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The following question based on average topic of quantitative aptitude

Questions : 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) $\text"x+y"/\text"2xy"$

(b) $\text"2xy"/\text"x+y"$

(c) $(x^2+y^2)/\text"x+y"$

(d) $\text"xy"/\text"x+y"$

The correct answers to the above question in:

Answer: (b)

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

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

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

If average of 20 observations x1,x2, ....., x20 is y, then the average of x1 – 101, x2 – 101, x3 –101, ....., x20 –101 is

a) y-20

b) y-101

c) 20y

d) 101y

Answer: (b)

${X_1 + X_2 + X_3 + X_4 + ..... + X_20}/20 = Y$

$X_1 + X_2 + X_3 + X_4 + ..... + X_20 = 20Y$

$= {X_1 - 101 + X_2 - 101 + X_3 - 101 + X_4 - 101 + ..... + X_20 - 101}/20$

$= (X_1 + X_2 + X_3 + X_4 + ..... + X_20) - 20 x 101/20$

$\text"20Y - 20 x 101"/20$

= Y - 101

Question : 2

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) $(x^3+y^3)/\text"x+y"$

b) $xy$

c) $(x^2+y^2)/\text"x+y"$

d) $xy^2+yx^2$

Answer: (b)

According to the question,

Average of x number is y2

∴ Sum of x number is = xy2

Average of y number is = x2

∴ Sum of y number is = yx2

Average of all number is

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

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

= xy

Question : 3

In a class, average height of all students is ‘a’ cms. Among them, average height of 10 students is ‘b’ cms and the average height of the remaining students is ‘c’ cms. Find the number of students in the class. (Here a > c and b > c )

a) $(a(b-c))/(a-c)$

b) $(b-c)/(a-c)$

c) $(b-c)/\text"10(a-c)"$

d) $\text"10(b-c)"/(a-c)$

Answer: (d)

Let the total number of students in the class be n.

According to the question,

an = 10 × b + (n – 10) c

⇒ an = 10b + nc – 10c

⇒ an – cn = 10b – 10c

⇒ n (a – c) = 10 (b – c)

$⇒ n =\text"10 (b - c )"/ ( a - c )$

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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
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