model 5 find new average from error Practice Questions Answers Test with Solutions & More Shortcuts

Question : 11

The average of 50 numbers is 38. If two numbers, namely 45 and 55 are discarded, the average of the remaining numbers is

a) 37.9

b) 37.5

c) 37.0

d) 36.5

Answer: (b)

Sum of 50 numbers = 50 × 38 = 1900

Sum of 48 numbers = 1900 – 45 – 55 = 1800

∴ Required average = $1800/48$ = 37.5

Question : 12 [SSC CGL Tier-I 2016]

The average of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. The correct average of the observations is :

a) 42

b) 28

c) 45

d) 32

Answer: (a)

Correct sum of 9 observations

= 9 × 35 – 18 + 81

= 315 + 63 = 378

∴ Required correct average = $378/9$ = 42

Question : 13 [SSC CGL Tier-1 2011]

The average of 25 observations is 13. It was later found that an observation 73 was wrongly entered as 48. The new average is

a) 14

b) 12.6

c) 13.8

d) 15

Answer: (a)

Difference of two observations = 73 – 48 = 25

∴ New average = 13 + $25/25$=14

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 25, m = 13

a = 73, b = 48

Correct Average = m + ${\text"(a-b)"}/ \text"n"n$

= 13 +${(73- 48)}/25$

= 13 + 1 = 14

Question : 14 [SSC Assistant Grade-III 2012]

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) 210

b) 200

c) 180

d) 240

Answer: (b)

Let the number of candidates be x, then

60x – 45x = 30 × 100

⇒ 15x = 3000

⇒ x = 200

Question : 15 [SSC CGL Tier-II 2015]

The average of 20 numbers is calculated as 35. It is dicovered later, that while calculating the average, one number, namely 85, was read as 45. The correct average is

a) 37

b) 36.5

c) 36

d) 37.5

Answer: (a)

Correct sum of 20 numbers

= 20 × 35 – 45 + 85

= 700 + 40 = 740

∴ Correct average = $740/20$ = 37

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 20, m = 35

a = 85, b = 45

Correct mean = m + ${\text"(a - b)"}/\text"n"$

= 35 +${(85 - 45)}/20$

= 35 + 2 = 37

IMPORTANT quantitative aptitude EXERCISES

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