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Erica has $460 in 5-and 10-dollar bills only. If she has

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Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post Updated on: 12 Sep 2013, 01:50
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

80% (02:08) correct 20% (02:19) wrong based on 308 sessions

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Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?

(A) 32
(B) 30
(C) 29
(D) 28
(E) 27

Originally posted by bmwhype2 on 28 May 2007, 22:32.
Last edited by Bunuel on 12 Sep 2013, 01:50, edited 1 time in total.
Edited the question and added the OA.
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Re: Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post 12 Sep 2013, 01:57
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New post 28 May 2007, 22:51
1
5x + 10y = 460
x > y

Pick an easy number to work with.
suppose x=30.
Then you have 150.
You need another 310 and that means you need y=31 which is greater than x.

Therefore, x > 30.

The only answer choice where x > 30 is if x = 32.

A is the answer.
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New post 31 May 2007, 18:05
ricokevin wrote:
5x + 10y = 460
x > y

Pick an easy number to work with.
suppose x=30.
Then you have 150.
You need another 310 and that means you need y=31 which is greater than x.

Therefore, x > 30.

The only answer choice where x > 30 is if x = 32.

A is the answer.


hmm, never struck me to use plugging in while solving it algebraically.
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New post 31 May 2007, 22:48
also, C and E won't work b/c when multiplied by 5 they give integer 5 in the units digit. subtracting xx5 from 460 will give you xx5. xx5/10 = integer.5.
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Re: PS Erica and bills  [#permalink]

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New post 01 Jun 2007, 06:14
bmwhype2 wrote:
Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?
(A) 32
(B) 30
(C) 29
(D) 28
(E) 27


If she had an equal number of each, , she could have 30 of each for a total of $450, and another 2 fivers to make $460. TOTAL = 32. Note that all other choices would give less than $450.
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Re: Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post 11 Sep 2013, 14:01
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I took way too long for this one, but should be pretty straightforward.

First thing I did, simplify x + 2y = 92.
Then we also have x>y

From the first equation, y= 46-x/2

So we replace in the inequality:

x>46-x/2
3x/2>46
x>92/3
x>30.6

Since 'x' has to be an integer then x=32.

Answer is (A)
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Re: Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post 04 Dec 2013, 15:44
OK so there is a 2:1 ratio between 5 and 10 dollar notes. This means that the answer choice can only be even.
so that's A, B or D.

Test b: number of 5 dollar notes equal 30. total revenue=150. To make up to 460 total, you need 31 ten dollar notes.
Since there is a greater number of $10 dollar notes than $5 dollar notes. Requirement not met.
We also know that D is wrong now...as it would give more $10 dollar notes than $5. Therefore A must be the right choice.
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Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post 10 Aug 2017, 13:59
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bmwhype2 wrote:
Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?

(A) 32
(B) 30
(C) 29
(D) 28
(E) 27


let x=least possible number of fives
5x+10(x-1)=460
x=31.3 :(
5x+10(x-2)=460
x=32 :-D
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Re: Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post 15 Aug 2017, 10:52
bmwhype2 wrote:
Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?

(A) 32
(B) 30
(C) 29
(D) 28
(E) 27


We can create the following equation in which f = the number of 5-dollar bills and t = the number of 10-dollar bills.

5f + 10t = 460

f + 2t = 92

If the number of 5-dollar bills and the number of 10-dollar bills are equal, i.e., f = t, then we have:

f + 2f = 92

3f = 92

f = 30⅔

Since f, the number of 5-dollar bills, must be a whole number, and f > t, f should be at least 31. Let’s check whether f can be 31t:

If f = 31, then 31 + 2t = 92 → 2t = 61 → t = 30.5. However, t must be whole number also, so f can’t be 31.

Now let’s check if f can be 32:

If f = 32, then 32 + 2t = 92 → 2t = 60 → t = 30. We see that t is a whole number and f > t. So, the least possible value of f is 32.

Answer: A
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Erica has $460 in 5-and 10-dollar bills only. If she has  [#permalink]

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New post 01 Oct 2018, 13:34
Bunuel wrote:
bmwhype2 wrote:
Erica has $460 in 5-and 10-dollar bills only. If she has fewer 10-than 5-dollar bills, what is the least possible number of 5-dollar bills she could have?

(A) 32
(B) 30
(C) 29
(D) 28
(E) 27


Given that:
5x+10y=460 --> 10y=460-5x;
x>y --> 10x>10y.

10x>460-5x --> 15x>460 --> x>30.something --> x=32.

Answer: A.


if there are 32 FIVES, 32*5 = 160 then since 10ns are fewer then 5, it means 10ns must be fewer.... :) iron logic :lol: if so then 31*10 = 310 +160 =470 WTM :? whats the matter :)
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Erica has $460 in 5-and 10-dollar bills only. If she has   [#permalink] 01 Oct 2018, 13:34
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