A merchant could sell one model of digital cameras at list price and receive $224 for all of them. ..

Question

A merchant could sell one model of digital cameras at list price and receive $224 for all of them. If he had eight more cameras, he could sell each one for $14 less and still receive $224. Find the list price of each camera.

Answer 1

Let Price = P and Quantity = Q

PQ = 224

(P-14)(Q + 8) = 224

From the first equation, Q = 224/P

Then, (P - 14)(224/P + 8) = 224

224 - 112 + 8P - 14*224/P = 224

8P - 112 - 14*224/P = 0

Multiply through by P/8

P^2 - 14P - 392 = 0

(P + 14)(P - 28) = 0

P = 28 Q = 224/28 = 8

Note that (28 - 14)(8 + 8) = 14 * 16 = 224

Answer 2

Meg

Suppose the list price is l, and he initially has c cameras.

Then c l = 224

If he has 8 more cameras, he has c + 8, and if the new price is 14 less each one sells for l - 14

Income is (c + 8) ( l - 14)

So

c l = 224 (1)

(c + 8) (l - 14) = 224 (2)

We need to find l

Well expanding (2)

cl - 14c + 8l - 112 = 224

cl = 224

so 224 - 14c + 8l - 112 = 224

So 8 l - 14 c = 112

c = 224/l

8l - 14 (224/l) = 112

8l² - 3136 = 112 l

So 8l² - 112l - 3136 = 0

Divide by 8

l² - 14l - 392 = 0

Solve this for l, giving l = 28.

Check. 224 with a list price of 28, gives 224 / 28 = 8 sold

If we sell 8+8 = 16 at a list price of 28 - 14 = 14 we get 16 x 14 = 224. So it is correct.

You could also experiment with numbers here. Obviously the list price has to be greater than 14, so we must sell at most 224/14 - 1 = 15 units.

So you could see what we happen if he has 15, 14, 13, 12 ... units, working out the list price from the first sentence then seeing what happens if you add 8 cameras and take 14 off the list price.

MIKE