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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Basic Average Questions Practice Test, Shortcuts, Tricks, PDF »
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New Average Aptitude MCQ Average of Consecutive Numbers »
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Top 99+ Average Aptitude: Twice, One Third of Numbers MCQ »
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New +99 Average Aptitude MCQ Find nth Average of Numbers »
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99+ Average Aptitude Find New Average from Error MCQ Test »
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Top 99+ Aptitude MCQ Test on Average With Excluded Number »
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Top 99+ Aptitude Average on Ages & Weights MCQ Test Quiz »
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Top Aptitude Average MCQ Find Monthly Income for Average »
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Top 149+ Aptitude Average on Cricket & Exam MCQ Test Quiz »
The following question based on average topic of quantitative aptitude
(a) 89.25 kg
(b) 88.95 kg
(c) 89.85 kg
(d) 89.55 kg
The correct answers to the above question in:
Answer: (c)
Difference in weight = 87 – 78 = 9 kg
∴ Correct average weight = 89.4 + $9/20$
= 89.4 + 0.45 = 89.85 kg
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 = 89.4
a = 87, b = 78
Correct Average = m + ${\text"(a-b)"}/ \text"n"n$
= 89.4+${(87-78)}/20$
= 89.4 + $9/20$
= 89.4 + 0.45 = 89.85 kg
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Read more new average from error Based Quantitative Aptitude Questions and Answers
Question : 1
The average weight of 15 students in a class increases by 1.5kg when one of the students weighing 40 kg is replaced by a new student. What is the weight (in kg) of the new student ?
a) 56 kg.
b) 64.5 kg.
c) 62.5 kg.
d) 60 kg.
Answer »Answer: (c)
Weight of the new student
= (40 + 15 × 1.5) kg
= (40 + 22.5) kg
= 62.5 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 = 15, T = 40, t = 1.5
Weight of new Person = T + Nt
= 40 + 15 × 1.5
= 40 + 22.5
= 62.5 kg.
Question : 2
The average weight of three men A, B and C is 84 kg. D joins them and the average weight of the four becomes 80 kg. If E whose weight is 3 kg more than that of D, replaces A, the average weight of B, C, D and E becomes 79 kg. The weight of A is
a) 70 kg.
b) 65 kg.
c) 80 kg.
d) 75 kg.
Answer »Answer: (d)
Total weight of A, B and C = 84 × 3 = 252 kg.
Again, A + B + C + D = 80 × 4 = 320 kg.
∴ D = (320 – 252) kg. = 68 kg.
E = 68 + 3 = 71 kg.
B + C + D + E = 79 × 4 = 316 kg.
Now, (A + B + C + D) – (B + C + D +E) = (320 – 316) kg.
∴ A – E = 4 kg.
or A = 4 + E = 4 + 71 = 75 kg
Question : 3
A student finds the average of ten 2-digit numbers. While copying numbers, by mistake, he writes one number with its digits interchanged. As a result his answer is 1.8 less than the correct answer. The difference of the digits of the number, in which he made mistake, is
a) 3
b) 2
c) 6
d) 4
Answer »Answer: (b)
Difference in average = 1.8
∴ Difference between the number and the number formed by interchanging the digits
= 1.8 × 10 = 18
(Since 53 – 35 = 18)
∴ Number = 35
∴ Difference of digits = 5 – 3 = 2
Question : 4
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 »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 : 5
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 »Answer: (a)
Marks obtained by eleventh candidate
= 22 × 45 – (10 × 55 + 11 × 40)
= 990 – (550 + 440)
= 990 – 990 = 0
Question : 6
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 »Answer: (b)
Let total number of candidates be x.
∴ 50x – 30 × 100 = 45x
⇒ 5x = 3000
⇒ x = $3000/5$=600
GET average PRACTICE TEST EXERCISES
model 1 basic average questions
model 2 average of consecutive numbers
model 3 twice, thrice, one third etc. of numbers
model 4 find nth average from 1st & last number
model 5 find new average from error
model 6 find average of excluded number
model 7 average on ages/weight
model 8 find monthly income
model 9 average on cricket/exam
average Shortcuts and Techniques with Examples
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model 1 basic average questions
Defination & Shortcuts … -
model 2 average of consecutive numbers
Defination & Shortcuts … -
model 3 twice, thrice, one third etc. of numbers
Defination & Shortcuts … -
model 4 find nth average from 1st & last number
Defination & Shortcuts … -
model 5 find new average from error
Defination & Shortcuts … -
model 6 find average of excluded number
Defination & Shortcuts … -
model 7 average on ages/weight
Defination & Shortcuts … -
model 8 find monthly income
Defination & Shortcuts … -
model 9 average on cricket/exam
Defination & Shortcuts …
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