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## solution

Cookbook: \$20 with 15% discount = \$17
Mystery: \$15 with 20% discount = \$12
Poetry: \$16 with 25% discount = \$12
Biography \$30 with 40% discount = \$18
Architecture \$46 with 50% discount = \$23

Total before discounts: \$127
Total after discounts: \$82

## detailed solution

From the first clue, we know that one book was \$46 before applying a discount. But only one discount will yield a whole number discounted price:

\$46 x (15%) = \$6.90 discount, \$39.10 book price
\$46 x (20%) = \$9.20 discount, \$36.80 book price
\$46 x (25%) = \$11.50 discount, \$34.50 book price
\$46 x (40%) = \$18.40 discount, \$27.60 book price
\$46 x (50%) = \$23 discount, \$23 book price

This means that the 50% discount is used on the \$46 book, leaving only four discounts left.

We could try using this same technique on the \$20 book, the other piece of information given in the first clue. Unfortunately, every remaining discount can be used on that book:

\$20 x (15%) = \$3 discount, \$17 book price
\$20 x (20%) = \$4 discount, \$16 book price
\$20 x (25%) = \$5 discount, \$15 book price
\$20 x (40%) = \$8 discount, \$12 book price

The second clue gives information about the relationship between the prices of two books before and after the discounts. If one book (A) is \$1 more than another book (B), prior to the discounts, then these books satisfy the equation:

A = B + \$1

After the discounts are applied (D1 and D2), the books are the same price:

A (D1) = B (D2)

Think about what numbers we will actually use in place of D1 and D2. We will not use .5, .4, .25, .2 and .15, because these numbers will give us the amount of money we saved, or in other words, the amount of the price reduction. We expect the amount we save to be different on each book, because they have different prices before the discounts and the same price after the discounts. Thus equating the amount we saved is not correct, and that’s not what we mean by the equation A (D1) = B (D2). Instead we will use one minus the percentages, which results in .5, .6, .75, .8, and .85, as these numbers, multiplied by the original dollar amount of the books, will give the new price.

One thing to keep in mind is that, even though we have assigned the 50% discount to the \$46 book, there is nothing in this clue that precludes the same discount (and thus the same \$46 book) to also be referred to here.

At this point we can start trying different combinations of discounts and see what values of A and B result. But before we do this, we can eliminate some of the possibilities. First, since we said that book A cost more initially, D1 must be a more significant discount than D2. Second, the same discount cannot be applied to two books (each book purchased was from a different category). Therefore, D1 < D2 (remember that we are using one minus the percentages, so a deeper discount results in a lower number when subtracted from one).

Thus we are left with the following combinations of discounts, which are displayed in the (empty for the moment) chart below:

Going back to the equations A = B + \$1 and A (D1) = B (D2), in order to solve for A and B given the values of the discounts, let us use some algebra to combine the two equations with two unknowns into one equation with one unknown.

If A (D1) = B (D2), then A = B (D2 / D1). Since both equations are now of the form A = (some expression), both of those expressions must be equal:

B (D2 / D1) = B + \$1
B (D2 / D1) – B = \$1

Factoring out the B in the left side of this equation yields:

B ((D2 / D1) – 1) = \$1

Then divide by (D2 / D1) – 1 to get the B by itself:

B = \$1 / ((D2 / D1) – 1).

In the chart, M is used to represent the expression (D2 / D1) – 1. Now all we have to do is plug in the two discount percentages, and this will immediately tell us what B is. To find A, we can use either of the two initial equations.

Using this method, we can find all ten of the possible discount combinations. Discount combinations that yield similar results will be discussed collectively below.

First, some of the discount possibilities do not produce whole number prices for the books A and B. These cannot be the correct solution.

Second, some of the discount possibilities, while producing whole number prices for books A and B, fail to produce whole number prices when the discounts are applied.

Going from left to right, when A = 3 and B = 2, the discounted price of both books would be \$1.50. When A = 4 and B = 3, the discounted price of both books would be \$2.40. And when A = 17 and B = 16, the discounted price of both books would be \$13.60.

This leaves us with just three remaining options.

Again going from left to right, one of those options uses the 50% discount, which we already have assigned to the \$46 book. Unfortunately, it does not come up with either A or B being \$46, so that discount possibility is out.

The second option is if A = 5 and B = 4, and using the 25% and 40% discounts, we get a discounted price of \$3 for each book. This means we have four book prices, \$46 and \$20 from the first clue, and \$4 and \$5 from the second clue. Since we know that the total cost of the books before the discounts was \$127, the last book must cost \$127 – (\$40 + \$20 + \$4 + \$5) = \$52. So far, we have used the 50%, 40%, and 25% discounts, leaving the 20% and the 15% discounts. Unfortunately, you cannot apply either one of those discounts to \$52 and get a whole number result.

Thus the final option must be the solution. Let’s check to be sure. If A = 16 and B = 15, we use the 25% and 20% discounts to get a discounted price of \$12 for each book. Thus we have \$46, \$20, \$16, and \$15 for four books, which totals \$97. Since the total for all five books is \$127, then the last book must cost \$30. We’ve used the 50%, 25%, and 20% discounts, which leaves the 40% and 15% discounts. Although the 15% discount doesn’t work with \$30, the 40% discount does, giving us a discounted price of \$18. Thus the 15% must be applied to the \$20 book.