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SSC CPO Quantitative Aptitude Quiz 7

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SSC CPO Quantitative Aptitude Quiz 7

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 7 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 7 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

1. How many seconds does Puja take to cover a distance of 500 m, if she runs at a speed of 30 km/hr?


    A. 60 sec
    B. 82 sec
    C. 95 sec
    D. 100 sec


Answer: Option A


Explanation:
Time = \(\frac{Distance}{Speed}\)
We see that the distance is given in metres while the speed is given in km/hr and the answer is asked in seconds.
So, convert km/hr into m/s by multiplying \(\frac{5}{18}\) m/s to the given value of speed.
30 \(\frac{km}{hr}\) = 30 \(\frac{5}{18}\)
i.e. Place these values in the formula:
Time = 500 x \(\frac{9}{75}\) = 60 sec


2. A cyclist covers a distance of 800 meters in 4 minutes 20 seconds. What is the speed in km/hr of the cyclist?


    A. 6.2 km/h
    B. 8.4 km/hr
    C. 11.05 km/hr
    D. 16.07 km/hr


Answer: Option C


Explanation:

Speed = \(\frac{Distance}{Time}\)
Time = 4 min 20 sec = 260 sec

Speed = \(\frac{800}{260}\) = 3.07 m/sec
Convert the speed from m/s to km/hr by multiplying with (5/18)
3.07 x \(\frac{18}{7}\) km/hr = 11.05 km/hr


3. Two towns P & Q are 275 km apart. A motorcycle rider starts from P towards Q at 8 a.m. at the speed of 25 km/hr. Another rider starts from Q towards P at 9 a.m. at the speed of 20 km/hr. Find at what time they will cross each other?


    A. 2.45 p.m.
    B. 2.30 p.m.
    C. 1.35 p.m.
    D. 1.15 p.m.


Answer: Option B


Explanation:

Assume, distance traveled by P in x hrs = 25 x km —–(1)
distance traveled by Q in (x-1) hrs = 20 (x-1) km —–(2)

Adding (1) & (2),

25 x + 20 (x -1) = 275

x = 6.5 hrs

(x -1) = (6.5 -1) = 5.5 hrs

Time at which they cross each other = 9 a.m. + 5.5hrs = 2.30 p.m.
The two motorcycle riders cross each other at 2.30 p.m.


4. An airplane flying 1000 km covers the first 200 km at the rate of 200 km/hr, the second 200 km at 400 km/hr, the third 200 km at 600 km/hr & the last 200 km at the rate of 800 km/hr. Determine the average speed of the airplane.


    A. 250 km/hr
    B. 300 km/hr
    C. 480 km/hr
    D. 600 km/hr


Answer: Option C


Explanation:

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

\(\frac {200}{200}\) + \(\frac {200}{400}\) + \(\frac {200}{600}\) + \(\frac {200}{800}\) = \(\frac {25}{12}\)
Average speed = \(\frac {1000}{25}\) x 12 = 480 km/hr


5. Jennifer travels the first 4 hours of her journey at a speed of 80 miles/hr and the remaining distance in 6 hours at a speed of 30 miles/hr. What is her average speed in miles/hr?


    A. 50 miles / hr
    B. 60 miles / hr
    C. 30 kmph
    D. 92 miles / hr
    E. None of these


Answer: Option A


Explanation:
i.e. Average speed = Total distance / Time

Distance =Time x Speed

Total distance covered by Jennifer = Distance covered in the first 4 hours + distance covered in next 6 hours

= (80 x 4) + (30 x 6)

= 500 miles / hr

Total time taken to complete the journey = 4 + 6 = 10 hrs

Therefore,
Average Speed = \(\frac {Total Distance}{Time}\) = \(\frac {500}{10}\)
= 50 miles/hr

1. Find the average of all numbers between 5 and 49 which are divisible by 5.


    A. 20
    B. 25
    C. 30
    D. 35


Answer: Option B


Explanation:

The numbers divisible by 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45.
Average = \(\frac {Sum of Quantities }{Number of Quantities }\)

\(\frac {(5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45) }{9}\)
= \(\frac {225}{9}\)


2. The average of 11 numbers is 30. If the average of the first six numbers is 17.5 and that of the last six is 42.5, then what is the sixth number?


