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In a set of 24 cards, each card is numbered with a different positive

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Joined: 02 Sep 2009
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In a set of 24 cards, each card is numbered with a different positive  [#permalink]

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25 Jun 2018, 05:51
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5% (low)

Question Stats:

87% (01:14) correct 13% (01:45) wrong based on 701 sessions

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In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

NEW question from GMAT® Official Guide 2019

(PS02127)

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Re: In a set of 24 cards, each card is numbered with a different positive  [#permalink]

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25 Jun 2018, 06:00
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Solution

Given:
• In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24
• One card will be drawn at random from the set

To find:
• The probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7

Approach and Working:
• From 1 to 24, inclusive, the number of cases where we can get a number which is divisible by both 2 and 3 = numbers divisible by 6 = 4 numbers (6, 12, 18, and 24)
• From 1 to 24, inclusive, the number of cases where we can get a number which is divisible by 7 = 3 numbers (7, 14, and 21)
• Total number of favourable cases = 4 + 3 = 7
• Therefore, the required probability = $$\frac{7}{24}$$

Hence, the correct answer is option C.

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Re: In a set of 24 cards, each card is numbered with a different positive  [#permalink]

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25 Jun 2018, 07:22
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Top Contributor
Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes)

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = number of integers that are divisible by both 2 and 3 OR are divisible by 7/24

number of integers that are divisible by both 2 and 3 OR are divisible by 7
The integers that meet this condition are: 6, 7, 12, 14, 18, 21, 24
There are 7 such numbers

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = 7/24

Cheers,
Brent
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Re: In a set of 24 cards, each card is numbered with a different positive  [#permalink]

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25 Jun 2018, 07:31
Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

NEW question from GMAT® Official Guide 2019

(PS02127)

Total number 24
Number div by 2 and 3 = multiple of LCM of 2 and 3 till 24
LCM of 2 and 3 =2*3=6
Multiple of 6 till 24=6,12,18,24...4 number
Number div by 7 =7,14,21...3 numbers
Total 3+4=7

Probability to pick these 7 out of 24 is 7/24
C
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Re: In a set of 24 cards, each card is numbered with a different positive  [#permalink]

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26 Jun 2018, 18:13
1
Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24

The numbers that are divisible by both 2 and 3 are 6, 12, 18, and 24.

The numbers that are divisible by 7 are 7, 14, and 21.

So the probability is 7/24.

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Re: In a set of 24 cards, each card is numbered with a different positive  [#permalink]

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03 Jan 2019, 20:11
how come 1 is not included? is one not divisible by 2,3, and 7?
Re: In a set of 24 cards, each card is numbered with a different positive   [#permalink] 03 Jan 2019, 20:11
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