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To find the total time taken by both working together, we use the LCM method.
Let the total units of work be the LCM of the individual days (12 and 15).
Total Work = LCM(12, 15) = 60 units.
Next, find the 1-day efficiency of each person:
A’s 1-day efficiency = 60 / 12 = 5 units/day.
B’s 1-day efficiency = 60 / 15 = 4 units/day.
Combined 1-day efficiency of A and B:
Combined Efficiency = 5 + 4 = 9 units/day.
Total time taken to complete 60 units together:
Time = Total Work / Combined Efficiency = 60 / 9 = 20 / 3 = 6 2/3 days.
Shortcut Formula: When A does a job in x days and B in y days, working together they take (xy) / (x + y) days.
Time = (12 * 15) / (12 + 15) = 180 / 27 = 20 / 3 = 6 2/3 days.
To find the time taken by B alone, we first find the total work using the LCM of the given days (8 and 12).
Total Work = LCM(8, 12) = 24 units.
Next, find the combined 1-day efficiency of A and B, and the 1-day efficiency of A alone:
(A + B)’s 1-day efficiency = 24 / 8 = 3 units/day.
A’s 1-day efficiency = 24 / 12 = 2 units/day.
Now, find B’s 1-day efficiency by subtracting A’s efficiency from the combined efficiency:
B’s 1-day efficiency = (A + B)’s efficiency – A’s efficiency = 3 – 2 = 1 unit/day.
Total time taken by B alone to complete 24 units:
Time = Total Work / B’s efficiency = 24 / 1 = 24 days.
Shortcut Formula: If A and B together take x days and A alone takes y days, then B alone takes (xy) / (y – x) days.
Time = (8 * 12) / (12 – 8) = 96 / 4 = 24 days.
To find the time taken by A, B, and C working together, we first find the total work using the LCM of the given combined days (12, 15, and 20).
Total Work = LCM(12, 15, 20) = 60 units.
Next, find the 1-day efficiencies of the pairs:
(A + B)’s 1-day efficiency = 60 / 12 = 5 units/day.
(B + C)’s 1-day efficiency = 60 / 15 = 4 units/day.
(C + A)’s 1-day efficiency = 60 / 20 = 3 units/day.
Adding all three equations gives twice the combined efficiency of A, B, and C:
2 * (A + B + C)’s efficiency = 5 + 4 + 3 = 12 units/day.
Therefore, the combined efficiency of A, B, and C together is:
(A + B + C)’s efficiency = 12 / 2 = 6 units/day.
Total time taken by all three to complete 60 units:
Time = Total Work / Combined efficiency = 60 / 6 = 10 days.
Shortcut Formula: If (A+B) takes x days, (B+C) takes y days, and (C+A) takes z days, then A, B, and C together take (2xyz) / (xy + yz + zx) days.
Time = (2 * 12 * 15 * 20) / ((12 * 15) + (15 * 20) + (20 * 12)) = 7200 / (180 + 300 + 240) = 7200 / 720 = 10 days.
Let the 1-day efficiency of B be 1 unit/day.
Since A is twice as efficient as B, A’s 1-day efficiency = 2 units/day.
Combined 1-day efficiency of A and B = 2 + 1 = 3 units/day.
Given that they finish the work together in 14 days, we can find the total work:
Total Work = Combined Efficiency * Total Days = 3 * 14 = 42 units.
Now, find the time taken by A alone to complete 42 units:
Time taken by A = Total Work / A’s efficiency = 42 / 2 = 21 days.
Shortcut Formula: If A is k times as efficient as B and they together take D days, then the time taken by A alone is D * (k + 1) / k days.
Time = 14 * (2 + 1) / 2 = 14 * 3 / 2 = 42 / 2 = 21 days.
Efficiency is inversely proportional to the time taken.
Since the ratio of efficiencies of A to B is 3 : 1, the ratio of time taken by A to B is 1 : 3.
Let the time taken by A be x days and the time taken by B be 3x days.
Given that A takes 60 days less than B:
3x – x = 60
2x = 60
x = 30 days.
So, A takes 30 days and B takes 3 * 30 = 90 days to complete the work individually.
Now, find the total work using the LCM of 30 and 90:
Total Work = LCM(30, 90) = 90 units.
A’s 1-day efficiency = 90 / 30 = 3 units/day.
B’s 1-day efficiency = 90 / 90 = 1 unit/day.
Combined efficiency of A and B = 3 + 1 = 4 units/day.
Total time taken together = Total Work / Combined efficiency = 90 / 4 = 45 / 2 = 22 1/2 days.
Shortcut Formula: If A is k times as efficient as B and takes D days less than B, the time taken together is given by (k * D) / (k^2 – 1) (when comparing individual time difference). Alternatively, using individual times x and y: (xy) / (x + y).
To solve problems involving alternate days, we first find the total work using the LCM of the individual days (10 and 15).
Total Work = LCM(10, 15) = 30 units.
Next, find the 1-day efficiency of each person:
A’s 1-day efficiency = 30 / 10 = 3 units/day.
B’s 1-day efficiency = 30 / 15 = 2 units/day.
Since they work on alternate days starting with A, a complete cycle consists of 2 days (A works on day 1, and B works on day 2).
Work done in 1 complete 2-day cycle = 3 + 2 = 5 units.
To complete 30 units of work, the number of cycles required is:
Number of cycles = 30 / 5 = 6 cycles.
Since each cycle takes 2 days, the total time taken is:
Total Days = 6 cycles * 2 days/cycle = 12 days.
