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

The correct answers to the above question in:

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

Practice average (model 1 basic average questions) Online Quiz

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

a, b, c, d, e, f, g are consecutive even numbers. j, k, l, m, n are consecutive odd numbers. The average of all the numbers is

a) $3(\text"a+n"/2)$

b) $(\text"l+d"/2)$

c) $\text"a+b+m+n"/4$

d) $\text"j+c+n+g"/4$

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 = ${d+l}/2$ 

Question : 2

The average of nine consecutive numbers is n. If the next two numbers are also included the new average w

a) increase by 2

b) remain the same

c) increase by 1.5

d) increase by 1

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

Question : 3

The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2) ?

a) $\text"2n+5"/2$

b) $n+2$

c) $n+3$

d) $\text"2n+9"/2$

Answer: (a)

According to the question let m = 1

∴ 5 consecutive integers are = 1,2,3,4,5

$\text"1+2+3+4+5"/5=n$

$n = 15/5=3$

6 consecutive integers starting with (m + 2) are = 3,4,5,6,7,8

∴$\text"3+4+5+6+7+8"/6=33/6=11/2$

Now check from option to put n = 3

While in Option :(a) $\text"2n+5"/2 = 11/2 $

Question : 4

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)

$⇒ n =\text"10 (b - c )"/ ( a - c )$

Question : 5

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 : 6

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

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