Hard fraction problem

Question

The question is below:

Rebecca has a collection of 45 books. The collection consists of novels and textbooks written in either Chinese or English. $\frac{4}{5}$ of the novels are in English and $\frac{3}{4}$ of the textbooks are in English. The total number of books written in English is 35. How many of her books are textbooks written in English?

ANSWER: 15 books

And here are my initial thoughts:

Initially I thought that since $\frac{4}{5}$ of novels are in English and $\frac{3}{4}$ of textbooks are in English, and there is a total of 35 books written in English. So I thought that if I add $\frac{4}{5}$ and $\frac{3}{4}$ which is equal to $\frac{31}{20}$.This will represent 35 English books, but $\frac{31}{20}$ is improper fraction so it cannot represent 35 English books. Am I interpreting the question wrong? Help would be greatly appreciated.

Also this question should not require any algebra since it's only Year 8 mathematics.

Answer 1

Let $T$ be the number of textbooks and $N$ the number of novels.

Then we have $T+N=45$ and $\frac34T+\frac45N=35$.

Do you know how to solve this system for $T$?

Then the answer to the question is $\frac34T$.

Answer 2

Pure logic, no algebra:

Since fractions of a book doesn't make sense, the number of novels must be a multiple of 5 and the number of textbooks a multiple of 4. Also the total number of books 45 is a multiple of 5, so we see the number of textbooks must be multiple of 5 too giving it a multiple of 20. So that leaves only 0, 20 or 40 textbooks and you can check individually which is correct, or use a similar argument for the number of English books to narrow that down.

Answer 3

You can't add the fractions together like that because the fractions do not represent a number of books of each type, but rather a proportion of books with a certain characteristic (in your case, written in English), among the total number of books of each type (novels and textbooks).

To illustrate with an example, suppose I have a bowl of candies, $50$ of which are red and $16$ of which are yellow. If $4/5$ of the red candies are sour flavored and $3/4$ of the yellow candies are sour flavored, then would it make sense to add $4/5 + 3/4$ to get the total number of sour candies in the bowl? Of course not, because when we say $4/5$ of the $50$ red candies are sour, that means $(4/5)(50) = 40$ of the red candies are sour; similarly, $(3/4)(16) = 12$ of the yellow are sour, and now we can add $40 + 12 = 52$ because now we are adding whole numbers of candies that share the same characteristic together.

Therefore, the same principle applies here with the books. However, we don't know how many of the books are novels and how many are textbooks, so this is why the question is a bit more challenging. Indeed, it may be a bit much to ask to avoid using algebra entirely. One thing we could try is to observe that the number of novels is a multiple of $5$, and the number of textbooks is a multiple of $4$; otherwise, we can't take $4/5$ and $3/4$ of these to get an integer number of English-language books of each type.

Since there are $45$ books in total, we can compare the multiples of $5$ and $4$ less than $45$ and see if there is a combination can add up to $45$:

$$\{0, 5, 10, 15, 20, 25, 30, 35, 40, 45\} \\ \{0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44\}.$$

Notice that there is no way to pick a number from the multiple of $4$s that does not end in a $0$, because there is no corresponding number in the multiple of $5$s that will make a sum of $45$. So the only possible choices in the second list are $\{0, 20, 40\}$. This means there are only three possible choices: either there are:

1. $45$ novels and $0$ textbooks,
2. $25$ novels and $20$ textbooks, or
3. $5$ novels and $40$ textbooks.

In the first case, $4/5$ of $45$ novels is $36$ English novels, which is too many, so this won't work.

In the second case, $4/5$ of $25$ is $20$ English novels, and $3/4$ of $20$ is $15$ English textbooks, resulting in $20 + 15 = 35$ English language books. This gives $15$ English language textbooks as the solution.

In the third case, $4/5$ of $5$ is $4$ English novels, and $3/4$ of $40$ is $30$ English textbooks, and this adds up to $4 + 30 = 34$ English language books, which is one short, so this is not a solution.

No algebra used!