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February 28th, 2014, 07:52 AM   #1
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Boys and some Sweets puzzle

Some boys share a bag of sweets.

If each boy has 6 sweets, there will be 5 sweets left in the bag.
If there were 3 more sweets in the bag, each boy could have exactly 7 sweets.

How many sweets were in the bag?
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Could this be a simultaneous equation?
Also possible if you showed how you got your answer as this boggled me a bit.

Have fun
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February 28th, 2014, 08:06 AM   #2
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Re: Boys and some Sweets puzzle

Let x be the number of boys.
Let B be the number of sweets in the bag.
6x = B - 5
7x = B + 3
x = 8, B = 53.
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February 28th, 2014, 08:08 AM   #3
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Re: Boys and some Sweets puzzle

The best way to start off is to let b = number of boys and s = number of sweets. Then, we have:

s = 6b + 5 (if every boy has 6 sweets there are 5 left)

s + 3 = 7b (if there were 3 more, every boy could have 7)

So, yes, that generates two simultaneous equations. Can you solve those?

... I see greg's done it for you!
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February 28th, 2014, 08:25 AM   #4
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Re: Boys and some Sweets puzzle

Still don't see exactly how you got there though. From forming the equations "6x = b-5" and "7x = b+3" how did you end up with 8 boys and 53 sweets?
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February 28th, 2014, 08:32 AM   #5
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Re: Boys and some Sweets puzzle

The trick is to subtract one equation from the other:

(s + 3) - s = 7b - (6b + 5)

3 = b - 5

b = 8

Then put b into one of the equations:

s = 6b + 5 = 48 + 5 = 53
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February 28th, 2014, 10:48 AM   #6
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Re: Boys and some Sweets puzzle

I still don't quite get it. could you do like step by step?
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February 28th, 2014, 10:55 AM   #7
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Re: Boys and some Sweets puzzle

That is step by step, I'm sorry to say!
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February 28th, 2014, 11:00 AM   #8
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Re: Boys and some Sweets puzzle

It may be but it's confusing.
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March 1st, 2014, 05:30 AM   #9
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Re: Boys and some Sweets puzzle

Another way of doing it would be to;

5 + 3 = 8 boys
6 x 8 = 48 + 5 = 53 sweets.
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March 1st, 2014, 05:53 AM   #10
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Re: Boys and some Sweets puzzle

6x = B - 5
7x = B + 3
7x - 6x = (B + 3) - (B - 5)
x = 8.
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