Basic Question
Q: A can do a piece of work in 20 days, and B alone can do it in 30 days. In how many days can A and B complete the work together?
Time Total work Efficiency
A -> 20 Days 60 Units 60/20 = 3 units
B -> 30 Days 60/30 = 2 units
LCM of days = 60
Assume total work = 60 Units
Next, we find efficiency. Efficiency is the work completed in a given time (days, minutes, hours)
according to the question.
Now, we get to know that A takes 20 days to complete 60 units of work.
Work done in one day by A = 60/20 = 3 units
Work done in one day by B = 60/30 = 2 units
Now, if both A and B work together, then 3 + 2 = 5 units of work will be completed in a day.
Therefore, 60 units of work at the rate of 5 units per day will take 60/5 = 12 days.
So, when A and B work together, the work will be done in 12 days.
Formula:
Time = Work / Efficiency
In this case, Work = 60 units, and the combined efficiency of both A and B is 5 units/day.
Therefore, Time = 60 / 5 = 12 days.
Example 2
Q: A and B together can do a piece of work in 7(1/2) days, and B alone can do it in 25 days. In how many days can A alone finish the work?
Time Total Work Efficiency
A + B = 15/2 days 75 units (75 * 2) / 15 = 10
B = 25 days 75 / 25 = 3
While taking the LCM ignore the denominator part.
Combined efficiency of A and B = 10 units/day
Efficiency of B alone = 3 units/day
Therefore, efficiency of A alone = 10 - 3 = 7 units/day
Formula:
Time = Work / Efficiency
Time taken by A alone = 75 / 7 = 105⁄7 days
Q: A, B, and C can do a piece of work in 8 days. If B alone can do the same work in 18 days and C alone can do the same work in 24 days, in how many days can A alone finish the same work?
Time Total Work Efficiency
A + B + C 8 days 9
B 18 days 72 units 4
C 24 days 3
Combined efficiency of A, B, and C = 9 units/day
Efficiency of B alone = 4 units/day
Efficiency of C alone = 3 units/day
Therefore, efficiency of A alone = 9 - 4 - 3 = 2 units/day
Formula:
Time = Work / Efficiency
Time taken by A alone = 72 / 2 = 36 days
Hence, A can complete the work in 36 days alone.
Q: A can do 1/2 of a piece of work in 5 days, B can do 3/5 of the same work in 9 days, and C can do 2/3 of the work in 8 days. In how many days can all three of them together do the work?
A takes to do 1/2 of the work in 5 days
for 1 unit it takes 5 * 2 = 10 days
B takes to do 3/5 of the work in 9 days
for 1 unit it takes (9 / (3/5)) = 15 days
C takes to do 2/3 of the work in 8 days
for 1 unit it takes (8 / (2/3)) = 12 days
Time Total Work Efficiency
A 10 days 60 units 6
B 15 days 4
C 12 days 5
Total efficiency (sum of A, B, and C) = 6 + 4 + 5 = 15 units/day
Therefore, if all work together, it will take 60 / 15 = 4 days to complete the work.
Q: A and B can do a piece of work in 8 days, B and C can do it in 24 days, and C and A can do it in 8(4/7) days. In how many days can C alone do the whole work?
Time Total Work Efficiency
A + B 8 days 15
B + C 24 days 120 units 5
C + A 8(4/7) days 14
(60/7 days)
Efficiency of A + B = 15 units/day
Efficiency of B + C = 5 units/day
Efficiency of C + A = 14 units/day
Let's find A, B and C's efficiencies
A + B = 15 --(1)
B + C = 5 --(2)
upon addition of (1) and (2) equation we will get 2B + A + C = 20 --(3)
we know A + C = 14, replace A + C with 14 in equation (3)
2B + 14 = 20
B = 3
C = 2
Efficiency of C alone = 2 units/day
Total work = 120 units
Time taken by C alone = Work / Efficiency
Time = 120 / 2 = 60 days
Hence, C alone can complete the work in 60 days.
Q: A can complete a piece of work in 20 days and B can complete the same work in 25 days. If they start the work together but after 5 days B left the work. In how many days will the total work be finished?
Time Total Work Efficiency
A 20 days 100 units 5 units/day
B 25 days 4 units/day
If A and B work together, they will complete 9 units of work in a day
In 5 days, both worked together, so:
5 * 9 = 45 units of work completed in 5 days
Remaining work which A has to do = 100 - 45 = 55 units
Using the formula T = W / E:
T = 55 / 5 (efficiency of A)
T = 11 days
Therefore, the total time to finish the work is:
5 days (when both worked together) + 11 days (when only A worked) = 16 days
Q: A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. Then after how many days did A leave the work?
