Thomas can process a pile of paperwork in 4 hours less time than Frank. Together they can process the pile in 2.1 hours. How long would it take Frank to do the job alone?

F = 7 hours

Step-by-step explanation:

Let the time taken by Frank to work alone be F.

Translating the word problem into an algebraic equation;

Thomas alone takes (F - 4) hours.

Combine time = 2.1 hours

Their individual work rate expressed in piles per hour are;

Frank = 1/F

Thomas = 1/(F - 4)

Combined rate = 1/F + 1/(F - 4)

Simplifying the equation, we have;

Combined rate = (F - 4 + F)/F(F - 4)

Combined rate = (2F - 4)/F(F - 4)

Combined rate = (2F - 4)/(F²- 4F)

Now to find the time taken when they work together is;

(2F - 4)/(F²- 4F) = 1/2.1

Cross-multiplying, we have;

2.1*(2F - 4) = F² - 4F

4.2F - 8.4 = F² - 4F

Rearranging the equation, we have;

F² - 4.2F - 4F + 8.4 = 0

F² - 8.2F + 8.4 = 0

Solving the quadratic equation by factorization;

Factors = -7 and -1.2

F² - 7F - 1.2F + 8.4 = 0

F(F - 7) - 1.2(F - 7) = 0

(F - 7)(F - 1.2) = 0

Therefore, F = 7 or 1.2 hours

The time taken by Frank alone would be 7 hours.

idk

step-by-step explanation: