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SSC CPO Quantitative Aptitude Quiz 18

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SSC CPO Quantitative Aptitude Quiz 18

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 18 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 18 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

1. In how many ways can the letters of the word INDIA be arranged, such that all vowels are never together?


    A. 48
    B. 42
    C. 28
    D. 36


Answer: Option B


Explanation:

First, we find out the number of times a particular letter occurs in the given word:

2 – I
1 – A
1 – D
1 – N

=> Total number of solutions minus the number of times all three vowels are together.

Now, using the concept of permutation and combination:

=> Divide the possible combination by 2! because it occurs twice and replacing one by another will cause no difference in the word.

=> Total number of words that can be formed: 5!/2!

=> Total number of words that can be formed keeping all the vowels together: 3! 3!/2!

=> 60 – 18 = 42


2. The income of A is 25% more than the income of B. What is the income of B in terms of income of A?


    A. 80%
    B. 75%
    C. 78.66%
    D. 71.25%


Answer: Option D


Explanation:

One of the most basic questions.

Let the income of B be 100. The income of A is 25% more than the income of B which means Income of A becomes 125

Now income of B in terms of A = 100/125 *100 = 80%


3. A starts walking at 4 kmph and 4 hours after his start B starts cycling at 10 kmph. After how much distance will B catch up with A?


    A. 26.2 km
    B. 25.7 km
    C. 23.2 km
    D. 26.67 km


Answer: Option D


Explanation:

Distance travelled by each of them is to be made equal.

=> Distance traveled walking = Distance covered cycling.

Let after 4 hours of walking, A walk for x hours more before B catches up with him.

=> Distance = Speed * Time

(4+x)4 = 10x

x = 8/3

Therefore, It takes B 8/3 hours to catch up with A. Distance: 8/3 x 10 = 80/3 km = 26.67


4. If one is added to the numerator of the fraction it becomes one. If one is added to the denominator of the fraction it becomes 1/2. The fraction is?


    A. \(\frac{1}{2}\)
    B. \(\frac{3}{5}\)
    C. \(\frac{2}{3}\)
    D. \(\frac{2}{5 }\)


Answer: Option C


Explanation:

Let the fraction be \(\frac{X}{Y }\).

=> \(\frac{X + 1}{Y}\) = 1

=> x+1 = y—(1)
\(\frac{X}{Y – 1 }\) = \(\frac{1}{2 }\)

=> 2x = (y+1)—(2)

=>Equating (1) and (2)

=> x+1 = 2x-1

=>x=2

Substituting the value of x in (1)

=>y=3

=>2/3


5. A boat with speed 15 km/hr in standing water goes 30 km downstream and returns in a total of 4.5 hours. What is the speed of current?


    A. 4 kmph
    B. 6 kmph

    C. 5 kmph

    D. 8 kmph


Answer: Option C


Explanation:
Speed of the boat in still water: 15 km/hr

Speed of boat downstream: (15+x)km/hr where x is the speed of the current.

The boat travels 30 km downstream and then 30 km upstream and takes \(\frac{9}{2 }\) hours.

Total time= Time taken to travel downstream + Time is taken to travel upstream

=> 4/5= (30/(15+x))+(30/(15-x))

=>x= 5

Directions ( 1- 5): Find the odd one out
1. 125, 106, 88, 76, 65, 58, 53


    A. 88
    B. 106

    C. 125
    D. 76


Answer: Option A


Explanation:

This sequence represents a series in which from the reverse order a prime number is added:

53+5=58

58+7=65

65+11=76

76+13=89

89+17=106

106 + 19 = 125

Hence 88 is the answer.


2. What is the CP of Rs 100 stock at 4 discount, with 1/5% brokerage?


    A. 99.6
    B. 96.2
    C. 97.5

    D. 98.25


Answer: Option B

Explanation:
Use the formula,
CP= 100 – discount + brokerage%
CP= 100 – 4 +\(\frac{1}{5 }\)
96.2

Thus the CP is Rs 96.2.


3. A pipe is 30 m long and is 45% longer than another pipe. Find the length of the other pipe.

    A. 20.68
    B. 20

    C. 20.12
    D. 20.5


Answer: Option B

Explanation:
Let length of other pipe be X
According to question,
30 = \(\frac{45}{100 }\) X + X
30 = 0.45X + X
30 = 1.45 X
X= \(\frac{30}{1.45 }\)
X = 20.68m

Thus the length of the other pipe is 20.68 meters.


4. 15th august 2010 was which day of the week?

    A. Thursday
    B. Friday
    C. Wednesday
    D. Sunday


Answer: Option D

Explanation:
15th August 2010 can be written as 2009 + days from 1st January 2010 to 15th August 2010.

=> Total number of odd days in 400 years = 0

Hence, the total number of odd days in 2000 years = 0 (as 2000 is a perfect multiple of 400)

Odd in days in the period 2001-2009:
7 normal years + 2 leap yeas

=> (7*1) + (2*2) = 11

=> Odd days will be 11- (7*1) = 4

Days from January 1 to August 15 in 2010: 31+28+31+30+31+30+31+15

= 227 days.

