SOLUTION: The total outing cost for a group of people amounted to Php. 8000. When an additional 4 persons decided to join the outing, the cost per person decreased by Php. 100. How many peop
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Question 393661: The total outing cost for a group of people amounted to Php. 8000. When an additional 4 persons decided to join the outing, the cost per person decreased by Php. 100. How many people actually joined the outing? (Show Your Solution) Found 2 solutions by mananth, stanbon:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! The total outing cost for a group of people amounted to Php. 8000. When an additional 4 persons decided to join the outing, the cost per person decreased by Php. 100. How many people actually joined the outing? (Show Your Solution)
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the total outing cost for a group of people amounted to P8000. when an additional 4 persons decided to join the outing, the cost per person decreased by P100. how many actually joined the outing?
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let there be x persons
cost per person = 8000/x
..
4 persons join
=x+4
cost reduces by 100
..
8000/x - 8000/(x+4)= 100
LCD = x(x+4)
(8000(x+4)-8000x)/x(x+4)=100
8000x+32000-8000x=100(x(x+4))
32000=100x^2+400x
/100
320=x^2+4x
x^2+4x-320=0
x^2+20x-16x-320=0
x(x+20)-16(x+20)=0
(x-16)(x+20)=0
x= 16 which is positive
Initially 16 persons joined the outing.
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m.ananth@hotmail.ca
You can put this solution on YOUR website! The total outing cost for a group of people amounted to Php. 8000.
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Original DATA:
Let number in the group be "x".
Average cost per person is 8000/x Php.
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When an additional 4 persons decided to join the outing, the cost per person decreased by Php. 100.
New DATA:
Number in the group is "x+4".
Average cost perperson is 8000/(x+4) Php.
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How many people actually joined the outing?
Equation:
original average - new average = 100 Php
8000/x - 8000/(x+4) = 100
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Divide thru by 100 to get:
80/x - 80/(x+4) = 1
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Multiply thru by x(x+4) to get:
80(x+4) - 80x = x(x+4)
80*4 = x^2+4x
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x^2+4x-320 = 0
(x-16)(x+20) = 0
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Positive solution:
x = 16 (original # of people in the group)
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Cheers,
Stan H.