Time and work MCQs with answers pdf for competitive exams. Time and work mcqs pdf download and basic mathematics mcqs with answers pdf for competitive exams.
Two pipes can fill a tank in 10 hrs. and 12 hrs. respectively while a third pipe empties the full tank in 20 hrs. If all the three pipes operate together , in how much time the tank will be filled?
A. 12 hrs
B. 7.5 hrs
C. 5 hrs
D. 7 hrsIf two pipes function together, the cistern will be filled in 6 hrs. One pipe fills the cistern 5 hrs. faster than the other. How many hours it take the second pipe to fill the cistern ?
A. 15 hrs
B. 20 hrs
C. 5 hrs
D. 10 hrsA can do a particular work in 6 days . B can do the same work in 8 days. A and B signed to do it for Rs. 3200. They completed the work in 3 days with the help of C. How much is to be paid to C ?
A. Rs. 420
B. Rs. 400
C. Rs. 380
D. Rs. 600Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24
Amount of work A + B + C can do = 1/3
Amount of work C can do in 1 day = 1/3 – 7/24 = 1/24
work A can do in 1 day: work B can do in 1 day: work C can do in 1 day
= 1/6 : 1/8 : 1/24 = 4 : 3 : 1
Amount to be paid to C = 3200 × (1/8) = 400P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 10 days. How many days more will P take to finish the work ?
A. 18
B. 12
C. 7
D. 3p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 10 days = 10 × (1/16) = 10/16 = 5/8
Remaining work = 1 = 5/8 = 3/8
Work done by P in 1 day = Work done by P,Q and R in 1 day – Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (3/8) / (1/48) = 18If X and Y complete a certain work in 10 days, Y and Z in 16 days and X and Z in 22 days, find the time required for each one to complete the work while working separately ?
A. 40, 30, 120 days
B. 30, 40, 60 days
C. 120, 40, 60 days
D. 120, 60, 80 daysA and B can finish a piece of work in 20 days .B and C in 30 days and C and A in 40 days. In how many days will A alone finish the job ?
A. 45
B. 48
C. 34(2/7)
D. 44(1/A)+(1/B) = (1/20) (1/B)+(1/C) = (1/30) (1/A)+(1/C) = (1/40) Solving the three equations,
A=48Bilal can do a work in 15 days and Jalal in 12 days. If they work on it together for 4 days, the fraction of work that is left is ?
A. 4/7
B. 1/4
C. 3/5
D. 2/5Part of work completed by Bilal in a day = 1/15 Part of work completed by Jalal in a day = 1/12 Part of work completed by them in a day = (1/15)+(1/12) = 3/20 Part of work completed by them in 4 days = (3/20)*4 = 3/5 Fraction of work left
= 1-(3/5)
= 2/5A and B can do a work in 8 days, 13, and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in ?
A. 12 days
B. 8 days
C. 4 days
D. 6 days(A + B + C)’s 1 day’s work = 1/6; (A + B)’s 1 day’s work = 1/8 (B + C)’s 1 day’s work = 1/12 (A + C)’s 1 day’s work = ( 2 X 1/6 ) – ( 1/8 + 1/12 ) = ( 1/3 – 5/24 ) = 3/24 = 1/8 So, A and C together will do the woirk in 8 days.
A can do a certain work-in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in ?
A. 25 days
B. 30 days
C. 15 days
D. 20 days(A + B)’s 1 day’s work = 1/10; C’s 1 day’s work = 1/50 (A + B + C)’s 1 day’s work = 1/10 + 1/50 = 6/50 = 3/25 Also, A’s 1 day’s work = (B + C)’s 1 day’s work
From (i) and (ii), we get : 2 x (A’s 1 day’s work) = 3/25 As 1 day’s work = 3/50 B’s 1 day’s work = 1/10 – 3/50 = 2/50 = 1/25 So, B alone could do the work in 25 days.A and B can do a piece of work in 72 days; B and C can do it in 120 days; A and C can do it in 90 days. In what time can A alone do it?
A. 120 days
B. 150 days
C. 80 days
D. 100 days