SOLUTION: An ice-cream parlor has one size of ice-cream cone for 7 cents and another for a cents. One afternoon when the smaller cones outsold the larger ones two to one, the sale totaled 6

Question 1138693: An ice-cream parlor has one size of ice-cream cone for 7 cents and another for a cents. One afternoon when the smaller cones outsold the larger ones two to one, the sale totaled 6.21 pesos. How many cones of each size were sold?
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

"... a cents..."???

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After seeing the response from the other tutor, I see that perhaps you really do mean a cents each for the other cones, and we are supposed to find all solutions for which the total sales is 6.21 pesos = 621 cents.

So let x be the number of cones sold at a cents each; then the number sold at 7 cents each is 2x. Then if the total sales was 621 cents,

To find solutions in integers, find the prime factorization of the total sales number.

So (14+a) has to be in integer that is a divisor of 621; with this prime factorization of 621, there are two possibilities:
(14+a) = 23 --> a = 9; or
(14+a) = 3*3*3 = 27 --> a = 13

Solution #1: If (14+a) = 23, making a = 9, then x = 3*3*3 = 27. This gives us x=27 cones at 9 cents each and 2x=54 cones at 7 cents each.

CHECK: 27(9)+54(7) = 243+378 = 621

Solution #2: If (14+a) = 3*3*3 = 27, making a = 13, then x = 23. This gives us x=23 cones at 13 cents each and 2x=46 cones at 7 cents each.

CHECK: 23(13)+46(7) = 299+322 = 621

Answer by josgarithmetic(39625) (Show Source): You can put this solution on YOUR website!
Let the a cents per cone be the larger cone.

2x of the 7 cent cones and x of the a cent cones were sold.

You can look for any whole number values of x and a which work, but better if you are GIVEN a value for a.
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