Time and Work is an important concepts in math aptitude. In this article, we will explore essential formulas on time and work, and present a series of examples with solutions, starting from simple to more complex.
Formulas for Time and Work
1) The basic formula for time and work is as follows
Work = Rate × Time
i.e Work Done = Time Taken × Rate of Work
Here, “Rate” represents the work efficiency of an individual or a group
“Time” refers to the duration taken to complete the work.
“Work” represents the total amount of work or tasks completed.
Hence Time and work are always in direct proportion
2) Inverse Proportionality: When multiple individuals work together, their combined rate is inversely proportional to the time required to complete the task. The formula can be expressed as:
Work 1 × Time 1 = Work 2 × Time 2
Here Work 1 & Work 2 represent the amount of work done by different individuals or groups, Time 1 & Time 2 represent the time taken by different individuals or groups to complete their respective tasks.
3) Work Ratio: Suppose two individuals, P and Q, can complete a task in Time P & Time Q, respectively. The ratio of their work rates is inversely proportional to the ratio of the times taken. The formula can be expressed as:
Rate P: Rate Q = Time Q: Time P
Here Rate P & Rate Q represent the work rates of individuals P and Q, respectively. Time P & Time Q represent the time taken by individuals P and Q, respectively, to complete the task.
i.e If the number of people engaged to do a piece of work is increased (or decreased) in a certain ratio the time required to do the same work will be decreased (or increased) in the same ratio.
4) If a person can do a piece of work in “X” days, then that person can do “1/X” of the work in one day.
Total Work Done = Number of Days × Efficiency
Hence Time and Efficiency are inversely proportional to each other
Time and Work questions with solutions
Example-1: If A can do a piece of work in 10 days and B can do the same work in 15 days, then how much time will they take if they work together?
Solution:
A’s one day work = 1/10
B’s one day work = 1/15
(A+B)’s one day work = (1/10)+(1/15) = 1/6
So, A and B together can complete the work in 6 days.
Example-2: If A can do a piece of work in x days and B can do the same work in y days, then how much time will they take if they work together?
Solution:
A’s one day work = 1/x
B’s one day work = 1/y
(A+B)’s one day work = (1/x)+(1/y)
So, A and B together can complete the work in (xy)/(x+y) days.
Example-3: If A can do a piece of work in ‘X’ days and B is twice as efficient as A, then how much time will they take if they work together?
Solution:
A’s one day work = 1/x
B’s one day work = 2/x
(A+B)’s one day work = (1/x)+(2/x) = 3/x
So, A and B together can complete the work in x/3 days.
Example-4: A pump can fill a tank in 12 hours, and another pump can fill the same tank in 6 hours. If both pumps are used, but the second pump is turned off after 3 hours, how long will it take to fill the tank completely?
Solution: Let’s: R1 = Rate of the first pump (tank filled per hour) & R2 = Rate of the second pump (tank filled per hour)
As per the given data,
The first pump fills the tank in 12 hours, so its rate is R1 = 1/12 tank/hour.
The second pump fills the tank in 6 hours, so its rate is R2 = 1/6 tank/hour.
When both pumps are used for 3 hours, the portion of the tank filled by each pump is as follows:
First pump: 3 hours x R1 = 3 x ( 1/12) = 1/4 of the tank filled
Second pump: 3 hours x R2 = 3 x (1/6) = 1/2 of the tank filled
Now, let’s calculate the remaining portion of the tank that needs to be filled after 3 hours:
Remaining portion = 1 – [(1/4) +( 1/2)] = 1 – (3/4 ) = 1/4 of the tank.
Finally, we can find the time total time (T) it takes to fill the remaining 1/4 of the tank using only the first pump
T = (The remaining portion of the tank to be filled) / (Rate of the first pump) = (1/4) / (1/12) = (1/4) x (12/1) = 3 hours.
Therefore, it will take an additional 3 hours to fill the remaining 1/4 of the tank using only the first pump.
In total, the time to fill the tank completely is 3 hours (when both pumps are used for 3 hours) + 3 hours (using only the first pump) = 6 hours.
