Summer Math Packet: How Many Books Does Ruby Buy?

Summer time means many of our students bring in their school assigned summer math packets. We provide help when needed, and sometimes we have a great workout ourselves. Try this middle school problem that had a student stumped, and briefly stumped us!

Ruby paid $36 for books  marked down $5 each from the original price. If she had bought them at the original price, she would have bought 5 fewer books. How many books did she buy?

The problem solving strategies suggested in the packet are:

Why don’t you try and solve it yourself first.

Let’s make an organized list in table form to systematically see how we can pay $36 for various combinations of prices and books. Assuming whole numbers (especially for books), let’s use the discounted price as the lowest possible price, and thus as the independent variable that drives the solution. We’re going to tabulate how we can pay $36 for both discounted books and full priced books; and check if the difference between the books bought is 5.

The last row answers that Ruby bought 9 books at the discounted price of $4. At the full price of $9, she only buys 4 books.

Did our instructors solve using a table? No, we harnessed the power of algebra. Our concern is to first derive the answer, then guide the student to a solution. We modeled the problem thusly:

Let n be the number of books bought at marked down price.
Let p be the original price. Then the marked down price must be (p – 5).
Hence Ruby pays:

 (1):  n(p – 5) = $36

At full price, Ruby buys 5 fewer books for the same amount, hence:

(2):  (n – 5)p = $36

The first and second equations can be equated:

n(p – 5) = (n – 5)p
np – 5n = np – 5p
n = p

So the algebra confirms that the number of books bought is also the value of the original price of each book as given by the prior solution. It makes visceral sense when looking at (1) and (2), with the wordy cruft of the word problem stripped away, that the problem  simultaneously models both side of the problem statement.  We left the hint of the partial solution that n = p to the student to complete. Why don’t you give it a try!

More interestingly, if we had modeled the problem after the table solution, then:

Let N be the number of books bought at discount.
Let P be the discount price, then the full price must be (P + 5).

(1b):  NP = 36

Then at full price:

(2b): (N – 5)(P + 5) = 36

Expand (2b) and substitute (1b)

NP + 5N – 5P – 25 = 36
36 + 5N – 5P – 25 = 36
5N – 5P = 25
N – P = 5

Humm… there is no simple algebraic suggestion about how to simply explain the relationship of N and P and we’d likely have given up finding an explanation and say, "solve it with the table!"

Now let’s use a guess and check approach to the first model. Since we now know that n = p, then rewriting equation (1) gives:

(3):   n(n – 5) = 36.

We can infer from the factor (n – 5) that n > 5.

Try the next whole number 6;  6(6 – 5) = 6 × 1 = 6, no too low!
Double up and try 12;  12(12 – 5) = 12 × 7 = 84, no too high!
Try midway between 6 and 12 = ^{(6 + 12)}/₂ = 9;  9(9 – 5) = 9 × 4 = 36, yes!

So it’s confirmed that the answer is Ruby bought 9 books at the discounted price of $4 each.

Let’s use the full power of algebra to elegantly arrive at the solution. Starting with (3), we expand using distribution, reorganize it into the quadratic equation, factor, and solve for its roots:

n² – 5n = 36
n² – 5n – 36 = 0
(n + 4)(n – 9) = 0

Giving the possible solutions of n = ^-4 and n = 9. Since only n = 9 makes sense, this absolutely confirms that Ruby buys 9 books and it’s indeed a whole number solution.

Is your child's summer math puzzling you? You still have time to bring them to Mathnasium for help!

Contact:

Ruby Yao and Benedict Zoe, Mathnasium of Fort Lee
201-969-6284 (WOW-MATH), [email protected]
246 Main St. #A
Fort Lee, NJ 07024

Happily serving communities of Cliffside Park, Edgewater, Fort Lee, Leonia, Palisades Park, North Bergen, West New York, and Fairview.