## Slush Shack

Math brain teasers require computations to solve.You have a small shack in a parking lot, out of which you sell slushes for a few hours a day. You sell them in 4 sizes: small (12 oz), medium (20 oz), large (32 oz), and x-large (40 oz), and you charge $1.50, $2.25, $3.75, and $4.50 for each size, respectively. The cups cost you 3 cents for small, 5 cents for medium, 8 cents for large, and 10 cents for x-large, and the slush ingredients cost you 6 cents per ounce. Today, you are having a special: with every slush purchased, the customer receives a coupon for a free small slush. You figure that this special will increase sales for the day, but that the losses (due to giving out free slushes in the future) will cut the profit down to a much-below-average day. Your hope, though, is that it will help by drawing more regular customers in the long run. Your goal is to profit $20 even after the coupon losses. Through the day, medium was the best-selling size -- they sold exactly 4 times as many as smalls! -- followed by large, then x-large, and finally small. At the end of the day, the profit (not accounting for the coupons) from your sales was $46.50. Did you make the $20 net profit goal?

(This is strictly profit of sale over purchase, not involving paying employees, bills, etc. Also, assume that the customers who use the coupons would not have returned if they didn't have the coupon. Therefore, the loss from a coupon is the cost for you to make the slush, not the price you would normally receive for it.)

(This is strictly profit of sale over purchase, not involving paying employees, bills, etc. Also, assume that the customers who use the coupons would not have returned if they didn't have the coupon. Therefore, the loss from a coupon is the cost for you to make the slush, not the price you would normally receive for it.)

### Answer

(1): m=4s(2): m > l > x > s

(3): 75s + 100m + 175l + 200x = 4650, divided by 25 gives...

(4): 3s + 4m + 7l + 8x = 186

(1) subbed into (4) gives (5): 19s + 7l + 8x = 186

Can s=2? m would be 8, the maximum for l would be 7, and the maximum for x would be 6. For these maximum values, (5) says 135 = 186. The left side, which is at its maximum, is less than the right side, so s must be at least 3.

Can s=5? The least x can be is 6, and the minimum for l is 7. This minimum substituted into (5) gives 192 = 186. So 5 is too high to be s.

Now we know that s must be either 3 or 4. Let's set s equal to 3 for now, and see what happens.

Substituting into (5) gives: 7l + 8x = 129, solved for l gives (6): l = (129 - 8x) / 7

We know x is at least 4, because s is 3. So let's try different values for x, starting at 4, and substitute into (6).

If x=4, then l=13 6/7, which is not a whole number, proving impossible.

If x=5, then l=12 5/7, also not a whole number. As x increases by 1, the remainder decreases by 1. So we have to increase x by 5 to get to a whole number l-value.

If x=10, then l = 7. Now we have a whole number for l...however, x is greater than l, so (2) is now false. We can also see that if we increase x further, l will be decreasing, so (2) will only continue farther and farther from truth.

This proves that s cannot be 3, and is now narrowed down to s=4.

Substituting s=4 into (5) gives: 7l + 8x = 110, solved for l gives (7): l = (110 - 8x) / 7.

Now we again test x-values (minimum of x=5) for valid results: If x=5, l=10. Because l is a whole number greater than x, this is a valid solution! Increasing x by 7 to find the next possibility gives x=12, l=2. x > l, so any greater x-value is impossible.

The solution is s=4, m=16, x=5, l=10. This is a total of 35 slushes, and therefore 35 coupons given out. 35 * $0.75 = $26.25 in losses from the coupons. $46.50 - $26.25 = $20.25 net profit when accounting for the coupons. You made your goal!

Hide

Back to Top