If 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.
Last updated date: 18th Mar 2023
•
Total views: 207.4k
•
Views today: 5.86k
Answer
207.4k+ views
Hint: Assume a variable x which will represent the number of days a man will take to complete the work alone and assume a variable y which will represent the number of days a boy will take to complete the work alone. Using a unitary method, calculate the work done by man and the boy in 1 day. Use the information given in the question and convert each data for 1 day and form two equations which can be solved to get x and y.
Complete step-by-step answer:
In the question, it is given that 8 men and 12 boys can finish a piece of work in 10 days. Also, it is given that 6 men and 8 boys can finish it in 14 days. We are required to find the time taken by one man alone and that by one boy alone to finish the work.
Let us assume that 1 man can finish the work alone in x days and 1 boy can finish the work alone in y days.
Let us write the above assumed data in arrow form.
1 man $\to $ 1 work $\to $ x days
By unitary method, we can say,
1 man $\to $ $\dfrac{1}{x}$ work $\to $ 1 day
This means that 1 man can do $\dfrac{1}{x}$ work in 1 day.
So, 8 men will do $\dfrac{8}{x}$ work in 1 day. $..............\left( 1 \right)$
Also, 6 men will do $\dfrac{6}{x}$ work in 1 day ……………….$\left( 2 \right)$
Similarly, we can write,
1 boy $\to $ 1 work $\to $ y days
By unitary method, we can say,
1 boy $\to $ $\dfrac{1}{y}$ work $\to $ 1 day
This means that 1 boy can do $\dfrac{1}{y}$ work in 1 day.
So, 12 boys will do $\dfrac{12}{y}$ work in 1 day. $..............\left( 3 \right)$
Also, 8 boys will do $\dfrac{8}{y}$ work in 1 day ……………….$\left( 4 \right)$
Using $\left( 1 \right)$ and $\left( 3 \right)$, we can say that 8 men and 12 boys together can
do $\dfrac{8}{x}+\dfrac{12}{y}$ work in 1 day.
It is given in the question that 8 men and 12 boys can do 1 work in 10 days. So, we can say
that 8 men and 12 boys can do $\dfrac{1}{10}$ work in 1 day. From this, we can say,
$\dfrac{8}{x}+\dfrac{12}{y}=\dfrac{1}{10}..............\left( 5 \right)$
Using $\left( 2 \right)$ and $\left( 4 \right)$, we can say that 6 men and 8 boys together can do $\dfrac{6}{x}+\dfrac{8}{y}$ work in 1 day.
It is given in the question that 6 men and 8 boys can do 1 work in 14 days. So, we can say that 6 men and 8 boys can do $\dfrac{1}{14}$ work in 1 day. From this, we can say,
$\dfrac{6}{x}+\dfrac{8}{y}=\dfrac{1}{14}..............\left( 6 \right)$
Multiplying equation $\left( 5 \right)$ by 4 and equation $\left( 6 \right)$ by 6, we get,
$\dfrac{32}{x}+\dfrac{48}{y}=\dfrac{4}{10}$
$\dfrac{36}{x}+\dfrac{48}{y}=\dfrac{6}{14}$
Subtracting the above two equations, we get,
$\begin{align}
& \left( \dfrac{32}{x}+\dfrac{48}{y} \right)-\left( \dfrac{36}{x}+\dfrac{48}{y} \right)=\dfrac{4}{10}-\dfrac{6}{14} \\
& \Rightarrow \dfrac{-4}{x}=\dfrac{56-60}{140} \\
& \Rightarrow \dfrac{-4}{x}=\dfrac{-4}{140} \\
& \Rightarrow x=140 \\
\end{align}$
Substituting x=140 in equation $\left( 5 \right)$, we get,
\[\begin{align}
& \dfrac{8}{140}+\dfrac{12}{y}=\dfrac{1}{10} \\
& \Rightarrow \dfrac{12}{y}=\dfrac{1}{10}-\dfrac{8}{140} \\
& \Rightarrow \dfrac{12}{y}=\dfrac{14-8}{140} \\
& \Rightarrow \dfrac{12}{y}=\dfrac{6}{140} \\
& \Rightarrow y=280 \\
\end{align}\]
Hence, a man can complete the whole work alone in 140 days and a boy can complete the whole work alone in 280 days.
