Kurt T. answered 02/04/18
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Math Tutoring and Test Prep
It's usually helpful to treat work rate problems in terms of work performed per unit time, not time required to complete the entire job.
Betty can paint a house in 14 hours, so she can paint 1/14 of a house in one hour.
Karen can paint a house in x hours, and the two women working together can paint the house in 2x/3 hours. So Karen can paint 1/x of a house in one hour and both can do the job in 3/2x hours.
Add the "work per hour" figures:
1/14 + 1/x = 3/2x
Common denominator is 14x.
x/14x + 14/14x = 21/14x
Disregard the denominators.
x + 14 = 21
x = 7
Now let's see if the answer makes sense.
We already knew Betty could paint a house in 14 hours, or 1/14 of a house in one hour.
We now know Karen can paint a house in 7 hours, or 1/7 of a house in one hour.
1/14 + 1/7 = 3/14. The two women can paint 3/14 of a house in an hour, or an entire house in 14/3 or 4 2/3 hours.
4 2/3 is 2/3 of 7.
Betty can paint a house in 14 hours, so she can paint 1/14 of a house in one hour.
Karen can paint a house in x hours, and the two women working together can paint the house in 2x/3 hours. So Karen can paint 1/x of a house in one hour and both can do the job in 3/2x hours.
Add the "work per hour" figures:
1/14 + 1/x = 3/2x
Common denominator is 14x.
x/14x + 14/14x = 21/14x
Disregard the denominators.
x + 14 = 21
x = 7
Now let's see if the answer makes sense.
We already knew Betty could paint a house in 14 hours, or 1/14 of a house in one hour.
We now know Karen can paint a house in 7 hours, or 1/7 of a house in one hour.
1/14 + 1/7 = 3/14. The two women can paint 3/14 of a house in an hour, or an entire house in 14/3 or 4 2/3 hours.
4 2/3 is 2/3 of 7.
Star M.
02/04/18