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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, 03:28
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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 5cent and 25cent stamps and twice as many 10cent stamps as 5cent stamps, what is the least number of 1cent stamps she could have bought? (A) 5 (B) 10 (C) 15 (D) 20 (E) 25
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Re: Kim bought a total of $2.65 worth of postage stamps in four [#permalink]
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29 Jun 2014, 03:57
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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 5cent and 25cent stamps and twice as many 10cent stamps as 5cent stamps, what is the least number of 1cent 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 10cent stamps as 5cent 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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Re: Kim bought a total of $2.65 worth of postage stamps in four [#permalink]
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29 Jun 2014, 14:14
manideepgre wrote: Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5cent and 25cent stamps and twice as many 10cent stamps as 5cent stamps, what is the least number of 1cent stamps she could have bought? (A) 5 (B) 10 (C) 15 (D) 20 (E) 25 Moved to the proper subforum. Please post your questions in the PS/DS/SC/CR/RC subforums and tag properly. Thank you.
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Kim bought a total of $2.65 worth of postage stamps in four [#permalink]
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19 Nov 2014, 01:56
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Answer = C = 15 ............... 1 ............... 5 ................ 10 ............... 25 ................ Total ............... b ................ a ................... 2a ............ a ...................... 265 b + 5a + 20a + 25a = 265 b + 50a = 265 To have b "minimum"; 50a has to be maximum Nearest to 265 is 250 which is divisibly by 50 265250 = 15 b = 15
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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, 07:25
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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 5cent and 25cent stamps and twice as many 10cent stamps as 5cent stamps, what is the least number of 1cent 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
Answer: C



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Kim bought a total of $2.65 worth of postage stamps in four [#permalink]
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07 May 2016, 04:30
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]
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22 Oct 2016, 17:47
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
265250 = 15



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Re: Kim bought a total of $2.65 worth of postage stamps in four [#permalink]
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12 Nov 2017, 08:46
manideepgre wrote: Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5cent and 25cent stamps and twice as many 10cent stamps as 5cent stamps, what is the least number of 1cent stamps she could have bought?
(A) 5 (B) 10 (C) 15 (D) 20 (E) 25 We can create the following variables: a = number of 1cent stamps b = number of 5cent stamps c = number of 10cent stamps d = number of 25cent 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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Re: Kim bought a total of $2.65 worth of postage stamps in four [#permalink]
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23 Feb 2018, 10:49
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 5cent and 25cent stamps and twice as many 10cent stamps as 5cent stamps, what is the least number of 1cent stamps she could have bought?
(A) 5 (B) 10 (C) 15 (D) 20 (E) 25 We can create the following variables: a = number of 1cent stamps b = number of 5cent stamps c = number of 10cent stamps d = number of 25cent 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 Dear JeffTargetTestPrep , where can I find a detailed general explanation about the topic? If it exists clearly




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