Practice Average - quantitative aptitude Online Quiz (set-2) For All Competitive Exams

Q-1)   A student finds the average of 10, 2 – digit numbers. If the digits of one of the numbers is interchanged, the average increases by 3.6. The difference between the digits of the 2-digit numbers is

(a)

(b)

(c)

(d)

Explanation:

Total increase = 3.6 × 10 = 36

∴ If the number be 10x + y,

then Number obtained after reversing the digits = 10y + x

∴ 10y + x – 10x – y = 36

⇒ 9y – 9x = 36

⇒ 9 (y – x) = 36

⇒ y – x = $36/9$ = 4


Q-2)   The average of two numbers is 8 and the average of other three numbers is 3. The average of the five numbers is :

(a)

(b)

(c)

(d)

Explanation:

Average of five numbers

= ${2×8+3×3}/{2+3}$

= ${16+ 9}/5$ = $25/5$ = 5


Q-3)   The average of nine numbers is 50. The average of the first five numbers is 54 and that of the last three numbers is 52. Then the sixth number is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

The sixth number

= 9 × 50 – 5 × 54 – 3 × 52

= 450 – 270 – 156 = 24


Q-4)   The average of 8 numbers is 20. The average of first two numbers is 15$1/2$ and that of the next three is 21$1/3$. If the sixth number be less than the seventh and eighth numbers by 4 and 7 respectively, then the eighth number is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

Sum of 8 numbers = 20 × 8 = 160

Sum of the first two numbers =$31/2$ ×2 = 31

Sum of next three numbers = $64/3$×3=64

Sum of the remaining three numbers = 160 – (31 + 64) = 160 – 95 = 65

Let 6th number = x

∴ 7th number = x + 4,

8th number = x + 7

⇒ x + x + 4 + x + 7 = 65

⇒ 3x = 65 – 11

⇒ x = $54/3$ =18

∴ Eighth number = 18 + 7 = 25


Q-5)   The average of three numbers is 135. The largest number is 195 and the difference between the other two is 20. The smallest number is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 1,

According to the question,

195 + x + x + 20 = 135 × 3

⇒ 2x + 215 = 405

⇒ 2x = 405 – 215 = 190

∴ x =$190/2$ = 95

x = Smallest number


Q-6)   A batsman has a certain average of runs for 12 innings. In the 13th innings he scores 96 runs thereby increasing his average by 5 runs. What will be his average after 13th innings?

(a)

(b)

(c)

(d)


Q-7)   A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is

(a)

(b)

(c)

(d)

Explanation:

If the average in 10 tests be x, then,

x × 10 + 100 = (x + 5) × 11

⇒ 11x – 10x = 100 – 55

⇒ x = 45

∴ Required average = 50


Q-8)   The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next inning so as to increase his average of runs by 4 ?

(a)

(b)

(c)

(d)

Explanation:

Let the batsman make x runs. Total runs in 10 innings = 10 × 32 = 320

∴ ${320+x}/11$ = 32+4

⇒ 320 + x = 36 × 11

⇒ x = 396 – 320 = 76

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 32, N = 11, t = 4

Required Run = T + Nt

[Here N is taken as (n + 1)]

= 32 + 11 × 4

= 32 + 44 = 76


Q-9)   The average run of a player is 32 out of 10 innings. How many runs must he make in the next innings so as to increase his average by 6 ?

(a)

(b)

(c)

(d)

Explanation:

Runs scored in the next innings = x (let)

According to the question,

10 × 32 + x = 11 × 38

⇒ 320 + x = 418

⇒ x = 418 – 320 = 98

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 32, N= (10 + 1) = 11, t = 6

Required Runs = T + Nt

= 32 + 11 × 6

= 32 + 66 = 98


Q-10)   A cricketer, whose bowling average was 12.4 runs/wicket takes 5 wickets for 22 runs in a match, thereby decreases his average by 0.4. The number of wickets, taken by him before this match was :

(a)

(b)

(c)

(d)

Explanation:

Let the number of wickets before the last match be x.

According to the question,

12.4x + 22 = (x + 5 ) × 12

⇒ 12.4x + 22 = 12x + 60

⇒ 12.4x – 12x = 60 – 22

⇒ 0.4x = 38

⇒ x = $38/{0.4}$ =$380/4$ = 95


Q-11)   The average salary, per head, of all the workers of an institution is Rs.60. The average salary of 12 officers is Rs.400; the average salary, per head, of the rest is Rs.56. The total number of workers in the institution is

(a)

(b)

(c)

(d)

Explanation:

Number of other workers except officers = x

∴ 12 × 400 + x × 56 = (x + 12) × 60

⇒ 4800 + 56x = 60x + 720

⇒ 60x – 56x = 4800 – 720

⇒ 4x = 4080

⇒ x = $4080/4$ = 1020

∴ Total number of workers = 1020 + 12 = 1032


Q-12)   Among three numbers, the first is twice the second and thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is

(a)

(b)

(c)

(d)

Explanation:

Let the second number be x.

