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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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 Some Further Analysis Comment 13 of 13 |
(In reply to Puzzle Resolution by K Sengupta)

If the fundamental tenets corresponding to the given problem had qualified the four items as A, B, C, D; then, denoting by P_a, P_b, P_c and P_d as the respective prices (in dollars) of A, B, C, D we would have all the permutations of (1.20, 1.25, 1.50, 3.16) as valid solutions and the said solutions are provided hereunder as follows:

(P_a, P_b, P_c, P_d)

= (1.20, 1.25, 1.50, 3.16); (1.20, 1.25, 3.16, 1.50);
(1.20, 1.50, 1.25, 3.16); (1.20, 1.50, 3.16, 1.25);
(1.20, 3.16, 1.25, 1.50); (1.20, 3.16, 1.50, 1.25)
(1.25, 1.50, 3.16, 1.20); (1.25, 1.50, 1.20, 3.16);
(1.25, 3.16, 1.50, 1.20); (1.25, 3.16, 1.20, 1.50);
(1.25, 1.20, 1.20, 3.16); (1.25, 1.20, 3.16, 1.20);
(1.50, 3.16, 1.20, 1.25); (1.50, 3.16, 1.25, 1.20);
(1.50, 1.20, 3.16, 1.25); (1.50, 1.20, 1.25, 3.16);
(1.50, 1.25, 3.16, 1.20); (1.50, 1.25, 1.20, 3.16);
(3.16, 1.20, 1.25, 1.50); (3.16, 1.20, 1.50, 1.25);
(3.16, 1.25, 1.20, 1.50); (3.16, 1.25, 1.50, 1.20);
(3.16, 1.50, 1.20, 1.25); (3.16, 1.50, 1.25, 1.20)

Edited on June 25, 2007, 12:25 pm
 Posted by K Sengupta on 2007-06-25 12:24:26

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