1200 men and 500 women can build a bridge in 2 weeks. 900 men and 250 women will take 3 weeks to build the same bridge. How many men will be needed to build the bridge in one week?
3600
This problem involves a classic concept in quantitative aptitude: work and time. To solve this, we need to determine the work rate of men and women relative to each other and then calculate the total work required to build the bridge. Finally, we can find out how many men are needed to complete this total work in one week.
Let's define the work rate of one man as 'M' units of work per week, and the work rate of one woman as 'W' units of work per week. The total work required to build the bridge remains constant.
The total work done is the product of the number of workers, their individual rates, and the time taken.
From Scenario 1:
Total Work $= (1200 \times \text{M} + 500 \times \text{W}) \times 2$ weeks
Total Work $= 2400 \text{M} + 1000 \text{W} \quad \text{(Equation 1)}$
From Scenario 2:
Total Work $= (900 \times \text{M} + 250 \times \text{W}) \times 3$ weeks
Total Work $= 2700 \text{M} + 750 \text{W} \quad \text{(Equation 2)}$
Since the total work is the same in both scenarios, we can equate Equation 1 and Equation 2:
$2400 \text{M} + 1000 \text{W} = 2700 \text{M} + 750 \text{W}$
Rearrange the terms to group M and W:
$1000 \text{W} - 750 \text{W} = 2700 \text{M} - 2400 \text{M}$
$250 \text{W} = 300 \text{M}$
Divide both sides by 50 to simplify:
$5 \text{W} = 6 \text{M}$
This means that the work done by 5 women is equivalent to the work done by 6 men. We can express the work rate of one woman in terms of a man's work rate:
$\text{W} = \frac{6}{5} \text{M}$
Now, substitute the value of 'W' from the relationship $W = \frac{6}{5} M$ into either Equation 1 or Equation 2 to find the total work in terms of men's work units. Let's use Equation 1:
Total Work $= 2400 \text{M} + 1000 \text{W}$
Total Work $= 2400 \text{M} + 1000 \left(\frac{6}{5} \text{M}\right)$
Total Work $= 2400 \text{M} + (200 \times 6) \text{M}$
Total Work $= 2400 \text{M} + 1200 \text{M}$
Total Work $= 3600 \text{M}$
This means the entire bridge building project requires 3600 "man-weeks" of work.
We need to find out how many men ('X') are required to build the bridge in 1 week. The total work done by X men in 1 week must be equal to the total work required for the bridge.
Work done by X men in 1 week $= \text{X} \times \text{M} \times 1$ week $= \text{X M}$
Equate this to the total work required:
$\text{X M} = 3600 \text{M}$
Divide both sides by M:
$\text{X} = 3600$
Therefore, 3600 men will be needed to build the bridge in one week.
| Scenario | Workers & Time | Total Work (Equation) |
|---|---|---|
| 1 | 1200 Men + 500 Women for 2 weeks | $2400M + 1000W$ |
| 2 | 900 Men + 250 Women for 3 weeks | $2700M + 750W$ |
| Key Relationship: $5W = 6M \implies W = \frac{6}{5}M$ | ||
| Total Work (in M units): $3600M$ | ||
The total work required is equivalent to 3600 units of a man's work for one week. To complete this in exactly one week, 3600 men would be needed.
The number of significant figures in $\frac{25 \times 0.524}{100.0} = 0.131$ is
The number of 3-digit numbers such that the digit 1 is never to the immediate right of 2 is
Students applying for hostel rooms are allotted rooms in order of seniority. Students already staying in a room will move if they get a room in their preferred list. Preferences of lower ranked applicants are ignored during allocation.
Given the data below, which room will Ajit stay in?
Names | Student seniority | Current | Room preference list |
Amar | 1 | P | R, S, Q |
Akbar | 2 | None | R, S |
Anthony | 3 | Q | P |
Ajit | 4 | S | Q, P, R |
There are 16 teachers who can teach Thermodynamics (TD), 11 who can teach Electrical Sciences (ES), and 5 who can teach both TD and Engineering Mechanics (EM). There are a total of 40 teachers. 6 cannot teach any of the three subjects, i.e. EM, ES or TD. 6 can teach only ES. 4 can teach all three subjects, i.e. EM, ES and TD. 4 can teach ES and TD. How many can teach both ES and EM but not TD?