SOLUTION: Mary bought 10 gift cards in The denominations of $20, $50,$100. She spent $370. She has twice as many $20 gift cards as $50 gift cards. How many of each denomination of gift car

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Question 1167121: Mary bought 10 gift cards in The denominations of $20, $50,$100. She spent $370. She has twice as many $20 gift cards as $50 gift cards. How many of each
denomination of gift card did she buy?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52864)   (Show Source): You can put this solution on YOUR website!
.
Mary bought 10 gift cards in The denominations of $20, $50,$100. She spent $370.
She has twice as many $20 gift cards as $50 gift cards. How many of each denomination of gift card did she buy?
~~~~~~~~~~~~~ 


Let x be the number of $50 cards.


Then the number of the $20 cards is 2x, and the number of the $100 cards is  (10-x-2x) = (10-3x).


Now you can write the money equation


    50x + 20*(2x) + 100*(10-3x) = 370  dollars.


Simplify and solve


    50x + 40x + 1000 - 300x = 370

    1000 - 370 = 300x - 50x - 40x

     630       = 210x

       x       = 630/210 = 3.


ANSWER.  3  $50-cards;  3*2 = 6  $20-cards  and  the rest  10- 3 - 6 = 1  $100-card.

Solved     //   using only one equation in one unknown.



Answer by greenestamps(13206)   (Show Source): You can put this solution on YOUR website!


Twice as many $20 gift cards as $50 gift cards; and 10 gift cards in all. So

x = # of $50 gift cards
2x = # of $20 gift cards
(10-3x) = # of $100 gift cards

Now write the equation that says the total cost was $370:



Solve for x using basic algebra; then use that value to answer the question.

Even if a formal algebraic solution was required, you can get some good brain exercise by solving this using logical reasoning and some simple mental arithmetic (easier arithmetic than required in the formal algebraic solution suggested above).

Here's one way to solve the problem informally.

Try different numbers of $100 gift cards and see if that limits the problem to only one possible solution.

3 $100 gift cards would mean there would have to be 7 other gift cards of $50 or $20 each to make the remaining $70. That is clearly not possible.

Even 2 $100 gift cards won't work; there would have to be 8 other gift cards of $50 or $20 each to make the remaining $170, and a bit of quick arithmetic shows that is not possible.

Then, since the problem says she bought gift cards of all three amounts, we can assume there was 1 $100 gift card, leaving 9 gift cards of $50 or $20 each to make the remaining $270. Some mental arithmetic and perhaps some trial and error will finish solving the problem quickly.

And here is an even quicker path to an informal solution using logical reasoning.

The total cost of the $100 and $50 gift cards will be a multiple of $50. With a total cost of $370, that means there must be either $20 in $20 gift cards (1 $20 card), or $120 in $20 gift cards (6 $20 cards).

A single $20 gift card would mean there would have to be 9 $50 or $100 gift cards to make the remaining $350; that is clearly impossible.

So this line of reasoning quickly tells us that there must be 6 $20 gift cards.

That leaves 4 $50 or $100 gift cards to make the remaining $250; again simple arithmetic and maybe some trial and error finds there must be 1 $100 gift card and 3 $50 gift cards.

ANSWER: 1 $100 gift card, 3 $50 gift cards, and 6 $20 gift cards.
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