- DI & DS
- English Language
- GK
-
Intelligence & CR
- Alphabet & Number Ranking
- Analytical Reasoning
- Blood Relations Test
- Coding - Decoding
- Comparision of Ranks
- Direction Sense Test
- Mathematical Operation / Number Puzzles
- Series
- Sitting Arrangement
- Statement and Arguement
- Statement and Conclusion
- Statement and Course of Action
- Statement-Assumption
- Syllogism
-
Mathematical Skills
- Average
- Calender
- Clocks
- Geometry
- Height and Distance
- Logarithms
- Mensuration
- Mixtures and Alligations
- Number System
- Percentage
- Permutation and Computation
- Probability
- Profit and Loss
- Ratio and Proportion
- Set Theory
- Simple calculations
- Simple Equations
- Simple Interest and Compound Interest
- Time and Work
- Time, Speed and Distance
-
31.
A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C, and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated by the number adjacent to the arrow representing the street.
Motorists travelling from point S to point T would obviously take the route for which the total cost of travelling is the minimum. If two or more routes have the same least travel cost, then motorists are indifferent between them. Hence, the traffic gets evenly distributed among all the least cost routes.
The government can control the flow of traffic only by levying appropriate toll at each junction. For example, if a motorist takes the route S-A-T (using junction A alone), then the total cost of travel would be Rs 14 (i.e. Rs 9 + Rs 5) plus the toll charged at junction A.[1] If the government wants to ensure that all motorists travelling from S to T pay the same amount (fuel costs and toll combined) regardless of the route they choose and the street from B to C is under repairs (and hence unusable), then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
(1) 2, 5, 3, 2
(2) 0, 5, 3, 1
(3) 1, 5, 3, 2
(4) 2, 3, 5, 1
(5) 1, 3, 5, 1[2] If the government wants to ensure that no traffic flows on the street from D to T, while equal amount of traffic flows through junctions A and C, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
(1) 1, 5, 3, 3
(2) 1, 4, 4, 3
(3) 1, 5, 4, 2
(4) 0, 5, 2, 3
(5) 0, 5, 2, 2
[3] If the government wants to ensure that all routes from S to T get the same amount of traffic, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
(1) 0, 5, 2, 2
(2) 0, 5, 4, 1
(3) 1, 5, 3, 3
(4) 1, 5, 3, 2
(5) 1, 5, 4, 2[4] If the government wants to ensure that the traffic at S gets evenly distributed along streets from S to A, from S to B, and from S to D, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
(1) 0, 5, 4, 1
(2) 0, 5, 2, 2
(3) 1, 5, 3, 3
(4) 1, 5, 3, 2
(5) 0, 4, 3, 2[5] The government wants to devise a toll policy such that the total cost to the commuters per trip is minimized. The policy should also ensure that not more than 70 per cent of the total traffic passes through junction B. The cost incurred by the commuter travelling from point S to point T under this policy will be:
(1) Rs 7
(2) Rs 9
(3) Rs 10
(4) Rs 13
(5) Rs 14asked in CAT
View Comments [0 Reply]
-
32.
Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth day it was priced at Rs 110. At the end of each day, the MCS share price either went up by Rs 10, or else, it came down by Rs 10. Both Chetan and Michael took buying and selling decisions at the end of each trading day.
The beginning price of MCS share on a given day was the same as the ending price of the previous day.
Chetan and Michael started with the same number of shares and amount of cash, and had enough of both. Below are some additional facts about how Chetan and Michael traded over the five trading days.
> Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other hand, each day if the price went down, he bought 10 shares at the closing price.
> If on any day, the closing price was above Rs 110, then Michael sold 10 shares of MCS, while if it was below Rs 90, he bought 10 shares, all at the closing price.[1] If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once during the five days, what was the price of MCS at the end of day 3?
(1) Rs 90
(2) Rs 100
(3) Rs 110
(4) Rs 120
(5) Rs 130[2] If Michael ended up with Rs 100 less cash than Chetan at the end of day 5, what was the difference in the number of shares possessed by Michael and Chetan (at the end of day 5)?
(1) Michael had 10 less shares than Chetan.
(2) Michael had10 more shares than Chetan.
(3) Chetan had 10 more shares than Michael.
(4) Chetan had 20 more shares than Michael.
(5) Both had the same number of shares.[3] If Chetan ended up with Rs 1300 more cash than Michael at the end of day 5, what was the price of MCS share at the end of day 4?
