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Q:

Amit can do a piece of work in 45 days, but Bharath can do the same work in 5 days less, than Amit, when working alone. Amit and Bharath both started the work together but Bharath left after some days and Amit finished the remaining work in 56 days with half of his efficiency but he did the work with Bharath with his complete efficiency. For how many days they had worked together?

A) 6	B) 8
C) 9	D) 12

Answer & Explanation Answer: B) 8

Explanation:

$\begin{matrix} A m i t & B h a r a t \\ N o . o f D a y s & 45 & 40 \\ E f f i c i e n c y & 2.22 P e r c e n t (= \frac{1}{45}) & 2.5 P e r c e n t (= \frac{1}{40}) \end{matrix}$

Amit did the work in 56 days = $56 \times \frac{1}{45 \times 2} = \frac{28}{45}$

Therefore, Rest work 17/45 was done by Amit and Bharath = $\frac{\frac{17}{45}}{\frac{17}{360}}$ = 8 days

( since Amit and Bharath do the work in one day = $\frac{1}{45} + \frac{1}{40} = \frac{17}{360}$ )

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Filed Under: Time and Work
Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

A) 5/7	B) 1/5
C) 2/7	D) 2/35

Answer & Explanation Answer: C) 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)]

= $P [(E \cap F) \cup (F \cap E)]$

= $P (E) P (F) + P (F) P (E)$

= $\frac{1}{7} \times \frac{4}{5} + \frac{1}{5} \times \frac{6}{7} = \frac{2}{7}$

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Filed Under: Probability
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

Q:

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A) 136/345	B) 17/87
C) 316/435	D) 158/435

Answer & Explanation Answer: C) 316/435

Explanation:

$n (s) = {}^{30}C_{2}$

Let A be the event of getting two oranges and

B be the event of getting two non-defective fruits.

and $(A \cap B)$ be the event of getting two non-defective oranges

$∴ P (A) = \frac{{}^{20}C_{2}}{{}^{30}C_{2}}, P (B) = \frac{{}^{22}C_{2}}{{}^{30}C_{2}} a n d P (A \cap B) = \frac{{}^{15}C_{2}}{{}^{30}C_{2}}$

$∴ P (A \cup B) = P (A) + P (B) - P (A \cap B)$

= $\frac{{}^{20}C_{2}}{{}^{30}C_{2}} + \frac{{}^{22}C_{2}}{{}^{30}C_{2}} - \frac{{}^{15}C_{2}}{{}^{30}C_{2}} = \frac{316}{435}$

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Filed Under: Probability

Q:

Two cards are drawn at random from a well - shuffled pack of 52 cards. what is the probability that either both are red or both are queens?

A) 17/112	B) 55/221
C) 55/121	D) 33/221

Answer & Explanation Answer: B) 55/221

Explanation:

n(S) = ${}^{52}C_{2}$ = 1326

Let A = event of getting both red cards

and B = event of getting both queens

then $(A \cap B)$ = event of getting two red queens

n(A) = ${}^{26}C_{2}$ = 325, n(B) = ${}^{4}C_{2}$ = 6

$n (A \cap B) = {}^{2}C_{2} = 1$

$∴ P (A) = \frac{325}{1326}, P (B) = \frac{6}{1326}$

$P (A \cap B) = \frac{1}{1326}$

$∴$ P ( both red or both queens) = $P (A \cup B)$

= $P (A) + P (B) - P (A \cap B) =$ $\frac{325}{1326} + \frac{1}{221} - \frac{1}{1326}$ = $\frac{55}{221}$

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Filed Under: Probability

Q:

The probability of occurance of two two events A and B are 1/4 and 1/2 respectively. The probability of their simultaneous occurrance is 7/50. Find the probability that neither A nor B occurs.

A) 25/99	B) 39/100
C) 61/100	D) 17/100

Answer & Explanation Answer: B) 39/100

Explanation:

P ( neither A nor B) = $P (A a n d B)$

= $P (A \cap B)$ = $P (A \cup B)$ = $1 - P (A \cup B)$

= $1 - \frac{61}{100} = \frac{39}{100}$

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Q:

A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.

A) 5/19	B) 3/20
C) 21/38	D) None of these

Answer & Explanation Answer: C) 21/38

Explanation:

n(S) = ${}^{20}C_{2}$ = 190

n(E) = ${}^{15}C_{2}$ = 105

Therefore, P(E) = 105/190 = 21/38

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Filed Under: Probability

Q:

One card is drawn from a pack of 52 cards , each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is neither a spade nor a king.

A) 0	B) 9/13
C) 1/2	D) 4/13

Answer & Explanation Answer: B) 9/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus n(E) = 13 + 3 = 16

Hence $n (\bar{E}) = 52 - 16 = 36$

Therefore, $P (E) = \frac{36}{52} = \frac{9}{13}$

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Filed Under: Probability

Q:

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28	B) 14
C) 21	D) 7

Answer & Explanation Answer: A) 28

Explanation:

Let x, y, z be the number of balls received by the three persons, then

$x \geq 5, y \geq 5, z \geq 5 a n d x + y + z = 21$

Let $u \geq 0, v \geq 0, w \geq 0 t h e n$

$∴$ x + y + z =21

$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\Rightarrow$ u + v + w = 6

$∴$ Total number of solutions = ${}^{6 + 3 - 1}C_{3 - 1} = {}^{8}C_{2} = 28$

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Filed Under: Permutations and Combinations

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