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SSC CPO Quantitative Aptitude Quiz 2

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SSC CPO Quantitative Aptitude Quiz 2

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


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1. Shivam travels 20% distance of the total journey by car and 50% of the remaining by train and taxi in the respective ratio of 5:3 and the remaining distance he covers on feet. If the sum of the distance which he travels by car and by Taxi is 126 km, then find the total distance which Shivam travels during his journey?

    A. 360 km
    B. 640 km
    C. 420 km
    D. 400 km


Answer: Option A

Explanation:
Distance covered by car = 20% of 100x = 20x
Distance covered by Train = \(\frac {5}{5 + 3}\)× 50% of (100x – 20x) = \(\frac {5}{8}\)× 40 x = 25x

Distance covered by Taxi = \(\frac {3}{5 + 3}\)× 50% of (100x – 20x) = \(\frac {3}{8}\)× 40 x = 15x
Distance covered on feet = 100x – 20x – 25x – 15x = 40x

Total distance covered by Car and Taxi = 20x + 15x = 35x = 126 km (Given)
Length of journey = 100 × \(\frac {126}{35}\)


2. A beats B by 15 sec in a 200 m race, B beats C by 25 sec in a 500 m race, C beats D by 32 sec in 800 m race and D beats E by 35 sec in a 1 km race. What must be the speed of A in order to beat E by 800 m in a 2 km race?

    A. 2.5 m/s
    B. 3.33 m/s

    C. 5 m/s

    D. 6.66 m/s


Answer: Option B

Explanation:
A beats B by 15 sec means A reaches the destination 15 sec ahead of B or B reaches 15 sec later than B. Let all of them compete in a 2 km or 2000 m race, now we can compare them as follows:

A beats B by 150 sec, B beats C by 100 sec, C beats D by 80 sec, D beats E by 70 sec, ⇒ E would finish the race 400 sec after A.

Now, if A has to beat E by 800 m then we can say that E will cover the remaining 800m in 400sec, and cover 2000m in 1000 sec. A has to reach 400s earlier i.e in 600 sec,

so A’s speed = \(\frac {2000}{600}\) = 3.33 m/s


3. There is a race of 1 km among Hiten, Vikash, and Priyank. Hiten and Vikash run 1 km and Hiten wins by 10 seconds. Hiten and Priyank run 1 km and Hiten wins by 125 mt. When Vikash and Priyank run a km, Vikash wins by 15 seconds. What is the ratio of time taken by Hiten and Vikash to run a km?

    A. 38: 31

    B. 33: 35

    C. 35: 37

    D. 13 : 47


Answer: Option C

Explanation:
If Hiten covers 1 km in x seconds, Vikash covers the same distance in (x +10) seconds.

If Hiten covers 1 km, then in the same time Priyank covers only 875 mt.

If Vikash covers 1 km in (x + 10) sec, then Priyank covers 1 km in (x + 25) sec

Thus in x seconds, Priyank covers the distance of 875 mt.

\(\frac {X}{875}\) × 1000 = x + 25

x = 175

Time is taken by Hiten to complete 1 km = 175 seconds

Time is taken by Vikash to complete 1 km = 185 seconds

Required ratio = 175 : 185 = 35 : 37


4. A train started from point A at a speed of 60 km/hr and after 2 hours another train of the same length started from A at a speed of 80 km/hr in the same direction as the first one. After how much time the second train will meet the first train?

    A. 5 hours
    B. 3 hours
    C. 6 hours
    D. 8 hours


Answer: Option C

Explanation:
∵Let after x hours the second train will meet the first train.

Because distance is the same,

\({s}_{1}\)\({t}_{1}\) = \({s}_{2}\)\({t}_{2}\)
60 (x + 2) = 80 × x

60x + 120 = 80x

80x – 60x = 120

20x = 120

x = 6 hours


5. Two trains of lengths 160 m and 200 m travel at the speeds of 48 m/s and 52 m/s respectively in opposite direction to each other. What is the total time is taken by them to cross each other?

    A. 3.6 sec
    B. 4 sec
    C. 5.2 sec
    D. 6.8 sec


Answer: Option A

Explanation:
Relative speed = 48 + 52 = 100 m/s

Total distance covered by both the trains = 160 + 200 = 360 m

Speed = \(\frac {Distance}{Time}\) = \(\frac {360}{Time}\)
Time = 3.6 seconds

1. The average weight of three friends Amar, Visera, and Daman is 70 kg. Another person Vishal joins the group and now the average is 66 kg. If another person Tahir whose weight is 6 kg more than Vishal, joins the group replacing Amar, then the average weight of Visera, Daman, Vishal, and Tahir becomes 75 kg. What is the weight of Amar (in kg)?


