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# Joseph has exactly 185 pesos in 50 peso coins, 25 peso

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Joseph has exactly 185 pesos in 50 peso coins, 25 peso [#permalink]

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01 Sep 2010, 16:22
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Question Stats:

67% (02:19) correct 33% (01:24) wrong based on 188 sessions

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Joseph has exactly 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins only, and he has at least one of each. How many 25 peso coins does he have?

(1) He has exactly two 50 peso coins.
(2) He has fewer than 40 1 peso coins.
[Reveal] Spoiler: OA

Last edited by zisis on 02 Sep 2010, 18:43, edited 1 time in total.
Manager
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Posts: 225
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Kudos [?]: 317 [0], given: 16

Re: The Quest for 700: Weekly GMAT Challenge peso [#permalink]

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01 Sep 2010, 16:24
zisis wrote:
Joseph has exactly 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins only, and he has at least one of each. How many 25 peso coins does he have?

1) He has exactly two 50 peso coins.

2) He has fewer than 40 1 peso coins.

IMO E

(1)if 2*50, then could have 1 or 3 25peso

(2) could have 1,3, or 5 25 peso

1 and 2: could ahev 1 or 3 25peso
Manager
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Posts: 225
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Kudos [?]: 317 [1] , given: 16

Re: The Quest for 700: Weekly GMAT Challenge peso [#permalink]

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02 Sep 2010, 18:43
1
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zisis wrote:
Joseph has exactly 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins only, and he has at least one of each. How many 25 peso coins does he have?

1) He has exactly two 50 peso coins.

2) He has fewer than 40 1 peso coins.

A: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B: Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C: BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D: EACH statement ALONE is sufficient.

E: Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

OA
[Reveal] Spoiler:
The easiest way to count the number of possibilities for how Joseph could have 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins is to count via the largest increment, 50. Since 3 is the max number of 50 peso coins he could have, here are all the possibilities:
50 peso 25 peso 1 peso
3 1 10
2 3 10
2 2 35
2 1 60
1 5 10
1 4 35
1 3 60
1 2 85
1 1 110

1) INSUFFICIENT. Joseph has exactly 2 50 peso coins. We now have three options, all with different numbers of 25 peso coins.
50 peso 25 peso 1 peso
2 3 10
2 2 35
2 1 60

2) INSUFFICIENT. He has fewer than 40 1 peso coins. We now have 5 options, with 5 different numbers for how many 25 peso coins he has.
50 peso 25 peso 1 peso
3 1 10
2 3 10
2 2 35
1 5 10
1 4 35

1+2) INSUFFICIENT. He has exactly 2 50 peso coins and fewer than 40 1 peso coins. Since the chart for statement 1 above is shorter, refer back to it, and eliminate the 1 option that has more than 40 1 peso coins. Two options remain:
50 peso 25 peso 1 peso
2 3 10
2 2 35

Joseph could have either 2 or 3 25 peso coins.

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Re: Joseph has exactly 185 pesos in 50 peso coins, 25 peso [#permalink]

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24 Oct 2013, 18:08
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Re: Joseph has exactly 185 pesos in 50 peso coins, 25 peso [#permalink]

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25 Oct 2013, 22:09
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Expert's post
zisis wrote:
Joseph has exactly 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins only, and he has at least one of each. How many 25 peso coins does he have?

(1) He has exactly two 50 peso coins.
(2) He has fewer than 40 1 peso coins.

We have that 50x + 25y + z = 185, where x, y, z are positive integers.
Question is for which positive integer values of x, y, and z will the equation hold true??

S1 :- INSUFFICIENT
x = 2 ---------> 100 + 25y + z = 185 -----------> 25y + z = 85
From the above equation we will get three possible solutions of y and z
y=1 and z=60
y=2 and z=35
y=3 and z=10

S2 :- INSUFFICIENT
Z < 40 ---------> Trying various values of z from 1 to 39, we will get multiple different values of x and y

S1 + S2 :- INSUFFICIENT
Despite knowing that Z can not be greater than 40, we still have two cases
y=2 and z=35
y=3 and z=10

Hope that helps!
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Re: Joseph has exactly 185 pesos in 50 peso coins, 25 peso [#permalink]

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19 May 2016, 19:01
zisis wrote:
Joseph has exactly 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins only, and he has at least one of each. How many 25 peso coins does he have?

(1) He has exactly two 50 peso coins.
(2) He has fewer than 40 1 peso coins.

Suppose:
50A+25B+C=185

1. A=2
25B+C=85
we can have:
B=1 -> C=65
B=2 -> C=35
B=3 -> C=10
not sufficient.

2. C<40
A=1
C=35
B=4
or
A=2
C=10
B=3

not sufficient.

1+2
A=2
C=10
B=3

or
A=1
C=35
B=4

2 outcomes - not sufficient
Re: Joseph has exactly 185 pesos in 50 peso coins, 25 peso   [#permalink] 19 May 2016, 19:01
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