I need to hand it in tomorrow and I have to go to sleep in 1 hour. Hurry please.
An easy problem, PLEASE answer quickly.
Andrew, Boris, Cedric and Dan are collecting old paper. Toghether they collected 288 kg of it. How much paper did each of them collect if we know that Andrew collected 36kg more then Boris which equals 3/4 the ammount which Boris and Cedric collected together and if we know that Dan collected twice as much paper as Cedric.
A complete explenations including all the algerbra stuff. Thanks to all od you
I can do this now because its Sunday morning here and no work for me, but I am slow at typing (one-finger typing), so it will be s-l-o-w-l-y.Originally Posted by diamondfox
Let
A = number of kg of old papers collected by Andrew
B = ....by Boris
C = ....by Cedric
D = ....by Dan
A +B +C +D = 288 --------(1)
A = B +36 -----------------(2)
A = (3/4)(B +C) ----------(3)
D = 2C ------------------(4)
4 equations, 4 unknowns, solvable.
One way to continue is this:
The C is shown on 3 of the 4 equations, so we use that C. We eliminate A, B, D.
A from (2) = A from (3),
B +36 = (3/4)(B +C)
Clear the fraction, multiply both sides by 4,
4B +144 = 3B +3C
4B -3B = 3C -144
B = 3C -144 -------------(5)
Substitute that into (2),
A = (3C -144) +36
A = 3C -108 --------------(6)
Substitute those, and D=2C, into (1),
(3C -108) +(3c -144) +C +2C = 288
3C +3C +C +2C = 288 +108 +144
9C = 540
C = 540/9 = 60 kg -----------answer.
Hence,
A = 3C -108 = 180 -108 = 72 kg ----------answer.
B = 3C -144 = 180 -144 = 36 kg ------answer.
D = 2C = 120 kg ---------------------answer.
The hardest part is translating the information into mathematics.Originally Posted by diamondfox
I'll call: Andrew = A, Boris = B, Cedric = C, Dan = D.
Toghether they collected 288 kg of it => A + B + C + D = 288
Andrew collected 36kg more then Boris => A = B + 36
which equals 3/4 the ammount which B&C collected together => A = 3/4(B+C)
Dan collected twice as much paper as Cedric => D = 2C
This gives the following system of four lineair equations in four unknowns: