# Problem solving question help...

• Apr 7th 2013, 09:12 PM
curt26
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
• Apr 7th 2013, 09:56 PM
earthboy
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.
• Apr 8th 2013, 04:24 PM
Soroban
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:. . . . $\displaystyle {\color{red}53} \;\le\;x\;<\;{\color{red}93}$

• Apr 8th 2013, 04:45 PM
Prove It
Re: Problem solving question help...
Except 65 + 75 + 87 = 227, not 230.
• Apr 8th 2013, 08:53 PM
curt26
Re: Problem solving question help...
Thank you so much you guys!