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Q:

A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident

A) 30/100	B) 35/100
C) 45/100	D) 50/100

Answer & Explanation Answer: B) 35/100

Explanation:

Let A = Event that A speaks the truth

B = Event that B speaks the truth

Then P(A) = 75/100 = 3/4

P(B) = 80/100 = 4/5

P(A-lie) = $1 - \frac{3}{4}$ = 1/4

P(B-lie) = $1 - \frac{4}{5}$ = 1/5

Now, A and B contradict each other =[A lies and B true] or [B true and B lies]

= P(A).P(B-lie) + P(A-lie).P(B)

= $(\frac{3}{5} * \frac{1}{5}) + (\frac{1}{4} * \frac{4}{5}) = \frac{7}{20}$

= $(\frac{7}{20} * 100)$ = 35%

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Filed Under: Probability

Q:

A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:

A) 0%	B) 5%
C) 7.5%	D) 10%

Answer & Explanation Answer: A) 0%

Explanation:

C.P = Rs.3000

S.P =Rs. [3600 x 10] / [100+(10 x 2)] = Rs.3000

Gain =0%

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Filed Under: True Discount

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 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:

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

A) 0.04	B) 0.043
C) 0.047	D) 0.05

Answer & Explanation Answer: B) 0.043

Explanation:

Vowels are A I A I O,

C A S T I G A T I O N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

$n (E) = \frac{5 P_{5}}{2! * 2! (A, I a r e 2 t i m e s)} * \frac{6 P_{6}}{2! (T i s 2 t i m e s)} = 21600$

$n (S) = \frac{11!}{2! * 2! * 2! (A, I a r e 2 t i m e s)} = 4989600$

$\frac{21600}{4989600} = 0.043$

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Filed Under: Probability
Exam Prep: GRE

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:

Which of the following statements is not correct?

A) log10 = 1	B) log (2 + 3) = log (2 x 3)
C) log1 = 0	D) log (1 + 2 + 3) = log 1 + log 2 + log 3

Answer & Explanation Answer: B) log (2 + 3) = log (2 x 3)

Explanation:

log (2 + 3) = log 5

log (2 x 3) = log 2 + log 3

log (2 + 3) $\neq$ log (2 x 3)

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Filed Under: Logarithms
Exam Prep: Bank Exams
Job Role: Bank PO

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