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In the 5-digit integer 54xy2, x and y are single digits. What is the p

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Math Revolution GMAT Instructor
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In the 5-digit integer 54xy2, x and y are single digits. What is the p  [#permalink]

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New post 15 Dec 2017, 08:29
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

82% (00:58) correct 18% (02:02) wrong based on 49 sessions

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[GMAT math practice question]

In the 5-digit integer \(54xy2\), x and y are single digits. What is the probability that \(54xy2\) is divisible by 4?

A. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)

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Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p  [#permalink]

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New post 15 Dec 2017, 11:14
MathRevolution wrote:
[GMAT math practice question]

In the 5-digit integer \(54xy2\), x and y are single digits. What is the probability that \(54xy2\) is divisible by 4?

A. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{4}\)


any even digit in the tens place will work
any digit in the hundreds place will work
1/2*1=1/2
C
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Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p  [#permalink]

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New post 15 Dec 2017, 23:20
A number is divisible by 4 if the last two digits form a number divisible by 4.
So 54xy2 is divisible by 4 when y= (1,3,5,7,9)= 5 possible cases

54xy2 is not divisible by 4 when y= (0,2,4,6,8)= 5 possible cases

Required probability= \(\frac{5}{10}\)= \(\frac{1}{2}\)

Answer: C.
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Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p  [#permalink]

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New post 18 Dec 2017, 00:29
=>
The divisibility of \(54xy2\) by \(4\) depends on the last two digits only. So, \(54xy2\) is divisible by \(4\) exactly when \(10y + 2\) is divisible by \(4\).
Out of the possible values for \(10y + 2, 12, 32, 52, 72, 92\) are divisible by \(4\), but \(02, 22, 42, 62, 82\) are not. Therefore, the required probability is \(\frac{1}{2}\).

Therefore, the answer is C.

Answer : C
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Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p &nbs [#permalink] 18 Dec 2017, 00:29
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