Puzzles! #3

[B_S_A]

Three boys (A, B, and C) are talking about how many sweets they each have.

A: B has the most!

B: If C gave me one sweet, I'd have twice as many as A does.

C: It'd be better if B gave me two sweets. Then we'd all have the same amount!

How many sweets are there in total?

[Hidden]

Not a scooby doo! Talking myself round in circles on this one Ben!

[StaceM8]

9 B has 5

A has 3 & C has 1 ;o)

[fuzzyhead]

Are we supposed to derive this without trial and error? I know what the answer is, but I don't know how to prove it algebraically. Ok here goes.

C+2 = B-2 = A.

B-4 = C.

B+1 = 2A.

4-1 = 3, but does that logically prove that A=3 and therefore that C=1 and B=5? Or do I just think it does because I tried a few values and worked out which of them was right?

I'll get my coat...

[B_S_A]

9 is correct!

The difference between the number of sweets A and B have is two. If B had one more sweet, he would have twice as many as A. Therefore, A + 3 = 2A. which means A has three sweets. B must then have five, as he has one less than twice what A has. If C had two more sweets he would have the same as A, so he must have one.

[Danslatete in reply to B_S_A]

Read this over and over but I just don't get it. They're kids so the sweets won't last long anyhow as they'll be eaten!

[RogerCMerriman in reply to B_S_A]

I got lost attemping to get the boys sweets equal, which they could if they had 3 each. but then the maths doesn't seem to work.

[Danslatete]

Oh ek! Give me a few years to get my head around that one...

They put logic puzzles in the classrooms and change them each week. I can do primary 4 most days which is about age 8/9 , further up the school it takes me an age, other times I get them as soon as I look at them. Maybe it's things I've seen before but don't realise.

This type of puzzle has me completely baffled. I'm better at word ones.

[Kirk5w7]

With logic problems it helps too practice and then you get better, all about making those pathways in your brain. Xx Janet

[barny1]

The total number of sweets is 9: A+B+C = 9

Here's why....

B + 1 sweet = A x 2

C + 2 sweet = B - 2 sweet

And since they'll all have the same number of sweets then it holds true that:

B - 2 sweet = A

So we have 3 equations:

Eq1: B+1 = 2A

Eq2: C+2 = B-2

Eq3: B-2 = A

Both Eq1 and Eq3 have the same terms (2 unknowns, A and B). So subtract them to eliminate one term, subtracting each term individually.

Eq1 - Eq3: (B-B) + (1--2 = 1+2) = (2A-A)

We're left with: 3 = A

We can substitute 3 = A into Eq1 to get B....

B+1 = 2A

B+1 = 6

B = 5

Now we only need to find C....

Substitute B = 5 into Eq2...

C+2 = B-2

C+2 = 3

C = 1

So....

A + B + C = 3 + 5 + 1 = 9 sweets

Ta-dah!

youtube.com/watch?v=HtdgCFe...(Not working as expected?)

[Hidden in reply to barny1]

Lol Barney

I am confused even when you explain it - don't worry about explaining again I still won't get it!

Rain man is such a good film - so moving

[barny1 in reply to Hidden]

Just what part didn't you get?

See, all that maths they taught you in school really is useful!