The solutions to time and work questions are given here. These solutions are detailed with proper explanations. Also, the best approach to solve the questions are given to help the CAT aspirants solve the related questions easily and accurately in the least possible time.
Time and Work: Detailed Solutions
1). Answer: Option (d)
After the first 12 days, in the next 12 days, Ram will again do 40% of the work. Hence, Ravi will do 20% of the work.
Therefore ram is twice as efficient as ravi. Or 100% more efficient. (since, (2-1)/1 x100 = 100%).
2). Answer: Option (c)
Let’s solve this using answer options. We can see that if B takes 40 days to complete the entire work then he’ll take 20 days to do half of it. But the sum of A and B is 20. Let’s try option c. B will do half the work in 15 days. Now using equation (1/a)+(1/b)=2/15, We see that a = 10. Hence half the job is done in 5 days. It’s the correct option.
3). Answer: Option (b)
Method 1- Using variables
Let the number of filling pipes be x and the number of empty pipes be y. Then y/6 – x/8 = 1/6.. 4(y-1)=3x Now go from answer options
We see that for x= y=4, the relation is getting satisfied. Hence the answer is 4.
Method 2- Using Percentage
Inlet- 8 hours => 12.5% per hour Outlet= 6 hours => 16.66% per hour The difference = 4.16% As the tank is full, the difference should account for 100% of the tank, as it takes 6 hours to empty 4.16 x 6 = 25% i.e. 25% work is done by one outlet pipe; therefore there has to be 4 outlet pipes therefore the number of inlet pipes = 8-4 = 4. Answer is option (b).
4). Answer: Option (a)
A can finish the work in 12 days. He can finish 100% of the work in 12 days. In 1 day he can finish (100/12) % = 8.33% of the work. Similarly, B can finish (100/15) % = 6.67% of the work in 1 day. When they both work together, they can finish (8.33 + 6.67) % = 15% of the work in 1 day So, to complete 100% of the work, they will take 100/15 = 6 2/3 days.
5). Answer: Option (c)
Arun can finish 100% of work in 12 days, he can finish (100/12)= 8.33% of the work in a day.
Ajit can finish 100% of work in 15 days, he can finish (100/15) = 6.67% of the work in a day.
Amit can finish 100% of the work in 20 days, he can finish (100/20) = 5% of the work in a day.
So, if all three work together, then they can finish (8.33 + 6.67 + 5) % = 20% of the work in a day. So, they can complete the work in 100/20 = 5 days.
6). Answer: Option (d)
Suppose Ajit can finish the work in ‘x’ days. In one day he can do (100/x) % of the work.
As Ajit is thrice as good as Dev, Dev will do 1/3rd of what Ajit can do in a day.
So, Dev can do (100/3x)% of work in a day.
Now they complete the work in 5 days, so in 1 day they must be doing 100/5 = 20% of the work.
So, (100/x) + (100/3x) = 20 ? x = 20/3.
So, Ajit can complete the work alone in 20/3 days.
7). Answer: Option (a)
Let a, b and c be the % of the work done by A, B, and C in one day respectively.
A & B can complete the work in 12 days ? a + b = 100/12% = 8.33% … (i)
B & C can complete the work in 15 days ? b + c = 100/15% = 6.67% … (ii)
A & C can complete the work in 20 days ? a + c = 100/20% = 5% … (iii)
Adding (i), (ii) and (iii) 2(a + b + c) = 20% ? (a + b + c) = 10%.
So, working together they finish 10% of work in a day. So, they can complete the work in 10 days.
8). Answer: Option (b)
A can finish the work in 16 days. In one day he can finish (100/16) %= 6.25% of the work.
In 4 days, working alone he will finish [(100/16)*4] = 25% of the work.
Now B joins him.Amount of work left = 100 – 25 = 75%. B can do (100/8)% = 12.33% of the work.
A and B together can do 12.33+ 6.25 = 18.75 % of the work in one day. Work left after A has finished 25% of the work in 4 days = 75%. Hence Time taken = [75/18.75] = 4 days. So, total Time taken = 4 + 4 = 8 days.
9). Answer: Option (d)
Amit takes 20 days to complete the work. So, in 1 day Amit completes 100/20 = 5% of the work.
Suraj is 25% more efficient than Amit. So, Suraj will do [5 + (5 *.25)] = 6.25% of the work in 1 day.
Suraj joined Amit 4 days before the work was completed. This means Suraj worked for 4 days.
In four days Amit and Suraj together complete [(5 + 6.25)*4] = 45% of the work.
