| 1. Andrew and Phil play a game in which there are three packs of cards lying on a table. Pack I has only one card; Pack II has only two cards and Pack III has only three cards. The game is played with the following rules: i. Each player in his turn can pick only one card or all cards from just one pack. ii. Andrew & Phil take alternate turns and the player who picks up the last card loses. iii. Both players are rational & intelligent and play to win. From which pack should Phil pick up a card in order to win, given that he makes the first move? |
Difficult |
| A. Pack I B. Pack II C. Pack III D. Can’t be determined |
View Answer
Answer: Option B
Explanation:
We consider different cases:
Case I: Phil picks up one card from pack I
If Phil (P) picks up the only card in pack I, the Andrew (A) will pick up both cards in pack II, leaving P to pick up 1 or 3 cards from pack III and in either case he loses.
Case II: Phil picks up both cards from pack II
If P picks up both cards in pack II then A will pick all 3 cards in pack III, leaving P to pick the last card from pack I and P again loses.
Case III: Phil picks up one card from pack II
If P picks up 1 card from pack II, A has to pick the other card from this pack (condition i). Then P can pick all 3 cards in pack III and A will have to pick the last card from pack I. Thus, P wins.
Case IV: Phil picks up all three cards from pack III
If P picks all 3 cards from pack III, then A can pick both the cards from pack II, and P will have to pick the last card from pack I, and hence P loses.
Case V: Phil picks up one card from pack III
If P picks 1 card from pack III, then A picks 1 card from the same pack, leaving P to pick up the last card from this pack. Now, A picks up both the cards from pack II, leaving P to pick the last card from pack I and hence P loses.
Thus, only in case III, P will win. So, P should pick 1 card from pack II to ensure a win.
| 2. In a military line-up, four officers – Rick, Sick, Tick & Wick – are standing in a row, such that: i. One man is tall, strong and unmasked. ii. Two officers who are not tall are each standing next to Rick. iii. Sick is the only officer standing next to exactly one strong man. iv. Tick is the only officer not standing next to exactly one masked man. Who among them is tall, strong and unmasked? |
Difficult |
| A. Rick B. Sick C. Tick D. Wick |
View Answer
Answer: Option C
Explanation:
From iii, there is only one officer who is strong; from iv there is only one officer who is masked; from ii, Rick is neighbored with two officers who are not tall.
Now, the man who is strong is standing on an extreme, and he is the one who is tall, strong and unmasked (TSU) and needs to be identified. Also, from iii, Sick is the person standing next to the TSU man.
TSU man. So, the line-up looks like this:
TSU Sick Rick ___
❶ ❷ ❸ ❹
Now, two possibilities arise:
Case I: The TSU man is Wick
If this is true, then the arrangement looks like:
Wick Sick Rick Tick
❶ ❷ ❸ ❹
But this is not possible because from iv, Tick is not standing next to a masked man, and Rick is the only man who is masked.
Case II: The TSU man is Tick
If this is true, then the arrangement looks like:
Tick Sick Rick Wick
❶ ❷ ❸ ❹
This arrangement satisfies all conditions. So. Tick is the officer who is tall, strong and unmasked
| 3. Three children – P, Q, R – each have some pencils and erasers, such that: i. P has at least one pencil and twice as many erasers as pencils. ii. Q has at least one pencil and three times as many erasers as pencils. iii. R has at least one pencil and three more erasers than pencils. One of P, Q and R says, “If I tell you the numbers of pencils and erasers we have altogether, which does not exceed 24, you will get to know how many pencils and erasers I have, but not how many pencils and erasers each of the other has.” Who makes the above statement? |
Difficult |
| A. P B. Q C. R D. Can’t be determined |
View Answer
Answer: Option A
Explanation:
From i, P has the following combinations of pencils & erasers: 3, 6, 9, 12, 15, 18…. —– ❶
From ii, Q has the following combinations of pencils & erasers: 4, 8, 12, 16, 20….. —– ❷
From iii, R has the following combinations of pencils & erasers: 5, 7, 9, 11, 13, 15 …. —– ❸
From ❶, ❷ & ❸, the total number of pencils & erasers is at least 12, and it is given that this total cannot exceed 24, so the maximum total is 24. If T is the total of pencils and erasers with the three of them, then:
12 ≤ T ≤ 24 —–❹
Now, we use the trial & error method to arrive at possible values of T within the range mentioned in ❹
• T cannot be 12, 14, 15, 16, 17 because in that case the number of pencils and erasers (total) with each of them would be known, which contradicts the condition in the statement made in the question.
For example: T = 12 is possible only when P = 3, Q = 4, R = 5; T = 14 is possible only when P = 3, Q = 4, R = 7; T = 15 is possible only when P = 6, Q = 4, R = 5; and so on.
• T cannot be 13 because no values of P, Q and R add to 13
• T cannot be 18, 20, 21, 22, 23, 24 because these values of T would correspond to multiple values for each P, Q and R, and it would not be possible to know the number of pencils and erasers with any of P, Q, R; which again contradicts the statement made in the question.
For example, if T = 18, then the following possibilities exist: P = 3, Q = 4, R = 11; P = 3, Q = 8, R = 7; P = 9, Q = 4, R = 5. Thus, it is not possible to know the number of pencils and erasers with any of P, Q and R.
• Thus, the only permissible value of T is 19. Now, since T = 19, the number of pencils and erasers with P should even, because from ❷ & ❸, the respective number of pencils & erasers with Q and R is even and odd. Even plus odd is an odd number and since T is finally odd, only an even number can be added to odd to get an odd number. From ❶ we conclude that the only possible value for P is 6. Now, if P = 6, then either Q = 4 and R = 9 OR Q = 8 and R = 5.
Thus, the statement is made by P.
From ii, Q has the following combinations of pencils & erasers: 4, 8, 12, 16, 20….. —– ❷
From iii, R has the following combinations of pencils & erasers: 5, 7, 9, 11, 13, 15 …. —– ❸
From ❶, ❷ & ❸, the total number of pencils & erasers is at least 12, and it is given that this total cannot exceed 24, so the maximum total is 24. If T is the total of pencils and erasers with the three of them, then:
12 ≤ T ≤ 24 —–❹
Now, we use the trial & error method to arrive at possible values of T within the range mentioned in ❹
• T cannot be 12, 14, 15, 16, 17 because in that case the number of pencils and erasers (total) with each of them would be known, which contradicts the condition in the statement made in the question.
For example: T = 12 is possible only when P = 3, Q = 4, R = 5; T = 14 is possible only when P = 3, Q = 4, R = 7; T = 15 is possible only when P = 6, Q = 4, R = 5; and so on.
• T cannot be 13 because no values of P, Q and R add to 13
• T cannot be 18, 20, 21, 22, 23, 24 because these values of T would correspond to multiple values for each P, Q and R, and it would not be possible to know the number of pencils and erasers with any of P, Q, R; which again contradicts the statement made in the question.
For example, if T = 18, then the following possibilities exist: P = 3, Q = 4, R = 11; P = 3, Q = 8, R = 7; P = 9, Q = 4, R = 5. Thus, it is not possible to know the number of pencils and erasers with any of P, Q and R.
• Thus, the only permissible value of T is 19. Now, since T = 19, the number of pencils and erasers with P should even, because from ❷ & ❸, the respective number of pencils & erasers with Q and R is even and odd. Even plus odd is an odd number and since T is finally odd, only an even number can be added to odd to get an odd number. From ❶ we conclude that the only possible value for P is 6. Now, if P = 6, then either Q = 4 and R = 9 OR Q = 8 and R = 5.
Thus, the statement is made by P.