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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
Joined: 16 Aug 2015
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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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15 Dec 2017, 08:29
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25% (medium)

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76% (01:44) correct 24% (02:13) wrong based on 60 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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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" VP Joined: 07 Dec 2014 Posts: 1160 Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p [#permalink] ### Show Tags 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 Senior Manager Joined: 07 Jul 2012 Posts: 376 Location: India Concentration: Finance, Accounting GPA: 3.5 Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p [#permalink] ### Show Tags 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. _________________ Kindly hit kudos if my post helps! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7107 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: In the 5-digit integer 54xy2, x and y are single digits. What is the p   [#permalink] 18 Dec 2017, 00:29
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# In the 5-digit integer 54xy2, x and y are single digits. What is the p

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