Aptitude Quiz 1 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

1. Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?

    A. 270 m
    B. 250 m
    C. 280 m
    D. 220 m


Answer: Option A

Explanation:
Relative speed = (72 – 36) * \(\frac{5}{18}\) = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m.


2. [50 – {20 + 60% of (\(\frac{26}{13}\) + \(\frac{24}{12}\) + 21) – 2} + 8] = \(\frac{625}{?}\)

    A. 25
    B. 125
    C. 24\(^ {1/5}\)
    D. 25\(^ {3/5}\)
    E. 5


Answer: Option A

Explanation:
[50 – {20 + 60% of (\(\frac{26}{13}\) + \(\frac{24}{12}\) + 21) – 2} + 8] = \(\frac{625}{x}\) => [50 – {20 + \(\frac{3}{5}\) of (25) – 2} + 8] = 625/x

=> [50 – 33 + 8] = 625/x => x = 25


3. In a 200 meters race A beats B by 35 m or 7 seconds. A’s time over the course is:

    A. 40 sec
    B. 47 sec
    C. 33 sec
    D. None of these


Answer: Option C

Explanation:
B runs 35 min 7 sec.
B covers 200 m in (\(\frac{7}{35}\)*200) = 40sec
B’s time over the course = 40 sec.
A’s time over the course (40 – 7) sec = 33 sec.


4. The value of \(\frac {1}{log_{3}60}\)+\(\frac {1}{log_ {4}{60}}\)+\(\frac {1}{log_{5}{60}}\)is:


    A. 0
    B. 1
    C. 5
    D. 60


Answer: Option B

Explanation:
\(log_{60}{3}\)+ \(log_{60}{4}\) + \(log_{60}{5}\)
= \(log_{60}(3 \times 4 \times 5)\)
= \(log_{60}{60}\)
= 1.

5. Calculate the number of bricks, each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar?

    A. 4000
    B. 5000
    C. 6000
    D. 7000


Answer: Option C

Explanation:
10 * \(\frac{4}{100}\) * 5 * \(\frac{90}{100}\) = \(\frac{25}{100}\) * \(\frac{15}{100}\) * \(\frac{8}{100}\)* x

10 * 20 * 90 = 15 * 2 * x => x = 6000

1. A sum of Rs. 125000 amounts to Rs. 15500 in 4 years at the rate of simple interest. What is the rate of interest?

    A. 3%
    B. 4%
    C. 5%
    D. 6%


Answer: Option D

Explanation:
S.I. = (15500 – 12500) = Rs. 3000
Rate =\(\frac{(100 * 3000) }{(12500 * 4)}\) = 6%.


2. Three flags each of different colors are available for a military exercise, Using these flags different codes can be generated by waving
I. Single flag of different colors
II. Any two flags in a different sequence of colors.
III. three flags in a different sequence of colors.
The maximum number of codes that can be generated is.


    A. 6
    B. 9
    C. 15
    D. 18


Answer: Option C

Explanation:
This type of question becomes very easy when we assume three colors are red(R) blue(B) and Green(G).
We can choose any color.
Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, the number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colors, the number of ways is six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15.


3. The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?

    A. 50
    B. 60
    C. 70
    D. 80


Answer: Option C

Explanation:
A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.


4. Rohan and Mohan invest in ratio 5: 2. 10% of the profit is donated to a hospital before dividing it between the two. Rohan gets Rs. 6840. What is Mohan’s share?

    A. Rs. 2736
    B. Rs. 3800
    C. Rs. 4788
    D. Rs. 6840


Answer: Option A

Explanation:
Ratio of investment = Ratio of Profit
Ratio of Profit of Rohan to Mohan = 5:2
Rohan’s Share = Rs. 6840 = \(\frac {5}{5+2}\) x (90% Total Profit) ———-> 10% given to hospital

∴ 6840 = \(\frac {5}{7}\) x \(\frac {90}{100}\) x Total Profit

∴ Total Profit = Rs. 10,640/-
Vijay’s Share = 10,640 – 6840 – 1064 = Rs. 2736/-


5. A gets 3 times as much money as B gets, B gets only Rs.25 more then what C gets. The three gets Rs.675 in all. Find the share of B?

    A. Rs.125
    B. Rs.115
    C. Rs.110
    D. Rs.140


Answer: Option D

Explanation:
A+B+C = 675

A = 3B

3B+B+B-25 = 675

5B = 700

B = 140

1. A drink vendor has 80 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn’t want to mix any two drinks in a can. What is the least number of cans required?

    A. 35
    B. 37
    C. 42
    D. 30


Answer: Option B

Explanation:
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = \(\frac {80}{16}\) = 5
Number of cans of Pepsi = \(\frac {144}{16}\) = 9
Number of cans of Sprite = \(\frac {368}{16}\) = 23
The total number of cans required = 5 + 9 + 23 = 37 cans.


2. The compound and the simple interests on a certain sum at the same rate of interest for two years are Rs.11730 and Rs.10200 respectively. Find the sum.

    A. Rs.18000
    B. Rs.17000
    C. Rs.18500
    D. Rs.17500
    E. None of these.


Answer: Option B

Explanation:
The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for the first year also is Rs.5100. The compound interest for the second year on Rs.5100 for one year
So rate of the interest = \(\frac {(100 * 1530)}{(5100 * 1)}\) = 30% p.a.
So P = \(\frac {(100 * 10200)}{(30 * 2)}\) = Rs.17000


3. The length and breadth of the playground are 36 m and 21 m respectively. Flagstaffs are required to be fixed on all along the boundary at a distance of 3 m apart. The number of flagstaffs will be:

    A. 37
    B. 38
    C. 39
    D. 40


Answer: Option B

Explanation:
Perimeter = 2×(36+21) m = 114 m
Number of flag staffs = (\(\frac {114}{3}\)) = 38


4. A and B can do a work in 5 days and 10 days respectively. A starts the work and B joins him after 2 days. In how many days can they complete the remaining work?

    A. 1 day
    B. 2 days
    C. 3 days
    D. 4 days


Answer: Option B

Explanation:
Work done by A in 2 days = \(\frac {2}{5}\)
Remaining work = \(\frac {3}{5}\)
Work done by both A and B in one day = \(\frac {1}{5}\) + \(\frac {1}{10}\) = \(\frac {3}{10}\)
Remaining work = \(\frac {3}{5}\) * \(\frac {10}{3}\) = 2 days.


5. 3 varieties of wheat were mixed at a warehouse. The rate of Type 1 wheat was Rs. 145/ Kg and rate of Type 2 wheat were Rs. 20 per kg more than Type 1. The quantities of 3 varieties of wheat were in ratio 2:1:3 respectively. The mix was finally sold at the rate of Rs. 180 per kg. Find the price of the 3rd type of wheat?


    A. Rs. 196.58
    B. Rs. 208.33
    C. Rs. 210
    D. Rs. 215.67


Answer: Option B

Explanation:
Let the rate of the 3rd type of wheat be Rs. W
Quantity ratio = 2 : 1 : 3
This means if we take 2kg of Type 1 and 1 kg of Type 2, then we must take 3kg of Type 3.
Also, the mix will have 2kg + 1kg + 3kg = 6 kg quantity.
∴ (145 x 2) + (165 x 1) + (3 x W) = 6 x 180
∴ W = Rs. 208.33 per kg = 3rd type price per kg


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