Study the following information and answer the question that follow ::
A flat wilderness area has four widely separated shelters - M,N,O and P that are connected by exactly four straight paths-D,E,F and G that equal to each other in length and connect the shelters in the following ways : 1). D connects M and O only 2). E connects N and O only 3). F connects M and N only 4). G connects N and P only. The shelters are at ends of the paths.
Q. No. 1:
If a person walks the full length of each path exactly once, which of the following lists all those shelters and only those shelters at which the person must be exactly twice ?
Answer: A On the basis of information routes are ::
Q. No. 2:
If, by taking shortcuts that stray from the paths, a person could travel from O to P over a shortest distance than the shortest between O and P by path alone, which of the following must be true ?
A :
The route composed of E and G is not a straight line
B :
The route composed of F and G is not a straight line
C :
E meets G at a right angle
D :
The shortest sequence of paths between M and P is the shortest distance between M and P.
Answer: D On the basis of information routes are ::
Q. No. 5:
If a person is at P and want's to reach M by a sequence of paths no longer than necessary, there are how many path sequence(s) of minimal length from which to choose?
Answer: B On the basis of information routes are :: P,N,M and P,N,O,M
Q. No. 6:
If a straight line distance between M and P is the same as the straight line between O and P, which of the following can result false if new straight paths are added between M and P and between O and P ?
A :
The shortest distance by path between M and O is less than the shortest distance between O and P
B :
The shortest distance by path between any shelter and any other shelter is the same.
C :
A person must travel fewer path to travel the shortest distance between M and N
D :
The number of paths required for the shortest possible travel by path between any shelter and any other shelter is one.
Answer: B On the basis of information routes are ::
Study the following information and answer the question that follow ::
Six musicians Ali, Baba, Giri, Julie, Mumtaz and Tarak - are planning to perform a programme consisting entirely; of three sessions. Each session requires two violins, one veena, and a piano. Each person must play in at least one session, and each person can play, at most, one instrument in session. No person can play the same type of instrument (violin, veena or piano) in two successive sessions. 1). Ali plays violin only, and must play in the first session. 2). Bala plays violin or piano. 3). Giri plays violin or veena. 4). Julie plays veena only. 5). Mumtaz plays violin or piano. 6). Tarak plays piano only.
Q. No. 1:
Which of the following groups of musicians includes all those and only those, who CANNOT be scheduled to play in all three sessions, no matter what scheduled is devised ?
Answer: D According to the directions no person can play the same type of instrument in two successive sessions. As per the directions only Ali, Julie and Tarak play one instrument and they cannot be scheduled to play in all three sessions.
Q. No. 2:
Any of the following musicians could play in the second session EXCEPT :
Answer: B As, Ali plays in the first session and he only plays violin. Hence, he cannot play in the second session.
Q. No. 3:
Unavailability of which of the following musicians would still permit scheduling the five remaining players so that the proposed programme could be performed ?
Answer: A In the directions, we see that Mumtaz and Bala are playing same instruments. In the two only one could be made unavailable. Since, Bala is not in options. So, Mumtaz is the answer.
Q. No. 4:
If Ali, Baba, Giri and Julie play in the first session, which of the following could be the group of musicians playing in the second session ?
Answer: A Veena is played by either Julie or Giri and If Julie plays in 1st session then Giri will play in 2nd session.
Study the following information and answer the question that follow ::
At the start of a game of cards, J and B together had four times as much money as T, while T and B together had three times as much as J. At the end of the evening, J and B together has three times as much money as T, while T and B together has twice as much as J. B lost Rs 200.
Q. No. 1:
What fraction of the total money did T have at the beginning of the game ?
Answer: D At the start of the game J and B together had 4 times the money that T had. Let the total money be m. Then, j+b+t = 4t+t = 5t = m Thus, the total money is divisible by 5. Similarly, from the other three statements in the data, we find that the total money is divisible by 4 and 3 as well. Hence, the total money is divisible by 60. Let the total money be 60y. At the start of the game, j+b = 4t j+b+t = 5t = 60y => t = 12y. t+b = 3j j+b = 48y j = 15y and b = 33y. The amounts with J,B and T at the start of the game were 15y, 33y and 12y respectively. Now, at the end of the game :: j+b = 3t => 4t = 60y t = 15y. j+b = 45y t+b = 2j Hence, The amounts with J , B and T at the end of the game were 20y, 25y and 15y respectively. Hence, the fraction of total money that T had at the beginning was 12y/60y = 1/5.