∴ First number = 2x

∴ Third number =${2x}/3$

∴ 2x+x+${2x}/3$ = 49.5×3

⇒ 6x + 3x + 2x =49.5×9 = 445.5

⇒ 11x = 445.5

⇒ x = ${445.5}/11$ = 40.5

∴ Required difference

= 2x - ${2x}/3$ = ${4x}/3$

= ${4×40.5}/3$ = 54

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

Here, a = 2, b = $3/2$, x = 49.5

First Number = $\text"3ab"/ \text"1+b+ab"$ 5x

= ${3×2×{3/2}}/{1+{3/2}+2}×{3/2}$×49.5

= ${18/2}/{11/2}$×49.5

= ${18×49.5}/11$ = 18×4.5

= ${18×45}/10$ = 81

Third Number = $3/\text"1+b+ab"$ ×x

= $3/{1+{3/2}+2×{3/2}}$×49.5

= $3/{11/2}$ ×49 5.

= 6 ×4.5 = 27

Difference = 81 – 27 = 54


Q-13)   The mean of 20 items is 47. Later it is found that the item 62 is wrongly written as 26. Find the correct mean.

(a)

(b)

(c)

(d)

Explanation:

Difference = 62 – 26 = 36

∴ Required average = 47 +$36/20$ = 47 + 1.8 = 48.8

Aliter : Using Rule 26,

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 20, m = 47

a = 62, b = 26

Correct Average = m + ${\text"(a - b)"}/\text"n"$

= 47 + ${(62 – 26)}/20$

= 47+$36/20$

= 47 + 1.8 = 48.8


Q-14)   The average of seven numbers is 18. If one of the number is 17 and if it is replaced by 31, then the average becomes :

(a)

(b)

(c)

(d)

Explanation:

Difference = 31 – 17 = 14

∴ Required average = 18 +$14/7$ = 20

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 7, m = 18

a = 31, b = 17

New Average = m + ${\text"(a - b)"}/\text"n"$

= 18 +${(31 – 17)}/7$

= 18 +$14/7$

= 18 + 2 = 20


Q-15)   The mean value of 20 observations was found to be 75, but later on it was detected that 97 was misread as 79. Find the correct mean.

(a)

(b)

(c)

(d)

Explanation:

Difference = 97 – 79 = 18

True average = 75 + $18/20$ = 75.9

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 20, m = 75

a = 97, b = 79

Correct mean = m + ${\text"(a - b)"}/\text"n"$

= 75 +${(97 -79)}/20$

= 75 +$18/20$

= 75 + 0.9 = 75.9


Q-16)   3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is :

(a)

(b)

(c)

(d)

Explanation:

Total age of 5 members, 3 years ago = 17 × 5 = 85 years

Total age of 5 members, now

= (85 + 3 × 5) = 100 years

Total age of 6 members, now

= 17 × 6 = 102 years

∴ Age of the baby

= 102 – 100 = 2 years

Aliter : Using Rule 17,

( 't’ years before, the average age of N members of a family was 'T’ years. If during this period 'n’ children increased in the family but average age (present) remains same, then,

Present age of n children = n.T – N.t )

Here, t = 3, N = 5, T = 17, n = 1

Present age of baby = nT – Nt

= 1 × 17 – 5 × 3

= 17 – 15 = 2 years


Q-17)   The average monthly income of P and Q is Rs.5, 050. The average monthly income of Q and R is Rs.6, 250 and the average monthly income of P and R is Rs.5,200. The monthly income of P is

(a)

(b)

(c)

(d)

Explanation:

Total monthly income of P and Q = 2 × 5050 =Rs.10100

Total monthly income of Q and R = 2 × 6250 =Rs.12500

Total monthly income of P and R = 2 × 5200 =Rs.10400

On adding all three,

Total monthly income of 2(P + Q + R)

=Rs.(10100 + 12500 + 10400)

=Rs.33000

∴ Total monthly income of (P + Q + R) = $33000/2$

=Rs.16500

∴ P’s monthly income

=Rs.(16500 – 12500)

=Rs.4000


Q-18)   A man spends in 8 months as much as he earns in 6 months. He saves Rs. 6000 in a year. His average monthly income is :

(a)

(b)

(c)

(d)

Explanation:

Let the average monthly income of man be Rs. x.

∴ Man’s annual income = Rs. 12x

∴ Man’s annual expenses

=Rs.$({6x×12}/8)$

=Rs.9x

∴ Savings = 12x – 9x = Rs. 3x

∴ 3x = 6000

⇒ x = $6000/3$ = Rs. 2000


Q-19)   The average monthly income of A and B is Rs.14000, that of B and C is Rs.15600 and A and C is Rs.14400. The monthly income of C is

(a)

(b)

(c)

(d)

Explanation:

A + B =Rs.28,000 ...(i)

B + C = Rs.31,200 ...(ii)

C + A = Rs. 28,800 ...(iii)

Adding,

2(A + B + C) = 88000

⇒ A + B + C = 44000

From equation (i),

28000 + C = 44000

⇒ C = 44000 – 28000 = Rs.16000


Q-20)   The average weight of a group of 20 boys was calculated to be 89.4 kg and it was later discovered that one weight was misread as 78 kg. instead of 87kg. The correct average weight is

(a)

(b)

(c)

(d)

Explanation:

Difference in weight = 87 – 78 = 9 kg

∴ Correct average weight = 89.4 + $9/20$

= 89.4 + 0.45 = 89.85 kg

Aliter : Using Rule 26,

If average of n numbers is m later on it was found that a number 'a’ was misread as 'b’.

The correct average will be = m +${\text"(a-b)"}/ \text"n"$.

Here, n = 20, m = 89.4

a = 87, b = 78

Correct Average = m + ${\text"(a-b)"}/ \text"n"n$

= 89.4+${(87-78)}/20$

= 89.4 + $9/20$

= 89.4 + 0.45 = 89.85 kg