alg. 2 math help?? eek!?

Question

a mechanical sorter can process a bag of mail in 18 minutes. After the sorter has been working for a time, it breaks down. The rest of the mail it was sorting is divided equally between two older machines, each of which would take 60 minutes to complete the job working alone. The mail is finished being sorted 20 minutes after the first machine started working. How long did the first machine work before breaking down??

work too please. thank you very much. appreciate it. :]

Answer 1

N/r1 = 18
N/r2 = 60
18r1 = 60r2
r2 = 18r1/60
(N-n)/2r2 + n/r1 = 20
60(N - n)/36r1 + n/r1 = 20
(60/36)(N/r1) - (60/36)(n/r1) + n/r1 = 20
(60/36)(18) + n/r1(1 - (60/36)) = 20
t = (20 - (60/36)(18))/(1 - (60/36))
t = (20*36 - 60*18)/(36 - 60)
t = (720 - 1080)/-24
t = 360/24
t = 15 min

Answer 2

****** NOTE THE ANSWER ABOVE IS INCORRECT ******
Notice that the problem says the older machine would take 60 minutes to COMPLETE the job. It does not say 60 minutes to sort a bag of mail.

T = Time that the first machine worked before breaking down.

1 - (T/18) = (18 - T) / 18 Amount of mail left after it breaks down.

((18 - T) /18) / 60 = (18 - T) / 1080 is the rate that one of the older machines can process the mail.

2*(18 - T) / 1080 = (18 -T) / 540 is the rate that both older machines process the mail

The two older machines work for (20 - T) minutes

( (18 - T) / 540 )(20 - T) = T/18
(18 - T)(20 - T) = 30T

360 - 38T + T^2 = 30 T

T^2 - 68T + 360 = 0

Use the quadratic formula:

T = [ 68 + or - sqrt (4624 - 1440) ] / 2
= [ 68 + or - sqrt (3184) ] / 2
= (68 + or - 56.43) / 2
= 5.79 Minutes

Answer 3

x = 18 min machine
y = 60 min machine
b = remainder of bag
f = first part of bag
b = minutes before breakdown

(y-20)/y = r
(60-20)/60 = r
40/60 = r
2/3 = r

1 - r = f
1- (2/3) = f
1/3 = f

x * f = b
18 * f = b
18* (1/3) = 6

6 min <---answer

In 6 min, the first machine at the rate of 18 min per bag would have done 1/3 of the job (6/18) which leaves 2/3 of the bag. The 60 min machines, working together could to a bag in half the time (30 min). 30 min times 2/3 of a bag equal 20 min. [30 * (2/3)] Therefore, the first machine worked for 6 minutes, completing 1/3 of the bag, then the 2 older machines finished the other 2/3 of the bag 20 minutes afterward.

Answer 4

I don't really understand it, but base on what i know from this problem is : the time for the older machine to finish the process is givien, 20 min. , which is 1/3 (20/60) times compare with the whole full job., so it means the first machine did 2/3 of the process, so take 2/3 times 18, you have 12 mins, the 1st machine work for 12 mins before it breaks down.

and you should do your homework earlier, should not wait until the last min. :)