Shortcut Approach: Calculate the work done per block of alternate turns. Multiply the number of blocks to get close to the total work, then count the remaining individual days sequentially. Here, 6 complete blocks finish the exact 30 units in 12 days without any remainder.
First, find the total work by taking the LCM of 20 and 30.
Total Work = LCM(20, 30) = 60 units.
Calculate the individual efficiencies:
A’s 1-day efficiency = 60 / 20 = 3 units/day.
B’s 1-day efficiency = 60 / 30 = 2 units/day.
Find the work done by both A and B together in 6 days:
Combined efficiency of A and B = 3 + 2 = 5 units/day.
Work done in 6 days = 5 * 6 = 30 units.
Calculate the remaining work:
Remaining Work = Total Work – Work done = 60 – 30 = 30 units.
Now, find the time taken by B alone to finish the remaining 30 units:
Time = Remaining Work / B’s efficiency = 30 / 2 = 15 days.
Shortcut Approach: Subtract the fraction of work completed by both in 6 days from the total work (1). Fraction done = 6 * (1/20 + 1/30) = 6 * (5/60) = 1/2. Remaining work = 1 – 1/2 = 1/2. Time for B to finish remaining work = (1/2) * 30 days = 15 days.
First, find the total work using the LCM of 12, 15, and 20.
Total Work = LCM(12, 15, 20) = 60 units.
Calculate the 1-day efficiency of each person:
A’s efficiency = 60 / 12 = 5 units/day.
B’s efficiency = 60 / 15 = 4 units/day.
C’s efficiency = 60 / 20 = 3 units/day.
Since A left 2 days before the completion, we can use the concept of adding hypothetical work: assume A did not leave and worked for those last 2 days.
Work done by A in 2 days = 2 * 5 = 10 units.
Add this work to the total work to treat it as if everyone worked until the very end:
Adjusted Total Work = 60 + 10 = 70 units.
Find the combined efficiency of all three (A, B, and C):
Combined Efficiency = 5 + 4 + 3 = 12 units/day.
Total time taken to complete the adjusted work:
Time = Adjusted Total Work / Combined Efficiency = 70 / 12 = 35 / 6 = 5 5/12 days.
Shortcut Approach: When someone leaves before completion, add their potential work for the remaining period to the total work and divide by the sum of all efficiencies. Formula: (Total Work + Work left by departing person) / (Sum of all efficiencies).
First, find the total work by taking the LCM of 15 and 10.
Total Work = LCM(15, 10) = 30 units.
Calculate the individual efficiencies:
A’s 1-day efficiency = 30 / 15 = 2 units/day.
B’s 1-day efficiency = 30 / 10 = 3 units/day.
Find the work done by both A and B together in 2 days before B leaves:
Combined efficiency of A and B = 2 + 3 = 5 units/day.
Work done in 2 days = 5 * 2 = 10 units.
Calculate the remaining work:
Remaining Work = Total Work – Work done = 30 – 10 = 20 units.
Now, find the time taken by A alone to finish the remaining 20 units:
Time = Remaining Work / A’s efficiency = 20 / 2 = 10 days.
Shortcut Approach: Calculate the fraction of work completed by B in 2 days, subtract it from 1 to find the remaining fraction, and multiply by A’s individual completion time.
Fraction by B = 2 / 10 = 1/5.
Remaining fraction = 1 – 1/5 = 4/5.
Time for A = (4/5) * 15 = 12 days total elapsed, meaning 12 – 2 = 10 days for A alone after B left.
Wages for completing a piece of work are distributed among workers in the ratio of their 1-day efficiencies (when they work for the same amount of time).
First, find the 1-day efficiencies of A and B:
A’s 1-day efficiency = 1 / 10
B’s 1-day efficiency = 1 / 15
Ratio of efficiencies of A to B = (1 / 10) : (1 / 15) = 3 : 2.
Total ratio parts = 3 + 2 = 5 parts.
Given total wages = Rs. 3000.
Value of 1 part = 3000 / 5 = Rs. 600.
A’s share = 3 parts * 600 = Rs. 1800.
Shortcut Formula: A’s share = Total Money * (B’s time) / (A’s time + B’s time) -> wait, efficiency ratio is inverse of time ratio. Time ratio is 10 : 15 = 2 : 3. Efficiency ratio is 3 : 2. A’s share = 3000 * (3 / (3 + 2)) = 3000 * (3 / 5) = Rs. 1800.
Given that 10 men equivalence equals 15 women, we first find the relation between their efficiencies:
10 * M = 15 * W, which simplifies to 2 * M = 3 * W (or 1 woman = 2/3 of a man).
Calculate the total work in terms of man-days:
Total Work = 10 men * 12 days = 120 man-days.
Next, convert the combined group of 4 men and 6 women into equivalent men.
Convert 6 women into men: 6 * (2 / 3) = 4 equivalent men.
Total workforce = 4 men + 4 equivalent men = 8 equivalent men.
Now, find the total time taken to complete 120 man-days of work:
Time = Total Work / Total Workforce = 120 / 8 = 15 days.
Shortcut Formula:
For “M1 men or W1 women = D days, find M2 men and W2 women”, the formula is D / ((M2 / M1) + (W2 / W1)).
Time = 12 / ((4 / 10) + (6 / 15)) = 12 / (0.4 + 0.4) = 12 / 0.8 = 15 days.
To solve problems involving men, days, and hours, we use the chain rule formula based on the constancy of total man-hours required for the same work.
