1. ## Cost Price For Second Radio Set

A man sells a radio set for 34.50 and makes a profit of 15 percent. He sells a second radio set for 35.10 at a loss of 10 percent. If on the whole, he neither gains nor losses, find the cost price of the second radio set.

Let CP = cost price

Let SP = selling price

CP = (SP • 100)/(100 + given percent)

Is this the needed formula to solve this problem?

2. ## Re: Cost Price For Second Radio Set

I would let:

$\displaystyle C_1$ = the cost of the first set

$\displaystyle C_2$ = the cost of the second set

These costs are in dollars.

Now, we are told:

$\displaystyle 1.15C_1=34.50$

$\displaystyle 0.9C_2=35.10$

Given that he sold both sets, for the gain/loss described, and broke even, I would write:

$\displaystyle 1.15C_1+0.9C_2=C_1+C_2$

This implies:

$\displaystyle 0.15C_1-0.1C_2=0$

Or:

$\displaystyle C_1=\frac{2}{3}C_2$

Now, we also know:

$\displaystyle C_1+C_2=69.6$

Can you proceed?

3. ## Re: Cost Price For Second Radio Set

C_1 = (2/3)(C_2)

(2/3)(C_2) + C_2 = 69.6

(5/3)(C_2) = 69.6

C_2 = 69.6 divided by (5/3)

C_2 = 4.64 or 4 dollars and 64 cents.

4. ## Re: Cost Price For Second Radio Set

Your algebra is correct, but the arithmetic at the end isn't:

$\displaystyle C_2=69.6\cdot\frac{3}{5}=41.76$

5. ## Re: Cost Price For Second Radio Set

Originally Posted by harpazo
C_1 = (2/3)(C_2)

(2/3)(C_2) + C_2 = 69.6

(5/3)(C_2) = 69.6

C_2 = 69.6 divided by (5/3)

C_2 = 4.64 or 4 dollars and 64 cents.
If you take this answer back to the question and check the "reasonableness" of your answer, it doesn't make much sense. That is, you are saying that the cost price for radio 2 is 4.64 dollars, but he makes a loss of 10% when he sells it for 35.10 dollars. Clearly incorrect!! A quick check of "reasonableness" can help you fix up your own mistakes (particularly arithmetic ones).

Also, to get your "answer" of 4.64 you calculated 69.6 / 5 / 3 instead of 69.6 / (5/3). Brackets are very important here.

6. ## Re: Cost Price For Second Radio Set

Thank you. I made a typo, of course.