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# Kim bought a total of \$2.65 worth of postage stamps in four

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

Ans=C
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Kim bought a total of \$2.65 worth of postage stamps in four [#permalink]
a * (0,05) + a * (0,25) + 2a * (0,10) + b * 0,01 = 2,65

So, a(0,05+0,25+0,20) + b*0,01 = 2,65 => a * 0,50 + b*0,01 = 2,65

Given that the least answer choice is 25, a>4 (4*0,5 = 2)

So, let's try a=5
5 * 0,50 + b*0,01 = 2,5 + b*0,01
b * 0,01 = 0,15
b=15
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Re: Kim bought a total of \$2.65 worth of postage stamps in four [#permalink]
Use the following equation:

1a+5b+10c+25d = 265
b=d
2b=c

Plug in values --> 1a+5b+20b+25b = 265
1a + 50b = 265

bmax = 5 --> 250

265-250 = 15
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Re: Kim bought a total of \$2.65 worth of postage stamps in four [#permalink]
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

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.

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

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.

Dear JeffTargetTestPrep , where can I find a detailed general explanation about the topic? If it exists clearly
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Re: Kim bought a total of \$2.65 worth of postage stamps in four [#permalink]
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

5-CENT : 25-CENT : 10-CENT : 1-CENT
X : X : 2X : X (Stem: twice as many 10-cent stamps as 5-cent stamps)
We need to minimize 1-cent denomination coins and we are not given total no of coins but we have denomination of coins
so we can approximate
5(x) + 25(x) + 10(2x) = 265
50x = 265
x is an integer at 5 so the remaining 265-250 = 15 are 1-cent denomination coins
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Re: Kim bought a total of \$2.65 worth of postage stamps in four [#permalink]
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Re: Kim bought a total of \$2.65 worth of postage stamps in four [#permalink]
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