Averages Questions for SSC CGL Set-2 PDF

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Averages Questions for SSC CGL set-2 PDF
Averages Questions for SSC CGL set-2 PDF

Averages Questions for SSC CGL Set-2 PDF

Download SSC CGL Averages Questions with answers set-2 PDF based on previous papers very useful for SSC CGL exams. Very important Averages Questions for SSC exams.

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Instructions

The table given below shows the runs scored by 5 players in four matches.

Question 1: What is the correct order of averages of the given players in the 4 matches?

a) E > A > C > D > B

b) E > A > D > C > B

c) A > E > C > D > B

d) A > E > B > C > D

Question 2: What is the average of the cubes of the first 13 natural numbers?

a) 196

b) 364

c) 485

d) 637

Question 3: The average of 50 results was calculated as 30 but later it was found that while calculating, 73 was taken as 33 by mistake, then what will be the correct average?

a) 29.2

b) 30.8

c) 31.6

d) 34

Question 4: The average runs scored by a batsman in 7 matches is 53 and in other 9 matches the average is 33. What is the average runs scored by the batsman in 16 matches?

a) 41.75

b) 44.25

c) 47

d) 49.175

Question 5: The average marks obtained by 150 students in an examination is 40. If the average marks of passed students is 60 and that of the failed students is 20, then what is the number of students who passed the examination?

a) 25

b) 50

c) 75

d) 100

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Instructions

The table given below shows the production (in ‘000 tonnes) of five companies from 2012 to 2016.

Question 6: What is the average (in ‘000 tonnes) of total production all companies in year 2013?

a) 371.1

b) 373.2

c) 378.4

d) 362.3

Question 7: The average of 11 results is 182. If the average of first 6 results is 199 and that of the last 6 results is 161, then what will be the 6th result?

a) 79

b) 118.5

c) 158

d) 237

Question 8: The average of 45 results was calculated as 27 but later it was found that while calculating 39 was taken as 93 by mistake, then what will be the correct average?

a) 25.8

b) 26.8

c) 27.2

d) 28.2

Instructions

The table give below shows the marks obtained by six students in 5 different subjects.

Question 9: What is the average of total marks obtained by all six students in subject R?

a) 77.5

b) 76.2

c) 93

d) 83.4

Instructions

The given pie chart shows the marks obtained (in degrees) by a student in different subjects. The total marks obtained by the student in the examination is 432.

 

Question 10: In how many subjects marks obtained are more than the average marks per subject?

a) 3

b) 1

c) 2

d) 4

Question 11: The average age of a husband and is wife was 23 years at the time of their marriage. After five years they have a one year old child. The average age of the family now is

a) 29.3 years

b) 19 years

c) 23 years

d) 28.5 years

Question 12: Six friends have an average height of 167 cms. A boy with height 162 cm leaves the group. Find the new average height.

a) 168 cm

b) 166 cm

c) 169 cm

d) 167 cm

Question 13: The average weight of 8 persons increases by 2.5 kg when a new persons comes in place of one of them weighing 65 kg. The weight of the new person is

a) 84 kg

b) 85 kg

c) 76 kg

d) 76.5 kg

Question 14: The average of 8 numbers is 27. If each of the numbers is multiplied by 8, find the average of new set of numbers.

a) 1128

b) 938

c) 316

d) 216

Question 15: In a prep school, the average weight of 30 girls in a class among 50 students is 16 kg and that of the remaining students is 15.5 kg. What is the average weight of all the students in the class ?

a) 15.2 kg

b) 15.8 kg

c) 15.4 kg

d) 15.6 kg

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Answers & Solutions:

1) Answer (A)

Decreasing order of average = Decreasing order of total runs scored.

Runs scored by :

A = 70 + 105 + 55 + 135 = 365

B = 40 + 35 + 95 + 72 = 242

C = 60 + 68 + 75 + 85 = 288

D = 95 + 45 + 55 + 60 = 255

E = 105 + 110 + 125 + 122 = 462

$\therefore$ Decreasing order = E > A > C > D > B

=> Ans – (A)

2) Answer (D)

Sum of the cubes of the first $n$ natural numbers = $[\frac{n(n+1)}{2}]^2$

=> Sum of the cubes of the first 13 natural numbers = $[\frac{13(13+1)}{2}]^2$

=> $(13)^2\times49$

$\therefore$ Required average = $\frac{(13)^2\times49}{13}$

= $13\times49=637$

=> Ans – (D)

3) Answer (B)

