Question

If 6 men and 8 boys can do a piece of work in 10 days, while 26 men and 48 boys can do the same work in 2 days, then what is the time taken by 15 men and 20 boys in doing the same type of work?
A) 3 days
B) 4 days
C) 5 days
D) 6 days

Answer 1

Hint: In this question it is given that if 6 men and 8 boys can do a piece of work in 10 days, while 26 men and 48 boys can do the same work in 2 days, then we have to find the time taken by 15 men and 20 boys in doing the same type of work. So to find the solution we have to consider that,
1 man's 1 day’s work = x and 1 boy's 1 day’s work = y
After finding the values of x and y we are able to find the time taken by 15 men and 20 boys in doing the same type of work, i.e the value of $$\dfrac{1}{15x+20y}$$.
Complete step-by-step solution:
 1man's 1 day’s work = x
$\therefore$ 6 mens 1 day’s work = 6x
Again,
 1 boy's 1 day’s work = y
$\therefore$ 8 boys 1 day’s work = 8y
So, it is given that 6 men and 8 boys 10 days work = 1,
Therefore, 6 men and 8 boys 1 day’s work = $$\dfrac{1}{10}$$
$$ \therefore 6x+8y=\dfrac{1}{10}$$.......(1)
Also 26 men and 48 boys can do the same work in 2 days.
So, similarly we can write,
$$\therefore 26x+48y=\dfrac{1}{2}$$
$$\Rightarrow 2(13x+24y)=\dfrac{1}{2}$$
$$\Rightarrow (13x+24y)=\dfrac{1}{4}$$......(2)
We have two linear equations, (1) and (2),
Now we are going to solve (1) and (2) by substitution method,
From equation (1),
$$8y=\dfrac{1}{10} -6x$$
$$\Rightarrow 3\times 8y=3\times \left( \dfrac{1}{10} -6x\right) $$
$$\Rightarrow 24y=\dfrac{3}{10} -18x$$
Now putting the value of 24y in equation (2), we get,
$$13x+24y=\dfrac{1}{4}$$
$$13x+\dfrac{3}{10} -18x=\dfrac{1}{4}$$
$$\Rightarrow 13x-18x=\dfrac{1}{4} -\dfrac{3}{10}$$
$$\Rightarrow -5x=\dfrac{5-6}{20}$$ [ since, LCM(4,10) = 20]
$$\Rightarrow -5x=-\dfrac{1}{20}$$
$$\Rightarrow x=\dfrac{1}{20\times 5}$$
$$\Rightarrow x=\dfrac{1}{100}$$
Now putting the value of ‘x’ in equation (1), we get,
$$6x+8y=\dfrac{1}{10}$$
$$\Rightarrow \dfrac{6}{100} +8y=\dfrac{1}{10}$$
$$\Rightarrow 8y=\dfrac{1}{10} -\dfrac{6}{100}$$
$$\Rightarrow 8y=\dfrac{1\times 10}{10\times 10} -\dfrac{6}{100}$$
$$\Rightarrow 8y=\dfrac{10}{100} -\dfrac{6}{100}$$
$$\Rightarrow 8y=\dfrac{10-6}{100}$$
$$\Rightarrow 8y=\dfrac{4}{100}$$
$$\Rightarrow y=\dfrac{4}{100\times 8}$$
$$\Rightarrow y=\dfrac{1}{200}$$

Therefore, time taken by 15 men and 20 boys in doing the same type of work
= $$\dfrac{1}{15x+20y}$$ days
= $$\dfrac{1}{\left( \dfrac{15}{100} +\dfrac{20}{200} \right) }$$ days
= $$\dfrac{1}{\left( \dfrac{30+20}{200} \right) }$$ days
= $$\dfrac{1}{\left( \dfrac{50}{200} \right) }$$ days
= $$\dfrac{200}{50}$$ days = 4 days.
Note: So to solve this type of question you need to know the relationship between the time, work and person, which are,
1) Work & Person: Directly proportional to each other, i.e. more work, more person required.
2) Time & Person: Inversely proportional to each other, i.e. more people, less time required.
3) Work & time: Directly proportional each other, i.e. more work, more time required.