Let M1, D1, H1 be the men, days, and daily working hours for the first group, and M2, D2, H2 for the second group.
The formula is (M1 * D1 * H1) / W1 = (M2 * D2 * H2) / W2. Since the amount of work (W1 and W2) is the same, it simplifies to:
M1 * D1 * H1 = M2 * D2 * H2
Substitute the given values into the equation:
12 * 15 * 8 = 20 * D2 * 6
1440 = 120 * D2
Solve for D2:
D2 = 1440 / 120 = 12 days.
Shortcut Formula: Use the inverse proportionality principle for compound proportions: D2 = D1 * (M1 / M2) * (H1 / H2).
D2 = 15 * (12 / 20) * (8 / 6) = 15 * (3 / 5) * (4 / 3) = 15 * (12 / 15) = 12 days.
To solve problems involving inlet and outlet pipes, we treat filling rates as positive efficiencies and emptying rates as negative efficiencies.
First, find the total capacity of the tank using the LCM of 20, 30, and 60.
Total Capacity = LCM(20, 30, 60) = 60 units.
Calculate the 1-minute work rate of each pipe:
Pipe A (inlet) = 60 / 20 = +3 units/minute.
Pipe B (inlet) = 60 / 30 = +2 units/minute.
Pipe C (outlet) = 60 / 60 = -1 unit/minute.
When all three pipes are opened together, the net 1-minute work rate is:
Net Efficiency = +3 + 2 – 1 = +4 units/minute.
Total time taken to fill the 60-unit tank with the net efficiency:
Time = Total Capacity / Net Efficiency = 60 / 4 = 15 minutes.
Shortcut Formula: When pipe A fills in x minutes, pipe B fills in y minutes, and pipe C empties in z minutes, the net time taken when all are open is (xyz) / (yz + xz – xy).
Time = (20 * 30 * 60) / ((30 * 60) + (20 * 60) – (20 * 30)) = 36000 / (1800 + 1200 – 600) = 36000 / 2400 = 15 minutes.
To solve this, we first find the total capacity of the cistern using the LCM of the given hours (12, 15, and 20).
Total Capacity = LCM(12, 15, 20) = 60 units.
Calculate the 1-hour work rate of each component:
Pipe 1 = 60 / 12 = +5 units/hour.
Pipe 2 = 60 / 15 = +4 units/hour.
Combined filling rate of both pipes = 5 + 4 = +9 units/hour.
With the leak active, the net filling rate becomes:
Net Rate = 60 / 20 = +3 units/hour.
Now, find the rate of the leak by setting up the equation:
Combined Rate + Leak Rate = Net Rate
9 + Leak Rate = 3
Leak Rate = 3 – 9 = -6 units/hour (negative indicates emptying).
Total time taken by the leak alone to empty the full 60-unit cistern:
Time = Total Capacity / Leak Rate = 60 / 6 = 10 hours.
Shortcut Formula: If two pipes fill a tank in x and y hours respectively, and with a leak they take t hours, the time taken by the leak alone to empty the tank is
(t * LCM) / (Net work – Combined work) or derived directly via individual combined efficiencies: Time = Total Work / |(1/x + 1/y – 1/t)^(-1)|.
First, find the total capacity of the tank using the LCM of 15 and 20.
Total Capacity = LCM(15, 20) = 60 units.
Calculate the 1-minute work rate of each pipe:
Pipe A (inlet) = 60 / 15 = +4 units/minute.
Pipe B (outlet) = 60 / 20 = -3 units/minute.
Since they operate alternately for 1 minute each, a complete cycle consists of 2 minutes (A works in minute 1, B works in minute 2).
Net work done in 1 complete 2-minute cycle = (+4) + (-3) = +1 unit.
In alternate filling and emptying problems, the inlet pipe fills the remaining units in its final turn without the outlet pipe emptying it. Therefore, we subtract the inlet pipe’s capacity (4 units) from the total capacity to find the threshold before the final turn:
Threshold = Total Capacity – A’s 1-minute work = 60 – 4 = 56 units.
To accumulate 56 units at a rate of 1 unit per 2-minute cycle:
Cycles required = 56 / 1 = 56 cycles.
Time taken for 56 cycles = 56 cycles * 2 minutes/cycle = 112 minutes.
In the 113th minute, pipe A opens and fills the remaining 4 units (60 – 56 = 4 units) in 1 minute.
Total time = 112 + 1 = 113 minutes.
Shortcut Approach:
For alternate inlet-outlet pipe problems, calculate the net work per cycle.
Subtract the last filling pipe’s capacity from the total work, find the time for the cycles up to that threshold, and add the final filling time:
Total Time = (Cycles to reach Threshold * Cycle Duration) + Final Inlet Time.
First, let’s determine the efficiency of B. Let B’s 1-day efficiency be 100 units/day (or 5 units/day for clean division).
Since B can complete the work in 21 days, let’s assume B’s efficiency = 5 units/day.
Total Work = B’s efficiency * Total Days = 5 * 21 = 105 units.
Given that A is 40% more efficient than B:
A’s 1-day efficiency = 100% + 40% of B’s efficiency = 140% of 5 = 1.4 * 5 = 7 units/day.
Combined 1-day efficiency of A and B = A’s efficiency + B’s efficiency = 7 + 5 = 12 units/day.
Total time taken by A and B working together:
Time = Total Work / Combined Efficiency = 105 / 12 = 35 / 4 = 8 3/4 days.
Shortcut Formula: If B takes y days and A is p percent more efficient than B, working together they take (y * 100) / (200 + p) adjusted proportionally.
