# model 1 basic average questions Practice Questions Answers Test with Solutions & More Shortcuts

#### average PRACTICE TEST [9 - EXERCISES]

Question : 1 [S.S.C. (CHSL) 2011]

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

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 : 2 [S.S.C. (CISF) 2011]

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

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

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

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

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

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

= Y - 101

Question : 3 [S.S.C. (CHSL) 2011]

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) $xy^2+yx^2$

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 : 4 [S.S.C. (CGL) 2016]

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

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)

$⇒ n =\text"10 (b - c )"/ ( a - c )$

Question : 5 [S.S.C. (CGL) 2015]

Average of n numbers is a. The first number is increased by 2, second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is

a) $a+(2^(n-1)-1)/n$

b) $a+2(2^n-1)/n$

c) $a+2^(n-1)/n$

d) $a+(2^n-1)/n$

Sum of new numbers

= na + (2 + 4 + 8 + 16 ..... to n terms)

Now, S = 2 + 4 + 8 + 16 + ..... to n terms

Here, a = first term = 2

r = common ratio = $4/2$ = 2

It is a geometric series.

∴ S =${a({r^n} - 1)}/{r-1}$= ${2({2^n} - 1)}/{2-1}$

= 2 ($2^n$ –1)

∴ Required average

= ${na +2({2^n}-1)}/n$

a +${2({2^n}-1)}/n$

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