stamps

Assume Kim bought x 5-cent and 25-cent stamps and 2x 10-cent stamps. Let y be the amount of 1-cent stamps bought.

.05(x) + 0.25(x) + 0.20(x) + 0.01(y) = 2.65

y = (2.65 - 0.5(x))*100 = 265 - 50x

In order to minimize y, we must maximize the value of x. The only value of x that works in that case is x = 5.

Therefore y = 265 - 250 = 15 stamps.

grandh01 wrote: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

To MINIMIZE the number of 1-cent stamps, we must MAXIMIZE the number of the other types of stamps.

She bought an equal number of 5-cent and 25-cent stamps.
She bought twice as many 10-cent stamps as 5-cent stamps.
In other words, for every 5-cent stamp, one 25-cent stamp and two 10-cent stamps must be bought:
5+25+2(10) = 50.

Thus, the sum of the other types of stamps must be a multiple of 50.
The greatest multiple of 50 less than 265 is 250.
Since 265-250 = 15, 15 cents must be spent on 1-cent stamps.
Thus, the number of 1-cent stamps = 15.

The correct answer is C.

Let the number of 5 & 25 cent stamps be x (since they are equal).
So, the no. of 10 cent stamps = 2x.

And let y denote 1 cent stamps.

So,

5x + 25x + 20x + y = 265
=> 50x + y = 265

For y to be minimum, 50x has to be maximized. The largest value in 265 which is a multiple of 50 is 250. Therefore, y = [spoiler]265 - 250 = 15. Ans: C[/spoiler]

grandh01 wrote: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

We can create the following variables:

a = number of 1-cent stamps

b = number of 5-cent stamps

c = number of 10-cent stamps

d = number of 25-cent stamps

Thus:

2.65 = 0.01a + 0.05b + 0.10c + 0.25d

and

b = d

and

c = 2b

Thus, we have:

2.65 = 0.01a + 0.05b + 0.10(2b) + 0.25b

2.65 = 0.01a + 0.50b

265 = a + 50b

265 - 50b = a

We want the value of b to be as large as possible and still maintain that the value of 265 - 50b is positive and that value will be the least value of a. We see that if b = 5, then a = 15 (if b > 5, then 265 - 5b, or a, will be negative). Thus, the least value of a is 15.

Answer: C