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10 Feb 2004, 13:31

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Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent stamps, what is the least number of 1-cent stamps she could have bought ?
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25

Please explain in lenght..

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10 Feb 2004, 13:55

Let X be the number of 5cents stamps.
We know though that number of 5cents stamps = number of 10cents stamps.
Then, we know that .05X + .25X + .10(2X) = .50X
2.65 / .5 = 5 remainder .15. Therefore least number of 1cent stamp is C)15
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Paul

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10 Feb 2004, 14:01

Let X be the no of 5 cent stamps

Let Y be the no of 1 cent stamps

The equation would be 5X+25X+10*2X+Y=265
50X+Y=265

Y=15.

Vivek.
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10 Feb 2004, 14:02

Minimum no of each denomination stamps he can buy is 1
Let us say he buys 1 5 cent stamp = 5c
then he buys 1 25 cent stamp as well = 25c
then he also buys 2 10 cent stamps = 20c

Total = 50c for three different stamps
We know that he spent $2.65 or 265cents on these stamps.
Since 265 is not divisible by 50 we chose the next highest number less than 265. That number is 250. So maximum he can spend 250 cents buying 5, 10 and 25 cent stamps.
265-250 = 15 are the least number of 1 cent stamps he can buy.

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10 Feb 2004, 14:34

Vivek
how did you figure that y=15..
you have two unknowns and 1 equation. Please explain
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10 Feb 2004, 14:35

Oops sorry,
didnt see that you mentioned x=5 cent stamps..

Thanks everyone..
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10 Feb 2004, 14:37

let me barge in

Vivek's equation is
50X + Y = 265 = 5*50 + 15
so he equates similar terms on both the sides
50X = 5*50
and Y = 15
This can be done in case of simple equations.

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11 Feb 2004, 01:40

15 is correct. Vivek's approach is the easiest and the best in this case.

[#permalink] 11 Feb 2004, 01:40

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