alegbra help w/ finding conveyor-belt time, rate of wind

[callie2291]

I have some questions guys and i can't seem to get it

Question 1: 1 conveyor belt can move 1000 boxes in 6mins. another can move 1000 boxes in 12mins. if another conveyor belt is added and all 3 are used, the boxes are moved in 3mins. How long wud it take the third conveyor belt alone to do the same job?

Question 2: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10miles w/the wind, & in the same amount of time runs 7 miles against wind what is the rate of the wind?

I cant even figure out where to start on these!
Please someone help me!

 

[Deleted member 4993]

I have some questions guys and i can't seem to get it

Question 1: 1 conveyor belt can move 1000 boxes in 6mins. another can move 1000 boxes in 12mins. if another conveyor belt is added and all 3 are used, the boxes are moved in 3mins. How long wud it take the third conveyor belt alone to do the same job?

I assume 1000 boxes are moved in 3 minutes when all three are running.

In 3 minutes, how many boxes can the conveyor belt 1 move? = m

In 3 minutes, how many boxes can the conveyor belt 2 move? = n

In 3 minutes, how many boxes can the conveyor belt 3 move? = 1000 - (m + n)

From this you can calculate, the time required for the conveyor belt #3 to move 1000 boxes.

Question 2: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10miles w/the wind, & in the same amount of time runs 7 miles against wind what is the rate of the wind?

Assume the speed of wind = w

Jim's Speed with wind = 5 + w
Jim's Speed against wind = 5 - w

Then use time = distance /speed

so

10/(5+w) = 7/(5-w).....(why?)

Solve for 'w'.

 

[soroban]

Hello, callie2291!

1) Conveyor belt \(\displaystyle A\) can move 1000 boxes in 6 mins; \(\displaystyle B\) can move 1000 boxes in 12 mins.
if a third conveyor belt \(\displaystyle C\) is added and all 3 are used, the boxes are moved in 3 mins.
How long would it take \(\displaystyle C\) to do the same job alone?

\(\displaystyle A\text{ does the job in 6 minutes.}\)
. . . \(\displaystyle \text{In one minute, }A\text{ can do }\frac{1}{6}\text{ of the job.}\)

\(\displaystyle B\text{ does the job in 12 minutes.}\)
. . . \(\displaystyle \text{In one minute, }B\text{ can do }\frac{1}{12}\text{ of the job.}\)

\(\displaystyle C\text{ does the job in }x\text{ minutes.}\)
. . . \(\displaystyle \text{In one minute, }C\text{ can do }\frac{1}{x}\text{ of the job.}\)

\(\displaystyle \text{Working together, in one minute, they can do: }\:\frac{1}{6} + \frac{1}{12} + \frac{1}{x}\text{ of the job.}\;\;{\bf[1]}\)


\(\displaystyle \text{But we are told: working together, it takes them 3 minutes to do the job.}\)
. . \(\displaystyle \text{Hence, in one minute, they can do }\frac{1}{3}\text{ of the job.}\;\;{\bf{[2]}\)

\(\displaystyle \text{Equate {\bf[1]} and {\bf[2]}: }\;\frac{1}{6} + \frac{1}{12} + \frac{1}{x} \;=\;\frac{1}{3}\quad\hdots \text{and solve for }x.\)

 

[Denis]

Different approach:
Together, A and B move 1000/6 + 1000/12 = 250 boxes per minute ; 1000/250 = 4 minutes.

C's contribution reduces time by 1 minute: so A and B moved 750, leaving 250 for C.

So C moved 250 in 3 minutes, hence 1000 will take 12 minutes.