Example-5: A construction team of 6 members can build a bridge in 30 days. Due to the delayed start, they have only 20 days to complete the bridge. How many more members should they hire to finish the bridge on time?
Solution: Let the required number of additional workers be N.
According to the work ratio formula: Rate P : Rate Q = Time Q : Time P
6 workers : (6 + N) workers = 20 : 30
6 / (6 + N) = 20 / 30
6 * 30 = 20 * (6 + N)
180 = 120 + 20N
20N = 180 – 120
20N = 60
N = 60 / 20
N = 3
They need to hire 3 more workers.
Example-6: A machine produces 400 units in 8 hours. If the production rate decreases by 25% after 6 hours, how many units will be produced in the next 4 hours?
Solution:
Let us: R1 = Production rate per hour for the first 8 hours (units per hour) R2 = Production rate per hour for the next 4 hours (units per hour) Total units produced in 8 hours = 400 units
As per the given data, the Production rate for the first 6 hours is the same as the total production rate in 8 hours: 400 units / 8 hours = 50 units per hour.
Calculate the production rate for the next 4 hours (R2) after the decrease of 25%
Production rate decrease = 25% New production rate = 100% – 25% = 75% of the original rate
R2 = 75% of R1 , R2 = 75% of 50 units / hour
R2 = 0.75 x 50 units/ hour
R2 = 37.5 units/hour
Hence the total number of units produced in the next 4 hours using the new production rate is R2
Total units produced in 4 hours = R2 x 4 hours Total units produced in 4 hours = 37.5 units/hour x 4 hours= 150 units
Therefore, the machine will produce 150 units in the next 4 hours after the production rate decreases by 25% at the 6th hour.
Example-7: A factory produces 2000 toys in 5 days working 8 hours a day. How many hours a day should they work to produce 5000 toys in 10 days?
Solution:
Let ‘X’ be the number of hours they need to work per day to produce 5000 toys in 10 days.
The number of toys produced is directly proportional to the number of hours worked and the number of days worked. So, the proportion can be set up as follows:
Combined work rate formula:
2000 toys / (5 days x 8 hours) = 5000 toys / (10 days x H hours)
2000 / 40 = 5000 / (10H)
50 = 5000 / (10H)
10H = 5000 / 50
H = 5000 / (50 x 10)
H = 10 hours
Example -8: Three workers can build a wall in 6 days. How many days will it take for two workers to build the same wall if they work at the same rate?
Solution:
To solve this problem, you need to find the rate of work for each worker and then add them together to get the total rate of work. The formula is:
where A and B are the times for each worker to do the job alone, and T is the time for them to do it together.
In this case, we know that three workers can build a wall in 6 days, so the rate of work for each worker is:
We want to find how long it will take for two workers to build the same wall, so we need to solve for T in the formula:
So, it will take two workers 9 days to build the same wall if they work at the same rate.
Example- 9: A pump can fill a water tank in 20 hours. If the tank is already 1/4 full, how long will the pump take to fill the remaining 3/4 of the tank?
Solution:
Let’s first find out how much of the tank the pump needs to fill:
If the tank is already 1/4 full, then the remaining portion that needs to be filled is 1 – 1/4 = 3/4 of the tank.
Now, we know that the pump can fill the entire tank in 20 hours. Let’s calculate the rate at which the pump fills the tank.
Rate = Amount of work ÷ Time
The pump fills the tank in 20 hours, so its rate of filling is 1 tank ÷ 20 hours = 1/20 tank per hour.
Now, we need to find out how long the pump will take to fill 3/4 of the tank.
Time = Amount of work ÷ Rate
The amount of work is 3/4 of the tank, and the pump rate is 1/20 tank per hour.
Time = (3/4) tank ÷ (1/20 tank per hour)
Now, let’s calculate the time:
Time = (3/4) ÷ (1/20) = (3/4) * (20/1) = 60/4 = 15 hours
So, the pump will take 15 hours to fill the remaining 3/4 of the tank.
