model 5 find new average from error Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 9 EXERCISES

Top 10,000+ Aptitude Memory Based Exercises

The following question based on average topic of quantitative aptitude

Questions : 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

The correct answers to the above question in:

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

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Read more new average from error Based Quantitative Aptitude Questions and Answers

Question : 1

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 : 2

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

Question : 3

A student, by mistake, wrote 64 in place of 46 as a number at the time of finding the average of 10 given numbers and got the average as 50. The correct average of the numbers is :

a) 48

b) 48.2

c) 49

d) 48.1

Answer: (b)

Correct sum of numbers

= 10 × 50 – 64 + 46

= 500 – 18 = 482

∴ Correct average = $482/10$ = 48.2

Question : 4

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 : 5

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 : 6

The average weight of 20 students in a class is increased by 0.75 kg when one of the students weighing 30 kg is replaced by a new student. Weight of the new student (in kg) is :

a) 40

b) 35

c) 50

d) 45

Answer: (d)

Required answer = 30 + 20 × 0.75

= 30 kg + 15 kg = 45 kg

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, N = 20, T = 30, t = 0.75

Weight of New student = T + Nt

= 30 + 20 × 0.75

= 30 + 15 = 45 kg

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