This means the rest 55% of the work is done by Amit alone. This 55% of work will be completed in (55/5) = 11 days
So, Amit worked alone for 11 days.
10). Answer: Option (a)
The ratio of efficiency is inversely proportional to the amount of time taken. Hence A and C will take Time in ratio 3: 5 to complete the same work.
So, in terms of numbers, days have taken to complete the work:
B: C = 2: 3 A: C = 3: 5
A: B: C = 9: 10: 15
If A is taking 9x days, B takes 10x days and C takes 15x days.
A takes 4 days less than C ? 9x = 15x – 4 ? x = 2/3.
A takes 6 days, B takes 20/3 days and C takes 10 days.
In one day A does (100/6)%, B does 15% and C does 10% of the work.
If A, B and C work for two days together. So, in 2 days, amount of work done = 2(100/6+ 15 + 10)=250/3 % Amount of work left = 100 – 250/3=50/3%. This can be completed by “A” in 1 day.
11). Answer: Option (b)
Suppose A takes x days to complete the job. Then B will take x + 4 days and C will take x – 2 days to do the same job.In one day, A does (1/x)%, B does [1/(x+4)]% and C does [1/(x-2)]%.
A & B will do [1/x]+ [1/(x+4)]% of the work in a day.
So, Time taken by A & B to complete the work when
working together = x(x+ 4)/(2x+4)
Time taken by C = (x – 2)
Now as given, x – 2 = x(x + 4)/(2x+4)
Solving we get x = 2(1 ± v3), we neglect 2(1 ± v3) because the number of days cannot be negative
So, days taken by A = 2(1 + v3) days taken by B = 2(3 + v3)
Ratio = (1 + v3): (3 + v3).
12). Answer: Option (d)
In one month GIL can do 100/5= 20% of the work. Along with NCC, it is able to do 100/4 = 25% of the work in one month. So, the amount of work done by NCC = 5% in a month. So, in the 4 months, NCC has done 5*4 = 20% of the work. So, the share of NCC = 20% of Rs. 35, 00, 000 = Rs. 7, 00, 000.Out of these as Sanjeev’s team is only 75% as efficient as Ramesh’s team, the money will be divided in the ratio of efficiency: Sanjeev’s team efficiency: Ramesh’s Team Efficiency = 75: 100 = 3: 4.
So, money earned by Sanjeev’s team = (3/7)× Rs.7,00,000 = Rs.3,00,000.
13). Answer: Option (c)
Method 1- Using man-days
Amount of work (man days) = 30 × 24 = 720 man-days.
So, the number of days taken by 40 people = 720/40= 18 days.
Method 2: Using the constant product rule
Increase in number of men = 1/3
Decrease in number of days =1/4=1/4of 24 = 6 24-6= 18 days.
14). Answer: Option (c)
Method 1- Using man-days
Amount of work in terms of man-days = 24 * 20 = 480 man-days
So, the number of men required to complete the work in 12 days = 480/12 = 40 men.
16 more men are required.
15). Answer: Option (d)
Amount of the work = 20 * 20 = 400 man days
Suppose the number of men is increased after ‘n’ days. Then for ‘n’ days amount of work done = 20n
Total Time taken = 75% of 20 = 15 days.
So, equating the man-days 20n + 30*(15 – n) = 400 (when no. of men is increased by 50%, it becomes 20 + .5*20 = 30)
20n + 450 – 30n = 400
10n = 50 ? n = 5. So, number of men should be increased after 5 days.
16). Answer: Option (a)
Equate the amount of work in terms of man-hours and volume of the platform.
15 workers, 4 hours a day, 25 days == 120 * 10 * 14
12 workers, 5 hours a day, N days = = 600 * 14 * 12 Taking the ratio,
(15*4*25)/( 12 *5*N)=(120 *10 * 14)/(600 *14 * 12)? N = 150.
They will take 150 days.
17). Answer: Option (b)
5 men can reap a field in 12 days and 8 women can also reap the same field in 12 days.
So, equating the amount of work done in both the cases: 5 * 12
man-days = 8 * 12 woman days, 5 man-days = 8 woman days, 1 man day = 1.6 woman days
Now when 3 men and 4 women are working then as 1 man day = 1.6 woman days, 1 man working for 1 that is equal to 1.6 women working for 1 day. Hence 3 men and 4 women = (3 * 1.6) + 4 = 8.8 women.
So, amount of days taken = 8 * 12/8.8 = 11 days (approx).
