761. P and Q can complete a work in 15 days and 10 days respectively. They started the work together and then Q left after 2 days. P alone completed the remaining work. The work was finished in__________days.
A. 12
B. 15
C. 22
D. 20

Explanation:
Work done by P in 1 day = 1/15
Work done by Q in 1 day = 1/10
Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6
Work done by P and Q in 2 days = 2 × (1/6) = 1/3
Remaining work = 1 – 1/3 = 2/3
Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days
Total time taken = 2 + 10 = 12 days

762. A and B can finish a work 30 days if they work together. They worked together for 20 days and then B left. A finished the remaining work in another 20 days. In how many days A alone can finish the work?
A. 70
B. 50
C. 40
D. 60

Explanation:
Amount of work done by A and B in 1 day = 1/30
Amount of work done by A and B in 20 days = 20 × (1/30) = 20/30 = 2/3
Remaining work – 1 – 2/3 = 1/3
A completes 1/3 work in 20 days
Amount of work A can do in 1 day = (1/3)/20 = 1/60
=> A can complete the work in 60 days

763. A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in___________days if they work together.
A. 18 days
B. 22 ½ days
C. 24 days
D. 26 days

Explanation:
If A completes a work in 1 day, B completes the same work in 3 days
Hence, if the difference is 2 days, B can complete the work in 3 days
=> if the difference is 60 days, B can complete the work in 90 days
=> Amount of work B can do in 1 day= 1/90
Amount of work A can do in 1 day = 3 × (1/90) = 1/30
Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45
=> A and B together can do the work in 45/2 days = 22 ½ days

764. Two taps X and Y can fill a tank in 10 hrs. and 15 hrs.respectively. If the both taps are opened together, the tank will be full in_________?
A. 3 hrs.
B. 4 hrs,
C. 5 hrs.
D. 6 hrs.

Explanation:
X’s 1 hour work = 1/10
Y’s 1 hour work = 1/15
(x+y)’s 1 hour work = 1/10+1/15=5/30=1/6
Both the taps can fill the tank in 6 hrs.

765. A, B and C together earn Rs.150 per day while A and C together earn Rs.94 and B and C together earn Rs.76. The daily earning of C is:__________?
A. 10 Rs,
B. 15 Rs,
C. 20 Rs.
D. 25 Rs.

Explanation:
B’s daily earning = Rs.(150-94)= Rs.56
A’s daily earning = Rs.(150-76)= Rs.74
C’s daily earning = Rs.[(150-(56+74)]=Rs.20

766. If 3 men or 4 women can construct a wall in 43 days, then the number of days that 7 men and 5 women take to construct it is :
A. 12 days
B. 14 days
C. 16 days
D. 18 days

767. If 6 men can make 10 sofas in 2 days, then 8 men can make 8 sofas in__________?
A. 1.8 days
B. 1.5 days
C. 1.2 days
D. 1 day

Explanation:
Time is
directly proportional to number of sofas
inversely proportional to number of people No. of days
= (6/8)*(8/10)*2
= 1.2 days

768. If 10 workers can make 10 tables in 10 days, then how many days would it take for 5 workers to make 5 tables?
A. 1
B. 5
C. 10
D. 25

Explanation:
10 workers make 10 tables in 10 days. 1 worker makes 1 table in 10 days. Similarly, 5 workers can make 5 tables in 10 days.

769. If 6 men take 9 days to complete a work, how many men can complete the work in 3 days?
A. 2 men
B. 12 men
C. 9 men
D. 18 men

Explanation:
For the work to be completed in 9 days,
6 men are required. Time and men are inversely proportional. For the work to be completed in 3 days,
6*9/3 = 18 men are required

770. Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days?
A. Rs.1104
B. Rs.1000
C. Rs.934
D. Rs.1210

Explanation:
Assume that cost of keeping a cow for 1 day = c ,
cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 —(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c=23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56×23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104

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