A certain number of men can complete a piece of work in 6k days, where k is a natural number.
By what percent should the number of men be increased so that the work can be completed in
5k days?
20%
In work and time problems, the total amount of work is usually considered constant. The key relationship is often between the number of workers and the time taken to complete the work. Assuming each worker does work at the same rate, the number of workers and the time taken are inversely proportional.
This means if you increase the number of workers, the time taken to complete the same amount of work will decrease, and vice versa. Mathematically, this can be represented as:
$$ \text{Number of Workers} \times \text{Time} = \text{Constant Work} $$
Let's apply this concept to the given problem involving a certain number of men completing a piece of work.
We are given two scenarios for completing the same piece of work:
Let's define the variables:
Since the amount of work is the same in both scenarios, we can use the relationship:
$$ M_1 \times T_1 = M_2 \times T_2 $$
Substitute the given values for the time:
$$ M_1 \times (6k) = M_2 \times (5k) $$
Assuming $k$ is a natural number, $k \ne 0$. We can cancel $k$ from both sides of the equation:
$$ 6 M_1 = 5 M_2 $$
Now, we can express $M_2$ in terms of $M_1$:
$$ M_2 = \frac{6}{5} M_1 $$
This equation tells us that the new number of men ($M_2$) is $\frac{6}{5}$ times the original number of men ($M_1$).
We need to find the percentage by which the number of men should be increased. The increase in the number of men is the difference between the final number of men and the initial number of men:
$$ \text{Increase in Men} = M_2 - M_1 $$
Substitute the expression for $M_2$ we found:
$$ \text{Increase in Men} = \frac{6}{5} M_1 - M_1 $$
To subtract $M_1$, we can write $M_1$ as $\frac{5}{5} M_1$:
$$ \text{Increase in Men} = \frac{6}{5} M_1 - \frac{5}{5} M_1 = \left(\frac{6}{5} - \frac{5}{5}\right) M_1 = \frac{1}{5} M_1 $$
The increase in the number of men is $\frac{1}{5}$ of the original number of men.
Now, to find the percentage increase, we use the formula:
$$ \text{Percentage Increase} = \frac{\text{Increase}}{\text{Original Number}} \times 100\% $$
Substitute the values:
$$ \text{Percentage Increase} = \frac{\frac{1}{5} M_1}{M_1} \times 100\% $$
Cancel out $M_1$ from the numerator and denominator:
$$ \text{Percentage Increase} = \frac{1}{5} \times 100\% $$
$$ \text{Percentage Increase} = 20\% $$
So, the number of men should be increased by 20% so that the work can be completed in $5k$ days instead of $6k$ days.
| Scenario | Number of Men | Time Taken |
|---|---|---|
| Initial | $M_1$ | $6k$ days |
| Final | $M_2 = \frac{6}{5} M_1$ | $5k$ days |
| Concept | Description | Formula |
|---|---|---|
| Total Work | Assumed constant for a specific task | Constant |
| Inverse Proportion | Number of workers and time taken are inversely related (if work rate is constant) | $M \times T = \text{Constant}$ |
| Percentage Change | Calculated as $(\text{Change} / \text{Original Value}) \times 100\%$ | $\frac{\text{Final} - \text{Initial}}{\text{Initial}} \times 100\%$ |
When solving work and time problems like this, we usually make certain assumptions unless stated otherwise:
Understanding these assumptions helps in correctly applying the inverse proportionality relationship between the number of workers and the time taken.
The compound interest on Rs. 75,000 for three years, at successive interest rates of 4%, 6% and 9% for each year respectively is:
The value of x in the following figures is :

When a child reaches adolescence, there is apt to be a conflict between the parents and the child, since
the latter considers himself to be by now quite capable of managing his own affairs, while the former
are filled with parental solicitude, which is often a disguise for love of power. Parents consider, usually,
that the various moral problems which arise in adolescence are peculiarly their province. The options
they express, however, are so dogmatic that the young seldom confide in them, and usually go their
own way in secret.