word problem?

Question

One pipe can fill a tank in 7 minutes less time than it takes a smaller pipe. Together they can fill the tank in 12 minutes.

Write an equation to answer this question: How long would it take the larger pipe to fill the tank? Let t = the number of minutes it takes the larger pipe to fill the tank alone.

Then, solve the equation, and write its answer.

Answer 1

p1t = p2(t + 7) = 12(p1 + p2)
p2 = p1t/(t + 7)
p1t = 12(p1 + p1t/(t + 7))
t = 12 + 12t/(t + 7)
t^2 +7t = 12t + 84 + 12t
t^2 - 17t - 84 = 0
t = (17 ± √(289 + 336))/2
t = (17 ± √625)/2
t = (17 ± 25)/2
t = 42/2
t = 21 min. for the larger pipe working alone
t + 7 = 28 min. for the smaller pipe working alone

21 =? 12 + 12*21/(21 + 7)
21 =?12 + 3*4*3*7/4*7
21 =?12 + 3*3
21 = 21

Answer 2

Let‘t’ be the time taken by larger pipe to fill the tank

Then‘t+7’ will be the time for smaller pipe to fill the tank

Let T be the time taken by both pipes to fill the tank.

If V is the volume of the tank,

In one minute the larger pipe will fill V/ t units.

In one minute the smaller pipe will fill V/ (t+7) units.

In one minute both pipes will fill V/ T units.

That is V/T = V/ t + V/ (t+7)

Canceling V through out,

1/T = (1/ t) + 1 / (t+7)

1/12 = (1/ t) + 1 / (t+7)

Multiplying both sides by 12* (t+7)* t

(t+7)* t = 12(t+7) + 12 t.

t*t + 7t = 12t + 84 + 12t

t*t - 17t - 84 = 0.

(t - 21)*(t+4) =0

t = 21 minutes or -4 minutes.

We take the +21 minutes as the answer.

If we have begun the filling processes 4 minutes earlier than now, the process will have the same effect

Answer 3

Let‘t’ be the time taken by larger pipe to fill the tank

Then‘t+7’ will be the time for smaller pipe to fill the tank

Let T be the time taken by both pipes to fill the tank.

If V is the volume of the tank,

In one minute the larger pipe will fill V/ t units.

In one minute the smaller pipe will fill V/ (t+7) units.

In one minute both pipes will fill V/ T units.

That is V/T = V/ t + V/ (t+7)

Canceling V through out,

1/T = (1/ t) + 1 / (t+7)

1/12 = (1/ t) + 1 / (t+7)

Multiplying both sides by 12* (t+7)* t

(t+7)* t = 12(t+7) + 12 t.

t*t + 7t = 12t + 84 + 12t

t*t - 17t - 84 = 0.

(t - 21)*(t+4) =0

t = 21 minutes or -4 minutes.

We take the +21 minutes as the answer.

If we have begun the filling processes 4 minutes earlier than now, the process will have the same effect.

Answer 4

Let t = # of minutes it takes the smaller pipe to fill the tank
t - 7 = # of minutes it takes the bigger pipe to fill the tank

Process:
In one minute, the smaller pipe had done 1/t of the job.
In one minute, the bigger pipe had done 1/(t-7) of the job
In one minute, both pipe had done 1/12 of the job.

When both pipe work together, we will have the equation:
1/t + 1/(t - 7) = 1/12
===> [(t - 7) + t] / (t^2 - 7t) = 1/12
===> (2t - 7) / (t^2 - 7t) = 1/12
===> 12(2t - 7) = t^2 - 7t
===> 24t - 84 = t^2 - 7t
===> 0 = t^2 - 31t + 84
===> 0 = t - 28 or 0 = t - 3
===> t = 28 or t = 3
===> the valid answer is t = 28 cause t - 7 = 21
It takes 28 minutes for the smaller pipe and 21 minutes for the bigger pipe to fill the tank.

to decide which answer is correct, read carefully the given problem and the representations made.

Answer 5

The lengths of time for the two pipes to fill the tank when working alone are t and t+7, respectively. Together it takes 12 minutes.

1/t + 1/(t+7) = 1/12

Clearing the denominators by multiplying by 12t(t+7) we get:

12(t + 7) + 12t = t(t + 7)
12t + 84 + 12t = t^2 + 7t
0 = t^2 - 17t - 84
t^2 - 17t - 84 = 0
(t - 21)(t + 4) = 0
t = 21, -4

We can eliminate -4 as a solution.

So the time for the large pipe working alone to fill the tank is:

Large pipe t = 21 minutes

Answer 6

if t is the larger pipes time then the smaller pipes time is t + 7 and together they fill it in 12 minutes means t + t + 7 = 12... so 2t + 7 = 12... so 2t = 5... so t = 2.5 minutes. so the large pipe takes 2.5 minutes and the small pipe takes 9.5 minutes.