SOLUTION: A woman bought 100 Christmas cards. For the ones that sing a song when you open them, she paid 30 cents each. For the rest she paid 5 cents each. If the cards cost $10.25 in all, h
Question 1028923: A woman bought 100 Christmas cards. For the ones that sing a song when you open them, she paid 30 cents each. For the rest she paid 5 cents each. If the cards cost $10.25 in all, how many of the expensive kind did she buy? Found 2 solutions by Edwin McCravy, stanbon:Answer by Edwin McCravy(20059) (Show Source):
Let the number of sing-a-songs be x
Let the number of ordinarys be y
Value Value
Type Number of of
of of EACH ALL
card cards card cards
-------------------------------------------
sing-a-songs x $0.30 $0.30x
ordinarys y $0.05 $0.05y
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TOTALS 100 ----- $10.25
The first equation comes from the second column.
x + y = 100
The second equation comes from the last column.
0.3x + 0.05y = 10.25
Get rid of decimals by multiplying every term by 100:
30x + 5y = 1025
So we have the system of equations:
.
We solve by substitution. Solve the first equation for y:
x + y = 100
y = 100 - x
Substitute (100 - x) for y in 30x + 5y = 1025
30x + 5(100 - x) = 1025
30x + 500 - 5x = 1025
25x + 500 = 1025
25x = 525
x = 21 = the number of sing-a-song cards.
Substitute in y = 100 - x
y = 100 - (21)
y = 79 ordinarys.
Checking: 21 sing-a-songs is $6.30 and 79 ordinarys is $3.95
That's 100 cards.
And indeed $6.30 + $3.95 = $10.25
Edwin
You can put this solution on YOUR website! A woman bought 100 Christmas cards. For the ones that sing a song when you open them, she paid 30 cents each. For the rest she paid 5 cents each. If the cards cost $10.25 in all, how many of the expensive kind did she buy?
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Equations:
s + r = 100 cards
30s + 5r = 1025 cents
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Modify for elimination::
s + r = 100
6s + r = 205
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Subtract and solve for "s"
5s = 105
s = 21 (# of song cards)
r = 100 -21 = 79 (# of other cards)
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Cheers,
Stan H.