Question

‘A’ can do a piece of work in x hours and B can do the same work in (x+5) hours. Working together they can do the same work in (x+1) hours, find x.

PLEASE HELP ME ASAP I DO NOT KNOW HOW TO SOLVE THIS QUESTION

Answer 1

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✦ Required Answer:

♦️ GiveN:

• A can do a piece of work in x hours
• B can do a piece of work in (x+5) hours
• Working together, they can do in (x+1) hours.

♦️ To FinD:

• Find the value of x....?

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✦ Explanation of Concept:

The above question can be solved by using unitary method or the time- work relation like If P can do a piece of work in 2 days, then his 1 day work would be 1/2. By using these kind of methods, we can solve it easily.

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✦ Solution:

A can do a work in x hours

Then, A's 1 hour work = 1/x

B can do a work in (x+5) hours

Then, B's 1 hour work = 1/x+5

Working together, the amount of work they can do in 1 hour is,

$\large{ \rm{ \dashrightarrow \: \frac{1}{x} + \frac{1}{x + 5} }} \\ \\ \large{ \rm{ \dashrightarrow \: \frac{x + 5 + x}{x(x + 5)}}} \\ \\ \large{ \rm{ \dashrightarrow \: \frac{2x + 5}{ {x}^{2} + 5x} }}$

So, they can complete the work in,

$\Large{ \rm{ \dashrightarrow \: \frac{ {x}^{2} + 5x }{2x + 5} \: hours }}$

It is given that, They take (x+1) hours to complete the works, that means we can say that,

$\large{ \rm{ \dashrightarrow \: \frac{ {x}^{2} + 5x }{2x + 5} = x + 1}} \\ \\ \large{ \rm{ \dashrightarrow \: {x}^{2} + 5x = (x + 1)(2x + 5)}} \\ \\ \large{ \rm{ \dashrightarrow \: {x}^{2} + 5x = 2 {x}^{2} + 5x + 2x + 5}} \\ \\ \large{ \rm{ \dashrightarrow \: {x}^{2} + 5x = 2{x}^{2} + 7x + 5}} \\ \\ \large{ \rm{ \dashrightarrow \: {x}^{2} + 2x + 5 = 0}} \\ \\ \rm{ \red{By \: using \: sreedhar \: acharya \: fomula}} \\ \dag{\boxed{ \rm{ \: x = \frac{ - b \: \pm \sqrt{ {b}^{2} - 4ac } }{2a}}}} \\ \\ \large{ \rm{ \dashrightarrow \: x = \frac{ - 2 \pm \: \sqrt{ {2}^{2} - 4 \times 1 \times 5 } }{2} }} \\ \\ \large{ \rm{ \dashrightarrow \: x = \frac{ - 2 \pm \: \sqrt{ - 16} }{2}}} \\ \large{ \rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \downarrow \: \green{(imaginary \: roots)}}}$

This means that the above condition is not possible, as there is no real roots of x. Hence, the above question is invalid because the hours can't be imaginary practically. May be the valued provided is wrong, So this is the method of solving, Try to understand from this.

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Answer 2

Correct Question -

‘A’ can do a piece of work in x hours and B can do the same work in (x+5) hours.

Working together they can do the same work in (x+1) hours, find x.

Solution -

‘A’ can do a piece of work in x hours and B can do the same work in (x+5) hours.

In one hour , A can complete ( 1 / x ) th of the work .

In one hour , B can complete ( 1 / x + 5 )th of the work.

Now ,

If they both work together -

In one hour , work done by both of them -

=> ( 1 / x ) + ( 1 / x + 5 )

=> ( 2x + 5 ) / ( x^2 + 5x )

So ,

Work done in ( x + 1 ) hrs

=> ( 2x + 5 ) / ( x^2 + 5x ) × ( x + 1 ) = 1

=> ( 2x + 5 ) × ( x + 1 ) = x^2 + 5x

=> 2x^2 + 7x + 5 = x^2 + 5x

=> x^2 + 2x + 5 = 0

In this given Quadratic equation ,

Discriminant , D

=> b^2 - 4ac

=> 4 - 4 × 5 × 1

=> -16

Now , as D is negative , only complex roots exist .

This is not possible as x has to be a natural number .

There are some mistakes in the values given in the question ..

Try to understand the method to solve similar questions ...