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. 50:
A vessel has three pipes connected with it, two to supply liquid and one to draw liquid. The first alone can fill the vessel in 9/2 hours, the second in 3 hour and the third can empty it in 3/2 hour. If all the pipes are opened simultaneously when the vessel is half-full, how soon will it be emptied ?
Answer: D 1st pipe can fill in 9/2 hour. 2nd pipe can fill in 3 hour. And, 3rd pipe can empty in 3/2 hour. The vessel is emptied in 2/9 + 1/3 - 2/3 = - 1/9 The vessel full emptied in 9 hours. The vessel half emptied in 9/2 hours.
Q. No. 51:
A started a work and left after working for 2 days. Then B was called and he finished the work in 9 days. Had A left the work after working for 3 days, B would have finished the remaining work in 6 days. In how many days can each of them, working alone, finish the whole work ?
Answer: C Suppose A can finish the work in x days and B can finish it in y days. 2/x + 9/y = 1......................(i) 3/x + 6/y = 1.....................(ii) On solving both the equations, we get x=5 and y = 15.
Q. No. 52:
12 men cam complete a piece of work in 36 days, 18 women can complete the same piece of work in 60 days. 8 men and 20 women work together for 20 days. If only women were to complete the remaining piece of work in 4 day, how many women would be required ?
Answer: A 12 men in 36 days can do a work. 1 man in a day can do 1/(12*36) work. 8 men in 20 days can do (8*20)/(12*36) = 10/27 work. Similarly, we find that 20 women in 20 days can do 10/27 work. Remaining work = 7/27 Now, because in 60 days a work is done by 20 women. In 1 day a work done by 20* 60 women. In 4 days 7/27 work is done by (20*60*7)/(27*4) = 70 women.
Q. No. 53:
Ten women can complete a piece of work in 15 days. Six men complete the same piece of work in 10 days. In how many days can five women and six men together complete the piece of work ?
Answer: B men will be denoted by M and women by W. 10W*15 = 6M*10 1M = 5W/2 Now, 5W+6M = (5+15)W = 20W 10*15 = 20*x days => x = 7.5 days.
Q. No. 54:
18 men can complete a piece of work in 63 days, 9 women take 189 days to complete the same piece of work. How many days will 4 men , 9 women and 12 children together take to complete the piece of work if 7 children alone can complete the piece of work in 486 days ?
Answer: D Men = 18*63 Women = 9*189 Children = 7*486 => 14M = 21W = 42 Children => 1M = 3 children and 1W = 2 children 4M+9W+12C = (12+18+12) children = 42 children 7*486 = 42*x => x= 81 days.