Algebra 2 Word Problem

Question

A fruit stand salesman had an equal number of papayas and apples for sale. he packed the papayas equally into 10 boxes and the apples equally into 6 boxes. By midday he had sold 6 boxes of papayas and 3 boxes of apples. "We have sold 66 fruits" exclaimed his wife. How many papayas and how many apples were left?

Answer 1

First lets break up the problem.

 

" A fruit stand salesman had an equal number of papayas and apples"

 

We should use variables to represent the papayas and apples:

 

P = total amount of papayas the salesman starts with at beginning of day

A = total amount of apples the salesman starts with at the end of the day

 and 

 

P = A

 

Next, " he packed the papayas equally into 10 boxes and the apples equally into 6 boxes"

 

P/10 = p 

A/6 = a 

where 

p = amount of papayas packed into 1 box 

a = amount of apples packed into 1 box

 

By midday he had sold 6 boxes of papayas and 3 boxes of apples

6p = amount of papayas he had sold by then

3a = amount of apples he had sold by then 

 

" We have sold 66 fruits"

 

So if we add how much apples and papayas that have been sold by now, we get 66. So,

 

6p + 3a = 66 

 

Lets plug in p = P/10 and a = A/6 that we saw earlier

 

6( P/10) + 3 ( A/6) = 66

 

which simplifies to 

 

(3/5) P + (1/2) A = 66

 

We saw earlier that P = A so we can use that too.

 

(3/5) A + (1/2) A = 66 

 

(11/10) A = 66

 

So, A = 60.

 

Because P = A, P = 60.

 

So, at the beginning of the day he started with 60 papayas and 60 apples.

 

Using relations from earlier, we know that there were 6 papayas per box and 10 apples per box. 

He sold 6 boxes of papayas and 3 boxes of apples,

 

So he sold 36 papayas and 30 apples. 

 

60 - 36 = 24 papayas left

60 - 30 = 30 apples left

 

So, the salesman was left with 24 papayas and 30 apples by mid-day.