Specifically, using the efficiency ratio 100 : (100 + p), time together is (B’s Days * B’s Efficiency) / (A’s Efficiency + B’s Efficiency).
To solve problems involving mixed groups of workers (men and boys), we first find the efficiency ratio between a man and a boy.
Let the 1-day work of a man be M and of a boy be B.
According to the problem, the total work done by the first group is equal to the total work done by the second group:
(2M + 3B) * 10 = (3M + 2B) * 8
20M + 30B = 24M + 16B
30B – 16B = 24M – 20M
14B = 4M
2M = 7B, which gives the efficiency ratio M / B = 7 / 2 (meaning 1 man is equivalent to 3.5 boys).
Next, substitute M = 3.5B into either equation to find the total work in terms of boy-days:
Total Work = (2 * 3.5B + 3B) * 10 = (7B + 3B) * 10 = 10B * 10 = 100 boy-days.
Now, find the combined workforce of 2 men and 1 boy in terms of boys:
Workforce = 2 * (3.5B) + 1B = 7B + 1B = 8 equivalent boys.
Total time taken by 2 men and 1 boy:
Time = Total Work / Workforce = 100 / 8 = 25 / 2 = 12 1/2 days.
First, find the total work using the LCM of 20, 30, and 60.
Total Work = LCM(20, 30, 60) = 60 units.
Calculate the 1-day efficiency of each person:
A’s 1-day efficiency = 60 / 20 = 3 units/day.
B’s 1-day efficiency = 60 / 30 = 2 units/day.
C’s 1-day efficiency = 60 / 60 = 1 unit/day.
The work is done in a 2-day repeating cycle:
Day 1 (A + B working together) = 3 + 2 = 5 units.
Day 2 (A + C working together) = 3 + 1 = 4 units.
Total work done in 1 complete 2-day cycle = 5 + 4 = 9 units.
To find how many full cycles are needed for 60 units:
60 / 9 = 6 full cycles (with a remainder of work).
Work completed in 6 cycles = 6 * 9 = 54 units.
Time taken for 6 cycles = 6 * 2 = 12 days.
Remaining work = Total Work – Work completed = 60 – 54 = 6 units.
Now, the next cycle starts with Day 13 (A + B working together, whose combined efficiency is 5 units/day):
Work done on Day 13 = 5 units.
Remaining work after Day 13 = 6 – 5 = 1 unit.
On Day 14, A and C work together with a combined efficiency of 4 units/day. The time required to finish the remaining 1 unit is:
Time = 1 / 4 days.
Total time taken = 12 days + 1 day + 1/4 day = 13 1/4 days.
First, find the total work using the LCM of 10, 12, and 15.
Total Work = LCM(10, 12, 15) = 60 units.
Calculate the 1-day efficiency of each person:
A’s 1-day efficiency = 60 / 10 = 6 units/day.
B’s 1-day efficiency = 60 / 12 = 5 units/day.
C’s 1-day efficiency = 60 / 15 = 4 units/day.
Let T be the total number of days taken to complete the work.
A works for the first 2 days. Work done by A = 2 * 6 = 12 units.
B leaves 3 days before completion, meaning B works for (T – 3) days. Work done by B = 5 * (T – 3) units.
C works for the entire duration T. Work done by C = 4 * T units.
The sum of work done by all individuals equals the total work:
12 + 5 * (T – 3) + 4 * T = 60
12 + 5T – 15 + 4T = 60
9T – 3 = 60
9T = 63
T = 63 / 9 = 7 days.
Shortcut Formula: Set up the algebraic equation equating individual work contributions to the total work: (A’s work days * A’s efficiency) + (B’s work days * B’s efficiency) + (C’s work days * C’s efficiency) = Total Work.
First, find the total work using the LCM of 12, 15, and 20.
Total Work = LCM(12, 15, 20) = 60 units.
Calculate the 1-day efficiency of each pair:
(A + B)’s efficiency = 60 / 12 = 5 units/day.
(B + C)’s efficiency = 60 / 15 = 4 units/day.
(C + A)’s efficiency = 60 / 20 = 3 units/day.
Add all three equations to find twice the combined efficiency of A, B, and C:
2 * (A + B + C)’s efficiency = 5 + 4 + 3 = 12 units/day.
Therefore, the combined efficiency of all three is:
(A + B + C)’s efficiency = 12 / 2 = 6 units/day.
Calculate the work done by all three working together for 4 days:
Work done in 4 days = 6 units/day * 4 days = 24 units.
Calculate the remaining work after C leaves:
Remaining Work = Total Work – Work done = 60 – 24 = 36 units.
Now, find the time taken by A and B together to finish the remaining 36 units (since (A + B)’s efficiency is already known to be 5 units/day):
Time = Remaining Work / (A + B)’s efficiency = 36 / 5 = 7 1/5 days.
Shortcut Formula: Calculate individual combined group efficiencies, find the total group efficiency, subtract the work done during joint operation, and divide the remaining units by the target sub-group’s combined efficiency: Time = (Total Work – Joint Work Done) / Target Group Efficiency.
First, find the total capacity of the tank using the LCM of 10 and 15.
Total Capacity = LCM(10, 15) = 30 units.
Calculate the 1-hour filling rate of each pipe:
Pipe A = 30 / 10 = 3 units/hour.
Pipe B = 30 / 15 = 2 units/hour.
Both pipes work together for the first 2 hours:
Combined rate = 3 + 2 = 5 units/hour.
Work done in 2 hours = 5 * 2 = 10 units.