Average of 50 results = 30

=> Sum of 50 results = $30\times50=1500$

After correcting the mistake new sum = $1500-33+73=1540$

=> Correct average = $\frac{1540}{50}=30.8$

=> Ans – (B)

4) Answer (A)

Average runs scored by a batsman in 7 matches = 53

=> Total runs scored in 7 matches = $53\times7=371$

Similarly, total runs scored in 9 matches = $33\times9=297$

$\therefore$ Average runs scored by the batsman in 16 matches = $\frac{(371+297)}{16}$

= $\frac{668}{16}=41.75$

=> Ans – (A)

5) Answer (C)

Let the number of students who passed the examination = $x$

=> Number of students who failed = $(150-x)$

Average marks of 150 students = 40

=> Total marks = $40\times150=6000$

Similarly, total marks scored by passed students = $60x$

According to ques,

=> $60x+20(150-x)=6000$

=> $60x+3000-20x=6000$

=> $40x=6000-3000=3000$

=> $x=\frac{3000}{40}=75$

$\therefore$ The number of students who passed the examination = 75

=> Ans – (C)

6) Answer (B)

Total production (in ‘000 tonnes) of all companies in year 2013

= 386 + 402 + 268 + 360 + 450 = 1866

=> Required average = $\frac{1866}{5}=373.2$

=> Ans – (B)

7) Answer (C)

Average of 11 results = 182

=> Sum of 11 results = $182\times11=2002$

Similarly, sum of first 6 results = $199\times6=1194$

And sum of last 6 results = $161\times6=966$

$\therefore$ 6th result = $(1194+966)-2002=158$

=> Ans – (C)

8) Answer (A)

Average of 45 results = 27

=> Sum of 45 results = $27\times45=1215$

After correcting the mistake new sum = $1215-93+39=1161$

=> Correct average = $\frac{1161}{45}=25.8$

=> Ans – (A)

9) Answer (A)

Total marks obtained by all six students in subject R

= 80 + 78 + 55 + 74 + 83 + 95 = 465

=> Required average = $\frac{465}{6}=77.5$

=> Ans – (A)

10) Answer (C)

Total marks in the examination = 432

Average marks per subject = $\frac{432}{5}=86.4$

Marks obtained in :

Hindi = $\frac{65^\circ}{360^\circ}\times432=78$

English = $\frac{75^\circ}{360^\circ}\times432=90$

Maths = $\frac{80^\circ}{360^\circ}\times432=96$

Science = $\frac{70^\circ}{360^\circ}\times432=84$

Social Science = $\frac{70^\circ}{360^\circ}\times432=84$

Thus, only in 2 subjects (English and Maths), marks obtained are more than the average marks per subject

=> Ans – (C)

11) Answer (B)

Sum of ages of husband and wife at the time of their marriage = $23\times2=46$ years

Sum of the family after 5 years = 5 years of husband + 5 years of wife + 1 year of child

=> Total age = $46+5+5+1=57$ years

=> Required average = $\frac{57}{3}=19$ years

=> Ans – (B)

12) Answer (A)

Average height of 6 friends = 167 cm

=> Sum of height of 6 friends = $167\times6=1002$ cm

When the boy with height 162 cm left, sum of height of remaining 5 friends = $1002-162=840$ cm

=> Required average = $\frac{840}{5}=168$ cm

=> Ans – (A)

13) Answer (B)

Let average weight of 8 persons = $x$ kg and weight of new person = $y$ kg

=> Sum of weights of persons = $8x$ kg

According to ques,

=> $\frac{8x-65+y}{8}=x+2.5$

=> $8x-65+y=8x+20$

=> $y=20+65=85$

$\therefore$ The weight of the new person = 85 kg

=> Ans – (B)

14) Answer (D)

Let the 8 numbers be $x_1,x_2,…..,x_8$

Sum of these 8 numbers = $x_1+x_2+…..+x_8=27\times8=216$ ————–(i)

If each number is multiplied by 8, => Numbers = $8x_1,8x_2,…..,8x_8$

Sum = $8(x_1+x_2+…..+x_8)$

= $8\times216$

$\therefore$ New average = $\frac{8\times216}{8}=216$

=> Ans – (D)

15) Answer (B)

Average weight of 30 girls = 16 kg

=> Total weight of 30 girls = $16\times30=480$ kg

Similarly, total weight of remaining $(50-30=20)$ boys = $15.5\times20=310$ kg

$\therefore$ Average weight of all the students in the class = $\frac{(480+310)}{50}$

= $\frac{79}{5}=15.8$ kg

=> Ans – (B)

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