# Averages Questions For SSC MTS

Download Top-20 SSC MTS Averages Questions PDF. Average questions based on asked questions in previous year exam papers very important for the SSC MTS exam.

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**Question 1: **A batsman scores 34, 36, 38, 40, 42 in his 5 innings respectively. Find the average runs in his 5 innings ?

a) 38

b) 38.5

c) 39

d) 39.5

**Question 2: **The average of 5 consecutive even numbers is 44. Find the largest number in that series ?

a) 44

b) 46

c) 48

d) 50

**Question 3: **The average weight of P, Q and R is 42 kg. If the average weight of P and Q be 35 kg and that of Q and R be 50 kg, then what is the weight (in kgs) of Q?

a) 26

b) 44

c) 56

d) 54

**Question 4: **The Average of three numbers is 6.66.The first number is 1/4th the sum of other two. What is the value of first number?

a) 8

b) 6

c) 4

d) 2

**Question 5: **The average of 5 consecutive natural numbers is 38. What is the largest of these 5 numbers?

a) 39

b) 40

c) 41

d) 42

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**Question 6: **The average weight of 50 students is 32 kg. The average weight of first 24 students is 30 kg and of last 25 students is 34 kg. What is the weight (in kg) of the 25th student?

a) 34

b) 32

c) 30

d) 28

**Question 7: **If the average score of MS Dhoni in his first 99 matches is 52. What would be his new average, if he had scored 183 runs in his 100th match ?

a) 55.51

b) 54.41

c) 52.21

d) 53.31

**Question 8: **If the average age of 10 students is 16. If the age of their teacher is added, then average becomes 117. Find the age of the teacher ?.

a) 25

b) 27

c) 29

d) 31

**Question 9: **The average age of four boys is 24 years. If the ratio of their ages is 7:9:3:5, what is the age of the youngest boy ?

a) 12

b) 15

c) 18

d) 21

**Question 10: **In a city the ratio of buses, cars and bikes is 3:5:13 respectively. If the average of all three types of vehicles is 6741. Then find the number of cars in the city ?

a) 4815

b) 4155

c) 4555

d) 4665

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**Question 11: **Ravi bought 2 mobiles, one for 3999/- and another for 29999/-. Find the average cost per mobile ?

a) 16999/-

b) 16998/-

c) 15999/-

d) 15998/-

**Question 12: **Find the average of 34, 56, 78, 90, 123, 45 ?

a) 73

b) 72

c) 71

d) 70

**Question 13: **Find x, if the average of 12, 23, 34, 45, x, 67 is 41 ?

a) 56

b) 65

c) 76

d) 67

**Question 14: **The average weight of 36 students is 18 Kg. If the candidate with the highest weight is removed, the average weight dropped by 0.5 kg. What is the weight of the omitted candidate?

a) 38 kgs

b) 43 kgs

c) 37.5 kgs

d) 35.5 kgs

**Question 15: **The average of 13 natural numbers is 64. If the highest and lowest numbers are removed, then the average will drop by 2. If the highest and lowest numbers are in the ratio of 2:3, what is the highest number?

a) 75

b) 100

c) 90

d) 85

**Question 16: **The average of 16 numbers is 15. The number that should be added to this group to increase the average by 3 is

a) 66

b) 68

c) 64

d) 62

**Question 17: **The average of 8 consecutive integers is 23/2. What is the average of first three integers?

a) 9

b) 19/2

c) 8

d) 10

**Question 18: **What is the average of first 7 multiples of 7?

a) 7

b) 14

c) 21

d) 28

**Question 19: **What is the average of first 11 multiples of 11?

a) 22

b) 44

c) 55

d) 66

**Question 20: **Average of 14 numbers is 32. If the average of last 5 numbers is 26, then what is the average of the remaining numbers?

a) 35.33

b) 41.33

c) 27.5

d) 44.5

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**Answers & Solutions:**

**1) Answer (A)**

Average = (34+36+38+40+42)/5 = 190/5 = 38

Shortcut:

34, 36, 38, 40, 42 are in AP

So average = middle number ( in case of odd count)

average = average of middle two numbers (in case of even count)

here we have 5 numbers, odd count

So average = middle number = 38

SO the answer is option A.

**2) Answer (C)**

Let x-4, x-2, x, x+2, x+4 are the 5 consecutive even numbers

Average = x = 44

Largest number = x+4 = 48

So the answer is option C.

**3) Answer (B)**

P+Q+R = 42*3 = 126 ——-(1)

Q+R = 50*2 = 100 ——-(2)

P+Q = 35*2 = 70 ——-(3)

(2)+(3)-(1)

Q+R+P+Q-P-Q-R = 100+70-126

Q = 44

So the answer is option B.

**4) Answer (C)**

Let those 3 numbers are a, b, c

a+b+c = 6.66*3 = 20 ——-(1)

Given that,

a = (1/4)(b+c)

4a = b+c

5a = a+b+c

5a = 20

a = 4

So the answer is option C.

