Question

A test is divided into 4 sets of problems with the same number pf problems in each set. Alice correctly solves 35 problems. How many problems are on the test if Alice solved more than 60 percent of all the problems, but less than 65 percent of all problems? Give all possible answers.

Answer 1

54, 55, 56, 57, 58

Explanation:

Answer 2

56 problems

Explanation:

Set up an equation.

`(3)/(5)x<35<(13)/(20)x`

Why do we do this? We are told that she solved MORE than 60%, or
`(3)/(5)`, and LESS than 65%, or
`(13)/(20)`. Therefore, if we set the TOTAL number of problems to x, we have an equation we can solve.

`(3)/(5)x<35<(13)/(20)x\\`

Multiply all parts of the inequality by 20 to get rid of the denominators.

`20*(3)/(5)x<20*35<20*(13)/(20)x\\ \\12x<700<13x`

Now we can solve TWO individual inequalities to isolate the x variable.

`12x<700\\x<(700)/(12)\\x < 175/3\\x<58`

We can approximate 175/3 to about 58 (rounding down). We will sometimes round down when we have to deal with whole numbers.

The second inequality is as follows.

`13x>700\\x>700/13\\x>53`

Therefore, we can combine the two inequalities.

`53<x<58`

There were in between 53 and 58 questions. Since the number of questions must be a whole number, there can be 54, 55, 56, 57, OR 58. Why does 58 also work? When you plug 58 back into the original equation, you get that it STILL works. This is due to the fact that inaccuracies in computations allow you to round UP.

However, the last thing to keep in mind is that there are four sections with an equal number of questions. Meaning, the final answer has to be a multiple of four. The only multiple of 4 is 56; therefore, the final answer is 56.