by Ron Angelo A. Gelacio
(Paranaque,Metro Manila,Philippines)
Ricky, Carl and Jerome have a total of n candies. Ricky ate 1/4 of the candies. Carl ate 1/3 of the remaining candies. Jerome ate 1/5 of the remaining candies.
Then they grouped the remaining candies into 4 groups. If there were 10 candies in each group,how many candies did the three have in first?
Solution
Ricky ate 1/4 of the candies
This means that Ricky ate 1/4 of n
1/4 of n = 1/4 × n = n/4
Carl ate 1/3 of the remaining candies
The remaining candies is n - n/4
n - n/4 = 4n/4 - n/4 = 3n/4
This means that Carl ate 1/3 of 3n/4
1/3 of 3n/4 = 1/3 × 3n/4 = 3n/12 = n/4
Jerome ate 1/5 of the remaining candies
The remaining candies is 3n/4 - n/4
3n/4 - n/4 = 2n/4 = n/2
This means that Jerome ate 1/5 of n/2
1/5 of n/2 = 1/5 × n/2 = n/10
Then they group n/10 into 4 groups.
This means that the remaining candies is divided by 4
n/10 divided by 4 = n/40
There are 10 candies in each group
n/40 = 10
Multiply both sides by 40
n = 10 × 40 = 400
The number of candies in each group is 400.
Since there are 4 groups, the total number of candies is 400 × 4 = 1600
Ricky ate 1/4 of 1600 or 1/4 × 1600 = 1600/4 = 400
1600 - 400 = 1200
Carl ate 1/3 of 1200 or 1/3 × 1200 = 1200/3 = 400
1200 - 400 = 800
Jerome ate 1/5 of 800 or 1/5 × 800 = 800/3 = 160
800 - 160 = 640
640 divided by 4 = 160
At the end each group had 160 candies
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Oct 24, 16 06:10 PM
Straightforward proof of the law of sines. Easy to follow and understand