    A. 18
    B. 20
    C. 22

    D. 24


Answer: Option D

Explanation:
Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg

Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg ……… (i)

∴ Weight of Vishal = 264 – 210 = 54 kg

∴ Weight of Tahir = 54 + 6 = 60 kg

Now, as per the question

Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg. ………………. (ii)

Subtracting (i) from (ii), we get

Tahir – Amar = 60 – Amar = 300 – 264 = 36

Therefore, weight of Amar = 60 – 36 = 24 kg


2. The average salary of a company increases by 100 when the salary of the manager, which is Rs. 9500, is included. If the number of employees excluding the manager is the smallest cube divisible by 16, what is the final average of the company?

    A. Rs. 4000
    B. Rs. 3400
    C. Rs. 3700
    D. Rs. 3100


Answer: Option D

Explanation:
The smallest cube divisible by 16 is 64.

Let’s assume the average salary before the manager’s salary is included is x

After addition of the Manager’s salary the average increases by 100

We can write down the above information in form of an equation as:

64x + 9500 = 65 × (x + 100)

Solving for x, we get x = 3000

The final average is 3000 + 100 = Rs. 3100


3. An exam was conducted in a state over 222 centers. The average number of applicants percenter was found to be 1560. However, it was later realized that in one center, the number of applicants was counted as 1857 instead of 1747. What was the correct average number of applicants percenter (up to two decimals)?


    A. 1557.87

    B. 1558.20

    C. 1558.92

    D. 1559.51


Answer: Option D

Explanation:

Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = 110/222 = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51


4. The sum of the ages of Kidambi and Srikanth 14 years hence will be equal to 2 times their present age. If at present Kidambi is 8 years elder to Srikanth, then what are their present ages?


    A. 22, 8
    B. 29, 12
    C. 18, 10
    D. 13, 6


Answer: Option C

Explanation:

Let the present ages of Kidambi and Srikanth be x years and y years respectively.

As per the question,

(x + 14) + (y + 14) = 2(x + y)

x + y + 28 = 2x + 2y

x + y = 28 …..(i)

Also,

y + 8 = x

x – y = 8 …..(ii)

Solving means (i) and (ii), we get

x = 18 and y = 10

Therefore, Present ages of Kidambi and Srikanth is 18 years and 10 years respectively


5. Mayank said to his friend “If you subtract 18 from my age the two digits of my age will reverse their positions. Also, my age is six less than 8 times the sum of digits of my age”. Find Mayank’s age.


    A. 46 years

    B. 37 years
    C. 56 years

    D. 42 years


Answer: Option D

Explanation:
Let Mayank’s age be (10x + y) years

Age by reversing the digits = (10y + x) yrs

Now, 10x + y − 18= 10y + x

9x – 9y = 18

x – y = 2………………(1)

Also,

10x+y = 8(x+y) – 6

2x – 7y = –6……….… (2)

Solving equations (1) and (2),

x = 4 , y = 2

Therefore, Mayank’s age = 10x + y

= 10(4) + 2

= 42 years

1. The least number which when divided by 6, 9, 12, 15, 18 leaves the same remainder 2 in each case, is:


    A. 176

    B. 178
    C. 180
    D. 182


Answer: Option D

Explanation:
Given numbers are: 6, 9, 12, 15, 18

LCM of given numbers = 180

So, 180 + 2 = 182 is the number that leaves 2 as a remainder.


2. The HCF of two numbers is 98 and their LCM is 2352. The sum of the numbers may be


    A. 1078
    B. 1398
    C. 1426
    D. 1484


Answer: Option A

Explanation:
Let two numbers are 98x and 98y. Then,

Product of number = Product of HCF and LCM

98x × 98y = 98 × 2352

xy = 24

Let x = 8 and y = 3 (As Co-prime factors of 24 be 8 and 3)

Then, Sum of number = 98 × 8 + 98 × 3 = 98 (11) = 1078


3. The product of HCF and LCM of 18 and 15 is


    A. 120

    B. 150
    C. 175
    D. 270


Answer: Option D

Explanation:
HCF of 18 and 15 = 3

LCM of 18 and 15 = 90

∴ Product of HCF and LCM of both numbers = 3 × 90 = 270.


4. Three planets revolve around the Sun once in 200, 250 and 300 days, respectively in their own orbits. When do they all come relatively to the same position as at a certain point of time in their orbits?


    A. After 3000 days
    B. After 2000 days
    C. After 1500 days
    D. After 1200 days


Answer: Option A


Explanation:

Given that, three planets revolve the Sun once in 200, 250 and 300 days, respectively in their own orbits.

∴ Required time = LCM of (200, 250 and 300) = 3000 days

Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits.


5. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.


    A. 0.21 and 6.30

    B. 1.05 and 6.30
    C. 2.1 and 0.63
    D. 0.63 and 1.05


Answer: Option A


Explanation:

Making the same number of decimal places, the given numbers are 0.63, 1.05 and 2.10.

Without decimal places, these numbers are 63, 105 and 210.

Now, H.C.F of 63, 105 and 210 is 21.

∴ H.C.F, of 0.63, 1.05 and 2.1 is 0.21.

L.C.M. 63, 105 and 210 is 630

∴ L.C.M. of 0.63, 1.05 and 2.1 is 6.30.


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