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Q:

What's the difference between L1 and L2 cache?

Answer

Level 1 cache is internal to the chip, Level 2 is external.

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Subject: Hardware

Q:

A certain factory employed 600 men and 400 women and the average wager was Rs. 25.50 per day, If a woman got Rs. 5 less than a man, then what are their daily wages ?

A) m:25.50 w:27.50	B) m:27.50 w:22.50
C) m:26.50 w:27.50	D) m: 24.50 w:26.50

Answer & Explanation Answer: B) m:27.50 w:22.50

Explanation:

Let the daily wage of a man be Rs. x.

Then, daily wage of a woman = Rs. (x - 5).

Now, 600x + 400 (x - 5) = 25.50 * (600 + 400) <=> 1000x = 27500 <=> x = 27.50.

Man's daily wages = Rs. 27.50; Woman's daily wages = (x - 5) = Rs. 22.50.

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Q:

The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then :

A) x = y + z	B) 2x = y + z
C) 3x = 2y - 2z	D) none of these

Answer & Explanation Answer: B) 2x = y + z

Explanation:

Clearly, we have : x = (3y + 3z ) / 6 or 2x = y + z.

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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:

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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Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

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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