Calculate the remaining capacity of the tank:
Remaining Work = Total Capacity – Work done = 30 – 10 = 20 units.
After 2 hours, pipe A is closed, so pipe B finishes the remaining work alone at its rate of 2 units/hour:
Time taken by B to finish remaining work = 20 / 2 = 10 hours.
Total time taken to fill the tank = Time working together + Time working alone = 2 hours + 10 hours = 12 hours.
Shortcut Formula: Total Time = Time worked together + (Remaining Work / Rate of remaining active pipe), where Remaining Work = Total Capacity – (Combined Rate * Time together).
Let the filling rates of pipes A, B, and C be compared using a common ratio.
Given that A is twice as fast as B and three times as fast as C, let us assign convenient rates:
If pipe C’s rate = 2 units/minute, then pipe A’s rate = 3 * 2 = 6 units/minute.
Since pipe A is twice as fast as pipe B, pipe B’s rate = A’s rate / 2 = 6 / 2 = 3 units/minute.
Thus, the individual rates are: A = 6 units/min, B = 3 units/min, C = 2 units/min.
Given that pipe C can fill the tank in 55 minutes at a rate of 2 units/minute:
Total Capacity = C’s rate * Time taken by C = 2 * 55 = 110 units.
Now, find the combined rate when all three pipes are opened together:
Combined Rate = A + B + C = 6 + 3 + 2 = 11 units/minute.
Total time taken by all three pipes together to fill the 110-unit tank:
Time = Total Capacity / Combined Rate = 110 / 11 = 10 minutes.
Shortcut Formula: When comparing pipe speeds, establish the ratio of their unit rates, compute the total tank capacity using any individual’s rate and time, and divide by the sum of all rates: Time = Total Capacity / (Sum of all individual rates).
First, find the total capacity in units using the LCM of the given hours (8 and 12).
Total Capacity = LCM(8, 12) = 24 units.
Calculate the 1-hour rate of the leak and the net rate when both are open:
Leak’s rate = 24 / 8 = -3 units/hour (emptying).
Net rate (leak + inlet pipe) = 24 / 12 = -2 units/hour (emptying).
Now, determine the filling rate of the inlet pipe alone:
Net Rate = Leak Rate + Inlet Rate
-2 = -3 + Inlet Rate
Inlet Rate = -2 + (-3) = +1 unit/hour (filling).
This means the inlet pipe fills 1 unit of the tank per hour.
We are given that the inlet pipe fills water at 6 liters per minute. Convert this into liters per hour:
Inlet rate per hour = 6 liters * 60 minutes = 360 liters/hour.
Since 1 unit of our assumed capacity corresponds to 360 liters:
Total Capacity = Total Units * Value of 1 Unit = 24 * 360 = 8640 liters.
Shortcut Formula: Find the net rate difference to determine the inlet pipe’s efficiency unit value: Capacity = LCM * (Actual Flow Rate per hour) / (Difference in unit rates).
Here, unit difference between leak alone (-3) and net (-2) is 1 unit/hour, which equals 360 liters/hour.
Total Capacity = 24 * 360 = 8640 liters.
First, find the total capacity of the cistern using the LCM of 12, 15, and 20.
Total Capacity = LCM(12, 15, 20) = 60 units.
Calculate the 1-hour filling rate of each pipe:
Pipe A = 60 / 12 = 5 units/hour.
Pipe B = 60 / 15 = 4 units/hour.
Pipe C = 60 / 20 = 3 units/hour.
Now, break down the work sequentially based on the schedule:
1. Hour 1: Only pipe A is open.
Work done in the 1st hour = 5 * 1 = 5 units.
Remaining capacity = 60 – 5 = 55 units.
2. Hour 2: Pipe B joins pipe A (both A and B are open).
Combined rate of A and B = 5 + 4 = 9 units/hour.
Work done in the 2nd hour = 9 * 1 = 9 units.
Remaining capacity = 55 – 9 = 46 units.
3. From Hour 3 onwards: Pipe C also joins, so all three pipes (A, B, and C) are open together.
Combined rate of all three pipes = 5 + 4 + 3 = 12 units/hour.
Time required to fill the remaining 46 units at 12 units/hour = 46 / 12 = 23 / 6 = 3 5/6 hours.
Now, add the time periods together to get the total time:
Total Time = 1 hour (A alone) + 1 hour (A + B) + 3 5/6 hours (A + B + C) = 2 + 3 5/6 = 5 5/6 hours.
Shortcut Approach: Calculate the work done step-by-step for the initial individual/partial intervals, subtract from the total capacity, and divide the remaining units by the final combined rate: Total Time = Time intervals before full group + (Remaining Work / Final Combined Rate).
First, find the total capacity of the tank using the LCM of 12, 15, and 20.
Total Capacity = LCM(12, 15, 20) = 60 units.
Calculate the 1-hour filling rate of each pipe:
Pipe A = 60 / 12 = 5 units/hour.
Pipe B = 60 / 15 = 4 units/hour.
Pipe C = 60 / 20 = 3 units/hour.
Since pipe A remains open continuously while B and C operate alternately every hour, the operation follows a 2-hour repeating cycle:
1. Hour 1: Pipe A and Pipe B are open together.
Combined rate = 5 + 4 = 9 units/hour.
Work done in Hour 1 = 9 * 1 = 9 units.
2. Hour 2: Pipe A and Pipe C are open together.
Combined rate = 5 + 3 = 8 units/hour.
Work done in Hour 2 = 8 * 1 = 8 units.
