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Q:

In the following question below are given some statements followed by some conclusions. Taking the given statements to be true even if they seem to be at variance from commonly known facts, read all the conclusions and then decide which of the given conclusion logically follows the given statements.

Statements:
I. All dens are spiral.
II. Some spirals are cards.

Conclusions:
I. Some spirals are not cards.
II. Some dens are not cards.
III. Some cards are den.

A) Only conclusion (I) and conclusion(II) follow.	B) Only conclusion (III) follows.
C) All conclusions follow.	D) No conclusion follows.

Answer & Explanation Answer: D) No conclusion follows.

Explanation:

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Filed Under: Statement and Conclusions
Exam Prep: Bank Exams

Q:

In the following question, two statements are given each followed by two conclusions I and II. You have to consider the statements to be true even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any, follows from the given statements.

Statements:
(I) Some notes are coins.
(II) No coin is a card.

Conclusion:
(I) All cards can be notes.
(II) Some notes are neither coins nor cards.

A) Conclusion I follows	B) Conclusion II follows
C) Neither I nor II follows	D) Both I and II follows

Answer & Explanation Answer: C) Neither I nor II follows

Explanation:

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Filed Under: Statement and Conclusions
Exam Prep: AIEEE , Bank Exams , CAT

Q:

A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king?

A) 15/52	B) 17/26
C) 13/17	D) 15/26

Answer & Explanation Answer: D) 15/26

Explanation:

Number of cards in a pack of cards = 52

Number of black cards = 26

Number of king cards = 4 (2 Red, 2 Black)

Required, the probability that if a card is drawn either card is black or a king =

$\frac{26}{52} + \frac{4}{52} = \frac{30}{52} = \frac{15}{26}$

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Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO

Q:

The SQL keyword(s) ________ is used with wildcards.

A) LIKE only	B) IN only
C) NOT IN only	D) IN and NOT IN

Answer & Explanation Answer: A) LIKE only

Explanation:

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Filed Under: Oracle Certification
Job Role: Database Administration , IT Trainer

Q:

A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is?

A) 1/26	B) 1/13
C) 2/13	D) 1/52

Answer & Explanation Answer: A) 1/26

Explanation:

Here in this pack of cards, n(S) = 52

Let E = event of getting a queen of the club or a king of the heart

Then, n(E) = 2

P(E) = n(E)/n(S) = 2/52 = 1/26

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

Q:

If two cards are taken one after another without replacing from a pack of 52 cards. What is the probability for the two cards be Ace ?

A) 51/1221	B) 42/221
C) 1/221	D) 52/1245

Answer & Explanation Answer: C) 1/221

Explanation:

Total Combination of getting a card from 52 cards = ${}^{52}C_{1}$

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= ${}^{51}C_{1}$

Total combination of getting 2 cards from 52 cards without replacement= ( ${}^{52}C_{1} \times {}^{51}C_{1}$ )

There are total 4 Ace in stack. Combination of getting 1 Ace is = ${}^{4}C_{1}$

Because there is no replacement, So number of cards after getting first Ace = 3

Combination of getting an another Ace = ${}^{3}C_{1}$

Total Combination of getting 2 Ace without replacement= ${}^{4}C_{1} \times {}^{3}C_{1}$

Now,Probability of getting 2 cards which are Ace = $\frac{{}^{4}C_{1} \times {}^{3}C_{1}}{{}^{52}C_{1} \times {}^{51}C_{1}}$ = 1/221.

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Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk , Analyst

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:

A, B, C, D and E play a game of cards. A says to B, "If you give me 3 cards, you will have as many as I have at this moment while if D takes 5 cards from you, he will have as many as E has." A and C together have twice as many cards as E has. B and D together also have the same number of cards as A and C taken together. If together they have 150 cards, how many cards has C got ?

A) 28	B) 29
C) 31	D) 35

Answer & Explanation Answer: A) 28

Explanation:

Clearly, we have :

A = B - 3 ...(i)

D + 5 = E ...(ii)

A+C = 2E ...(iii)

B + D = A+C = 2E ...(iv)

A+B + C + D + E=150 ...(v)

From (iii), (iv) and (v), we get: 5E = 150 or E = 30.

Putting E = 30 in (ii), we get: D = 25.

Putting E = 30 and D = 25 in (iv), we get: B = 35.

Putting B = 35 in (i), we get: A = 32.

Putting A = 32 and E = 30 in (iii), we get: C = 28.

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Filed Under: Arithmetical Reasoning

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