Homework Question

[xud]

Hi all, I was helping my son with homework tonight and had a total brain freeze.
Here's the question:

Mina has 5 more marbles than Kirsty
Kirsty has 2 more marbles than Seb
Altogether they have 30 marbles.
How many does each child have?

I just can't figure out how to work it out.

[Mamyalynne]

Seb - 7
Kirsty - 9
Mina - 14

[Captain2]

Kirsty has X marbles
Mina has X + 5 marbles
Seb has X - 2 marbles, so

X + (X+5) + (X-2) = 30

that help?

[Prudie]

You can see Seb has the least, let that be X
So Kirsty has X + 2
So Mina has X + 2 + 5
Add them up x+x+2+x+2+5 to get 3x + 9
3x +9 = 30
So 3X = 21
x = 7 so Seb has 7, Kirsty has 9 and Mina 14

[xud]

Thanks a bunch!

[sherrardk]

I would add the 2 and 5, subtract from the 30 and divide by 3 then add the extra marbles for the Mina and Kirsty,

[Mamyalynne]

I didn't add working out because I can't, just do it in my head.

That's why my Maths teacher took early retirement.

[TheChair]

^ //That's why my Maths teacher took early retirement.//
Made me laugh.

[wildwood]

The obvious misconception is that Mina may have 5 more than Kersty, but she does in fact have 7 more than Seb, so 7+2=9.

Divide the remaining 21 by 3, so each has 7 plus the extras, as per Mamyalynne answer.

[gingejbee]

This is the type of question for which algebra was invented....and Prudie's using it properly.

[Captain2]

and errr, cough ;0))

[fiction-factory]

If you don't like the algebraic approach you can sometimes solve these iteratively in a few steps.
Start with Mina having half of the marbles- i.e. Mina has 15 , Kirsty 10 and Seb 8: the total is 33 which is 3 too high.
So reduce each by 1 to 14, 9 and 7.
That's 30 so is the solution

[Tilly2]

I have a headache now.

[gingejbee]

....well, yes, Captain....you are using it, but not in quite as "tidy" a way, imo.

[Captain2]

I must admit I've no idea how maths/algebra is taught these days, but we were always taught that if you have the method correct you could still achieve 8/10 even if you had the wrong answer; whereas if you took an educated guess or just wrote down the answer after doing it in your head you would only attain a 5/10 even with the correct answer.
Does that still apply today?

[fiction-factory]

Captain's also seems fine to me- he was just pointing xud in the right direction. Prudie's is perhaps a little easier to follow up in that the equation to solve contains only additions.

As for the working out, it's always better to include it as you can gain method marks (provided the question is worth more than one mark, and this would be worth 2 or 3) , but you would be given full marks for a correct answer even if you showed no working- unless the question says you must show your working and explain your solution.

[mibn2cweus]

M = K + 5
K = S + 2
therefore
M = S + 2 + 5
so then
M = S + 7
K = S + 2
and
M + K + S = 30

Solving for S
S = 30 - (M + K)
S = 30 - (S + 2 + S + 7) = 30 - 23 = 7
S = 30 - 2S - 9
3S = 30 - 9
S = (30 - 9) / 3
S = 21 / 3

S = 7
K = S + 2 = 9
M = S + 7 = 14

Plugging 7 into the value of S in the above equation
S = 30 - (S + 2 + S + 7) = 30 - 23 = 7
confirms our value for S

Interestingly, as it so happens it turns out that M = 2S . . . whatever?

[mibn2cweus]

M = K + 5
K = S + 2
therefore
M = S + 2 + 5
so then
M = S + 7
K = S + 2
and
M + K + S = 30

Solving for S
S = 30 - (M + K)
S = 30 - (S + 2 + S + 7)
S = 30 - 2S - 9
3S = 30 - 9
S = (30 - 9) / 3
S = 21 / 3

S = 7
K = S + 2 = 9
M = S + 7 = 14

Plugging 7 into the value of S in the above equation
S = 30 - (S + 2 + S + 7) = 30 - 23 = 7
confirms our value for S

Interestingly, as it so happens it turns out that M = 2S . . . whatever?

[Mamyalynne]

My way was quicker.

[murraymints]

Prud the best...

[murraymints]

no offence Mamy you right but prud explained xx