Question

Together, Juanita and Mark can make preparations to cater a dinner in 6 hours. Alone, Juanita can do the job 2 hours faster than Mark. Find the time that each person takes to prepare to cater a meal.

Answer 1

If it takes them 6 hours to do it together, then Juanita can do it in 2 hours, and Mark can do it in 4 hours (since 2 + 4 = 6).

Answer 2

Time taken by Juanita and mark to cater meals is 11.08 hours or 11 hours 4 minutes and 13.08 hours or 13 hours 4 minutes respectively.

Solution:

Given, Together, Juanita and Mark can make preparations to cater a dinner in 6 hours.

Alone, Juanita can do the job 2 hours faster than Mark.

We need to find the time that each person takes to prepare to cater a meal.

Let the time taken by Juanita be x hours.

Then time taken by mark becomes x + 2 hours.

Now, work done by Juanita in 1 hour =
`(1)/(x)` [by assuming total work done as 1 unit]

And, work done by mark in 1 hour =
`(1)/((x+2))`

Together, in one hour they can complete

`(1)/(x) + (1)/((x+2))` work, which equals to
`(1)/(6)`

[ as given that together they completed in 6 hours ]

`\begin{array}{l}{(1)/(x)+(1)/(x+2)=(1)/(6)} \\\\ {(x+2+x)/(x(x+2))=(1)/(6)} \\\\ {6(2 x+2)=x^(2)+2 x} \\\\ {12 x+12=x^(2)+2 x} \\\\ {x^(2)+2 x-12 x-12=0} \\\\ {x^(2)-10 x-12=0}\end{array}`

Now, by using quadratic formula,

`\begin{array}{l}{x=\frac{-(-10) \pm \sqrt{(-10)^(2)-4 * 1 *(-12)}}{2 * 1}} \\\\ {x=(10 \pm √(100+48))/(2)} \\\\ {x=(10+√(148))/(2), (10-√(148))/(2)} \\\\ {x=(10+2 √(37))/(2^)}, (10-2 √(37))/(2)}\end{array}`

`x=5+√(37), 5-√(37)`

We can neglect
`5-√(37)` as time cannot be negative.

`x = 5+√(37)=5+6.0827`

= 11.08 approximately

So, Juanita completes work alone in 11.08 hours or 11 hours 4 minutes approximately.

So, mark completes work in 11.08 + 2 = 13.08 hours or 13 hours 4 minutes approximately.

Hence, time taken by Juanita and mark to cater meals is 11.08 hours or 11 hours 4 minutes and 13.08 hours or 13 hours 4 minutes respectively.