Wheels of diameter of 7cm and 14cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make same number of revolutions per second. If both of them meet after 10sec, the speed of the smaller wheel is
Answer: C Let number of revolutions be x Then, (2 * 22/7 * 7/2 + 28 22/7 *14/2) *10 * x = 1980 => x = 3. Rate of smaller wheel = 3*2*22/7 * 7/2 = 66 cm/s
Q. No. 14:
A rectangular tank is 225m by 162 m at the base. With what speed must water flow into it through an aperture 60cm by 45cm that the level may be raised 20 cm in 5 hour?
Answer: B Required rate = (225*162*20)/(5*100) = 60/100 * 45/100 * h => h= 5400 m/hr
Q. No. 15:
A can do 50% more work as B can do in the same time. B alone can do a piece of work in 20hr. A, with the help of B, can finish the same work in how many hours?
Answer: D A alone can do the work in = 20 * 100/150 = 40/3 hour A and B together can do the work in 40/(2+3) = 8 hour.
Q. No. 16:
4 men and 10 women were put on a work. They completed 1/3 of the work in 4 days. After this 2 men and 2 women were increased. They completed 2/9 more of the work in 2 days. If the remaining work is to be completed in 3 days, then how many more women must be increased?
Answer: B Let one man takes x days to complete the work and one woman takes y days to complete the work independently. Then, (4*4)/x +(10*4)/y = 1/3.........(4 men + 10 women for 4 days) (6*2)/x +(12*2)/y = 2/9...................(6 men + 12 women for 2 days) Solving above two equation we get, x= 198 and y=216 Let Z women be added to complete the work in 3 days Then, (6*3)/108 + 3(12+Z)/216 = 1-(1/3 + 2/9) = 4/9 or 3Z = 24 => Z = 8
Q. No. 17:
C is twice efficient as A. B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e AB,BC,CA) starting with AB on the first day, BC on the second day and CA on the third day and so on, then how many days are required to finish the work?
Answer: C Efficiency (A:B:C) =3:2:1 No. of days (A:B:C) = 2:3:1 Number of days taken by A =12, number of days taken by B =18 and by C = 6. 1 day's work of (A+B) = 5/36 1 day's work of (C+B) = 8/36 1 day's work of (A+C) = 9/36 In 5 days total work done = 35/36. Now, rest of the work (i.e 1/36) is done by AC Number of days taken by AC for the rest of the work = (1/36)/(9/36) =1/9 Therefore, total time taken to complete the work = 5+ 1/9 = 46/9 days.
Q. No. 18:
The work done by a woman in 8 hour is equal to the work done by a man in 6 hour and by a boy is 12 hour. If working 6 hour per day 9 men complete a work in 6 days, then in how many days can 12 men, 12 women and 12 boys together finish the same work, working 8 hour per days?
Answer: D Let the total work be 24 units (LCM of 6,8,12 ) Then, men =4, women =2 and children =2 Total work = 4*9*6*6 = 1296 Number of days taken = 1296/(9*12*8) = 1.5 days.