Lauren has two tests
test A is marked out of 40 and contributes to 40%of her total grade
test B is marked out of 60 and contributes to 60%of her total grade
if lauren get 32 out of 40 on test A
how many marks does lauren have to get on test B to acheive a 50% pass
Maybe I’ve oversimplified this, but it seems that the way you set it up each point = 1% of her grade. So, 50-32=18. She needs 18/60 on her other test.
Did I just do your homework?
What’s more fun would be if Test A was worth 60% and Test B was worth 40%, in which case each point on test A would be worth 3/2 of percentage point and test B would be worth 2/3 a percentage point. Under these conditions she’s just need 5/60 on the second test. (rounded from 4.5/60)
I think I did that math right.
Wait. Is this a trick question? Is Mark her tutor or prof or something?
Yes it’s 18, but what a bizzare and pointless question, the only actual maths you have to do is 50-32.
Yes, but nowadays you get marks for knowing the method and not just getting the answer right 
.
yes that’s been true for several decades, but there is very little mathematical method in the question
you guyz r all idiots. u have to use complex numbers, then ther are infinite ansers…i dont know how 2 do it, but its tru.
u take the eleventeenth root ov 47, tims it buy 9, invurt the #s and ad pie to infinity
u git 25,043 az the mane anser, but you haf to ad al the othur numburz dividable buy 1.6 for the otherz
The question never specifies which test she took first, so assuming she took test B first, then she needs a 50.
Why not just fail her?
If anybody actually cares about the generalized solution to these types of problems when they’re not so straight forward, it’s simply a matter of summing the products of your test scores and their weightings.
-Make sure all your weights add up to 100, then write:
(Test#1_Score/Test#1_Total)*Test#1_Weight + (Test#2_Score/Test#1_Total)*Test#1_Weight + … = Desired_Mark
You can add as many tests as you want as long as the weights add up to 100.
So for this question, you would write:
(32/40)*40 + (x/60)*60 = 50
Then solve for x.
That actually seriously helps me with a math quiz I forgot to study for. Thanks!
EDIT: How do you study for a math quiz, anyway?
Do questions until you understand the concepts…what’s actually going on in the questions.
Keep your calculator off. Seriously.
Interesting. The way I’ve generally seen it done is certainly simpler. Assuming your grade in a semester is “how many points you got”/“total possible points” in order to weigh a test higher you just make each problem worth 2 or 3 points and lower weighted assignment problems worth 1/2 or 1 point each.