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SSC CPO Averages Quiz 5

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SSC CPO Averages Quiz 5

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Average is a straight- forward concept and it can be solved easily by equal distribution method. Average is the sum of all the elements in a given data set divided by the total number of elements in the data set. The most commonly denoted term for average is Arithmetic Mean, simply termed as mean. The article SSC CPO Averages Quiz 5 very useful for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO, IBPS RRB Exams and etc. SSC CPO Averages Quiz 5 is very useful to crack the Quantitative Aptitude sections in several exams.


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1. The average of first 10 even numbers is?

    A. 37
    B. 37.8
    C. 39
    D. 39.8


Answer: Option D

Explanation:
Prime numbers between 30 and 50 are:
31, 37, 41, 43, 47

Average of prime numbers between 30 to 50 will be

Average = \(\frac{31+37+41+43+47}{5}\) = \(\frac{199}{5}\) = 39.8


2. Reeya obtained 65, 67, 76, 82 and 85 out of 100 in different subjects, What will be the average.

    A. 70
    B. 75
    C. 80
    D. 85


Answer: Option B

Explanation:
\(\frac{65+67+76+82+85}{5}\) = 75


3. Find the sum of first 30 natural numbers

    A. 470
    B. 468
    C. 465
    D. 463


Answer: Option C

Explanation:
Sum of n natural numbers

Average = \(\frac{158 }{10}\) = 15.8


4. The average of first 10 natural numbers is?

    A. 5
    B. 5.5
    C. 6.5
    D. 6


Answer: Option B

Explanation:
Sum of 10 natural no. = \(\frac{n(n+1) }{2}\) = 55

= \(\frac{30(30+1) }{2}\) = \(\frac{30(31) }{2}\) = 465


5. Find the average of first 10 multiples of 7

    A. 35.5
    B. 37.5
    C. 38.5
    D. 40.5


Answer: Option C

Explanation:
Sum of 10 odd no. = 100

Average = \(\frac{100 }{10}\) = 10

1. The average of first 10 prime numbers is?

    A. 10
    B. 12.5
    C. 12.9
    D. 15.5


Answer: Option C

Explanation:

= \(\frac{7(1+2+3+…+10) }{10}\) = \(\frac{7(10(10+1)) }{20}\)
\(\frac{7(110) }{20}\) = 38.5


2. The average of four consecutive odd numbers is 24. Find the largest number.

    A. 25
    B. 27
    C. 29
    D. 31


Answer: Option B

Explanation:
Let the numbers are x, x+2, x+4, x+6, then
\(\frac{4x+12)}{4}\) = 24
[x + 3 = 24 =>x = 21


3. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students

    A. 55
    B. 56
    C. 57
    D. 58


Answer: Option C

Explanation:
Let the average weight of the 59 students be A.
So the total weight of the 59 of them will be 59*A.
The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs
\(\frac{59A+45}{60}\) = a – 0.2
59 A + 45 = 60 A – 12
45 + 12 = 60 A -59 A
A = 57


4. A motorist travel to a place 150 km away at an avearge speed of 50 km/hr and returns ar 30 km/hr.His average speed for the whole jouney in km/hr is

    A. 36.5 km/hr
    B. 37.5 km/hr
    C. 35.5 km/hr
    D. 34.5 km/hr


Answer: Option B

Explanation:
Average speed will be 2xy(x+y) km/hr
= {2(50)(30)}{50+30} km/hr
= 37.5 km/hr


5. The average of six numbers is X and the average of three of these is Y.If the average of the remaining three is z, then

    A. x = y + z
    B. 2x = y + z
    C. x = 2y + z
    E. x = y + 2z


Answer: Option B

X =((3y+3z)/6)
or
2X= y + z

1. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is

    A. 40
    B. 35
    C. 45
    D. 55


Answer: Option A

Explanation:
23 + Sum of the present ages of husband, wife and child
= (27 * 3 + 3 * 3) years
= 90 years.
Sum of the present ages of wife and child
= (20 * 2 + 5 * 2) years
= 50 years.
Husband’s present age
= (90 – 50) years
= 40 years


2. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.

    A. 70
    B. 76

    C. 78
    D. 80


Answer: Option B

Explanation:
Average after 11 innings should be 36
So, Required score = (11 * 36) – (10 * 32)
= 396 – 320 = 76


3. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is

    A. 35
    B. 45
    C. 55

    D. 65


Answer: Option A

Explanation:
Number is (5*27) – (4*25) = 135-100 = 35


4. Average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then weight of the new man is

    A. 50

    B. 55
    C. 60
    D. 65


Answer: Option C

Explanation:
Total weight increased is 1.5 * 10 = 15.
So weight of new person is 45+15 = 60


5. A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17th match

    A. 36
    B. 37
    C. 38
    D. 39


Answer: Option D

Explanation:
Let the average after 17th match is x
then the average before 17th match is x-3
so, 16(x-3) + 87 = 17x
=> x = 87 – 48 = 39

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