Average Questions for SSC CGL Tier 2 PDF
Download SSC CGL Tier 2 Average Questions PDF. Top 15 SSC CGL Tier 2 Average questions based on asked questions in previous exam papers very important for the SSC exam.
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Question 1: The average of odd numbers upto 100 is
a) 50.5
b) 50
c) 49.5
d) 49
Question 2: The average of three consecutive odd numbers is 12 more than one third of the first of these numbers. What is the last of the three numbers?
a) 15
b) 17
c) 19
d) Data inadequate
Question 3: The average of the first nine integral multiples of 3 is
a) 12
b) 15
c) 18
d) 21
Question 4: A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is
a) 45
b) 40
c) 50
d) 55
Question 5: The average of 6 consecutive natural numbers is K. If the next two natural numbers are also included, how much more than K will the average of these 8 numbers be?
a) 3
b) 1
c) 2
d) 1.8
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Question 6: The average of 7, 11, 15, x, 14, 21, 25 is 15, then the value of x is
a) 13.3
b) 12
c) 3
d) 14.5
Question 7: The average salary, per head, of all the workers of an institution is 60. The average salary of 12 officers is = 400; the average salary, per head, of the rest is 56. The total number of workers in the institution is
a) 1030
b) 1035
c) 1020
d) 1032
Question 8: The average of 50 numbers is 38. If two numbers, namely 45 and 55 are discarded, the average of the remaining numbers is
a) 37.5
b) 37.9
c) 36.5
d) 37.0
Question 9: Average age of A, B and C is 84 years. When D joins them the average age becomes 80 years. A new person, E, whose age is 4 years more than D, replaces A and the average of B, C, D and E becomes 78 years. What is the age of A ?
a) 50 years
b) 60 years
c) 70 years
d) 80 years
Question 10: The average of 20 numbers is 15 and the average of first five is 12. The average of the rest is
a) 16
b) 15
c) 14
d) 13
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Question 11: The average age of Ram and his two children is 17 years and the average age of Ram’s wife and the same children is 16 years. If the age of Ram is 33 years, the age of his wife is (in years):
a) 31
b) 32
c) 35
d) 30
Question 12: The average age of 14 girls and their teacher’s age is 15 years. If the teacher’s age is excluded, the average reduces by 1. What is the teacher’s age?
a) 32 years
b) 30 years
c) 29 years
d) 35 years
Question 13: The average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, the average of the remaining numbers is :
a) 35
b) 32.5
c) 37.5
d) 36
Question 14: In a family of 5 members, the average age at present is 33 years. The youngest member is 9 years old. The average age of the family just before the birth of the youngest member was
a) 30 years
b) 29 years
c) 25 years
d) 24 years
Question 15: The average marks of 50 students in a class was found to be 64. If the marks of two students were incorrectly entered as 38 and 42 instead of 83 and 24, respectively, then what is the correct average?
a) 61.24
b) 64.54
c) 62.32
d) 61.86
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Answers & Solutions:
1) Answer (B)
Require sum of 1+3+5+7+9….99
Applying formula for summation of n digits with a as first digit and d is the difference
sum = $\frac{n}{2} (2a + (n-1) d)$
or this formula can be reduced to $\frac{n}{2} \frac{a + l}{2} $hence for calculating avg. it will be
$\frac{a + l}{2}$ (where $l$ is last term)
so $\frac{1 + 99}{2}$ = 50
2) Answer (C)
Let’s say numbers are $a, a+2, a+4$
So avg. will be $\frac{(a+a+2+a+4)}{3} = \frac{a}{3} +12$
Or a = 15
So numbers will be 15, 17, 19
3) Answer (B)
As we know average of numbers which are in A.P. is = $\frac{a+l}{2}$ (where a is first term and l is last term)
Here a=3
and l= 27
Hence average will be 15
4) Answer (C)
let his average of 10 matches = x
total runs in 10 matches = 10x
in 11th match he scored 100 runs, so
total runs after 11th match = 10x + 100
average of 11 matches = (10x+100)/11
after 11th match the average of his runs is increased by 5, so
(10x+100)/11 = x+5
10x+100 = 11x + 55
x = 45
but present average = x+5 = 45+5 = 50
so the answer is option C.
