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Q:

A family consists of six members P, Q, R, X, Y and Z. Q is the son of R but R is not mother of Q. P and R are a married couple. Y is the brother of R. X is the daughter of P. Z is the brother of P. Which is a pair of brothers ?

A) P and X	B) P and Z
C) Q and X	D) R and Y

Answer & Explanation Answer: D) R and Y

Explanation:

Q is the son of R but R is not the mother. So, R is the Father of Q. P is married to R

So, P is the wife of R and the mother of Q, X is the daughter of P and hence of R and so she is the Sister of Q, Y is the brother of R and Z is the brother of P.

Clearly. Y is brother of R who is a male. So, Y and R are a pair of brothers

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Filed Under: Blood Relations

Q:

I. A. B, C. D, E and F are six members of a family.

II. One couple has parents and their children in the family.

III. A is the son of C and E is the daughter of A.

IV. D is the daughter of F who is the mother of E.

Which of the following pairs is the parents of the couple ?

A) AB	B) BC
C) AF	D) CF

Answer & Explanation Answer: B) BC

Explanation:

E is the daughter of A and F is the mother of E So, A is the father of E and hence the husband of F. Now,

D is the daughter of F So. D and E are the daughters of A and F Also. A is the son of C.

Now, only B remains Thus. B and C are the parents of A.

Clearly. B and C are the parents of the couple

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Filed Under: Blood Relations

Q:

Which of the following is true ?

A) A is brother of F.	B) A is sister of F.
C) D has two grandsons.	D) None of these.

Answer & Explanation Answer: D) None of these.

Explanation:

The sex of A is not known. So. neither (a) nor (b) is definitely true. Clearly, D is the grandmother of A and F.

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Filed Under: Blood Relations

Q:

If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: D) 1/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is $7 C_{5}$ .

So,required probability = $7 C_{5}$ / $10 C_{5}$ = 1/12.

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Filed Under: Probability

Q:

A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: A) 5/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is $3 C_{1} * 7 C_{4}$

So,required probability = $3 C_{1} * 7 C_{4}$ / $10 C_{5}$ = 5/12.

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Filed Under: Probability

Q:

A is the brother of B. B is the brother of C. D is the father of A. Based on these three statements, which of the following statements cannot be definitely true ?

A) B is the brother of A.	B) B is the son of D.
C) A is the brother of C.	D) C is the brother of A.

Answer & Explanation Answer: D) C is the brother of A.

Explanation:

A is the brother of B and B is the brother of C. So, C may be the brother or sister of A.

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Filed Under: Blood Relations

Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134	B) 1/2
C) 552/15379	D) 1/9

Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = $50 C_{5}$
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = $29 C_{2}$ ways and remaining two in $20 C_{2}$ ways.

Therfore,favourable number of events = $29 C_{2} * 20 C_{2}$

Hence,required probability = $29 C_{2} * 20 C_{2}$ $/ 50 C_{5}$ =551 / 15134

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Filed Under: Probability

Q:

The market value of a 10.5% stock, in which an income of Rs. 756 is derived by investing Rs. 9000, brokerage being 1/4%, is:

A) 108.25	B) 112.20
C) 124.75	D) 125.25

Answer & Explanation Answer: C) 124.75

Explanation:

For an income of Rs. 756, investment = Rs. 9000.

For an income of $R s . \frac{21}{2}$ , investment = $R s . \frac{9000}{756} \times \frac{21}{2}$ = Rs. 125.

For a Rs. 100 stock, investment = Rs. 125.

Market value of Rs. 100 stock = $R s . (125 - \frac{1}{4})$ = Rs. 124.75

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Filed Under: Stocks and Shares

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