Question
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours ?
✓
Answer
Let, $n$ represent the number of workers building the wall and t represent the time required.
Since, the number of workers varies inversely with the time required to build the wall.
$
\begin{aligned}
& \therefore n \propto \frac{1}{\mathrm{t}} \\
& \therefore n=\mathrm{k} \times \frac{1}{\mathrm{t}}
\end{aligned}
$
where $\mathrm{k}$ is the constant of variation
$
\therefore \mathrm{n} \times \mathrm{t}=\mathrm{k........(i)}
$
15 workers can build a wall in 48 hours,
i.e., when $n=15, t=48$
$\therefore$ Substituting $n=15$ and $t=48$ in (i), we get
$
\begin{aligned}
& \mathrm{n} \times \mathrm{t}=\mathrm{k} \\
& \therefore 15 \times 48=\mathrm{k} \\
& \therefore \mathrm{k}=720
\end{aligned}
$
Substituting $k=720$ in (i), we get
$
\begin{aligned}
& n \times t=k \\
& \therefore n \times t=720 \text {...(ii) }
\end{aligned}
$
This is the equation of variation.
Now, we have to find number of workers required to do the same work in 30 hours.
i.e., when $t=30, \mathrm{n}=$ ?
$\therefore$ Substituting $\mathrm{t}=30$ in (ii), we get
$\mathrm{n} \times \mathrm{t}=720$
$\therefore \mathrm{n} \times 30=720$
$\therefore n=\frac{720}{30}$
$\therefore \mathrm{n}=24$
$\therefore 24$ workers will be required to build the wall in 30 hours.
Since, the number of workers varies inversely with the time required to build the wall.
$
\begin{aligned}
& \therefore n \propto \frac{1}{\mathrm{t}} \\
& \therefore n=\mathrm{k} \times \frac{1}{\mathrm{t}}
\end{aligned}
$
where $\mathrm{k}$ is the constant of variation
$
\therefore \mathrm{n} \times \mathrm{t}=\mathrm{k........(i)}
$
15 workers can build a wall in 48 hours,
i.e., when $n=15, t=48$
$\therefore$ Substituting $n=15$ and $t=48$ in (i), we get
$
\begin{aligned}
& \mathrm{n} \times \mathrm{t}=\mathrm{k} \\
& \therefore 15 \times 48=\mathrm{k} \\
& \therefore \mathrm{k}=720
\end{aligned}
$
Substituting $k=720$ in (i), we get
$
\begin{aligned}
& n \times t=k \\
& \therefore n \times t=720 \text {...(ii) }
\end{aligned}
$
This is the equation of variation.
Now, we have to find number of workers required to do the same work in 30 hours.
i.e., when $t=30, \mathrm{n}=$ ?
$\therefore$ Substituting $\mathrm{t}=30$ in (ii), we get
$\mathrm{n} \times \mathrm{t}=720$
$\therefore \mathrm{n} \times 30=720$
$\therefore n=\frac{720}{30}$
$\therefore \mathrm{n}=24$
$\therefore 24$ workers will be required to build the wall in 30 hours.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Construct a kite ABCD in which AB = 4cm, BC = 4.9cm and AC = 7.2cm.
The sides of a rectangle are in the ratio 2 : 3 and its perimeter is 20cm. Draw the rectangle.
In the following figure RISK and CLUE are parallelograms. Find the measure of x.
Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
$\text{x}=\frac{1}{2},\text{y}=\frac{2}{3},\text{z}=-\frac{1}{5}$
→$\text{x}=\frac{1}{2},\text{y}=\frac{2}{3},\text{z}=-\frac{1}{5}$
Solve the following equation and verify your answer:
$\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=2$
→$\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=2$
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
The marks obtained (out of 20) by 30 students of a class in a test are given below:
7, 10, 8, 16, 13, 14, 15, 11, 18, 11, 15, 10, 7, 14, 20, 19, 15, 16, 14, 20, 10, 11, 14, 17, 13, 12, 15, 14, 16, 17.
Prepare a frequency distribution table for the above data using class intervals of equal width in which one class interval in 3-8 (including 2 and excluding 8). From the frequency distribution table so obtained, draw a histogram.
7, 10, 8, 16, 13, 14, 15, 11, 18, 11, 15, 10, 7, 14, 20, 19, 15, 16, 14, 20, 10, 11, 14, 17, 13, 12, 15, 14, 16, 17.
Prepare a frequency distribution table for the above data using class intervals of equal width in which one class interval in 3-8 (including 2 and excluding 8). From the frequency distribution table so obtained, draw a histogram.
Draw a pie-diagram of the areas of continents of the world given in the following table:
|
Continents
|
Asia
|
U.S.S.R
|
Africa
|
Europe
|
Noth America
|
South America |
Australia
|
|
Area (in million sq.km)
|
26.9
|
20.5
|
30.3
|
4.9
|
24.3
|
17.9 |
8.5
|
It x and y vary inversely, fill in the following blanks:
|
x
|
16
|
32
|
8
|
128
|
|
y
|
4
|
...
|
...
|
0.25
|
In each pair of triangles in the following figures, parts bearing identical marks are congruent. State the test and correspondence of vertices by which triangles in each pair are congruent.
ii.
iii.
ii.
iii.
iv