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?
Not a scooby doo! Talking myself round in circles on this one Ben!
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...
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.
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!
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.
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.
With logic problems it helps too practice and then you get better, all about making those pathways in your brain. Xx Janet
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!
I got bush flower cup development going on
Very conflicted/confused
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Another downside of Christmas
