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GMAT 1: 760 Q51 V42
Posts: 10,057
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (02:31) correct
55%
(02:59)
wrong
based on 47
sessions
History
Date
Time
Result
Not Attempted Yet
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. \(\frac{9}{25}\)
A. \(\frac{1}{5}\)
B. \(\frac{6}{25}\)
C. \(\frac{7}{25}\)
D. \(\frac{8}{25}\)
E. \(\frac{9}{25}\)
GMAT 1: 760 Q51 V42
Posts: 10,057
Official Solution:
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
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
GMAT 1: 670 Q48 V34
Posts: 4
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
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
MathRevolution
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.
pulak1988
Joined: 15 Jan 2019
Last visit: 08 Sep 2020
Posts: 47
Location: India
Schools: ISB '20 IIMA Rotman '22 GMBA '21 (S)
GMAT 1: 650 Q42 V38
GPA: 4
which is the correct method??
Posts: 1
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
Very vauge and incomplete question and explaination. How can you put this question on the forum?
Posts: 38
ArnoldMagoto
@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.
I think this is a high-quality question and I agree with explanation.
_________________
_________________
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.
Frankly, i hate all those so called ''Math Revolution'' questions. They are everyting except gmat like and are often incomplete and vauge.
Rebaz
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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Bunuel
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!
GMAT-Club-Forum-so5na6lv.png [ 73.34 KiB | Viewed 655 times ]
Bunuel
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!
Attachment:
GMAT-Club-Forum-so5na6lv.png [ 73.34 KiB | Viewed 655 times ]
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