Question 1140528: A basket of oranges were divided among Tina, Rick and Ken. Tina received 5 more than half of the oranges. Rick received 4 more than half of the remaining oranges. Ken received half of the remainder, plus the last 3 oranges. What percentage of the oranges did Ken receive?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A basket of oranges were divided among Tina, Rick and Ken.
A = t + r + k
Tina received 5 more than half of the oranges.
t = .5A + 5 Tina's amt
then
A - (.5A+5)
(.5A-5) = amt remaining
Rick received 4 more than half of the remaining oranges.
r = .5(.5A - 5) + 4
r = .25A - 2.5 + 4
r = .25A + 1.5 Ricks amt
then
(.5A-5)-(.25A+1.5)
.5A - .25A - 5 - 1.5
.25A - 6.5 amt remaining
Ken received half of the remainder, plus the last 3 oranges.
k = .5(.25A-6.5) + 3
k = .125A - 3.25 + 3
k = .125A - .25 Kens amt
Find A, the sum of all 3 amts
A = (.5A + 5)+(.25A+1.5)+(.125A-.25)
A = .5A + .25A + .125A + 5 + 1.5 - .25
A = .875A + 6.25
A - .875A = 6.25
.125A = 6.25
A = 6.25/.125
A = 50 total oranges
:
What percentage of the oranges did Ken receive?
k = .125(50)-. 25
k = 6 oranges
Find the percentage: 6/50 * 100 = 12%
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