Time Total Work Efficiency
A 45 days 360 units 8 units/day
B 40 days 9 units/day
In 23 days, B completed 23 * 9 = 207 units of work
That means 360 - 207 = 153 units of work they might have worked together
Combined efficiency of A and B = 17 units/day
So, A + B combined can complete 153 units of work in 153 / 17 = 9 days
Therefore, after 9 days, A left the work.
Q: A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
Time Total Work Efficiency
A + B 30 days 4 units/day
B + C 24 days 120 units 5 units/day
C + A 20 days 6 units/day
Total sum of efficiencies:
2(A + B + C) = 15 units/day
A + B + C = 15/2 = 7.5 units/day
If they worked together for 10 days, they completed:
10 * 7.5 = 75 units of work together
The remaining work left = 120 - 75 = 45 units
Now we need to find the efficiency of A:
A + B + C efficiency = 7.5 units/day
B + C efficiency = 5 units/day
A + 5 = 7.5
A = 7.5 - 5
A = 2.5 units/day
So, 45 units of work would take:
45 / 2.5 = 18 days to complete
Hence, A will take 18 more days to finish the work.
Q: A can do a work in 16 days, B can do it in 36 days and C can do it in 48 days. A, B and C start working together. After working 4 days, A left. Then after 5 days, B also left. In how many days can C complete the remaining work?
Time Total Work Efficiency
A 16 days 9 units/day
B 36 days 144 units 4 units/day
C 48 days 3 units/day
A, B, and C's combined efficiency = 16 units/day
They all work together for 4 days, completing:
4 * 16 = 64 units of work
Remaining work left = 144 - 64 = 80 units
Now A left, which means B and C are working together.
B and C's combined efficiency = 7 units/day
They worked together for 5 days, completing:
5 * 7 = 35 units of work
Remaining work left = 80 - 35 = 45 units
Now, 45 units of work have to be completed by C alone.
C's efficiency = 3 units/day
It will take:
45 / 3 = 15 days for C to complete the remaining work.
Q: A, B, and C can do a work in 10 days, 12 days, and 15 days respectively. They start the work together, but A left after working 3 days and C left 4 days before completion of the work. In how much time was the whole work completed?
Time Total Work Efficiency Work Done
A 10 days 60 units 6 units/day 6 * 3 = 18 units (worked for 3 days)
B 12 days 60 units 5 units/day 5 * x = 5x units (worked for all days)
C 15 days 60 units 4 units/day 4 * (x - 4) = 4x - 16 units (left 4 days before completion)
A worked for only 3 days, completing 18 units of work.
Assuming the total work took 'x' days:
B worked for all 'x' days, completing 5x units of work.
C worked for (x - 4) days, completing 4(x - 4) = 4x - 16 units of work.
Total work = 60 units.
Adding up the work done by A, B, and C:
18 + 5x + 4x - 16 = 60
2 + 9x = 60
9x = 58
x = 6(4/9)
Hence, the whole work was completed in 6(4/9) days.
Q: P takes 10 days to complete 40% of a task while Q takes 18 days to complete 75% of the same work. P and Q start a work together but P leaves the work after a few days, and Q continued to work alone for 3 days. If 63% of work has still left, how many days did P leave the work?
P 10 days
so in 10 days 40% = 2/5 of the work is done, which means
2 units of work in 10 days
1 unit of work in 5 days, so 5 units of work in 25 days
Q 18 days
so in 18 days 75% = 3/4 of the work is done, which means
3 units of work in 18 days
1 unit of work in 6 days, so 4 units of work in 24 days
Time Total Work Efficiency
P 25 days 600 24
Q 24 days 25
63% of 600 = 378
This means 600 - 378 = 222 units of work were done.
Q worked alone for 3 days = 25 * 3 = 75 units
Now, 222 - 75 = 147 units of work were done by both
147 / 49 (efficiencies of both) = 3
Hence, after 3 days, P left the work.
Q: A can complete a piece of work in 6 days and B in 7 days respectively. If they work for a day alternately, in how many days will the work be finished if A begins the work?
Concept:
Time T.W Efficiency
A 6 days 42 units 7 units/day
B 7 days 6 units/day
day-1 day-2 day-3 day-4 day-5 day-6 ...
A B A B A B ...
7 6 7 6 7 6
\ / \ / \ /
13 13 13
In two consecutive days from the beginning, 13 units of work are done.
13 units -> 2 days
Total work = 42 units; we don't have to exceed it.
The multiple of 13 which is under 42 is 13 * 3 = 39.
39 units -> 6 days
But our work is not completed.
42 - 39 = 3 units of work are left.
An important point: This remaining work will be assigned to the person who started the work, in this
case, A.
