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SSC CPO Quantitative Aptitude Quiz 8

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SSC CPO Quantitative Aptitude Quiz 8

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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 SSC CPO Quantitative Aptitude Quiz 8 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 8 will assist the students to know the expected questions from Quantitative Aptitude.


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1. If \(\frac{x}{y}\) = \(\frac{3}{4}\) and 8x + 5y = 22, then find the value of x.


    3.2
    B. 2.1
    C. 2.3
    D. 1.5


Answer: Option D


Explanation:
Y = \(\frac{4}{3}\) X
Substitute this value in 8x + 5y = 22

8x + 5 \(\frac{4}{3}\) x = 22
44x = 66
x = 1.5


2. The population of city A which is 68000 decreases at the rate of 1200/year. The population of city B which is 42000, increases at the rate of 800 per year. Find in how many years, the population of cities A and B are equal?


    A. 9 years
    B. 10 years
    C. 13 years
    D. 15 years


Answer: Option C


Explanation:

We have to find the population of cities A and B after x years.

Step 1: Population of city A = 68000, decreases at the rate of \(\frac {1200}{year}\)
68000 – 1200x

Step 2: Population of city B = 42000, increases at the rate of \(\frac {800}{year}\)
42000 + 800x

Step 3: Find how many populations of cities A and B are equal.

Population of city A = Population of city B
68000 – 1200x = 42000 + 800x
68000 – 42000 = 1200x + 800x
26000 = 2000x
x = 13


3. A contractor pays Rs. 20 to a worker for each day and the worker forfeits Rs. 10 for each day if he is idle. At the end of 60 days, the worker gets Rs. 300. Find for how many days the worker was idle?


    A. 28 days
    B. 30 days
    C. 34 days
    D. 40 days


Answer: Option B


Explanation:

Step 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x)

Step 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20

Step 3: After subtracting the amount which he forfeited, he receives Rs. 300.

Therefore,
(60 – x) 20 – 10x = 300
1200 – 20x – 10x = 300
900 = 30x
x = 30 days


4. On a farm, along with 50 hens, there were 45 goats and 8 horses and some farmers. If a total number of feet be 224 more than the number of heads, then find the number of farmers.


    A. 11
    B. 15
    C. 16
    D. 18


Answer: Option B


Explanation:

Let’s the number of farmers be y.

Step 1: Find number of heads
= (50 hens + 45 goats + 8 horses + y farmers)
= (103 + y)

Step 2: Number of feet
= [(Hens 2 × 50) + (45 × 4) + (8 × 4) + (y × 2)]
= [100 + 180 + 32 + 2y]
= 312 + 2y

Step 3: Find number of farmers
(312 + 2y) – (103 + y) = 224
312 + 2y – 103 – y = 224
y = 15

Number of farmers = 15


5. An express train runs at an average speed of 27 km/hr including the time of stoppage at stations. Another train runs at an average speed of 41 km/hr excluding the stoppage time at stations. Find how many minutes does a train stop in 1 hour.


    A. 20.52 min
    B. 15.23 min
    C. 12.50 min
    D. 10.75 min
    E. None of these


Answer: Option A


Explanation:
Train 1: Travels at an average speed of 27 km/hr
Train 2: Travels at an average speed of 41 km/hr

Therefore, train 1 lags train 2 by (41 – 27) km i.e. 14 km.

Now, we have to find the time, train 2 stops in 1 hour.

We know, Speed = \(\frac {Distance}{Time }\)

We know, Distance = 14 km, speed = 41 km/hr

Therefore, Time = \(\frac {Distance}{Speed}\)

= \(\frac {14}{41}\) = 0.342 hr

Answer is in minutes, hence multiply by 60
0.342 hr = 0.342 x 60 = 20.52 min

1. John’s marks were wrongly entered as 83 instead of 63. If the average marks calculated for the whole class increased by half, then find the number of students in the class.


    A. 30
    B. 35
    C. 40
    D. 45


Answer: Option C


Explanation:

Assume number of students in the class be x

As the average increases by half, find the total increase in marks for x students.
Total increase in marks = (x) x (\(\frac {1}{2}\) ) = \(\frac {x}{2}\)

Therefore,
Total increase in marks = False value – true value
x/2 = 83 – 63
x = 40 students


2. In a school, average marks of three batches of 40, 50 and 60 students respectively is 45, 55 and 70. Find the average marks of all the students.