18). Answer: Option (d)
1 man = 2 children.So, 6 children + 2 men = 3 men + 2 men = 5 men
So, 5 men complete a job in 6 days. So, amount of work in man days = 5 * 6 = 30 man days. Now, when 4 men are working, then number of days taken = 30/4 = 7.5 days.
19). Answer: Option (c)
Let ‘w’ be the percent of work done by a woman in 1 day and ‘m’ be the percentage of work by a man in 1 day. Then:
3w + 4m = 25 (100/4= 25%)
5w + 2m = 20
(100/5= 20%) Solving we get w = 15/7 m = 65/14 ? m = 5w/2
Money given will be in the ratio of work done. So, if a woman gets Rs. 60, then a man should get 13/6×60= Rs. 130.
20). Answer: Option (d)
(x – 2) men can do a job in x days: amount of work = x(x-2) man days
(x + 7) men can do ¾ of the job in (x- 10) days. So, they will do 100% of the job in 4/3(x – 10) days: amount of work done = (4/3)(x – 10)(x + 7)As the work done is same: x(x – 2) = 4/3(x – 10)(x + 7)
3(x2 – 2x) = 4(x2 – 3x – 70)
3×2 – 6x = 4×2 – 12x – 280
x2 – 6x – 280 = 0
(x – 20)(x + 14) = 0 X= 20.So, amount of work = 20 * 18 = 360 man days.
So, x + 4 = 24 men will finish the work in 360/24= 15 days.
21).Answer: Option (b)
Suppose the Japanese complete x % of the work in a day. Then the Chinese will do 3x % of the work in a day, and Indians will do 6x % of work in a day.
The money will be divided in the ratio of the work done. As the amount of time for which they do the work is the same: the money will be distributed in the ratio Japanese: Chinese: Indian = 1: 3: 6.
So, money received by Chinese and Japanese together = Rs. 4, 00,000.
22).Answer: Option (a)
In 1 day 4 men and 2 boys can do [100/(20/3)]% = 15% of the job. … (i)
In 1 day 3 women and 4 boys can do 100/5 = 20% of the job. … (ii)
In 1 day 2 men and 3 women can do 100/4 = 25% of the job. … (iii)
adding (i), (ii), (iii) 6 men, 6 women and 6 boys can do 15 + 20 + 25 = 60% of the job in 1 day.
So, 1 man, 1 woman, and 1 boy can do 10% of the job in a day.
All double their efficiency, they can do 20% of the job in a day.
They can complete the job in 100/20 = 5 days.
23).Answer: Option (c)
Pipe A can fill a tank completely in 4 hours, in 1 hour, it can fill 100/4= 25% of the tank.
Pipe B can empty the tank in 5 hours, in 1 hour, it can empty 100/5= 20% of the tank.
So, when both are simultaneously working, then in 1 hour 5% of the tank will be filled. Filling the tank completely will take 100/5= 20 hours.
24). Answer: Option (b)
Approach 1: Using percentages
The tap can fill the tank in 12 hours. So, in one hour (100/12)= 8.33 % of the tank is filled.
But because of the leak, it is filling the tank in 15 hours.
In one hour only (100/15)= 6.66 % of the tank is getting filled.
So, the percentage of water that is getting leaked in 1 hour = 100/12–100/15= (100/60)= 1.66 %.
Now when the tank is completely filled and the tap is closed, then the water will leak out at the rate of (100/60) % in 1 hour. So the time taken to empty the tank = 100/(100/60)= 60 hours.
Approach 2: Using the constant product rule
Increase in Time taken = 3/12=1/4 Decrease in efficiency = 1/5 Since efficiency a 1/(time taken), Time it will take to empty the tank = 12 x 5 = 60 hours.
25).Answer: Option (d)
Method 1: Conventional approach
Suppose in one hour A can fill 100 litres.
As it fills the tank in 10 hours, the capacity of the tank is 100 * 10 = 1000 litres. Now 4 pipes each with efficiency 20% as that of Pipe A? Each of the 4 pipes will fill 20 litres in 1 hour. So, all together they will fill 80 litres in one hour. So, to fill the tank completely, they will take 1000/80 = 12.5 hours.
Method 2: Using the constant product rule
The 4 pipes have an efficiency =20% that of A. Together they have an efficiency = 80% that of A. Thus a decrease in efficiency = 1/5. Increase in Time = ¼ = 25% more than 10 hours = 12.5 hours.
Method 3: Shortcut
A fills the tank in 10 hours. Thus, with 20% efficiency, it will take 50 hours. Now there are 4 such pipes. Thus, the time taken = 50/4=12.5.
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