Question

Pinku takes six days more than those of Nishu to complete certain work. If they work together they finish it in four days. How many days would it take to complete the work if they work alone?

Answer 1

Hint: Here we use a unitary method to find the amount of work completed by each Pinku and Nishu separately. Assume the number of days to complete the work by Nishu as a variable and form an equation of number of days to complete work by Pinku. Use a unitary method to find the amount of work done by each in 1 day and make an equation with the sum of amount of work done by each in a day to the given number of days.
* Unitary method helps us to find the value of a single unit when we are given the value of multiple units by dividing the value of multiple units by the number of units.

Complete step-by-step answer:
Let us assume number of days taken by Nishu to complete the work as x
Then using unitary method, we can write
\[ \Rightarrow \]Work done by Nishu in 1 day is \[\dfrac{1}{x}\]
We know Pinku takes six days more than those of Nishu to complete certain work
Then number of days taken by Pinku to complete the work will be \[(x + 6)\]
Then using unitary method, we can write
\[ \Rightarrow \]Work done by Pinku in 1 day is \[\dfrac{1}{{x + 6}}\]
Now we are given Nishu and Pinku work together.
Work done by Nishu and Pinku together in one day will be the sum of work done by Nishu and work done by Pinku individually in one day i.e. \[\dfrac{1}{x} + \dfrac{1}{{x + 6}}\]
Take LCM to calculate the value
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{x + 6}} = \dfrac{{x + 6 + x}}{{x \times (x + 6)}}\]
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{x + 6}} = \dfrac{{2x + 6}}{{{x^2} + 6x}}\]
Since we are given that when they work together they complete the work in 4 days.
So the sum of work done will be equal to one-fourth of the work
\[ \Rightarrow \dfrac{{2x + 6}}{{{x^2} + 6x}} = \dfrac{1}{4}\]
Cross multiply the values
\[ \Rightarrow 4(2x + 6) = {x^2} + 6x\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow 8x + 24 = {x^2} + 6x\]
Shift all values to one side of the equation
\[ \Rightarrow {x^2} + 6x - 8x - 24 = 0\]
\[ \Rightarrow {x^2} - 2x - 24 = 0\]
We use factorization method to find the value of x
\[ \Rightarrow {x^2} + 4x - 6x - 24 = 0\]
Take ‘x’ common from first two terms and -6 common from last two terms
\[ \Rightarrow x(x + 4) - 6(x + 4) = 0\]
\[ \Rightarrow (x - 6)(x + 4) = 0\]
Equate both factors to zero
If \[x - 6 = 0\]
Then \[x = 6\]
If \[x + 4 = 0\]
Then \[x = - 4\]
Since x is the number of days so it cannot be negative.
So, number of days taken by Nishu to complete work is 6
Number of days taken by Pinku to complete work is \[6 + 6 = 12\]

\[\therefore \]Pinku takes 12 days and Nishu takes 6 days to complete the work individually.

Note: Alternative method:
Another method to find the roots of quadratic equation \[{x^2} - 2x - 24 = 0\]
General form of quadratic equation is \[a{x^2} + bx + c = 0\] and the roots are given by the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
On comparing the equation \[{x^2} - 2x - 24 = 0\]with general equation
\[a = 1,b = - 2,c = - 24\]
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Put the values
\[ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4 \times 1 \times ( - 24)} }}{{2 \times 1}}\]
Write multiplication of two negative signs as positive signs
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 96} }}{2}\]
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {100} }}{2}\]
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {{{10}^2}} }}{2}\]
Cancel square root by square power
\[ \Rightarrow x = \dfrac{{2 \pm 10}}{2}\]
Either \[x = \dfrac{{2 + 10}}{2}\] or \[x = \dfrac{{2 - 10}}{2}\]
Either \[x = \dfrac{{12}}{2}\] or \[x = \dfrac{{ - 8}}{2}\]
Cancel same factors from numerator and denominator
Either \[x = 6\] or \[x = - 4\]
Since x cannot be negative,
\[ \Rightarrow x = 6\]
So, number of days taken by Nishu to complete work is 6
Number of days taken by Pinku to complete work is \[6 + 6 = 12\]
\[\therefore \]Pinku takes 12 days and Nishu takes 6 days to complete the work individually.