Total work completed in 1 complete 2-hour cycle = 9 + 8 = 17 units.
Now, find how many full cycles are needed to approach 60 units:
60 / 17 = 3 full cycles (with a remainder).
Work completed in 3 cycles = 3 * 17 = 51 units.
Time taken for 3 cycles = 3 * 2 = 6 hours.
Calculate the remaining work:
Remaining Work = Total Capacity – Work completed = 60 – 51 = 9 units.
Now, the 4th cycle begins (Hour 7), where Pipe A and Pipe B are open again with a combined rate of 9 units/hour.
Time taken to finish the remaining 9 units = 9 units / 9 units/hour = 1 hour.
Total time taken = 6 hours + 1 hour = 7 hours.
This introduces an advanced concept in Pipes and Cisterns: flow rate is proportional to the cross-sectional area (square of the diameter or radius) of the pipe, meaning Rate is proportional to d^2.
1. Find the ratio of efficiencies based on their diameters:
– Diameter of Pipe A = 2 cm implies Efficiency is proportional to (2)^2 = 4 units/min.
– Diameter of Pipe B = 4 cm implies Efficiency is proportional to (4)^2 = 16 units/min.
– Simplifying the rates, let Pipe A’s rate = 1 unit/min, and Pipe B’s rate = 16 / 4 = 4 units/min.
2. Calculate the total capacity of the tank using Pipe A’s given time:
– Pipe A takes 40 minutes at a rate of 1 unit/min.
– Total Capacity = 1 * 40 = 40 units.
3. Find the combined rate when both pipes are opened together:
– Combined Rate = Rate of A + Rate of B = 1 + 4 = 5 units/min.
4. Calculate the total time taken together:
– Time = Total Capacity / Combined Rate = 40 / 5 = 8 minutes.
Shortcut Approach: For cross-sectional area flow problems, convert diameters to square-proportional rates (d^2), find the total capacity using one pipe’s given time, and divide by the sum of the squared diameter proportions.
This introduces a variable efficiency concept where a worker’s rate changes partway through the project.
1. Assume the total work units based on A’s original completion time:
– Let A’s original 1-day efficiency = 1 unit/day.
– Total Work = 1 unit/day * 20 days = 20 units.
2. Calculate the work done in the initial period before the efficiency change:
– A works at his original efficiency (1 unit/day) for the first 4 days.
– Work done in 4 days = 1 * 4 = 4 units.
– Remaining work = Total Work – Work done = 20 – 4 = 16 units.
3. Calculate the new efficiency after a 20% drop:
– Original efficiency = 1 unit/day.
– Drop = 20% of 1 = 0.2 units/day.
– New efficiency = 1 – 0.2 = 0.8 units/day (or 4/5 units/day).
4. Calculate the time taken to finish the remaining work at the new efficiency:
– Time for remaining work = Remaining Work / New Efficiency = 16 / 0.8 = 20 days.
5. Calculate the total time taken:
– Total Days = Initial Days + Days at New Efficiency = 4 + 20 = 24 days.
This combines mixed-group efficiency conversion with a multi-stage work schedule.
1. Find the individual efficiency ratio between a man and a woman:
– Total work by 12 men in 8 days = 12 * 8 = 96 man-days.
– Total work by 16 women in 12 days = 16 * 12 = 192 woman-days.
– Since total work is constant, equate them: 96 man-days = 192 woman-days, which simplifies to 1 man = 2 women.
– Let 1 man’s efficiency = 2 units/day and 1 woman’s efficiency = 1 unit/day.
2. Calculate the total work units:
– Total Work = 12 men * 8 days * 2 units/day = 192 units.
3. Calculate the work done in the first 4 days by 8 men and 8 women:
– Combined efficiency of (8 men + 8 women) = (8 * 2) + (8 * 1) = 16 + 8 = 24 units/day.
– Work done in 4 days = 24 units/day * 4 days = 96 units.
– Remaining work = 192 – 96 = 96 units.
4. Calculate the new workforce and time for the remaining work:
– After 4 days, all men leave, and 4 more women join the existing 8 women (8 + 4 = 12 women).
– Efficiency of 12 women = 12 * 1 = 12 units/day.
– Time taken to complete the remaining 96 units = 96 / 12 = 8 days.
Shortcut Approach: Convert all workers to a unified baseline unit (e.g., women-units), find total work, subtract initial multi-group output, and divide the remaining units by the newly adjusted group’s combined efficiency.
In time and work problems involving wages, earnings are always distributed in the direct ratio of the individual work efficiencies, provided all workers work for the same duration.
1. Find the efficiency ratio of A, B, and C:
– Let C’s efficiency = 1 unit/day.
– Since B is 3 times as efficient as C, B’s efficiency = 3 * 1 = 3 units/day.
– Since A is twice as efficient as B, A’s efficiency = 2 * 3 = 6 units/day.
– Therefore, the efficiency ratio of A : B : C = 6 : 3 : 1.
2. Calculate the total proportion of efficiency units:
– Total units = 6 + 3 + 1 = 10 units.
3. Calculate B’s share of the total earnings (Rs. 9600):
– B’s Share = (B’s efficiency ratio / Total ratio) * Total Earnings
– B’s Share = (3 / 10) * 9600 = 3 * 960 = Rs. 2880.
Shortcut Formula: When workers complete a job together over the same time period, wages are divided strictly according to their relative efficiency ratio: Individual Share = (Individual Efficiency / Sum of All Efficiencies) * Total Amount.
