My Math Forum Boys and some Sweets puzzle

 Algebra Pre-Algebra and Basic Algebra Math Forum

 March 1st, 2014, 06:33 AM #11 Newbie   Joined: Feb 2014 Posts: 12 Thanks: 0 Re: Boys and some Sweets puzzle That still looks kind of confusing. Always a different way to approach things though.
March 1st, 2014, 07:11 AM   #12
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Boys and some Sweets puzzle

Hello, Puzzled16!

I'll give it a try . . .

Quote:
 Some boys share a bag of sweets. (a) If each boy has 6 sweets, there will be 5 sweets left in the bag. (b) If there were 3 more sweets in the bag, each boy could have exactly 7 sweets. How many sweets were in the bag?

$\text{Let }N\text{ = number of sweets in the bag.}
\text{Let }b\text{ = number of boys.}$

$\text{From (a): }\:N \:=\:6b\,+\,5\;\;[1]
\text{From (b): }\:N\,+\,3 \:=\:7b \;\;\;\Rightarrow\;\;\;N \:=\:7b\,-\,3\;\;[2]$

$\text{Equate [2] and [1]: }\:7b\,-\,3 \:=\:6b\,+\,5 \;\;\;\Rightarrow\;\;\;b \:=\:8$

$\text{Therefore: }\a)\;N \:=\:6(\,+\,5 \:=\:53" />

 March 6th, 2014, 02:12 PM #13 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 After 3 sweets are added to the five left in the bag, there are 8 in the bag, and these are just enough to let each boy have one more sweet, so there are 8 boys. Hence the bag originally held 8 × 6 + 5 = 53 sweets.
 March 6th, 2014, 04:05 PM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra Re: Boys and some Sweets puzzle Number of sweets = 6b + 5 = 7b -3 => b = 8 => Number of sweets = 6 x 8 + 5 = 7 * 8 - 3 = 53

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