Shortcut: 6 whole days + remaining work / A's efficiency
= 6 (3/7) days
Now, how did this 3/7 come from?
A can do 7 units in one day.
So, for 1 unit, he will take 1/7 day.
Now, for 3 units, he will take 3 * 1/7 = 3/7 days.
So, the answer will be 6(3/7) days.
Q: A, B, and C do a work in 20 days, 25 days, and 30 days respectively. In how many days will the work be finished if they do it alternately, starting with A?
Time T.W Efficiency
A 20 days 300 units 15 units/day
B 25 days 12 units/day
C 30 days 10 units/day
In three consecutive days, the total work done = 15 + 12 + 10 = 37 units.
The largest multiple of 37 that is less than 300 is 296.
This means 296 units of work will be completed in 8 cycles of 3 days each, which is 8 * 3 = 24
days in total.
Now, the remaining work = 300 - 296 = 4 units.
This remaining 4 units of work will be assigned to A.
A will take 4/15 = 4/15 day to complete it.
Hence, the work will be finished in 24 + 4/15 days.
Q: A, B, and C can do a piece of work in 11 days, 20 days, and 55 days respectively. If A works continuously and B and C work alternatively with A, in how many days will the total work be finished?
Time T.W Efficiency
A 11 days 220 units 20 units/day
B 20 days 11 units/day
C 55 days 4 units/day
As it is not mentioned which out of B and C starts first, we take the one given first, which is B.
day-1 day-2 day-3 day-4 ...
A+B A+C A+B A+C ...
20+11 20+4 20+11 20+4 ...
31 24 31 24
\ / \ /
55 55
In two consecutive days from the beginning, 55 units of work are done.
So in two consecutive days from the beginning, 55 units of work are done.
Now, to find the total time, we look for the largest multiple of 55 that is less than or equal
to 220.
The largest multiple of 55 that is less than or equal to 220 is 220, which means 4 complete
cycles of A+B and A+C.
Each cycle takes 2 days, so the total time taken is 2 * 4 = 8 days.
Hence, the work will be finished in 8 days.
Q: A can do a piece of work in 30 days, B can do it in 45 days, and C can do the same work in 60 days. If on the first day A worked alone, on the second day A and B worked together, and on the third day A and C worked together, repeating this cycle, then in how many days can the total work be completed?
Time T.W Efficiency
A 30 days 180 units 6 units/day
B 45 days 4 units/day
C 60 days 3 units/day
day-1 day-2 day-3 day-4 day-5 day-6
A A+B A+C A A+B A+C
6 10 9 6 10 9
\ / \ /
25 25
In one cycle of 3 days, 25 units of work are completed.
The largest multiple of 25 that is less
than 180 is 175.
This means 25 units in 3 days will be completed 7 times in 21 days.
Now, the
remaining work is 180 - 175 = 5 units, which will be done by A.
So, it will take 5/6 days for A
to complete this remaining work.
Therefore, the total time to complete the work is 21 days + 5/6
days, which is approximately 21(5/6) days.
Q: If the working efficiency of A is 50% more than B. If B alone can do a piece of work in 36 days, in how many days can A alone finish the same work?
Method 1: Work and time ratio
We should know that Efficiency ∝ (since work is constant).
According to the question, if A's efficiency is 150 and B's efficiency is 100 (as A's efficiency is
50% more than B), the ratio of their efficiencies will be 150:100.
Simplifying the ratio of efficiencies, we get 3:2.
Since efficiency is inversely proportional to time, the ratio of time taken will be the inverse of
the efficiency ratio, which is 2:3.
If B takes 36 days to complete the work (3 units of time), then:
1 unit of time = 36 days / 3 = 12 days
So, 2 units of time = 2 * 12 days = 24 days
Therefore, A will take 24 days to complete the work.
Method 2: Common Sense Approach
A B
E -> 150 : 100
E -> 3 : 2
Now, B alone can complete the work in 36 days, and his efficiency is 2 units per day. So, in 36
days, he will complete:
2 units/day * 36 days = 72 units of work (total work).
Since A's efficiency is 3 units per day, the total work (72 units) will be completed by A in:
72 units / 3 units/day = 24 days.
Therefore, A will take 24 days to complete the work.
Q: A is thrice as good as B in work. A is able to finish a work in 24 days less than B. In how many days will they finish the whole work together?
A B
E -> 3 : 1
T -> 1 : 3
It's given that A works 24 days less, so the difference between A's work and B's work is 24
days.
Now, according to the ratio of time, the difference between them is 2 units.
This means 2 units of time = 24 days.
So, 1 unit of time = 12 days.
Now, we calculate the total work = A's efficiency (3) * Time taken to complete the work (12 days) =
36 units.
Combined efficiency = 4 units/day, so 36 / 4 = 9 days.
