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Kim bought a total of $2.65 worth of postage stamps in four [#permalink]

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29 Jun 2014, 02:28

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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?

Re: Kim bought a total of $2.65 worth of postage stamps in four [#permalink]

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29 Jun 2014, 02:57

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manideepgre 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

Let us covert all the denominations in terms of 5 cents. 25 cent=5 * 5 cent stamp 10 cent=2* 5 cent stamp

Kim bought equal no of 5 cent and 25 cent stamps and twice as many 10-cent stamps as 5-cent stamps. Therefore every lot of purchase must vary in multiple 1+5+4(2*2)=10 purchases of 5 cent stamp. Therefor every lot is worth 50 cents. Closest multiple of 50 to 265 is 250 (50*5). Therefore Kim has to purchase atleast 15 stamps of 1 cent.

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

Moved to the proper sub-forum. Please post your questions in the PS/DS/SC/CR/RC sub-forums and tag properly. Thank you.
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Re: Kim bought a total of $2.65 worth of postage stamps in four [#permalink]

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11 Jan 2015, 06:25

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Question: 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?

Let: # of 5 and 25 cents stamps = n # of 10 cent stamps = 2n # of 1 cent stamps = m

Therefore: 5n + 25n + 10(2n) + m = 265 cents 50n + m = 265

For least number of 1 cents stamps, maximize 50n. Hence,

50*5 + m = 265 (n max has to be 5. For 6, the cost becomes 300 cents that exceeds his purchase of 265 cents) m = 15 i.e. 15 stamps 1 cents each

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
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