model 1 basic average questions 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 : The average of x numbers is $y^2$ and the average of y numbers is $x^2$ . So the average of all the numbers taken together is

(a) ${x^2 + y^2}/{x + y}$

(b) x$y^2$ + y$x^2$

(c) xy

(d) ${x^3 + y^3}/{x + y}$

The correct answers to the above question in:

Answer: (c)

Total sum of x numbers = x$y^2$

Total sum of y numbers = y$x^2$

∴ Required average

= ${xy^2+yx^2}/{x+y}$

=${xy(y+x)}/{x+y}$ = xy

Aliter : Using Rule 10,

Here, $n_1$ = x, $a_1$ = $y^2$

$n_2$ = y, $a_2$ = $x^2$

∴ Average = ${n_1a_1+n_2a_2}/{n_1+n_2}$

= ${xy^2+yx^2}/{x+y}$

= xy$({x+y}/{x+y})$ = xy

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

The average of the marks obtained in an examination by 8 students was 51 and by 9 other students was 68. The average marks of all 17 students was :

a) 60

b) 60.5

c) 59.5

d) 59

Answer: (a)

Sum of total number of 8 students in exam = 8 × 51 = 408

Sum of total number of 9 students in exam = 9 × 68 = 612

∴ Required average

= ${408+ 612}/17$ =${1020}/17$ = 60

Aliter : Using Rule 10,

Here, $n_1$ = 8, $a_1$ = 51

$n_2$ = 9, $a_2$ = 68

∴ Average = ${n_1a_1+ n_2a_2}/{n_1+ n_2}$

${8×51+9×68}/{8+9}$

${408+612}/17$ = $1020/17$

= 60 marks.

Question : 2

There are 100 students in 3 sections A, B and C of a class. The average marks of all the 3 sections was 84. The average of B and C was 87.5 and the average marks of A is 70. The number of students in A was

a) 20

b) 25

c) 35

d) 30

Answer: (a)

Number of students in section A = x

∴ Number of students in sections B and C

= (100 – x)

∴ x × 70 + (100 – x) × 87.5

= 84 × 100

⇒ 70x + 87.5 × 100 – 87.5x = 8400

⇒ 8750 – 17.5x = 8400

⇒ 17.5x = 8750 – 8400 = 350

⇒ x = $350/{17.5}$ = 20

Question : 3

The average of x numbers is y and average of y numbers is x. Then the average of all the numbers taken together is

a) ${x^2 +y^2}/ {x+y}$

b) ${xy}/{x + y}$

c) ${2xy}/{x + y}$

d) ${x+y}/{2xy}$

Answer: (c)

Sum of x numbers = xy

Sum of y numbers = xy

∴ Required average = ${xy+xy}/{x+y}$ = ${2xy}/{x+y}$

Aliter : Using Rule 10,

Here, $n_1$ = x, $a_1$ = y

$n_2$ = y, $a_2$ = x

∴ Average = ${n_1a_1+n_2a_2}/{n_1+n_2}$

= ${xy+yx}/{x+y}$ = ${2xy}/{x+y}$

Question : 4

The average weight of first 11 persons among 12 persons is 95 kg. The weight of 12th person is 33 kg more than the average weight of all the 12 persons. The weight of the 12th person is

a) 131 kg

b) 97.45 kg

c) 128 kg

d) 128.75 kg

Answer: (a)

Weight of 12th person = x kg (let).

∴ Average weight of 12 persons

= $({11×95+x}/12)$kg

According to the question,

${11×95+x}/12$ +33 = x

⇒ 1045 + x + 396 = 12x

⇒ 1441 = 11x

⇒ x = $1441/11$ = 131 kg.

Question : 5

Total weekly emoluments of the workers of a factory is Rs.1534. Average weekly emolument of a worker is Rs.118. The number of workers in the factory is :

a) 13

b) 12

c) 14

d) 16

Answer: (a)

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

Number of workers in the factory = $1534/118$ = 13

Question : 6

A librarian purchased 50 story– books for his library. But he saw that he could get14 more books by spending Rs. 76 more and the average price per book would be reduced by Re. 1. The average price (in Rs.) of each book he bought, was :

a) 25

b) 20

c) 10

d) 15

Answer: (c)

Let the average cost of each book bought (of 64 books) be Rs.x.

According to the question,

64 × x – 50(x + 1) = 76

⇒ 64x – 50x – 50 = 76

⇒ 14x = 76 + 50 = 126

⇒ x = $126/14$ = 9

∴ Required average price

= 9 + 1 = Rs. 10

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