Puzzles! #3: Three boys (A, B, and C) are talking... - Headway

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Puzzles! #3

B_S_A profile image
11 Replies

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?

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B_S_A
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11 Replies

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

StaceM8 profile image
StaceM8

9 B has 5

A has 3 & C has 1 ;o)

fuzzyhead profile image
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 profile image
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 profile image
Danslatete in reply toB_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 profile image
RogerCMerriman in reply toB_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 profile image
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 profile image
Kirk5w7

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

barny1 profile image
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...

in reply tobarny1

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 profile image
barny1 in reply to

Just what part didn't you get? ;)

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

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