model 1 basic average questions Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

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

Answer: (b)

According to the question consecutive even numbers = a, b, c, d, e, f, g

Consecutive odd numbers = j, k, l, m, n

Consecutive even number = 2, 4, 6, 8, 10, 12, 14

$\text"2+4+6+8+10+12+14"/7=56/7=8$ middle term

Consecutive odd numbers 1, 3, 5, 7, 9

$\text"1+3+5+7+9"/5=25/5$ middle term

Same as in the above situation.

Average of a, b, c, d, e, f, g = d

Average of j, k, l, m, n, = l

∴ Required average = ${d+l}/2$ 

Answer: (d)

Given,

$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8)"/9 = n$

$=> \text"(9a + 36)"/9 = n$

=> a + 4 = n -------------------- ( 1 )

If the next 2 numbers are included , let new average = k

$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8 + a + 9 + a + 10)"/11 = k$

$\text"(11a + 55)"/11 = k$

a + 5 = k -------------------- ( 2 )

Subtracting ( 1 ) from ( 2 ) , we get :

a + 5 - a - 4 = k - n

=> k - n = 1

=> k = n + 1

Therefore the new average is n + 1

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

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

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

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