Note: There is a possibility that one may commit a mistake while using unitary method. To avoid this mistake, one should think logically while applying a unitary method. For example, if 1 man is completing the $\dfrac{1}{x}$ work in 1 day, then it can be noticed that the more will be the men, the more will be the work done. That is why we multiplied the work by 8 when we wanted to find the work done by 8 men in 1 day.
Complete step-by-step answer:
In the question, it is given that 8 men and 12 boys can finish a piece of work in 10 days. Also, it is given that 6 men and 8 boys can finish it in 14 days. We are required to find the time taken by one man alone and that by one boy alone to finish the work.
Let us assume that 1 man can finish the work alone in x days and 1 boy can finish the work alone in y days.
Let us write the above assumed data in arrow form.
1 man $\to $ 1 work $\to $ x days
By unitary method, we can say,
1 man $\to $ $\dfrac{1}{x}$ work $\to $ 1 day
This means that 1 man can do $\dfrac{1}{x}$ work in 1 day.
So, 8 men will do $\dfrac{8}{x}$ work in 1 day. $..............\left( 1 \right)$
Also, 6 men will do $\dfrac{6}{x}$ work in 1 day ……………….$\left( 2 \right)$
Similarly, we can write,
1 boy $\to $ 1 work $\to $ y days
By unitary method, we can say,
1 boy $\to $ $\dfrac{1}{y}$ work $\to $ 1 day
This means that 1 boy can do $\dfrac{1}{y}$ work in 1 day.
So, 12 boys will do $\dfrac{12}{y}$ work in 1 day. $..............\left( 3 \right)$
Also, 8 boys will do $\dfrac{8}{y}$ work in 1 day ……………….$\left( 4 \right)$
Using $\left( 1 \right)$ and $\left( 3 \right)$, we can say that 8 men and 12 boys together can
do $\dfrac{8}{x}+\dfrac{12}{y}$ work in 1 day.
It is given in the question that 8 men and 12 boys can do 1 work in 10 days. So, we can say
that 8 men and 12 boys can do $\dfrac{1}{10}$ work in 1 day. From this, we can say,
$\dfrac{8}{x}+\dfrac{12}{y}=\dfrac{1}{10}..............\left( 5 \right)$
Using $\left( 2 \right)$ and $\left( 4 \right)$, we can say that 6 men and 8 boys together can do $\dfrac{6}{x}+\dfrac{8}{y}$ work in 1 day.
It is given in the question that 6 men and 8 boys can do 1 work in 14 days. So, we can say that 6 men and 8 boys can do $\dfrac{1}{14}$ work in 1 day. From this, we can say,
$\dfrac{6}{x}+\dfrac{8}{y}=\dfrac{1}{14}..............\left( 6 \right)$
Multiplying equation $\left( 5 \right)$ by 4 and equation $\left( 6 \right)$ by 6, we get,
$\dfrac{32}{x}+\dfrac{48}{y}=\dfrac{4}{10}$
$\dfrac{36}{x}+\dfrac{48}{y}=\dfrac{6}{14}$
Subtracting the above two equations, we get,
$\begin{align}
& \left( \dfrac{32}{x}+\dfrac{48}{y} \right)-\left( \dfrac{36}{x}+\dfrac{48}{y} \right)=\dfrac{4}{10}-\dfrac{6}{14} \\
& \Rightarrow \dfrac{-4}{x}=\dfrac{56-60}{140} \\
& \Rightarrow \dfrac{-4}{x}=\dfrac{-4}{140} \\
& \Rightarrow x=140 \\
\end{align}$
Substituting x=140 in equation $\left( 5 \right)$, we get,
\[\begin{align}
& \dfrac{8}{140}+\dfrac{12}{y}=\dfrac{1}{10} \\
& \Rightarrow \dfrac{12}{y}=\dfrac{1}{10}-\dfrac{8}{140} \\
& \Rightarrow \dfrac{12}{y}=\dfrac{14-8}{140} \\
& \Rightarrow \dfrac{12}{y}=\dfrac{6}{140} \\
& \Rightarrow y=280 \\
\end{align}\]
Hence, a man can complete the whole work alone in 140 days and a boy can complete the whole work alone in 280 days.
Note: There is a possibility that one may commit a mistake while using unitary method. To avoid this mistake, one should think logically while applying a unitary method. For example, if 1 man is completing the $\dfrac{1}{x}$ work in 1 day, then it can be noticed that the more will be the men, the more will be the work done. That is why we multiplied the work by 8 when we wanted to find the work done by 8 men in 1 day.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
