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

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

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New post 25 Jun 2018, 05:51
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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



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(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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New post 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.

Answer: 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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New post 25 Jun 2018, 07:22
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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

Answer: C

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

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New post 26 Jun 2018, 18:13
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.

Answer: C
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Re: In a set of 24 cards, each card is numbered with a different positive &nbs [#permalink] 26 Jun 2018, 18:13
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