Answer: A At the start of the game J and B together had 4 times the money that T had.
Let the total money be m.
Then, j+b+t = 4t+t = 5t = m
Thus, the total money is divisible by 5.
Similarly, from the other three statements in the data, we find that the total money is divisible by 4 and 3 as well.
Hence, the total money is divisible by 60.
Let the total money be 60y.
At the start of the game, j+b = 4t
j+b+t = 5t = 60y => t = 12y.
t+b = 3j
j+b = 48y
j = 15y and b = 33y.
The amounts with J,B and T at the start of the game were 15y, 33y and 12y respectively.
Now, at the end of the game ::
j+b = 3t => 4t = 60y
t = 15y.
j+b = 45y
t+b = 2j
Hence, The amounts with J , B and T at the end of the game were 20y, 25y and 15y respectively. J won 20y - 15y = 5y Required fraction = 5y/60y = 1/12.
Answer: C At the start of the game J and B together had 4 times the money that T had.
Let the total money be m.
Then, j+b+t = 4t+t = 5t = m
Thus, the total money is divisible by 5.
Similarly, from the other three statements in the data, we find that the total money is divisible by 4 and 3 as well.
Hence, the total money is divisible by 60.
Let the total money be 60y.
At the start of the game, j+b = 4t
j+b+t = 5t = 60y => t = 12y.
t+b = 3j
j+b = 48y
j = 15y and b = 33y.
The amounts with J,B and T at the start of the game were 15y, 33y and 12y respectively.
Now, at the end of the game ::
j+b = 3t => 4t = 60y
t = 15y.
j+b = 45y
t+b = 2j
Hence, The amounts with J , B and T at the end of the game were 20y, 25y and 15y respectively. As, B lost Rs 200. 33y - 25y = 200 y = 25. 33y = 825. B started with Rs 825.
Study the following information and answer the question that follow ::
Seven persons A,B,C,D,E,F and G are planning to enjoy boating. There are only two boats, and the following conditions are to be kept in mind. i). A will go in the same boat in which E is to go. ii). F cannot go in the boat in which C is, unless D is also accompanying. iii). Neither B nor C can be given the boat in which G is. iv). The maximum number of persons in one boat can be four only.
Q. No. 1:
If F and B are in one boat, which of the following statements is true ?
Answer: D From condition (iv), the maximum number of persons in one boat can be four only. Hence, there must be at least three people in the complete list of people who must be sitting in either boat.
Q. No. 77:
Abdul, Mala and Chetan went bird watching. Each of them saw one bird that none of the others did. Each pair saw one bird that the third did not. And one bird was seen by all three. Of the birds Abdul saw, two were yellow. Of the birds Mala saw, three were yellow. Of the birds Chetan saw, four were yellow. How many yellow birds were seen in all ? How many non-yellow birds were seen in all ?
Answer: B Abdul + Mala + Chetan = 1 yellow bird Abdul + Chetan = 1 yellow bird Chetan = 1 yellow bird Mala = 1 yellow bird Abdul+Mala = Non-yellow bird Abdul = Non-yellow bird Hence, Total yellow bird = 5 and Non-yellow = 2.
Study the following information and answer the question that follow ::
(i). There is a group of 5 persons A,B,C,D and E. (ii). In the group there is one badminton player, one chess player and one tennis player. (iii). A and D are unmarried ladies and do not play any games. (iv). No lady is a chess player or a badminton player. (v). There is a married couple in the group of which E is the husband. (vi). B is the brother of C and is neither a chess player nor a tennis player.
Answer: D B(Male)----------C(female)----------E (Male) B and C are brother and sister. C and E are husband and wife. A and D are female. B- Badminton player. C- Tennis player E- Chess player.
Answer: B B(Male)----------C(female)----------E (Male)
B and C are brother and sister.
C and E are husband and wife.
A and D are female.
B- Badminton player.
C- Tennis player
E- Chess player.
Answer: C B(Male)----------C(female)----------E (Male)
B and C are brother and sister.
C and E are husband and wife.
A and D are female.
B- Badminton player.
C- Tennis player
E- Chess player.