    A. 54.78
    B. 55.23
    C. 50.36
    D. 58.33


Answer: Option C

Explanation:
Average = \(\frac {Sum of Quantities}{Number of Quantities}\)
Here,
Number of quantities = Number of students in each batch

As average marks of students are given, calculate total marks of each batch first. So total marks for
Batch 1 = (40 x 45) = 1800
Batch 2 = (50 x 55) = 2750
Batch 3 = (60 x 70) = 4200

Sum of marks = (1800 + 2750 + 4200) = 8750

Therefore,
Required Average = \(\frac {(Sum of Works) }{(Total No. of Students in each batch) }\)
\(\frac {(8750) }{(40 + 50 + 60) }\) = 58.33


3. There are two batches A and B of a class. Batch A consists of 36 students and batch B consists of 44 students. Find the average weight of the whole class, if the average weight of batch A is 40 kg and that of batch B is 35 kg.

    A. 29.23 kg
    B. 32.56 kg
    C. 35.66 kg
    D. 37.25 kg


Answer: Option D

Explanation:
Given: Average weight of batch A = 40 kg , average weight of batch B = 35 kg

1) First find the total weight of all students
– Weight of batch A = (36 x 40) = 1440
– Weight of batch B = (44 x 35) = 1540

Total weight of all students = (1440 + 1540) = 2980 kg

2) Find average weight of whole class
(Batch A + Batch B) students = (36 + 44) = 80 students

Average Weight = \(\frac {Total weight of all the students }{No. of Students }\)
\(\frac {2980}{80}\) = 37.25 kg


4. An employer pays Rs. 30 for each day a worker works and forfeits Rs. 5 for each day he is idle. At the end of 60 days, a worker gets Rs. 500. For how many days did the worker remain idle?

    A. 35
    B. 48
    C. 52
    D. 58


Answer: Option C

Explanation:
Suppose the worker remained idle for m days. Then, he worked for (60 – m) days.
30 (60 – m) – 5m = 500
1800 – 25m = 500
25m = 1300
m = 52
So, the worker remained idle for 52 days.


5. If 10 bulls can plow 20 identical fields in 3 days working 10 hours a day, then in how many days can 30 bulls plow 32 same identical fields working 8 hours a day?


    A. 2
    B. 4
    C. 8
    D. 10


Answer: Option B

Explanation:
M1*D1*W2 = M2*D2*W1
10*3*10*32 = 30*d*8*20
d = 2 days

1. A pack of people can complete a certain amount of work in 12 days. Two times the same number of persons will complete half of the work in?


    A. 12 days
    B. 3 days
    C. 4 days
    D. 6 days


Answer: Option B

Explanation:

More no of people: Less days (Inverse Relationship)

More work: More days (Direct Relationship)

The ratio is given:

Persons: 1:: 2
Work 1:: \(\frac {1}{2}\)
Time 12:: x

The solution of this can be explained by solving this ratio by the sort of relationships they possess with time.

1*\(\frac {1}{2}\)*12 = 2*1*x

=> x = 3 days


2. 12 men and 18 boys working 7.5 minutes an hour complete a particular piece of work in 60 hours. If a boy’s work is half as efficient as a man’s work, how many boys will be needed to help 21 men to achieve twice the work in 50 hours working 9 minutes in an hour?

    A. 42
    B. 46
    C. 48
    D. 38


Answer: Option A

Explanation:
1 man is equivalent to 2 boys in work capacity. 12 men + 18 boys = 12 x 2 + 18 = 42 boys.

Let the required number of boys be x. So a Total number of people doing work = 21 men x 2 (according to work capacity) + x boys.

Taking respective ratios as required:

Given:

=> Hours 50:60
=> Minutes per hour 9: \(\frac {15}{2}\)
=> Work 1:2
=> People: 42: 42 + x
=> 50 x 9 x 1 x (42+x) = 60 x \(\frac {15}{2}\) x 2 x 42
=> 42 +x = 84

x = 42


3. 5/8th of a job is completed in 10 days. If a person works at the same pace, how many days will he take to complete the job?


    A. 4
    B. 5
    C. 6
    D. 7


Answer: Option C

Explanation:
It is given that 5/8th of the work is completed in 10 days.

=> Remaining work = 3/8th of total

Applying the unitary method:

Total work will be completed in \(\frac {10 * 8 }{5}\)days

=> It takes 16 days to complete total work

=> Hence, remaining work days = 16 – 10 = 6 days


4. What day will be 61 days after today, if today is Monday?

    A. Tuesday
    B. Thursday
    C. Saturday
    D. Friday


Answer: Option C

Explanation:

The basic concept in a problem involving days is finding the number of odd days. In 61 days:

=> 7 x 8 + 5 = 5 odd days

=> Monday + 5 days = Saturday

Hence 61 days from today will be a Saturday.


5. \({15}^{th}\) august 2010 was which day of the week?


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


Answer: Option D

Explanation:

\({15}^{th}\) August 2010 can be written as 2009 + days from \({1}^{st}\) January 2010 to \({15}^{th}\) 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.


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