Answer
Verified
402k+ views
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.
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.
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE