# SSC Topic: Average with Solutions

Q1. The average of seven numbers is 18. If one of the number is 17 and if it is replaced by 31, then the average becomes:
(a) 21.5
(b) 19.5
(c) 20
(d) 21
Q2. The mean of 100 items was 46. Later on it was discovered that an item 16 was misread as 61 and another item 43 was misread as 34. It was also found that the number of items was 90 and not 100. Then what is the correct mean?
(a) 50
(b) 50.7
(c) 52
(d) 52.7

Q3. In an exam, the average marks obtained by the students was found to be 60. After omission of computational errors, the average marks of 100 candidates had to be changed from 60 to 30 and the average with respect to all the examinees came down to 45 marks. The total number of candidates who took the exam, was
(a) 200
(b) 210
(c) 240
(d) 180
Q4. The average to 10 items was found to be 80 but while calculating, one of the items was counted as 60 instead of 50. Then the correct average would have been:
(a) 69
(b) 79.25
(c) 79
(d) 79.5
Q5. The average of 9 integers is found to be 11. But after the calculation, it was detected that, by mistake, the integer 23 was copied as 32, while calculating the average. After the due correction is made, the new average will be
(a) 10
(b) 9
(c) 10.1
(d) 9.5

Q6. A student finds the average of 10, 2 digits numbers. If the digits of one of the numbers is interchanged, the average increases by 3.6. The difference between the digits of the 2 digits number is
(a) 4
(b) 3
(c) 2
(d) 5
Q7. The average marks secured by 36 students was 52. But it was discovered that an item 64 was misread as 46. What is the correct mean of marks?
(a) 54
(b) 53.5
(c) 53
(d) 52.5
Q8. The average marks obtained by the students in 6 subject is 88. On subsequent verification it was found that the marks obtained by him in a subject was wrongly copied as 86 instead of 68. The correct average of the marks obtained by him is:
(a) 87
(b) 86
(c) 85
(d) 84
Q9. Average age of 5 boys is 16 yr, of which that of 4 boys is 16 yr 3 months. The age of the 5th boy is
(a) 15 yr
(b) 15 yr 6 months
(c) 15 yr 4 months
(d) 15 yr 2 months
Q10. The average of marks of 14 students was calculated as 71. But if was later found that the marks of two students has been wrongly entered as 42 instead of 56 and of another as 74 instead of 32. The correct average is:
(a) 67
(b) 68
(c) 69
(d) 71
Q11. The average of a collection of 20 measurements was calculated to be 56 cm. But later it was found that a mistake had occurred in one of the measurements which was recorded as 64 cm., but should have been 61 cm. The correct average must be:
(a) 53 cm
(b) 54.5 cm
(c) 55.85 cm
(d) 56.15 cm

Q12. The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected (new) mean is:
(a) 35.2
(b) 36.1
(c) 36.5
(d) 39.1
Q13. The average of marks in Mathematics for 5 students was found to be 50: Later, it was discovered that in the case of one student the marks 48 were misread as 84. The correct average is:
(a) 40.2
(b) 40.8
(c) 42.8
(d) 48.2
Q15. The mean of 9 observation is 16. One more observation is included and the new mean becomes 17. The 10th observation is:
(a) 9
(b) 16
(c) 26
(d) 30
Q14. The average age of eleven cricket players is 20 years. If the age of the coach is also included, the average age increased by 10%. The age of the coach is:
(a) 48 years
(b) 44 years
(c) 40 years
(d) 36 years

SOLUTIONS

S1. Ans.(c)
Sol. Actual number = 17
New number = 31
Difference = 14
Difference ‘14’ effect the seven number = 14/7 = 2
∴ Present average = 18
New average = 18 + 2 = 20
S2. Ans.(b)
Sol.
mean of 100 item are = 46
Sum of 100 items are = 46 × 100 = 4600
Misread 61 instead of 16 and 34 instead of 43
∴ Difference = (61 + 34) – (16 + 43)
= 95 – 59
= 36 (more)
∴ Actual sum = 4600 – 36 = 4564
Now total observation are = 90
Actual average = 4564/90 = 50.7
S3. Ans.(a)
Sol.
Let the number of student = x
∴(60x-(60×100)+(30×100))/x=45
60x – 3000 = 45x
15x = 3000
x = 200
S4. Ans.(c)
Sol. Incorrect number = 60
Correct number = 50
Difference = 60 – 50 = 10 (more)
Difference ‘10’ effect the all 10 items = 10/100 = 1
old average = 80
new average = 80 – 1 = 79
S5. Ans.(a)
Sol.
Incorrect number = 32
Correct number = 23
Difference = 32 – 23 = 9 (more)
Difference ‘9’ effect the 9 integers = 9/9 = 1
∴ Old average = 11
New average = 11 – 1 = 10

S6. Ans.(a)
Sol. Let as consider by mistake he writes 10th number with its digits
∴(10x+y-(10y+x))/10=3.6
∴ In this remaining nine numbers are same and they concel out
(10x+y-10y-x)/10=3.6
9x – 9y = 36
x – y = 4
S7. Ans.(d)
Sol. Incorrect mark = 46
Correct mark = 64
Difference = 64 – 46 = 18 (more)
Difference ‘18’ effect the 36 students = 18/36 = 0.5
Old average = 52
New average = 52 + 0.5 = 52.5
S8. Ans.(c)
Sol. Correct average of the marks obtained by him.
⇒ 88 – ((86 – 68))/6
⇒ 88 – 18/6 = 88 – 3 = 85
S9. Ans.(a)
Sol. Given, average age of 5 boys = 16 yr
Sum of ages of 5 boys = 16 × 5 = 80 yr
∴ Average ages of 4 boys
= 16 yr 3 months
= 16 yr + 3/12
= 16 yr + 1/4 yr
Sum of ages of 4 boys
= 4 × (16+1/4) yr
= 64 + 1 = 65 yr
∴ Age of 5th boy
= (80 – 65) = 15 yr

S10. Ans.(c)
Sol. Right value = 56 + 32 = 88
Wrong value = 42 + 74 = 116
Difference = 116 – 88 = 28
Difference average = 28/14 = 2
Correct average = 71 – 2 = 69
S11. Ans.(c)
Sol.
Correct average = (20 × 56 – 64 + 61)/20
=(1120-3)/20=1117/20
= 55.85 cm.
S12. Ans.(c)
Sol. Correct observation
= (50 × 36 + 48 – 23)/50
=(1800+25)/50=1825/50
= 36.5
S13. Ans.(c)
Sol. correct average
=(5×50+48-84)/5
=(250-36)/5=214/5=42.8
S14. Ans.(c)
Sol. Mean of 9 observation is = 16
sum of a observation is = 16 × 9 = 144
when one more observation include the new mean = 17
Sum of 10 observation = 10 × 17 = 170
∴ 10th observation = 170 – 144 = 26
S15. Ans.(b)
Sol. Average age of eleven cricket players is 20 years
Total age of eleven cricket players is = 20 × 11 = 220
If the age of coach include then the average age increase by 10%
= 20 + 10/100 × 20 = 22 years
∴ Total age of eleven player and coach
= 22 × 12 = 264 year
∴ Age of coach
= 264 – 220
= 44 years