model 1 basic average questions 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 20 numbers is 15 and the average of first five is 12. The average of the rest is

(a) 14

(b) 13

(c) 15

(d) 16

The correct answers to the above question in:

Answer: (d)

If the average of remaining numbers be x, then

20 × 15 = 5 × 12 + 15x

⇒ 300 = 60 + 15x

⇒ 15x = 300 – 60 = 240

⇒ x = $240/15$ = 16

Aliter : Using Rule 13,

Here, m = 20, x = 15

n = 5, y = 12

Average of remaining Numbers = $({mx-ny}/{m-n})$

= $({20×15-5×12}/{20-5})$

= $({300-60}/15)$ = $240/15$ = 16

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

The average of marks obtained by 100 candidates in a certain examination is 30. If the average marks of passed candidates is 35 and that of the failed candidates is 10, what is the number of candidates who passed the examination?

a) 80

b) 90

c) 70

d) 60

Answer: (a)

Number of successful students in the exam = x

∴ Number of unsuccessful students = 100 – x

According to the question,

30 = ${35x+10(100-x)}/100$

⇒ 3000 = 35x + 1000 – 10x

⇒ 3000 = 25x + 1000

⇒ 25x = 3000 – 1000 = 2000

⇒ x = $2000/25$ = 80

Question : 2

The average of 30 numbers is 15. The average of the first 18 numbers is 10 and that of the next 11 numbers is 20. The last number is

a) 60

b) 50

c) 52

d) 56

Answer: (b)

Let the last number be x.

According to the question,

18 × 10 + 11 × 20 + x = 30 × 15

⇒ 180 + 220 + x = 450

⇒ 400 +x = 450

⇒ x = 450 – 400 = 50

Question : 3

A student was asked to find the arithmetic mean of the following 12 numbers : 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x He found the mean to be 12. The value of x will be :

a) 17

b) 31

c) 7

d) 3

Answer: (c)

Using Rule 1,

Average of two or more numbers/quantities is called the mean of these numbers, which is given by

$\text"Average(A)" = \text"Sumof observation / quantities"/\text"No of observation / quantities"$

∴ S = A × n

Mean = ${3 +11+ 9+7+15+ 13+ 8 +19 +17 +21 +14+x}/12$

According to question,

${137+x}/12$ = 12

∴ 137 + x = 144

∴ x = 144 – 137 = 7

Question : 4

The average marks of 32 boys of section A of class X is 60 whereas the average marks of 40 boys of section B of class X is 33. The average marks for both the sections combined together is

a) 46$1/2$

b) 45$1/2$

c) 45

d) 44

Answer: (c)

Required average

${32×60+33×40}/72$

${1920+1320}/72$ = $3240/72$ = 45

Question : 5

The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. The weight of the new person is

a) 76 kg

b) 76.5 kg

c) 85 kg

d) 84 kg

Answer: (c)

Weight of new person

= (65 + 8 × 2.5) kg

= (65 + 20) kg

= 85 kg

Aliter : Using Rule 23,

Here, x = 2.5, n = 8

Weight of new person

= weight of replaced boy + x × n

Question : 6

If average of 20 observations $x_1 , x_2 , ....., x_20$ is y, then the average of $x_1 – 101, x_2 – 101, x_3 – 101, ....., x_20 –101$ is

a) 20y

b) 101y

c) y – 101

d) y – 20

Answer: (c)

Using Rule 2,

If the given observations (x) are occuring with certain frequency (A) then,

$Average = {A_1x_1+A_2x_2+...............+A_nx_n}/{x_1+x_2+......+x_n}$

where, $A_1, A_2, A_3, .......... A_n$ are frequencies

Required average

${x_1+x_2+…+x_20}/20$ = ${101×2}/20$

= y-101

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