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Q:

After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member ?

A) 12	B) 13
C) 14	D) 15

Answer & Explanation Answer: D) 15

Explanation:

i) Let the ages of the five members at present be a, b, c, d & e years.

And the age of the new member be f years.

ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1)

iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years

So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2)

==> a + b + c + d + e = 5x + 15

==> a + b + c + d = 5x + 15 - e ------ (3)

iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above,

The new average is: (5x + 15 - e + f)/5

Equating this to the average age of x years, 3 yrs, ago as in (2) above,

(5x + 15 - e + f)/5 = x

==> (5x + 15 - e + f) = 5x

Solving e - f = 15 years.

Thus the difference of ages between replaced and new member = 15 years.

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Q:

The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years, The age of the 15th student is :

A) 11	B) 12
C) 13	D) 14

Answer & Explanation Answer: A) 11

Explanation:

Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.

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Q:

The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers :

A) 4	B) 8
C) 12	D) 16

Answer & Explanation Answer: B) 8

Explanation:

Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) - 57 = 8

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Q:

The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is :

A) 35	B) 36.5
C) 40	D) 42

Answer & Explanation Answer: B) 36.5

Explanation:

Correct Sum = (36 * 50 + 48 - 23) = 1825.

Correct mean = = 1825/50 = 36.5

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Q:

If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of $\frac{a^{2} + b^{2} + c^{2}}{3}$ is :

A) M	B) 3 M^2
C) 2 (M/3)	D) M^2

Answer & Explanation Answer: B) 3 M^2

Explanation:

We have : ( a + b + c) / 3 = M or (a + b + c) = 3M.

Now. ${(a + b + c)}^{2} = {(3 M)}^{2} = 9 M^{2}$ .

$< = > a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 9 M^{2}$

$a^{2} + b^{2} + c^{2} = 9 M^{2}, (a b + b c + c a = 0)$

Required mean = $(\frac{a^{2} + b^{2} + c^{2}}{3}) = \frac{9 M^{2}}{3} = 3 M^{2}$ .

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Q:

A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x ?

A) 6	B) 7
C) 8	D) 9

Answer & Explanation Answer: B) 7

Explanation:

Clearly, we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x) / 12 = 12.

or 137 + x = 144 or x = 144 - 137 = 7.

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Q:

The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what ia the pice of each of these two books ?

A) 16,12	B) 10,16
C) 23,9	D) 8,7

Answer & Explanation Answer: B) 10,16

Explanation:

Total pice of the two books = Rs. [(12 x 10) - (11.75 x 8)]

= Rs. (120 - 94) = Rs. 26.

let the price of one book be Rs. x

Then, the price of other book = Rs. (x + 60% of x) = Rs.(x+(3/5)x) = Rs. (8/5)x

So, x+(8/5)x =26 <=> x =10

The prices of the two books are Rs. 10 and Rs. 16

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Q:

The average price of three items of furniture is Rs. 15000. If their prices are in the ratio 3:5:7, the price of the cheapest item is :

A) 6000	B) 7000
C) 8888	D) 9000

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