Puzzle:
There are three kids and each one have some pencils. The first kid gave the second and third as many pencils as they each already had. Then, the second gave the first and third as many as they each had. Still later, the third gave the first and second as many as they each already had. Each kid now have 24 pencils. How many pencils did each kid have originally?
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Solution:
You can solve the problem easily backward
The first had 39 pencils, the second had 21 and the third had 12.
You can solve the problem easily backward
First+Second+Third
24 + 24 + 24 = 72
12 + 12 + 48 = 72
6 + 42 + 24 = 72
39 + 21 + 12 = 72
The first had 39 pencils, the second had 21 and the third had 12.
4 comments:
Kid1 -> 39
Kid2 -> 21
Kid3 -> 12
Kool, I worked out independently and got the same result!
39-21-12 (initially)
6-42-24 (aft x gives)
12-12-48 (aft y gives)
24-24-24 (finally aft z gives)!
x+ y+z = 72
(x-y-z) + 2y +2z = 72
2(x-y-z) +(2y -((x-y-z)+2z))+4z =72
4(x-y-z)+ 2(2y-((x-y-z)+2z))+
(4z-(2(x-y-z)+(2y -((x-y-z)+2z)))= 72
and each term is equals 24
4(x-y-z) = 24 => x-y-z = 6 --->1
2(2y-((x-y-z)+2z)) = 24
=> (2y -x+y -z) = 12
=> -x+3y-z = 12 ----->2
(4z-(2(x-y-z)+(2y -((x-y-z)+z)))=24
=> -x-y+7z = 24---->3
solve 1, 2, 3
x = 39, y = 21, z = 12
work in reverse direction and get result easily
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