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 7-11 (Posted on 2003-07-11)
A customer at a 7-11 store selected four items to buy, and was told that the cost was \$7.11. He was curious that the cost was the same as the store name, so he enquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four individual items. The customer protested that the four prices should have been ADDED, not MULTIPLIED. The clerk said that that was OK with him, but, the result was still the same: exactly \$7.11. What were the four prices? (Do NOT count sales tax.)

 See The Solution Submitted by luvya Rating: 3.6250 (8 votes)

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 Puzzle Resolution | Comment 12 of 13 |
(In reply to Answer by K Sengupta)

At the outset, we note that 711 is a positive integer multiple of 79 is a prime number. Accordingly, the cost of one of the items (in dollars) must be a positive integer multiple of 0.79. Let this cost be 0.79*n dollars and let the proice of the other three items be A, B and C ( ll in dollars).

Then, we have A+B+C = 7.11 - 0.79n and ABC = 9/n

Accodingly, we set up the following table:

n.......A+B+C .........ABC
1....... 6.32..........9.00
2....... 5.53..........4.50
3....... 4.74..........3.00
4....... 3.95..........2.25
5....... 3.16..........1.80
6....... 2.37..........1.50

Now, each of the other pairs with the exception of (3.95, 2.25)
(for n=4)would force one to create a round product from a non-round sum.

With a little trial and error, with (n, A+B+C, ABC) = (4, 3.95, 2.25), we observe that, taking A = 1.50 gives(B+C, BC) = (2.45, 1.50), so that: (B, C) = (1.25, 1.20)

The price of the fourth item for n=4 is 0.79n = 3.16

Consequently, the cost of the four items are \$ 1.20, \$ 1.25, \$ 1.50 and \$ 3.16.

 Posted by K Sengupta on 2007-06-19 05:25:24

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