This introduces an advanced variant where workers operate on unequal duration repeating blocks rather than simple 1-to-1 alternate days.
1. Find the total work using the LCM of 10 and 15:
– Total Work = LCM(10, 15) = 30 units.
2. Calculate the 1-day efficiency of each person:
– A’s efficiency = 30 / 10 = 3 units/day.
– B’s efficiency = 30 / 15 = 2 units/day.
3. Define the repeating cycle structure (3 days total per cycle):
– Days 1 and 2: A works alone for 2 days implies Work done = 3 units/day * 2 days = 6 units.
– Day 3: B works alone for 1 day implies Work done = 2 units/day * 1 day = 2 units.
– Total work completed in 1 complete 3-day cycle = 6 + 2 = 8 units.
4. Calculate full cycles needed to approach the 30-unit target:
– 30 / 8 = 3 full cycles (with a remainder).
– Work completed in 3 cycles = 3 * 8 = 24 units.
– Time taken for 3 cycles = 3 cycles * 3 days/cycle = 9 days.
5. Calculate remaining work and finish the final steps:
– Remaining Work = 30 – 24 = 6 units.
– A new cycle starts with A working for the next 2 days. On day 10, A works and completes 3 units (remaining = 6 – 3 = 3 units). On day 11, A works and completes the final 3 units (remaining = 0).
– Total time = 9 days + 2 days = 11 days.
This is a classic advanced Time and Work problem involving relative time differences from the combined working period.
1. Let the total time taken by A and B working together be T days.
– According to the problem, A takes T + 4 days alone.
– B takes T + 9 days alone.
2. Set up the fractional work equation for 1 day:
– Rate of A + Rate of B = Combined Rate
– 1 / (T + 4) + 1 / (T + 9) = 1 / T
3. Solve the equation:
– Cross-multiply the left side: ((T + 9) + (T + 4)) / ((T + 4)(T + 9)) = 1 / T
– (2T + 13) / (T^2 + 13T + 36) = 1 / T
– T * (2T + 13) = T^2 + 13T + 36
– 2T^2 + 13T = T^2 + 13T + 36
– T^2 = 36
– T = 6 days.
Shortcut Formula: If worker A takes a extra days and worker B takes b extra days compared to their combined working time, the combined time T is always equal to the square root of the product of the extra days: T = sqrt(a * b).
Here, T = sqrt(4 * 9) = sqrt(36) = 6 days.
This introduces the work equivalence chain concept, where relationships between different workers’ time spans are converted into a unified efficiency ratio.
1. Set up the work equivalence equations based on the given conditions:
– 2 * A’s daily work = 3 * B’s daily work, which means A / B = 3 / 2.
– 3 * B’s daily work = 4 * C’s daily work, which means B / C = 4 / 3.
2. Combine the ratios to find the efficiency ratio of A : B : C:
– A : B = 3 : 2 (multiply by 2 to align B’s term) = 6 : 4.
– B : C = 4 : 3.
– Combined ratio A : B : C = 6 : 4 : 3.
3. Calculate the total work using C’s given completion time and efficiency:
– Let C’s efficiency = 3 units/day.
– Since C takes 39 days alone, Total Work = 3 units/day * 39 days = 117 units.
4. Find the combined efficiency and total time:
– Combined efficiency of A, B, and C = 6 + 4 + 3 = 13 units/day.
– Time = Total Work / Combined Efficiency = 117 / 13 = 9 days.
This introduces the waste pipe capacity matching concept, where a combined net filling time helps isolate an unknown emptying rate.
1. Find the total capacity of the tank using the LCM of 15, 20, and 10:
Total Capacity = LCM(15, 20, 10) = 60 units.
2. Calculate the individual rates:
Pipe A rate = 60 / 15 = 4 units/hour.
Pipe B rate = 60 / 20 = 3 units/hour.
Combined net rate of (A + B + C) = 60 / 10 = 6 units/hour.
3. Determine the rate of the waste pipe C:
Rate(A) + Rate(B) + Rate(C) = Net Rate
4 + 3 + Rate(C) = 6
7 + Rate(C) = 6
Rate(C) = 6 – 7 = -1 unit/hour (meaning pipe C empties 1 unit per hour).
4. Relate the unit rate to the actual volume:
We are given that waste pipe C empties 30 liters per hour. Therefore, 1 unit of our assumed capacity corresponds to 30 liters.
Total Capacity = 60 units * 30 liters/unit = 1800 liters.
1. Find the total work using the LCM of 8, 12, and 6.
Total Work = LCM of 8, 12, and 6 = 24 units.
2. Calculate the combined rates of the groups:
Combined rate of A and B equals 24 divided by 8, which is 3 units per day.
Combined rate of B and C equals 24 divided by 12, which is 2 units per day.
Combined rate of A, B, and C equals 24 divided by 6, which is 4 units per day.
3. Determine individual rates:
Rate of A equals the combined rate of all three workers minus the combined rate of B and C, giving 4 minus 2 equals 2 units per day.
Rate of C equals the combined rate of all three workers minus the combined rate of A and B, giving 4 minus 3 equals 1 unit per day.
4. Calculate the combined rate and time for A and C:
Combined rate of A and C equals 2 plus 1, which is 3 units per day.
Time equals Total Work divided by combined rate, giving 24 divided by 3 equals 8 days.
Shortcut Approach: Determine individual efficiencies by subtracting overlapping group rates from the total combined rate, then divide the total work by the sum of the target individuals rates.
First, find the total capacity of the cistern using the least common multiple of 20 and 30.
Total Capacity = 60 units.
