# Math Help - Math Fraction Help

1. ## Math Fraction Help

When Jane asked about her quiz average thus far. Dr Sullivan remembered that she intended for each quiz to have a value of 10 points. Her record of Janes scores appears in the table.

1 = 11/20
2 = 12/25
3 = 3/5

She wondered the best way to compute Jane's average. If you were Jane, how would you want Dr. Sullivan to compute the quiz average?

2. What a rubbish question!

Turn each fraction into a decimal (by doing the division), add them up and divide by 3 to get the mean.

Arrange them in order of size and take the middle one to get the median.

You can't get the mode because all numbers are different.

"If you were Jane, how would you want Dr. Sullivan to compute the quiz average?"

I would answer: "Accurately and honestly."

3. Originally Posted by Matt Westwood
What a rubbish question!

Turn each fraction into a decimal (by doing the division), add them up and divide by 3 to get the mean.

Arrange them in order of size and take the middle one to get the median.

You can't get the mode because all numbers are different.

"If you were Jane, how would you want Dr. Sullivan to compute the quiz average?"

I would answer: "Accurately and honestly."
That honestly kind of didnt help...

I did

11/20=.55
12/25=.48
3/5 =.60

.55+.48+.60 = 1.63

1.63/3 = .54

That still doesnt answer "If you were Jane, how would you want Dr. Sullivan to compute the quiz average" or what other way there is to get the value of 10 for each quiz

4. Originally Posted by WaynePD
When Jane asked about her quiz average thus far. Dr Sullivan remembered that she intended for each quiz to have a value of 10 points. Her record of Janes scores appears in the table.

1 = 11/20
2 = 12/25
3 = 3/5

She wondered the best way to compute Jane's average. If you were Jane, how would you want Dr. Sullivan to compute the quiz average?

You can get the average but they way you are doing it you only get the percentage. Each quiz is out of 10 points according to the problem. So...
Code:
  11/20 = .55
12/25 = .48
3/5 = .60
+____________
1.63  / 3  = .543 * 10 (points) = 5.43 average score on the quizes
You need to do this because the .54 is out of 1, just like 54% is 54 out of 100. You need the score which is out of 10.

I dont know what they are looking for on the how would you want them to average it part. It's not like a different method would get a different result without rounding. Unless you say you would rather him round to the tenths place after each calculation. But it would only get you another 3 points in the end so it's not like it would help.

5. ## my idea of the answer

Since the teacher intended to mark each quiz out of ten.You must change each quiz to be out of 10 marks. Looking at the total marks she gave for each quiz you can see that they are all factors of 100. So convert the fraction to fractions with denominators of 100 then to factions out of 10.
11/20 = 55/100 =5.5/10
12/25 = 48/100 =4.8/10
3/5 = 60/ 100 = 6/10
Jane marks out of 10 are:
5.5
4.8
6
To find the average add 5.5 +4.8 +6 then divide the answer by 3.
=16.3 ÷ 3
=5.4
Therefore the average = 5.4/10

6. Hello, WaynePD!

When Jane asked about her quiz average thus far. Dr. Sullivan remembered
that she intended for each quiz to have a value of 10 points.
Her record of Jane's scores appears in the table.

. . $\begin{array}{c|c}
\text{Quiz} & \text{Correct} \\ \hline
1 & \text{11 out of 20} \\
2 & \text{12 out of 25} \\
3 & \text{3 out of 5} \end{array}$

She wondered the best way to compute Jane's average.
If you were Jane, how would you want Dr. Sullivan to compute the quiz average?
These two statements imply that there is more than one way to compute Jane's average.

(A) The obvious method would be:

. . $\begin{array}{c|c|c}
\text{Quiz} & \text{Correct} & \text{Percent}\\ \hline
1 & \text{11 out of 20} & 55\%\\
2 & \text{12 out of 25} & 48\%\\
3 & \text{3 out of 5} & 60\%\end{array}$

$\text{Average} \:=\:\dfrac{55\% + 48\% + 60\%}{3} \;=\;\dfrac{163\%}{3} \;=\;54\frac{1}{3}\%$

(B) Another method would be:

. . $\begin{array}{c|c}
\text{Quiz} & \text{Correct} \\ \hline
1 & \text{11 out of 20} \\
2 & \text{12 out of 25} \\
3 & \text{3 out of 5} \\ \hline
\text{Total} & \text{26 out of 50} \end{array}$

$\text{Average} \;=\;\dfrac{26}{50} \;=\;52\%$

Given these two choices, Jane would prefer method (A).