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doughnut

by baller12q » Tue Mar 31, 2009 3:20 pm
At the bakery lew spent a total of $6 for one kind od cupcake and one kind of doughnut. How many doughnuts did he buy?

1)the price of 2 doughnuts was 10 less than the price of 3 cupcakes

2) the average price of 1 doughnut and 1 cupcake was .35
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by cramya » Tue Mar 31, 2009 4:21 pm
PM Response:

Let p1 be the price of 1 donut
Let d be the number of donuts

Let p2 be the price 1 cupcake
Let c be the number of cupcakes

We need to find d

Given : p1*d + p2*c = 6 $ or 600 cents

Stmt I

2* p1 = 3*p2-10

3*p2-2*p1= 10

let say p2=4 p1=1

For 4 cents we can buy 100 cupcakes= 400 cents
For 1 cent we can 200 donuts = 200 cents
400+200=600 cents

For 4 cents we can buy 50 cupcakes= 200 cents
For 1 cent we can 400 donuts = 400 cents
400+200=600 cents

INSUFF

Stmt II


p1+p2/2 = 35

p1+p2 = 70

p1=40 p2=30

Just like above many different combinations for d and c possible.

INSUFF

Together:

3*p2-2*p1= 10
p1+p2 = 70

We can obtain p1 and p2 but still many different combinations for d and c possible.

Choose E


Hopt this helps! I have tried to explain it as detailed as possible but let me know if u still hv questions.


Good luck.

Regards,
CR
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by GID09 » Wed Apr 01, 2009 1:34 am
  • 3*p2-2*p1= 10
    p1+p2 = 70

    We can obtain p1 and p2 but still many different combinations for d and c possible.
Cramya, Could you explain why we cannot find two unknowns using two equations?

Thanks for the explanation.
GID09[/list][/quote]
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