    A. 30
    B. 36
    C. 45
    D. 47


Answer: Option C

Explanation:
Given: Average of 11 numbers = 30

Step 1: Calculate total of 11 numbers by multiplying it by average value 30 = 11 x 30 = 330
Step 2: Calculate total of first six members by multiplying it by average value 17.5 = 17.5 x 6 = 105
Step 3: Calculate total of last six members by multiplying it by average value 42.5 = 42.5 x 6 = 255

Therefore, we can find the sixth number by adding the value of the first six and the last six numbers and subtracting it from the total value of 11 numbers.

Sixth number = (105 + 255)- 330 = 30


3. The average of 15 numbers is 15. If the average of first five numbers is 14 and that of other 9 numbers is 16, then find the middle number.

    A. 12
    B. 11
    C. 10
    D. 9


Answer: Option B

Explanation:
Given: Average of 15 numbers = 15, Average of 5 numbers = 14, Average of 9 numbers = 16
Average = \(\frac {Total Numbers}{No. of Numbers}\)
15 = \(\frac {Total Numbers}{15}\)
Therefore, total numbers = 15 x 15 = 225
Middle number = (Total numbers) – [(Average of 5 num x no of num) + ( Average of 9 num x no of num)]
= (225) – [(14 x 5) + (16 x 9)]
= (225) – [214]
= 11

Therefore, the middle number is 11


4. The average of four consecutive even numbers is 27. Find the largest of these numbers.

    A. 28
    B. 30
    C. 32
    D. 34


Answer: Option B

Explanation:
Consider the consecutive even numbers as : x, (x + 2), (x + 4) and (x+ 6)

Average = \(\frac {Sum of Quantities}{Number of Quantities}\)
= \(\frac {x + (x + 2) + (x + 4) + (x + 6) }{4}\) = 1
= 1 \(\frac {(4x + 12) }{4}\) = 27
Simplifying we get, x = 24

Therefore,
Largest number = (x + 6) = (24 + 6) = 30
Smallest number = 24


5. 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 A

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
x = 9 days

1. 26th January, 1996 was a Friday. What day of the week lies on 26th January, 1997?


    A. Saturday
    B. Sunday
    C. Monday
    D. Thursday


Answer: Option B

Explanation:

The year 1996 was a leap year. Hence, it adds 2 odd days. As 26th January, 1996 was a Friday, 26th January, 1997 was a Sunday.


2. The calendar for the year 2001 is same for which of the following year?

    A. 2005
    B. 2007
    C. 2011
    D. 2006


Answer: Option B

Explanation:
The total number of odd days from 2001 onwards should be zero.
Now, as an ordinary year adds 1 odd day and a leap year adds 2 odd days, we have:
2001, 2002, 2003, 2005, 2006 – 1 odd day each
2004 – 2 odd days
Hence, at the end of 2006 the total number of odd days = 7 or 0
Therefore, the calendar for the year 2001 is repeated in the year 2007.


3. The last day of the century cannot be:


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


Answer: Option D

Explanation:
100 years have 5 odd days. Hence the last day of the 1st century is a Friday.
200 years have 10 odd days or 1 week + 3 odd days. Hence, the last day of the 2nd century is a Wednesday.
300 years have 15 odd days or 2 weeks + 1 odd day. Hence, the last day of the 3rd century is a Monday.
400 years have 0 odd days. Hence, the last day of the 4th century is a Sunday.


4. Which of the following year is not a leap year?

    A. 1960
    B. 2080
    C. 2024
    D. 2100


Answer: Option D

Explanation:

The two conditions that decide that a year is a leap year or not is:
• For a year to be a leap year, it should be divisible by 4.
• No century is a leap year unless it is divisible by 400.
Hence, the year 2100 is not a leap year as it is not divisible by 400.


5. What will be the simple interest on Rs. 80,000 at 16(\(\frac {2}{3}\) ) % per annum for 9 months?


    A. 8,000
    B. 9,000
    C. 10,000
    D. 11,000


Answer: Option D

Explanation:

1) Principal = Rs. 80,000

2) Rate of interest = 16 \(\frac {2}{3}\)%
3) Time = 9 months
Rate of interest = 16 \(\frac {2}{3}\)% = \(\frac {50}{3}\)
Time = \(\frac {9}{12}\) = \(\frac {3}{4}\) years

Simple Interest = \(\frac {(P × R × T)}{100}\)
Substituting the given values, we get

Simple Interest = \(\frac {80,000 }{100}\) X 80,000 \(\frac {50}{3}\) X \(\frac {3}{4}\)
Simple Interest = Rs.10,000


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