Quantitative Aptitude - SPLessons

SSC CPO Quantitative Aptitude Quiz 10

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

SSC CPO Quantitative Aptitude Quiz 10

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 10 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 10 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

1. Three unbiased coins are tossed. What is the probability of getting at least 2 tails?


    0.75
    B. 0.5
    C. 0.25
    D. 0.2


Answer: Option B


Explanation:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
E = {HTT, THT, TTH, TTT}
n(S) = 8
n(E) = 4
P(E) = \(\frac{n(E) }{n(S)}\) = \(\frac{4}{8}\) = 0.5


2. Tickets numbered 1 to 50 are mixed and one ticket is drawn at random. Find the probability that the ticket was drawn has a number which is a multiple of 4 or 7?


    A. \(\frac{9 }{25}\)
    B. \(\frac{9 }{50}\)
    C. \(\frac{18 }{25}\)
    D. None of these


Answer: Option A


Explanation:

S = {1, 2, 3, … , 49, 50}
E = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 7, 14, 21, 35, 42, 49}
n(S) = 50
n(E) = 18
P(E) = \(\frac{n(E) }{n(S)}\) = \(\frac{18 }{50}\)
= \(\frac{9 }{25}\)

Time is taken to travel 60 km at 10 km/hr = \(\frac {60}{10}\) hrs = 6 hrs
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = \(\frac {60}{5}\) kmph. = 12 kmph


3. From a pack of 52 cards, one card is drawn at random. Find the probability that the drawn card is a club or a jack?


    A. \(\frac{17 }{52}\)
    B. \(\frac{8 }{13}\)
    C. \(\frac{4 }{13}\)
    D. \(\frac{1 }{13}\)


Answer: Option C


Explanation:

n(S) = 52
n(E) = 16
P(E) = \(\frac{n(E) }{n(S)}\) = \(\frac{16 }{52}\)
= \(\frac{4 }{13}\)


4. Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?


    A. \(\frac{8 }{35}\)
    B. \(\frac{34 }{35}\)
    C. \(\frac{27 }{35}\)
    D. None of these


Answer: Option A


Explanation:

P(A) = \(\frac{2 }{5}\)
P(B) = \(\frac{4 }{7}\)
E = {A and B both get selected}
P(E) = P(A)*P(B)
= \(\frac{2 }{5}\) * \(\frac{4 }{7}\)
= \(\frac{8 }{35}\)


5. Two pipes A & B can fill the tank in 12 hours and 36 hours respectively. If both the pipes are opened simultaneously, how much time will be required to fill the tank?


    A. 6 hours
    B. 9 hours

    C. 12 hours

    D. 15 hours


Answer: Option B


Explanation:
If a pipe requires ‘x’ hrs to fill up the tank, then part filled in 1 hr = \(\frac{1 }{X}\)

If pipe A requires 12 hrs to fill the tank, then partly filled by pipe A in 1 hr = \(\frac{1 }{12}\)

If pipe B requires 36 hrs to fill the tank, then partly filled by pipe B 1 hr = \(\frac{1 }{36}\)

Hence, part filled by (A + B) together in 1 hr = \(\frac{1 }{12}\) + \(\frac{1 }{36}\)

= \(\frac{48 }{432}\) = \(\frac{1 }{9}\)

In 1 hr both pipes together fill 1/9th part of the tank. This means, together they fill the tank in 9 hrs.

1. Two pipes can fill a tank in 10 and 14 minutes respectively and a waste pipe can empty 4 gallons per minute. If all the pipes working together can fill the tank in 6 minutes, what is the capacity of the tank?


    A. 120 gallons
    B. 240 gallons

    C. 450 gallons

    D. 840 gallons


Answer: Option D


Explanation:

Work done by waste pipe in 1 min = (Part filled by total pipes together) -(part filled by first pipe + part filled by second pipe)
= 1/6 – [(\(\frac{1 }{10}\)) + (\(\frac{1 }{14}\))] = \(\frac{1 }{6}\) – (\(\frac{24 }{140}\)) = –\(\frac{1 }{210}\)
Here, negative (-) sign indicates emptying of tank.

To find the capacity, we need to determine the volume of \(\frac{1 }{210}\) part.
Therefore, volume of \(\frac{1 }{210}\) part = 4 gallons ———(given condition)
Hence, the capacity of tank = volume of whole = 4 x 210 = 840 gallons.


2. Two pipes fill a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?


    A. 35 hrs
    B. 70 hrs
    C. 180 hrs

    D. 300 hrs


Answer: Option C

Explanation:
Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.

The time taken by the leak to empty the full tank = \(\frac{xy }{y – x}\) hrs

But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60

Hence, the time taken by these pipes to fill the tank = \(\frac{60 }{7}\) hrs = 8 .57 hrs = 8 hrs 34 min ——–( by multiplying ‘0.57’ hrs x 60 = 34 minutes)

Due to leakage, time taken = 8 hrs 34 min + 26 min = 8 hrs 60 min = 9 hrs ———-( because 60 min = 1 hr)

Thus, work done by (two pipes + leak) in 1 hr = \(\frac{1 }{9}\)
Hence, work done by leak in 1 hr = work done by two pipes – \(\frac{1 }{9}\)
= \(\frac{7 }{60}\) – \(\frac{1 }{9}\) = \(\frac{3 }{540}\) = \(\frac{1 }{180}\)

Therefore, the leak will empty the full cistern in 180 hours.


3. If two pipes function simultaneously, the reservoir will be filled in 24 hrs. One pipe fills the reservoir 20 hours faster than the other. How many hours does it take for the second pipe to fill the reservoir?