= 32 weeks and 3 days, this gives additional 3 odd days.

=> Total odd days= 3+4=7

=> 7 odd days=1 week= 0 odd days

=> 0 odd days= Sunday

Thus, 15th August 2010 was a Sunday.


5. A boatman rows 96 km downstream in 8 hours with a stream speed of 4kmph. How much time will he take to cover 8km upstream?


    A. 4 hours
    B. 6 hours
    C. 2 hours
    D. 1 hours


Answer: Option C

Explanation:
Speed = distance/time

Speed downstream: \(\frac{96}{8 }\) km/hr = 12kmph

Speed of stream = 4kmph

Effective speed of boat = (12-4) kmph
= 8kmph

Distance to be traveled upstream= 8 km

Speed upstream = boat speed-current speed

= 8-4 kmph

= 4 kmph

Time taken = \(\frac{Distance}{Speed }\) = \(\frac{8}{4 }\) hours
= 2 hours

Thus it will take 2 hours to go upstream.

1. While calculating the edge of a square, a worker makes an error of 2% in excess. What % error does he make in calculating area? (%)


    A. 4.04

    B. 4
    C. 40.4

    D. 4.004


Answer: Option A

Explanation:
Given Error = 2% while measuring the side of a square.

If the correct value of the side of the square is 100, the measured value:

=> 100 + 2% *100

= 100+2=102

The area of a square with edge 100 = side*side

=> 100*100

=> 10000

The area of a square with side 102 = 102*102= 10404

Error in area calculation = 10404-1000 = 404

% error= \(\frac{404}{10000 }\)*100
= 4.04%


2. Akash and Akshay are business partners. Akash invests Rs 35,000 for a period of 8 months and Akshay invests Rs 42,000 for a period of 10 months. Out of a total profit of Rs 31,570 what is Akash’s share?

    A. Rs 12,420
    B. Rs 18,040
    C. Rs 18,942

    D. Rs 12,628


Answer: Option D

Explanation:
For a given business profit is directly dependent upon the capital invested and the time of investment.

=> Ratio of shares of Akash and Akshay becomes: (35,000*8)/(42,000*10) = \(\frac{2}{3 }\)
=>% of profit belonging to Akash: \(\frac{2}{3 + 2 }\)*(31,570)
=>Rs 12,768


3. In what ratio must one add water to milk so as to gain 16.666% on the selling this mixture at the cost price?


    A. 1:6
    B. 1:3
    C. 6:1
    D. 3:1


Answer: Option A

Explanation:

To start off this question let us assume that cost price of 1 litre milk is Rs 1

No need to make a mixture and sell this mixture at 1 Rs per liter such that the total gain on the mixture is 16.667%.

Therefore, CP of 1 liter of the mixture becomes (quantity of milk)/ (quantity of mixture containing 1 L milk)*(the price of 1-liter milk).

=>\(\frac{(100}{(100+50/3) }\)*1

=>CP of 1-litre milk of mixture: Rs \(\frac{5}{6 }\)

As the price of any amount of water is zero, and as 1-liter milk costs Rs 1.5/6litre of the mixture will comprise entirely of cost of milk which means,1 liter of the mixture will

contain 5/6th amount of milk.

=>Water is added in the ratio of (1-\(\frac{5}{6 }\))= \(\frac{1}{6 }\)


4. Ram can finish a puzzle in 3 hours and Shyam can do the same in 2 hours. Both of them finish the puzzle and get 15 candies. What is Ram’s share?

    A. 3
    B. 6
    C. 11
    D. 12


Answer: Option B

Explanation:

The question is based on efficiency to do work.

Ram can finish a puzzle in 3 hours. Shyam can finish the puzzle in 2 hours.

=>In one hour Ram can finish 1/3rd of a puzzle

=>In one hour Shyam can finish half the puzzle

A total of 15 candies are to be shared amongst both of them

Hence Ram’s share must be = ((Work done by Ram in 1 hour)/(Work is done by Shyam in one hour)+(Work done by Ram in an hour))*15

=> 1/3/((1/2)+(1/3))*15

=> (1/3/(5/6))*15

=> \(\frac{6}{15 }\)*(15)

=>6 candies


5. A man sells 45 lemons for Rs 40 and loses 20%. At how much price should he sell 24 lemons to the next customer to make a 20% profit?


    32
    B. 20
    C. 24
    D. 16


Answer: Option A

Explanation:

Let cost price of lemons be x.

Selling price of lemons becomes CP – loss.

=>SP=x-(20/100)x

=>40=x(80/100)

=>50

So 45 lemons cost Rs 50
Cost of 1 lemon = \(\frac{50}{45 }\) Rs
Cost of 24 lemons = (\(\frac{50}{45}\))*24 =80/3 Rs
Selling Price of 24 lemons = (\(\frac{80}{3 }\))+(\(\frac{20}{100 }\))(\(\frac{80}{3 }\))=Rs 32


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