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M61-04

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MathRevolution
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M61-04 [#permalink] New post   18 Jun 2018, 00:49
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In the 6-digit integer \(543,2xy\), \(x\) and \(y\) are chosen from the digits 0, 2, 4, 6 and 8. What is the probability that \(543,2xy\) is divisible by 8?

A. \(\frac{1}{5}\)
B. \(\frac{6}{25}\)
C. \(\frac{7}{25}\)
D. \(\frac{8}{25}\)
E. \(\frac{9}{25}\)
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Re M61-04 [#permalink] New post   18 Jun 2018, 00:49
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In the 6-digit integer \(543,2xy\), \(x\) and \(y\) are chosen from the digits 0, 2, 4, 6 and 8. What is the probability that \(543,2xy\) is divisible by 8?

A. \(\frac{1}{5}\)
B. \(\frac{6}{25}\)
C. \(\frac{7}{25}\)
D. \(\frac{8}{25}\)
E. \(\frac{9}{25}\)


An integer with three or more digits is divisible by 8 if and only if its last three digits from a number that is divisible by 8.

The values of \(2xy\) that are divisible by 8 are 200, 208, 224, 240, 248, 264, 280, and 288.

The total number of 6-digit integers of the form \(543,2xy\) is equal to the number of ways of choosing two digits from 0, 2, 4, 6 and 8, which is 5*5 = 25.

Thus, the probability that \(543,2xy\) is divisible by 8 is \(\frac{8}{25}\).


Answer: D
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ArnoldMagoto
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Re: M61-04 [#permalink] New post   01 Aug 2019, 09:51
Max value = 288
Min value = 200

Both these numbers are divisible by 8, therefore

>>> 288 - 200 = 88

> 88/8 = 11

Meaning there are a total of 11+1 = 12 numbers and the probability would be 12/25!

In this set "200, 208, 224, 240, 248, 264, 280, and 288" you have missed: 216, 232, 256 and 272 making a total of 12 numbers that are divisible by 8
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Re: M61-04 [#permalink] New post   18 Sep 2019, 16:46
MathRevolution wrote:
In the 6-digit integer \(543,2xy\), \(x\) and \(y\) are chosen from the digits 0, 2, 4, 6 and 8. What is the probability that \(543,2xy\) is divisible by 8?

A. \(\frac{1}{5}\)
B. \(\frac{6}{25}\)
C. \(\frac{7}{25}\)
D. \(\frac{8}{25}\)
E. \(\frac{9}{25}\)


Guys you actually missed some multiples of 8; in this range they are 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288 and 196 making the probability value actually equal to 13/25.
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Re: M61-04 [#permalink] New post   19 Sep 2019, 09:13
which is the correct method??
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Re: M61-04 [#permalink] New post   26 Oct 2019, 20:20
i got tripped up by the 'missing' multiples of 8 too, but if you look at the question, the 10's digit also has to be even, so those others are tossed
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Re: M61-04 [#permalink] New post   06 Jan 2020, 05:29
Very vauge and incomplete question and explaination. How can you put this question on the forum?
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Re: M61-04 [#permalink] New post   02 Feb 2020, 15:13
ArnoldMagoto wrote:
Max value = 288
Min value = 200

Both these numbers are divisible by 8, therefore

>>> 288 - 200 = 88

> 88/8 = 11

Meaning there are a total of 11+1 = 12 numbers and the probability would be 12/25!

In this set "200, 208, 224, 240, 248, 264, 280, and 288" you have missed: 216, 232, 256 and 272 making a total of 12 numbers that are divisible by 8

@ARNLODMAGOTO

In your explanation how did you choose 216, 232, 256, 272? since 1, 3,5,7 cannot be selected. As per question only 0,2,4,6,8 can be selected.
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Re M61-04 [#permalink] New post   12 Aug 2023, 09:51
I think this is a high-quality question and I agree with explanation.
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Re: M61-04 [#permalink] New post   13 Apr 2024, 06:38
This question doesnot make sense in any way and is incomplete and vauge! You will never ever get such incomplete question on your actual gmat test,meaning donot waste 1 second of your valuable time on this question.

Frankly, i hate all those so called ''Math Revolution'' questions. They are everyting except gmat like and are often incomplete and vauge.
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Re: M61-04 [#permalink] New post   13 Apr 2024, 12:45
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Rebaz wrote:
This question doesnot make sense in any way and is incomplete and vauge! You will never ever get such incomplete question on your actual gmat test,meaning donot waste 1 second of your valuable time on this question.

Frankly, i hate all those so called ''Math Revolution'' questions. They are everyting except gmat like and are often incomplete and vauge.

­How do even find these questions? They are no longer a part of GMAT Club Tests, the company, Math Revolution, does no exist anymore, and the topic is archived.  
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M61-04 [#permalink] New post   19 Apr 2024, 06:09
Bunuel
Bunuel wrote:
Rebaz wrote:
This question doesnot make sense in any way and is incomplete and vauge! You will never ever get such incomplete question on your actual gmat test,meaning donot waste 1 second of your valuable time on this question.

Frankly, i hate all those so called ''Math Revolution'' questions. They are everyting except gmat like and are often incomplete and vauge.

­How do even find these questions? They are no longer a part of GMAT Club Tests, the company, Math Revolution, does no exist anymore, and the topic is archived.  

­ 
2 Math Revolution quizes used to be part of the free tests along with two other quizes of Exam Pal and the 4 free Manhattan tests long time a go. I made these quizes/tests, and still when you make a quize, you can find it by going to Dashboard and then you can find these attempted quizes in your Custom Tests History. As long as your subscription is valid(paid subscription), you can select those old quizes, open them and still review these questions.­
 
See below my Custom tests History!

 
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