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

MOST IMPORTANT quantitative aptitude - 9 EXERCISES

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The following question based on average topic of quantitative aptitude

Questions : 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

The correct answers to the above question in:

Answer: (b)

Correct sum of numbers

= 10 × 50 – 64 + 46

= 500 – 18 = 482

∴ Correct average = $482/10$ = 48.2

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

Question : 1

In an examination, the average of marks was found to be 50. For deducting marks for computational errors, the marks of 100 candidates had to be changed from 90 to 60 each and so the average of marks came down to 45. The total number of candidates, who appeared at the examination, was

a) 300

b) 600

c) 150

d) 200

Answer: (b)

Let total number of candidates be x.

∴ 50x – 30 × 100 = 45x

⇒ 5x = 3000

⇒ x = $3000/5$=600

Question : 2

The average marks obtained by 22 candidates in an examination are 45. The average marks of the first 10 candidates are 55 and those of the last eleven are 40. The number of marks obtained by the eleventh candidate is

a) 0

b) 45

c) 47.5

d) 50

Answer: (a)

Marks obtained by eleventh candidate

= 22 × 45 – (10 × 55 + 11 × 40)

= 990 – (550 + 440)

= 990 – 990 = 0

Question : 3

The average marks of 100 students were found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct average corresponding to the correct score.

a) 39

b) 38.7

c) 41

d) 39.7

Answer: (d)

Total of correct marks = 100 × 40 – 83 + 53 = 3970

∴ Correct average marks = $3970/100$ = 39.70

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 = 100, m = 40

a = 53, b = 83

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

= 40 +$({53-83}/100)$

= 40- $30/100$ = 39.70

Question : 4

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 : 5

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 : 6

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

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