another one please?

needhelpmom

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Oct 5, 2005
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A bagger with 2 years experience at supermarket starts work at 8:00am @ 6.75$ hour. A cashier with 3 years starts work at 10:00 am @8.90 an hour. They finished at the same time and earned the same amount that day. How many hours did each work?

(round to the nearest 100th of an hour)
 
Hope I don't over do it again.


Let the time they finish work = t.

Hours worked = (t - start time).

Wage = hours worked * hourly rate.


Wage = (t-start time)*hourly rate


They earn the same


(t-8)*6.75 = (t-10)*8.9


6.75t - 54 = 8.9t - 89


35 = 2.15t

3500/215 = t

700/43 = t

The hours the bagger worked = 700/43 - 8.


The hours worked by the cashier = 700/43 - 10.


Multiply these by there respective hour rate to check they did earn the same.

Perhaps you can wrestle with this until someone posts a simpler way.
 
Thanks. My daughter is in 7th grade and she is coming home with theseword problems. I really appreciatte your help. I don't know how to help her. My problem is trying to figure out how to set up the problem.
any suggestions for me.
 
needhelpmom said:
Thanks. My daughter is in 7th grade and she is coming home with theseword problems. I really appreciatte your help. I don't know how to help her. My problem is trying to figure out how to set up the problem. any suggestions for me.

Hard to give "suggestions", but maybe if you look at this way of doing it,
it may help you:

Let h = hours by cashier earning $6.75 per hour;
then hours by other cashier = h - 2 : follow that, mom?

so $6.75 cashier earned: 6.75 * h,
and other cashier earned: 8.90 * (h - 2);
still with me?

And, since what they earned is the SAME, then:
8.90(h - 2) = 6.75(h)
8.90(h) - 17.80 = 6.75(h)
8.90(h) - 6.75(h) = 17.80
2.15(h) = 17.80
h = 17.80 / 2.15 = 8.279...

So they worked 8.28 and 6.28 hours.

Don't be shy if you have questions on above.
 
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