Word Problem: Roger bought some pencils and erasers at the stationery store

Question

Roger bought some pencils and erasers at the stationery store. If he bought more pencils than erasers, and the total number of the pencils and erasers he bought is between 12 and 20 (inclusive), which of the following statements must be true?

Select all that apply.

a. Roger bought no fewer than 7 pencils.

b. Roger bought no more than 12 pencils.

c. Roger bought no fewer than 6 erasers.

d. Roger bought no more than 9 erasers.

To me, all of them are correct. Here is why:

With p = pencil and e = eraser.

a. p ≥ 7 ⇒ (p,e): (7,5), (8,4), (9,3), (10,2), (11,1), ect.

b. p ≤ 12 ⇒ (p,e): (12,8), (11,9)

c. e ≥ 6 ⇒ (p,e): (7,6), (8,7), (9,8), (10,9)

d. e ≤ 9 ⇒ (p,e): (10,9), (9,8), (8,7), (7,6), (7,5), ect.

However the correct answers according to the book are (a) and (d). Please help me understand why, thank you!

Answer 1

For a, if Roger had only bought $6$ pencils, he can have bought at most $5$ erasers, and he can no way make $12$ items, so he must have bought at least $7$ pencils.

Similarly for d, if Roger bought more than $9$ erasers, he must have bought at least $10$ pencils, and now he has too many items.

So a and d are absolutely true (this is the Pigeonhole principle btw).

b and c both have exceptions ($(11,0)$ is an exception to b, and $(11,10)$ is an exception to c), although they can possibly be true, as your answer suggests.

Answer 2

The solution $(p,e)=(13,7)$ is also viable, so $(b)$ is not correct as Roger could very well have bought $13$ pencils and not violated the statement given in the actual problem. Similarly $(p,e)=(15,5)$ is a viable solution, so $(c)$ is not correct, because Roger could very well have bought $5$ erasers.

Having a few examples to support your claim doesn't make that claim true, because there may still be examples that do not support your claim, and any one of the latter is enough to appear as a counterexample to your claim, hence rendering it false.