9 replies to this topic

[CyberMew]

Tom had an average score of 87 after 5 tests. Assuming he can get 100 for the rest of his tests, how many more tries does he need in order to increase his average score to 93?


i would appreciate it if someone can explain the steps in detail... kind of mind boggling for me

[Waser Lave]

You're starting with 5 tests with an average of 87 so that's:

87
87
87
87
87

Then you just need to keep adding an extra 100 and averaging the new total until you're over 93.

87,87,87,87,87,100: Mean = 89.17
87,87,87,87,87,100,100: Mean = 90.7
87,87,87,87,87,100,100,100: Mean = 91.875
87,87,87,87,87,100,100,100,100: Mean = 92.78
87,87,87,87,87,100,100,100,100,100: Mean = 93.5

[alphadark]

The way the question is worded I'd prefer waser's method but cody's way is more efficient if there are more variables

[ShadowLink64]

Tom had an average score of 87 after 5 tests. Assuming he can get 100 for the rest of his tests, how many more tries does he need in order to increase his average score to 93?


i would appreciate it if someone can explain the steps in detail... kind of mind boggling for me

The average is in the form (A + B + C + D + ...) / N, where N equals the number of total tests.

After five tests, we know that he has an average of 87. This means that (A+B+C+D+E)/5 must equal 87. This allows us to find A+B+C+D+E by multiplying 87 by 5. Therefore: A+B+C+D+E = 435.

Now, assuming that subsequent tests (F, G, H, etc..) are all 100, we would like:

(A+B+C+D+E+ N(100))/(N + 5) = 93

where N is the number of ADDITIONAL tests. Notice that the denominator has N + 5 because we are factoring in the 5 original tests already, and in the numerator N(100) tells us how many additional 100% scores we are adding up.

Replacing A+B+C+D+E = 435:

(435 + N(100))/(N+5) = 93
435 + N(100) = 93(N+5)
435 + N(100) = 93N + 465
Move stuff around:
7N = 30
N = (30/7) = 4.28 tests, but we can't write a partial test, so we need to write at least 5 tests to get the min. average of 93%.

Try it out (5 tests):
(435 + 100 + 100 + 100 + 100 + 100)/(5+5) = 93.5%

But if we only write 4 tests:
(435 + 100 + 100 + 100 + 100)/(5+4) = 92.8%

Hopefully I've made the method painfully obvious.

[Waser Lave]

Mine is the way a normal person would do it.

[ShadowLink64]

Mine is the way a normal person would do it.

What if they had 20 more tests to write? I don't want to average 20 times.

But you're right, normal people don't write that many tests.

[Waser Lave]

What if they had 20 more tests to write? I don't want to average 20 times.

But you're right, normal people don't write that many tests.

Obviously you wouldn't do it my way with lots of tests to do but when you know it's going to be only a few I would go my way.

[Sweeney]

When the question is so easy that no one can help you without actually giving you the answer, you're probably a lazy fuck.

[WingswithBlade]

What if they had 20 more tests to write? I don't want to average 20 times.

But you're right, normal people don't write that many tests.

lol most tests dont even give that much space unless its on a scrap piece of paper xD

[ShadowLink64]

lol most tests dont even give that much space unless its on a scrap piece of paper xD

Well don't write all the steps then. I wrote that much to make it painfully obvious to CyberMew what to do, lol.