It will take 9 days to complete the work when they both work together.
Q: Sam and Ravi together can paint a house in 5 days. If Ravi is 20% less efficient than Sam and Saran who is 20% more efficient than Sam, then find the number of days taken by Ravi and Saran together to paint the house.
Ravi Sam Saran
E -> 80 100 120
E -> 4 5 6
Sam and Ravi together can work in 5 days.
So, the combined efficiency of Sam and Ravi = 9 units/day.
Total work = 9 * 5 = 45 units.
Now we have to find how many days Ravi and Saran together can finish the work.
Combined efficiency of Ravi and Saran = 10 units/day.
45 / 10 = 4.5 days.
So, it will take 4.5 days for Ravi and Saran to paint the house together.
Q: A and B can complete a work in 18 days. Whereas B and C together can complete the same work in 15 days. If C is 50% more efficient than A, then find the time taken by A alone to complete the whole work?
Time T Ratio E
A + B 18 days 6 5
B + C 15 days 5 6
A : C
1 : 2
A + B = 5 units/day
B + C = 6 units/day
Subtraction gives us:
A - C = 1 unit
Now, the difference between A's efficiency and C's efficiency is 50%.
So, 50% = 1 unit
100% = 2 units
So, A's efficiency = 2 units and C's efficiency = 3 units.
Therefore, B's efficiency will be 2 + B = 5 units/day
B = 3 units/day
Now, total work = 18 * 5 = 90 units
90 / 2 = 45 days will be taken by A alone.
Q: The efficiency of A is 200% more than the efficiency of C and the efficiency of D is 25% less than the efficiency of C. If A and B together can complete the work in 14 days and C and D together can complete the work in 36 days, in how many days can B alone complete the work?
T T Ratio E
A + B 14 days 7 18
C + D 36 days 18 7
A : C : D
300 : 100 : 75
12 : 4 : 3
A + B = 18 units/day
And we know A's efficiency = 12 units/day
12 + B = 18
B = 6 units/day
Now, total work = 18 * 14 = 252 units
B alone will do it in 252 / 6 = 42 days
Q: 15 men working 8 hours/day make a dam in 10 days. In how many days will 12 men working 10 hours/day make it?
Concept:
M1 * D1 * T1 = M2 * D2 * T2
15 * 10 * 8 = 12 * D2 * 10
15 * 80 = 12 * 10 * D2
1200 = 120 * D2
D2 = 10
So, it will take 10 days to complete the same work with 12 men working 10 hours a day.
Q: 24 laborers working 8 hours/day make a road in 15 days. 48 laborers working 6 hours/day in how many days will make the triple long road?
Concept:
3(M1 * D1 * H1) = M2 * D2 * H2
3(24 * 15 * 8) = 48 * D2 * 6
Calculate the total work:
3 * (24 * 15 * 8) = 3 * 2880 = 8640
Set up the equation:
8640 = 48 * D2 * 6
Solve for D2:
8640 = 288 * D2
D2 = 8640 / 288
D2 = 30
So, in 30 days, 48 laborers working 6 hours a day will finish the triple long road.
Q: 18 Women complete a work in 16 days, they started the work together & after x days, 6 women left the work. Remaining work is completed by 12 women in 12 days. Find the value of x.
Q: A contractor contracted to finish a work in 62 days. He employed 32 men to work. But 1/3rd of the work was done in 30 days. How many additional men would be employed so that the remaining work is finished on time?
32 (men) * 30 (days) = 960 units of work done by them
Total work = 960 * 3 (since 1/3 of the work is 960 units)
Total work = 2880 units
Remaining work = 2880 - 960 = 1920 units
Remaining days = 62 - 30 = 32 days
We need to find the additional men required to finish the remaining work in the remaining days
(32 + x) * 32 (days) = 1920 units
(32 + x) = 1920 / 32
(32 + x) = 60
x = 60 - 32
x = 28
So, 28 additional men will be required to finish the remaining work on time.
Q: 1 Man or 2 Boys or 3 Girls can do a piece of work in 88 days. In how many days can 1 man, 1 boy, and 1 girl together do the same work?
We got to know that the efficiencies of 1 man = 2 boys = 3 girls.
M : B : G
E -> ( 1/1 : 1/2 : 1/3 ) * 6
6 : 3 : 2
The reciprocal becomes the ratio here.
To remove the denominator, we multiply the ratio by the LCM (6).
Total work = 6 (units/day) * 88 (days) = 528 units
Now, this work has to be done by 1 man, 1 boy, and 1 girl.
Combined efficiency of all = 6 (man) + 3 (boy) + 2 (girl) = 11 units/day
So, it will take (528 units) / (11 units/day) = 48 days.
Reference