SOLUTION: In the latest class reunion of HS '70, a certain number of succesful alumni promised to donate equally for a new school project amounting to $240 000.00. Four of them failed to pro
Question 1041185: In the latest class reunion of HS '70, a certain number of succesful alumni promised to donate equally for a new school project amounting to $240 000.00. Four of them failed to provide the certain amount of money. Hence, each of the remaining alumni had to pay $3 000.00 more to produce the sum. How many alumni had promised to pay originally?
Please state the solution! I really need to review this before our exams on Monday. Thanks alot. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! In the latest class reunion of HS '70, a certain number of successful alumni promised to donate equally for a new school project amounting to $240 000.00.
Four of them failed to provide the certain amount of money.
Hence, each of the remaining alumni had to pay $3 000.00 more to produce the sum.
How many alumni had promised to pay originally?
:
let a = the no. of alums that promised to pay
then
(a-4) = no. that actually paid
then = amt that each originally agreed to pay
and = amt that actually was paid
:
the equation, lets see if we can do this in "thousands of dollars" - = 3
multiply by a(a-4)
a(a-4)* - a(a-4) = 3a(a-4)
cancel the denominators
240a - 240(a-4) = 3a^2 - 12a
240a - 240a + 960 = 3a^2 + 12a
combine on the right to form a quadratic equation
0 = 3a^2 - 12a - 960
simplify divide by 3
a^2 - 4a - 320 = 0
You can us the quadratic formula a=1; b=-4; c=-320 but this will factor to:
(a+16)(a-20) = 0
the positive solution is what we want here
a = 20 alums the original no. that promised to pay
:
:
let's see if this checks out, (16 alums actually paid). - =
15 thousand - 12 thousand = 3 thousand