    A. 12 hrs
    B. 30 hrs

    C. 44 hrs
    D. 60 hrs


Answer: Option D

Explanation:
P + Q + R + S = (30 x 4) P + Q + R + S = 120 …. (i)
Assume that the reservoir is filled by first pipe in ‘x’ hours.
So, the reservoir is filled by the second pipe in ‘x + 20’ hours.
Now, from these above conditions, we can form the equations as,

\(\frac{1 }{x}\) + \(\frac{1 }{(x + 20)}\) = \(\frac{1 }{24}\)

\(\frac{[x + 20 + x] }{[x(x + 20)] }\) = \(\frac{1 }{24}\)

x2 – 28x – 480 = 0

By solving this quadratic equation , we get the factors (x – 40) (x+12) = 0
Hence, we get two values : x = 40 & x = -6
Since filling of reservoir is positive work , we can neglect the negative value of ‘x’.
Thus, x = 40

This means that the second pipe will take (x+ 20) hrs = 40 + 20 = 60 hrs to fill the reservoir.


4. Two pipes can fill a tank in 8 hrs & 6 hrs respectively. If they are opened on alternate hours and if pipe A gets opened first, then in how many hours, the tank will be full?

    A. 6 hrs
    B. 7 hrs
    C. 8 hrs
    D. 14 hrs


Answer: Option B

Explanation:
Pipe A’s work in 1 hr = 1/8
Pipe B’s work in 1 hr = 1/6

Pipes (A+B)’s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24

Now,

In 4 hrs they fill : 2 X (\(\frac{7 }{24}\)) = \(\frac{7 }{12}\)
In 6 hrs they fill : 3 X (\(\frac{7 }{24}\)) = \(\frac{7 }{8}\)

After 6 hrs, part left empty = \(\frac{1 }{8}\)

Now it is A’s turn to open up.

In one hr it fills \(\frac{1 }{8}\) of the tank.

So, the tank will be full in = 6 hrs + 1 hr = 7 hrs


5. It is observed that the pipe A can fill the tank in 15 hrs and the same tank is filled by pipe B in 20 hrs. The third pipe C can vacant the tank in 25 hrs. If all the pipes get opened initially and after 10 hrs, the pipe C is closed, then how long will it take to fill the tank?


    A. 3 hrs
    B. 7 hrs
    C. 12 hrs
    D. 20 hrs


Answer: Option C

Explanation:
Assume that the tank will be full in ’10 + x’ hrs.

Part filled by pipe A in 1 hr = \(\frac{1 }{15}\)
Part filled by pipe B in 1 hr = \(\frac{1 }{20}\)
Part emptied by pipe C in 1 hr = \(\frac{1 }{25}\)

(A+ B)’s part filled in 1 hr = \(\frac{1 }{15}\) + \(\frac{1 }{20}\) = \(\frac{7 }{60}\)

As Pipe C is closed after 10 hours, let us find the part of the tank filled in 10 hrs.
Tank filled in 10 hrs = 10 (part filled by A in 1 hr + part filled by B in 1 hr– part emptied by C in 1 hr)
= 10 [\(\frac{1 }{15}\) + \(\frac{1 }{20}\) – \(\frac{1 }{25}\)] = \(\frac{23 }{30}\)

Remaining part = 1 – part filled in 10 hours
= 1 – \(\frac{23 }{30}\) = \(\frac{7 }{30}\)
We know that, \(\frac{Part filled by (A + B) in 1 hr }{Remaining part }\) = \(\frac{Total part filled}{Time taken}\)
So, we get the ratio as ,
\(\frac{7 }{60}\) : \(\frac{7 }{30}\) :: 1 : x

So, the value of x = \(\frac{2 }{30}\) x 1 x \(\frac{60 }{7}\) = 2 hrs.

1. 6, 12, 20, 30, 42, 54, 72, 90, 110


    A. 20

    B. 42
    C. 54
    D. 90


Answer: Option C

Explanation:

2 × 3 = 6, 3 × 4 = 12, 4 × 5 = 20, 5 × 6 = 30, 6 × 7 = 42, 7 × 8 = 56, 8 × 9 = 72
The numbers in the series are in order as shown above.
After 42, the next number in series should be 56 instead of 54.
Hence, the odd number is 54


2. 34, 105, 424, 2123, 12756

    A. 12756
    B. 2123
    C. 424
    D. 34


Answer: Option B

Explanation:
34 × 3 = 102, but here the next number is 105.
Hence, 102 + 3 = 105

Try this concept for next numbers in sequence.
105 × 4 = 420, 420 + 4 = 424
It’s working! Check all the numbers in series
424 × 5 = 2120, 2120 + 5 = 2125
2125 × 6 = 12750, 12750 + 6 = 12756
Hence, correct answer is 2123. The term after 424 should be 2125 instead of 2123.


3. 9, 12, 17, 20, 25, 28, 34, 36, 41


    A. 25
    B. 28
    C. 34

    D. 41


Answer: Option C

Explanation:
Here, the differences between two successive numbers from the beginning are 3, 5, 3, 5…
34 is the wrong term because the next number should have the difference of 5 i.e. the number should be 33.
Hence, the odd number is 34


4. 6, 9, 11, 12, 14, 15, 25

    A. 9
    B. 11
    C. 14
    D. 25


Answer: Option B

Explanation:

Each number is composite except 11
Hence, the odd number is 11


5. 3, 5, 7, 14, 17, 19


    A. 3
    B. 7
    C. 14
    D. 17


Answer: Option C

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10 x – y = 20 …. (i)

and x + 20 = 2(y – 20) x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.


IBPS RRB PO – Related Information
IBPS RRB Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 2
Book for Quantitative Aptitude