Problem solving question help...

A students scores on the first three tests in a math course are 65,75 and 87. What range of scores on a fourth test will give the student an average that is less than 80 but no less than 70 for the four tests?

65+75+87+x/4 < 79.9, x<92.6

65+75+87+x/4 >(and equal to) 70, x>(and equal to) 53

So the range of scores would be 53 to 92.6? Is this correct and am I writing this down correctly?

Thank you very much

Re: Problem solving question help...

now for the conditions you have stated, the inequalities are:

$\displaystyle 80>\frac{65+75+87+x}{4}>70$ where x is the marks in the fourth subject.

$\displaystyle \implies 320>65+75+87+x $and $\displaystyle 65+75+87+x>70 \implies 93>x$ and $\displaystyle x>53$ (after solving the two inequalities).

therefore the range of marks to satisfy the stated condition are $\displaystyle (93,53)$, the open interval from 53 to 93,i.e every number possible between 53 and 93 but without 53 and 93 itself.

Re: Problem solving question help...

Hello, curt26!

Quote:

A student's scores on the first three tests in a math course are 65, 75 and 87.

What range of scores on a fourth test will give the student an average

that is less than 80 but no less than 70 for the four tests?

Let $\displaystyle x$ = his score on the fourth test.

His average will be: .$\displaystyle \frac{65 + 75 + 87 + x }{4} \:=\:\frac{x+230}{4}$

He wants his average to be *at** *__least__ 70 and less than 80.

Hence, we have: .$\displaystyle 70 \;\le\;\frac{x+{\color{red}227}}{4}\;<\;80$

Multiply by 4: .$\displaystyle 280 \;\le\;x+{\color{red}227} \;<\;320$

Subtract 227:. . . . $\displaystyle {\color{red}53} \;\le\;x\;<\;{\color{red}93}$

Re: Problem solving question help...

Except 65 + 75 + 87 = 227, not 230.

Re: Problem solving question help...

Thank you so much you guys!