5) Answer (B)
Let the 6 consecutive numbers be a-3,a-2,a-1,a,a+1,a+2
average = $\frac{SumofElements}{NumberofElements}$
It is given that average of 6 consecutive numbers be k and hence
k = $\frac{a-3+a-2+a-1+a+a+1+a+2}{6}$ = $\frac{6a-3}{6}$ = a – $\frac{1}{2}$
now next two numbers (a+3, a+4) are also added
Sum of 8 numbers = a-3+a-2+a-1+a+a+1+a+2+a+3+a+4 = 8a +4
average of 8 numbers = $\frac{8a+4}{8}$ = a + $\frac{1}{2}$ = k + 1
so average of 8 numbers is more than average of 6 numbers by = k+1 – K = 1
6) Answer (B)
we know that average =$\frac {sum of elements}{numberof elements} $
Number of elements =7
Sum of elements = 7+11+15+14+21+25+x = 93+x
=> Average = $\frac {93+x}{7} $
=> 15 = $\frac {93+x}{7} $
=> x = 12
7) Answer (D)
Let the total number of members in the institution be z
average = $\frac{SumofElements}{NumberofElements}$
Average salary of institution = Rs 60
total salary of Institution =Rs 60z
Given that out of z persons , there are 12 officers and there average salary is = Rs 400
and so total salary of 12 officers = 12 x 400 =Rs 4800
So total salary of other (z-12) members =Rs ( 60z – 4800 )…………..(1)
It is given that average salary of (z-12) persons = Rs 56
and hence from here the total salary of (z-12) people = Rs 56(z-12)……….(2)
Equation 1 and 2 are equal
60z – 4800 = 56z – 672
4z = 4128
z = 1032
8) Answer (A)
Average = $\frac{Sum of Elements}{Number of Elements}$
Given that Initially Number of Elements = 50
Initial Average = 38
Sum of Elements = 50 x 38 = 1900
Now as two numbers are discarded, hence number of elements left = 48
Sum of elements after discarding numbers = 1900 – 55 – 45 = 1800
Hence New Average = $\frac{1800}{48}$ = 37.5
9) Answer (D)
avg age of a,b,c = 84
$\frac{A+B+C}{3}$ = 84
A+B+C = 84*3 =252 ……………………(1)
similarly , $\frac{A+B+C+D}{4}$ = 80
A+B+C+D = 80*4 = 320
so using (1)
252 + D = 320
D = 68 i.e, E = 72 (as mentioned in the question)…………….(2)
$\frac{B+C+D+E}{4}$ = 78
B+C+D+E = 78*4 = 312
using (2)
B+C+68+72 =312
B+C = 172 ……………………………(3)
put (3) in (1)
A = 80
10) Answer (A)
Given that average of 20 numbers = 15
using average = $\frac{Sum}{Numberof Elements}$
Sum of all numbers = 20 x 15 = 300
Average of first five numbers = 12
So sum of first five numbers = 12×5 = 60
Sum of numbers left = 300 – 60 = 240
Number of numbers left = 20 – 5 = 15
So average of left numbers = $\frac{240}{15}$ = 16
11) Answer (D)
let the present age of Ram , Ram’s wife ,and his two children be R ,W,S1,S2 respectively
Average =$\frac{SumofAges}{Number of Ages}$
Given that average age of Ram and his childrens is =17 years
R+S1+S2= 17 × 3 = 51
Given R = 33 years .So, S2+S2 = 51-33 = 18 years
Now given that average age of Rams wife and two children = 16 years
So ,
W+S2+S1= 16×3=48
W= 48-18 =30 years
12) Answer (C)
the average age of 14 girls and teacher is given as 15 and as we know
Average = $\frac {sum of ages}{number ofpersons}$
Sum of ages of girls and teachers = 15×15 =225 years
Now the age of teacher is excluded and as a result the average reduced by 1 .So
New average = 14
New number of persons = 14
Hence new sum of ages after exclusion of teacher = 14x 14 = 196
Hence age of teacher =225-196 = 29 years
13) Answer (C)
Given that average age of 50 members = 38 years
So using , average = $\frac{SumofAge}{number of persons}$
Sum of ages = 50 x 38 = 1900
Now 45 and 55 are discarded and hence number of term left = 48
and Total Sum left = 1900 – 45 – 55 = 1800
New Average = $\frac{1800}{48}$ = 37.5
14) Answer (A)
average age of 5 members = 33×5 = 165 years
given that youngest member has age = 9 years
so 9 years ago , youngest member was not present and for other 4 people, 9 years will be reduced from each person’ age and hence
total age of 4 members apart from the youngest one 9 years ago = 165- 9 – 36 = 120 years
average age when the youngest member was born = $\frac{120}{4}$
= 30 years
15) Answer (B)
Given,A
Average of 50 students in a class = 64
Marks of two members were wrongly copied as 38 and 42 instead of 83 and 24.
Mean = $\frac{\text{Sum of all observations}}{\text{Number of Observations}}$
let the sum of all observations be x
According to the problem,
64 = $\frac{x}{50}$
$64×50$ = $x$
$x$ = 3200
Subtract the wrongly copied numbers from the total sum
= $3200 – (38+42)$
= $3200 – 80$
= $3120$
Now add the new numbers to the total sum
= $3120 + (83+24)$
= 3227
Thus, new sum is 3227
New Mean = $\frac{3227}{50}$ = 64.54
Therefore, new Mean = 64.54
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