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Q:

A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:

A) gains Rs. 55	B) gains Rs. 50
C) loses Rs. 30	D) gains Rs. 30

Answer & Explanation Answer: B) gains Rs. 50

Explanation:

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Filed Under: True Discount

Q:

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

A) 3/5	B) 1/3
C) 2/5	D) 3/4

Answer & Explanation Answer: A) 3/5

Explanation:

Let A be the event of A will tell truth. B be the event of B tell truth

$P (A) = \frac{2}{3}, P (A^{c}) = 1 - P (A) = \frac{1}{3}$

$P (B) = \frac{4}{5}, P (B^{c}) = 1 - P (B) = \frac{1}{5}$

When both agree then they say true or they say false together, that is

$A \cap B o r A^{c} \cap B^{c}$

Also these events will be mutually exclusive :

$P (A \cap B) + P (A^{c} \cap B^{c}) = P (A) P (B) P (C) = \frac{2}{3} * \frac{4}{5} + \frac{1}{3} * \frac{1}{5} = \frac{3}{5}$

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Filed Under: Probability

Q:

The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5, 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?

A) 3/70	B) 4/70
C) 9/70	D) 1/50

Answer & Explanation Answer: C) 9/70

Explanation:

Let A be the event of driver A drive safely after consuming liquor.

Let B be the event of driver B drive safely after consuming liquor.

Let C be the event of driver C drive safely after consuming liquor.

P(A)=2/5,P(B)=3/7,P(C)=3/4

The events A, B and C are independent . Therefore,

$P (A \cap B \cap C) = P (A) P (B) P (C) = \frac{2}{5} * \frac{3}{7} * \frac{3}{4} = \frac{9}{70}$

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

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Filed Under: Probability

Q:

The odds favouring the event of a person hitting a target are 3 to 5. The odds against the event of another person hitting the target are 3 to 2. If each of them fire once at the target, find the probability that both of them hit the target.

A) 1/20	B) 4/20
C) 1/20	D) 3/20

Answer & Explanation Answer: D) 3/20

Explanation:

Let A be the event of first person hitting the target,

$P (A) = \frac{3}{3 + 5} = \frac{3}{8} (o d d i n f a v o u r)$

Let B be the event of Second person hitting a target.

$P (B) = \frac{2}{3 + 2} = \frac{2}{5} (o d d a g a i n s t)$

Since both events are independent and both will hit the target so,

$P (A \cap B) = P (A) P (B) = \frac{3}{8} \times \frac{2}{5} = \frac{3}{20}$

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Filed Under: Probability

Q:

A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is divisible by 5 ?

A) 1/4	B) 1/5
C) 1/2	D) 3/4

Answer & Explanation Answer: B) 1/5

Explanation:

If a number ends with 5, 0 then the number will be divisible by 5
Here only 5 is present, end place will be fixed by 5 so,

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Filed Under: Probability

Q:

Course materials are sent to students by a distance teaching institution. The probability that they will send a wrong programme’s study material is 1/5. There is a probability of 3/4 that the package is damaged in transit, and there is a probability of 1/3 that there is a short shipment. What is the probability that the complete material for the course arrives without any damage in transit?

A) 4/5	B) 2/3
C) 1/2	D) 2/15

Answer & Explanation Answer: D) 2/15

Explanation:

Probability of sending the correct material is 4/5

Probability of the material not being damaged in transit is 1/4

Probability that there is no short shipment = 2/3

∴ Required probability =4/5 x 1/4 x 2/3 = 2/15

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Filed Under: Quantitative Aptitude - Arithmetic Ability

Q:

A coin is tossed 5 times. What is the probability that head appears an odd number of times?

A) 1/2	B) 1/3
C) 2/3	D) 1

Answer & Explanation Answer: A) 1/2

Explanation:

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

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Filed Under: Probability

Q:

A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?

A) 1/5	B) 3/4
C) 2/5	D) 3/5

Answer & Explanation Answer: C) 2/5

Explanation:

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)

= $(\frac{1}{5} * \frac{2}{3}) + (\frac{4}{5} * \frac{1}{3})$ = $\frac{2}{5}$

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Filed Under: Probability

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