A problem on typing speed

[burgess]

Need help to solve algebra 1 homework

X, Y and Z are three typists who working simultaneously can type 216 pages in 4 hours. In one hour, Z can type as many pages more than Y as Y can type more than X. During a period of five hours, Z can type as pages as X can during seven hours. How many pages does each of them type hour?

Please tell me the procedure to solve these type of problem

Thanks

[Guest]

216 / 4 = 54 pages per hour all together working.

Consider 1 hour working....

(Y-X) = diff = same as diff (Z-Y)

Z completes (X + (Y-X) + (Y-X))

Y completes (X + (Y-X))

X completes X

Add them to get all working together for 1 hour = 54

Gives.... 3Y = 54

So Y = 18

And..... 5Z = 7X .....told this....

5(2Y - X) = 7X
10Y - 5X = 7X
10Y = 12X
180 = 12X

X = 15


Z = 54 - 18 - 15 = 54-33 = 21

So Z = 21

[burgess]

Thank you for your detailed explanation

216 / 4 = 54 pages per hour all together working.

Consider 1 hour working....

(Y-X) = diff = same as diff (Z-Y)

Z completes (X + (Y-X) + (Y-X))

Y completes (X + (Y-X))

X completes X

Add them to get all working together for 1 hour = 54

Gives.... 3Y = 54

So Y = 18

And..... 5Z = 7X .....told this....

5(2Y - X) = 7X
10Y - 5X = 7X
10Y = 12X
180 = 12X

X = 15


Z = 54 - 18 - 15 = 54-33 = 21

So Z = 21