Ramesh has two examinations on Wednesday - Engineering mathematics in the morning and Engineering Drawing in the afternoon. He has a fixed amount of time to read the textbooks of both these subjects on Tuesday. During this time he can read 80 pages of Engineering Mathematics and 100 pages of Engineering drawing. Alternatively, he can also read 50 pages of Engineering Mathematics and 250 pages of Engineering drawing. Assume that the amount of time it takes to read one page of the textbook of either subject is constant. Ramesh is confident about Engineering Drawing and wants to devote full time to reading Engineering Mathematics. The number of Engineering Mathematics text book pages he can read on Tuesday is :
Answer: C Let, Ramesh read x pages of engineering math and y pages of engineering drawing. Total time = 80/x + 100/y = 50/x + 250/y => y = 5x Since, he has to read only engineering maths in a day, putting y=5x in above equation, we get 50/x + 250/5x = 100/x He can read 100 pages of maths in total time.
Q. No. 8:
Three professors Dr. Gupta, Dr. Sharma, Dr. Singh are evaluating answer script of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work ?
Answer: A As per the information given, the ratio of efficiencies of Dr. Gupta, Dr. Sharma and Dr. Singh are 42:30:25. hence, the ratio of time taken by Dr. Gupta and Dr. Sharma = 5:7 Dr. Gupta takes 10 days less than Dr. Sharma, time taken by them will be 25 days and 35 days respectively. Hence, time taken by Dr. Singh will be 42 days. Part of work completed by Dr. Gupta = 2/5 Part of work completed by Dr. Sharma = 3/7 Remaining work = 1 - (2/5 + 3/7) = 6/35 will be completed by Dr. Singh He can complete it in 42* 6/35 = 7.2 days.
Q. No. 9:
Three pipes A,B and C are connected to a tank. These pipes can fill the tank separately in 5hr, 10 hr and 15hr respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at 3/4th of their normal rates for the 1st hour after which they supplied water at normal rate. Pipe C supplied water at 2/3rd of its normal rate for 1st 2 hour, after which it supplied at its normal rate. In how much time, tank would be filled ?
Answer: C The part of the tank filled by A and B in first two hour => 3/4 * (1/5 + 1/10) + (1/5 + 1/10) The part of tank filled by C in first two hours = 2 * 2/3 * 1/15 Remaining part = 139/360 In 1 hour, all the three pipes together will fill = 11/30 hence, the time taken to fill the remaining tank = 139/360 * 30/11 = 1.0530 hour. Thus, the total time taken to fill the remaining tank = 3.05 hour.
Q. No. 10:
A cylindrical overhead tank is filled by two pumps P1 and P2. P1 can fill the tank in 8 hr while P2 can fill the tank in 12 hr. There is a pipe P3 which can empty the tank in 8 hr. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that the tank is exactly filled up by the time he is back. Due to technical fault P3 opens when the tank is one-third filled. If the supervisor comes back as per the plan what percent of the tank is till empty ?
Answer: C P1 and P2 can fill the tank in 24/5 hr.[In one hour these fill (1/8+ 1/12) part of tank]. It takes 12/5 hr in filling half the tank. For remaining half of the tank P3 will open and this will takes 6 hour. Supervisor has gone out for (12/5 + 6) hr. Now, 1/3rd tank will fill in 8/5 hr. In remaining 42/5 hr only 33/60th part of the tank will fill. Empties part of tank = 1- (1/3 + 33/60) = 1/10 Which is 10% of tank.