(1) Rs 90
(2) Rs 100
(3) Rs 110
(4) Rs 120
(5) Not uniquely determinable[4] What could have been the maximum possible increase in combined cash balance of Chetan and Michael at the end of the fifth day?
(1) Rs 3700
(2) Rs 4000
(3) Rs 4700
(4) Rs 5000
(5) Rs 6000[5] If Michael ended up with 20 more shares than Chetan at the end of day 5, what was the price of the share at the end of day 3?
(1) Rs 90
(2) Rs 100
(3) Rs 110
(4) Rs 120
(5) Rs 130asked in CAT
View Comments [0 Reply]
-
33.
Mathematicians are assigned a number called Erdös number, (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below:
Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity.
> In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F.
> On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3.
> At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other.
> On the fifth day, E co-authored a paper with F which reduced the group‘s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper.
> No other paper was written during the conference.[1] The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time):
(1) 5
(2) 7
(3) 9
(4) 14
(5) 15[2] How many participants in the conference did not change their Erdös number during the conference?
(1) 2
(2) 3
(3) 4
(4) 5
(5) Cannot be determined
[3] The Erdös number of C at the end of the conference was:
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5[4] The Erdös number of E at the beginning of the conference was:
(1) 2
(2) 5
(3) 6
(4) 7
(5) 8[5] How many participants had the same Erdös number at the beginning of the conference?
(1) 2
(2) 3
(3) 4
(4) 5
(5) Cannot be determinedasked in CAT
View Comments [0 Reply]
-
34.
Let S be the set of all pairs (i, j) where 1 ≤ i ≤ j < n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1,2) and (2, 3) are also friends, but (1,4) and (2, 3) are enemies.
[1] For general n, how many enemies will each member of S have?
(1) n – 3
(2)1/2 * (n2 -3n-2)
(3) 2n – 7
(4)1/2 * (n2 -5n+6)
(5) 1/2 * (n2 -7n+14)
[2] For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members?
1). 1/2 * (n2 -5n+6)
2). 2n-6
3). 1/2 * [n(n-3)]
4). n-2
5). 1/2 * (n2 -7n+16)
asked in CAT
View Comments [0 Reply]
-
35.
In a sports event, six teams (A, B, C, D, E and F) are competing against each other. Matches are scheduled in two stages. Each team plays three matches in Stage-I and two matches in Stage-II. No team plays against the same team more than once in the event. No ties are permitted in any of the matches. The observations after the completion of Stage-I and Stage-II are as given below.
Stage-I:
> One team won all the three matches.
> Two teams lost all the matches.
> D lost to A but won against C and F.
> E lost to B but won against C and F.
> B lost at least one match.
> F did not play against the top team of Stage-I.
Stage-II:
>The leader of Stage-I lost the next two matches.
>Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.
>One more team lost both matches in Stage-II.[1] The team(s) with the most wins in the event is (are):
(1) A
(2) A & C
(3) F
(4) E
(5) B & E[2] The two teams that defeated the leader of Stage-I are:
(1) F & D
(2) E & F
(3) B & D
(4) E & D
(5) F & D[3] The only team(s) that won both the matches in Stage-II is (are):
(1) B
(2) E & F
(3) A, E & F
(4) B, E & F
(5) B & F[4] The teams that won exactly two matches in the event are:
(1) A, D & F
(2) D & E
(3) E & F
(4) D, E & F
(5) D & Fasked in CAT
View Comments [0 Reply]
-
36.
Answer the following questions based on the statements given below:
(i) There are three houses on each side of the road.
(ii) These six houses are labeled as P, Q, R, S, T and U.
(iii) The houses are of different colours, namely, Red, Blue, Green, Orange, Yellow and White.
(iv) The houses are of different heights.
(v) T, the tallest house, is exactly opposite to the Red coloured house.
(vi) The shortest house is exactly opposite to the Green coloured house.
(vii) U, the Orange coloured house, is located between P and S.
(viii) R, the Yellow coloured house, is exactly opposite to P.
(ix) Q, the Green coloured house, is exactly opposite to U.
(x) P, the White coloured house, is taller than R, but shorter than S and Q.[1] What is the colour of the tallest house?
(1) Red
(2) Blue
(3) Green
(4) Yellow
(5) none of these[2] What is the colour of the house diagonally opposite to the Yellow coloured house?
(1) White
(2) Blue
(3) Green
(4) Red
(5) none of these[3] Which is the second tallest house?
(1) P
(2) S
(3) Q
(4) R
(5) cannot be determinedasked in CAT
View Comments [0 Reply]