# Jameson received four grades on his algebra tests, bringing his average to an 88 for that class. What grade would he have to have on his last test in order to bring his average up to 90?

Nov 16, 2016

Jameson needs to make $98$ on his fifth test to bring his average up to $90$.

#### Explanation:

Let's let each of the test grades be represented by

${x}_{1} , {x}_{2} , {x}_{3} , \ldots$

The average of the first four tests is

$\frac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}}{4} = 88$

Jameson would like the final average of the five tests to be $90$ therefore

$\frac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} + {x}_{5}}{5} = 90$

We notice that this average contains the sum of ${x}_{1}$ through ${x}_{4}$, just like the first average we did, except that it's not divided by 4. Lets solve for ${x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} + {x}_{5}$ in our first equation giving

${x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} + {x}_{5} = 88 \cdot 4 = 352$

We can now substitute this into the average of the five tests

$\frac{352 + {x}_{5}}{5} = 90$

Solving for ${x}_{5}$ we get

${x}_{5} = 98$

So Jameson needs to make $98$ on his fifth test to bring his average up to $90$.

Nov 16, 2016

Jameson needs 98 on the next test.

#### Explanation:

$\text{Mean" = "Total"/"Number of values}$

From this we can can see that $\text{Total = Mean" xx "Number}$

The mean of 4 tests is $88 \Rightarrow \text{Total} = 4 \times 88 = 352$

If the mean of 5 tests is $90 \Rightarrow \text{Total} = 45 \times 90 = 450$

The difference between the two totals is how much Jameson needs to score on the next test to average 90.

$450 - 352 = 98$