Question 134354: These word problems were on my test, please help me solve them.
A) A boy earned $1.50 per day more than his sister. The boy worked 5 days and the girl worked 6 days. Together they earned $35. What did each earn per day?
B) A woman is now 20 years older than her daughter. In 6 years she will be twice as old as her daughter. How old is she now?
C) A woman has money in two accounts. One account pays 7% annual interest, while the other pays 9% annual interest. If she has $600 more invested at 9% than she does at 7% and her total interest for a year is $182, how much does she have in each account?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A) A boy earned $1.50 per day more than his sister. The boy worked 5 days and the girl worked 6 days. Together they earned $35. What did each earn per day?
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Let sister's wage be "x" dollars per day;
Boy's wage will be "x+1.50" dollars per day:
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EQUATION:
5(x+1.50) + 6x = 35
5x+7.50+6x = 35
11x = 27.5
x = $2.50 (sister's wage per day)
x+1.50 = $4.00 (boy's wage per day)
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B) A woman is now 20 years older than her daughter. In 6 years she will be twice as old as her daughter. How old is she now?
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Age NOW DATA:
Let daughter's age be "x"; mother's age is "x+20"
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Six yrs. from now DATA:
Daughter will be "x+6" ; Mother will be "x+26"
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EQUATION:
x+26 = 2(x+6)
x+26 = 2x+12
x = 14 (daughter's age Now)
x+20 = 34 (mother's age Now)
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C) A woman has money in two accounts. One account pays 7% annual interest, while the other pays 9% annual interest. If she has $600 more invested at 9% than she does at 7% and her total interest for a year is $182, how much does she have in each account?
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Let amt. invested at 7% be "x" ; Interest on this is 0.07x dollars
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Amt invested at 9% is "x+600" ; Interest is 0.09(x+600) = 0.09x+ 54 dollars
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EQUATION:
interest + interest = 182
0.07x + 0.09x+54 = 182
0.16x = 128
x = $800 (amount invested at 7%)
x+600 = $1400 (amount invested at 9%)
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Cheers,
Stan H.
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