Calculate the hourly filling rate of each pipe:
Pipe A rate equals 60 divided by 20, which is 3 units per hour.
Pipe B rate equals 60 divided by 30, which is 2 units per hour.
The pipes operate in a repeating two hour cycle:
Hour 1 involves pipe A filling 3 units.
Hour 2 involves pipe B filling 2 units.
Total work completed in one complete two hour cycle equals 3 plus 2, which is 5 units.
To find how many full cycles are needed for 60 units:
Divide 60 by 5, resulting in 12 full cycles with no remainder.
Total time equals 12 cycles multiplied by 2 hours per cycle, which gives 24 hours.
First, find the total capacity of the tank using the least common multiple of 10, 15, and 12.
Total Capacity = 60 units.
Calculate the hourly rate of each pipe:
Pipe A rate equals 60 divided by 10, which is 6 units per hour.
Pipe B rate equals 60 divided by 15, which is 4 units per hour.
Pipe C rate equals 60 divided by 12, which is negative 5 units per hour because it empties the tank.
When all three pipes operate together, the combined net rate equals 6 plus 4 minus 5, which is 5 units per hour.
Calculate the work done in the first 4 hours while all pipes are open:
Work done equals 5 units per hour multiplied by 4 hours, resulting in 20 units.
Calculate the remaining capacity of the tank:
Remaining work equals 60 minus 20, which is 40 units.
After 4 hours, pipe C is closed, so only pipes A and B continue working.
Combined rate of pipes A and B equals 6 plus 4, which is 10 units per hour.
Calculate the time required to finish the remaining 40 units:
Time equals 40 divided by 10, which is 4 hours.
Calculate the total time taken to fill the tank:
Total time equals 4 hours of joint operation plus 4 hours of remaining operation, giving a total of 8 hours.
Shortcut Approach: Calculate the initial net output of all active inlet and outlet pipes combined, subtract that work from the total capacity, and divide the remaining units by the new combined rate of the active pipes.
1. Use the man days proportionality principle: (Men1 multiplied by Days1) divided by Work1 equals (Men2 multiplied by Days2) divided by Work2.
2. First phase parameters:
– Initial men equals 40 men
– Days worked equals 30 days
– Work completed equals 1/2 of the total work
3. Second phase parameters:
– Total planned days equals 50 days. Remaining days equals 50 less 30, which is 20 days.
– Remaining work equals 1 less 1/2, which is 1/2 of the total work.
4. Calculate total man days required:
– 40 men working for 30 days complete 1/2 of the work, which equals 1200 man days for half the work.
– Total man days required for the entire work equals 1200 multiplied by 2, which is 2400 man days.
– Remaining man days needed equals 2400 less 1200, which is 1200 man days.
5. Calculate required workforce for the remaining 20 days:
– Required total men equals 1200 man days divided by 20 remaining days, which is 60 men.
– Additional men to be employed equals Total required men (60) less initial men (40), which gives 20 men.
This is a logical and conceptual problem examining staged operational changes where the active workforce increases midway.
1. Determine the filling rate of a single tap:
Since one tap fills the entire tank in 6 hours, its rate equals 1/6 of the tank per hour.
2. Calculate the time taken to fill the first half of the tank:
The first half of the tank (1/2 work) is filled by the first tap working alone.
Time for the first half equals (1/2) divided by (1/6), which equals 3 hours.
3. Calculate the combined rate when the second identical tap opens:
With two identical taps operating together, their combined rate equals 1/6 plus 1/6, which equals 2/6 or 1/3 of the tank per hour.
4. Calculate the time taken to fill the remaining half of the tank:
Remaining work equals 1/2 of the tank.
Time for the second half equals (1/2) divided by (1/3), which equals 3/2 or 1.5 hours.
5. Determine the total time taken:
Total time equals 3 hours plus 1.5 hours, giving a total of 4.5 hours.
1. Establish the efficiency ratio: Since A is twice as efficient as B, the ratio of efficiencies of A to B is 2 to 1.
2. Determine the time ratio: Because time is inversely proportional to efficiency, the ratio of time taken by A to B is 1 to 2.
3. Calculate individual time values using the given difference: The difference in time ratio parts equals 2 less 1, which is 1 part. Since 1 part equals 30 days, A takes 30 days alone, and B takes 2 multiplied by 30, which is 60 days alone.
4. Calculate the combined time: Using the formula for combined work, multiply 30 by 60 and divide by the sum of 30 and 60, resulting in 1800 divided by 90, which equals 20 days.
Logical Approach: Use the inverse proportionality between efficiency and time to find individual durations from the given time difference, then apply the standard product over sum formula for combined work.
1. Find the total capacity of the reservoir using the least common multiple of 4, 6, and 12, which is 12 units.
2. Calculate the hourly rate of each pipe:
– First pipe rate equals 12 divided by 4, which is 3 units per hour.
– Second pipe rate equals 12 divided by 6, which is 2 units per hour.
– Third pipe rate equals 12 divided by 12, which is 1 unit per hour.
3. Calculate the combined rate when all three pipes operate together:
– Combined rate equals 3 plus 2 plus 1, which is 6 units per hour.
4. Calculate the work done in 1 hour:
– Work completed equals 6 units per hour multiplied by 1 hour, resulting in 6 units.
5. Determine the remaining empty fraction:
– Remaining units equal total capacity of 12 less completed work of 6, leaving 6 units.
– Fraction remaining equals remaining units divided by total capacity, giving 6 divided by 12, which simplifies to 1/2.