Perhaps a dumb question, but how do you get \(\frac{c+h}{hc}\) from summing 1/h and 1/c

???

This is a basic algebra operation: lowest common divisor

1/h and 1/c= hc which divided to the first fraction become c in the numerator and for the second is h in the numerator

h+c/hc

Look at here https://www.purplemath.com/modules/fraction4.htm

Regards

it would take the first one 1/h and the second one 1/c .. together they add up to (c+h)/hc per unit. reverse that to get the amount of time : hc/(c+h). since c<h .. then hc < hh. Therefore, A is the right alternative.

Time taken to fill it together is (c*h)/(c+h). Since,h>c thus h^2>ch and so Quantity A > Quantity B.

This can be solved easily with Work = Rate x Time formula and plugging in some values

Given: one hose takes h mins and another hose takes c mins. Assume total work to be done is 1.
h > c so let's take simple values of 3 and 2 for h and c respectively

Quantity A simplifies to 9/5

Quantity B
The number of minutes required to fill the hot tub using the hot water hose and the cold water hose simultaneously

h does 1/3 of the work in a minute and c does 1/2 of the work in a minute
Adding to get the combined rate 1/3 + 1/2 = 5/6

W = RT
T = W/R
T = 1 / ( 5/6 ) = 6/5

Quantity A is 9/5 and Quantity B is 6/5. A is bigger.

What if h is less than 1? Doesn't that reverse the inequality and make the answer D?

We have two quantities as:

Quantity A: h^2 / h + c
Quantity B: hc / h + c

We can eliminate the denominator. So we have to compare h^2 to hc.

We know c < h.
Before multiplying both the sides with h, we need to know whether h is less than or greater than or equal to 1. If it's less than 1, the inequality flips to
ch < h^2
Otherwise it stays the same ch > h^2


Can someone clarify why A is unequivocally greater than B?

Since h and c denotes time, they cannot be negative. You don't say "it takes -5 minutes to fill a tub"

i dont understand. in the option B there is not written that for rate. then why wouldnot it be c+H

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