**5) Answer (B)**

Let those numbers are x-2, x-1, x, x+1, x+2

Average of these 5 numbers = $\frac{x-2+x-1+x+x+1+x+2}{5}$ = x = $38$

Largest number = x+2 = 38+2 = 40

So the answer is option B.

**6) Answer (C)**

Total wt of 50 students = wt of 1st 24 students + wt of 25th student + wt of last 25 students

50*32 = 24*30 + wt of 25th student + 25*34

1600 = 720 + wt of 25th student + 850

1600 = 1570 + wt of 25th student

wt of 25th student = 1600 – 1570 = 30

So the answer is option C.

**7) Answer (D)**

Total score in 99 matches = 99*52 = 5148

Total score in 100 matches = 5148+183 = 5331

New average = 5331/100 = 53.31

So the answer is option D.

**8) Answer (B)**

Sum of ages of students = 10*16 = 160

Let k be the age of their teacher

New average = 17

$\frac{160+k}{10+1} = 17$

$\frac{160+k}{11} = 17$

$160+k = 187$

$ k = 27$

So the answer is option B.

**9) Answer (A)**

Total age of 4 boys = 24*4 = 96

Age of youngest guy = $\frac{3}{7+9+3+5}\times 96 = \frac{3}{24}\times 96 = 12$

So the answer is option A.

**10) Answer (A)**

let the number of buses = 3k

no.of cars = 5k

no.of bikes = 13k

Total no.of vehicles = 3*(6741)

==>3k+5k+13k = 20223

==>21k = 20223

==>k = 963.

No.of cars = 5k = 5(963) = 4815/-

So the answer is option A.

**11) Answer (A)**

Average = $\frac{3999+29999}{2} = \frac{33998}{2} = 16999$/-

So the answer is option A.

**12) Answer (C)**

Average of 34, 56, 78, 90, 123, 45 = $\frac{34+56+78+90+123+45}{6} = \frac{426}{6} = 71$

So the answer is option C.

**13) Answer (B)**

Average of 12, 23, 34, 45, x, 67 = 41

$\frac{12+23+34+45+x+67}{6} = 41$

$181+x = 246$

$x = 246-181 = 65$

So the answer is option B.

**14) Answer (D)**

Total weight of 36 students = 18 * 36 = 648 kg.

New average after omitting the candidate who weighs highest = 18 -0.5 = 17.5 Kg.

Total weight of the 35 remaining students = 17.5 * 35= 612.5 kg.

So, the weight of omitted candidate = 648 – 612.5 = 35.5 kg.

Therefore, option D is the right answer.

**15) Answer (C)**

The sum of all natural numbers = 13 * 64 = 832.

New average after omitting the highest and lowest numbers = 64 – 2 = 62 .

The sum of remaining 11 natural numbers = 11 * 62 = 682.

Therefore, sum of the lowest and highest number = 832 – 682 = 150.

The lowest and highest numbers are in the ratio of 2:3.

Therefore, the highest number will be =$ \frac{3}{5} \times 150$ = 90.

Hence, option C is the right answer.

**16) Answer (A)**

Sum of all 16 numbers = 16* 15 = 240.

After adding the new number, the total number count will be = 16 + 1 = 17.

We want the average to go up by 3 points. Therefore, the new average is 15 +3 = 18.

Sum of all the numbers including the newly added number = 17 * 18 = 306.

Hence, the newly added number will be 306 – 240 = 66.

Therefore, option A is the right answer.

**17) Answer (A)**

Let the 8 consecutive integers be = $(x),$ $(x+1),$ $(x+2),$ $(x+3),$ $(x+4),$ $(x+5),$ $(x+6)$ and $(x+7)$

Sum of integers = $(x)+$ $(x+1)+$ $(x+2)+$ $(x+3)+$ $(x+4)+$ $(x+5)+$ $(x+6)+(x+7)$ $=\frac{23}{2}\times8$

=> $8x+28=92$

=> $8x=92-28=64$

=> $x=\frac{64}{8}=8$

$\therefore$ Average of first three integers = $\frac{(8)+(8+1)+(8+2)}{3}$

= $\frac{27}{3}=9$

=> Ans – (A)

**18) Answer (D)**

First 7 multiples of 7 = 7,14,21,28,35,42,49

=> Sum = $7(1+2+3+4+5+6+7)=7\times28$

=> Average = $\frac{7\times28}{7}=28$

=> Ans – (D)

**19) Answer (D)**

First eleven multiples of 11 will form an arithmetic progression : 11,22,33,…….,121

First term, $a=11$ and common difference, $d=11$

Sum of $n$ terms = $\frac{n}{2}(a+l)$

=> $S_{11}=\frac{11}{2}(11+121)$

= $11\times66$

Now, average of 11 terms = $\frac{11\times66}{11}=66$

=> Ans – (D)

**20) Answer (A)**

Average of 14 numbers = 32

=> Sum of 14 numbers = $32\times14=448$

Average of last 5 numbers = 26

=> Sum of last 5 numbers = $26\times5=130$

Thus, sum of remaining (14-5) 9 numbers = $448-130=318$

$\therefore$ Required average = $\frac{318}{9}=35.33$

=> Ans – (A)