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Of a set of 25 consecutive integers beginning with 4, what is the prob

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Bunuel
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Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   13 Dec 2016, 07:06
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Of a set of 25 consecutive integers beginning with 4, what is the probability that a number selected at random will be divisible by 3?

A. 13/25
B. 9/25
C. 8/25
D. 6/25
E. 3/8
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vitaliyGMAT
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   13 Dec 2016, 08:27
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We need to find # of multiples of 3 within given set. Our set starts with 4 and has 25 elements (consecutive numbers). Last element will be 4+25 = 29

Fist multiple of 3 is 6, last is 27

\(\frac{27-6}{3} + 1 = 8\)

Our probability is (# multiples of 3) / (total # of elements in the set) \(= \frac{8}{25}\)

Answer C
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   08 Jan 2017, 01:51
Can any one explain these??..
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Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   Updated on: 08 Jan 2017, 08:54
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rikson wrote:
Can any one explain these??..


Yes sir. The question says starting with 4 and 25 consecutive integers. So the set you're working with contains numbers 4-29 inclusive. Factors of 3 between 4 and 29 are 6, 9, 12, 15, 18, 21, 24, 27. Answer you want is number of outcomes wanted/total possible outcomes.

So answer is 8/25.

Originally posted by rishit1080 on 08 Jan 2017, 02:34.
Last edited by rishit1080 on 08 Jan 2017, 08:54, edited 1 time in total.
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   08 Jan 2017, 07:17
rikson wrote:
Can any one explain these??..



The 25 consecutive numbers starting from 4 will end at 28 .
To calculate the number of consecutive integers , the formula is (last - first +1) .

What numbers that are divisible by 3 . You can use the formula or manually count it .
Doing manually , numbers are : 6 , 9, 12, 15, 18, 21, 24, and 27 (8 in total )

probability(that a number is divisible by 3) is = 8 / 25

Answer is C .
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   09 Jan 2017, 06:37
Thanks to both of you(Rishit and Sb),,,actually I took these easy question in different format,,,now it's clear,,,thanks,,,,
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   11 Jan 2017, 07:58
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Bunuel wrote:
Of a set of 25 consecutive integers beginning with 4, what is the probability that a number selected at random will be divisible by 3?

A. 13/25
B. 9/25
C. 8/25
D. 6/25
E. 3/8


We are given that there is a set of 25 consecutive integers, starting with the number 4. Thus, the set of integers is from 4 to 28, inclusive. We need to determine how many multiples of 3 fall within that range. To do so, we can use the following formula:

# of multiples of 3 = [(largest multiple of 3 in the set - smallest multiple of 3 in the set)/3] + 1

# of multiples of 3 = [(27 - 6)/3] +1 = 21/3 + 1 = 8

Thus, the probability of selecting a multiple of 3 is 8/25.

Answer: C
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   06 Feb 2017, 08:19
25 consecutive integers are 4, 5, 6 ...... 28, 29
First multiple of 3 is 6
Last multiple of 3 is 27
Number of multiples of 3 is (27 - 6)/3 + 1 = 8
So, the required probability is 8/25

Answer : C
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   13 Oct 2020, 11:44
ScottTargetTestPrep wrote:
Bunuel wrote:
Of a set of 25 consecutive integers beginning with 4, what is the probability that a number selected at random will be divisible by 3?

A. 13/25
B. 9/25
C. 8/25
D. 6/25
E. 3/8


We are given that there is a set of 25 consecutive integers, starting with the number 4. Thus, the set of integers is from 4 to 28, inclusive. We need to determine how many multiples of 3 fall within that range. To do so, we can use the following formula:

# of multiples of 3 = [(largest multiple of 3 in the set - smallest multiple of 3 in the set)/3] + 1

# of multiples of 3 = [(27 - 6)/3] +1 = 21/3 + 1 = 8

Thus, the probability of selecting a multiple of 3 is 8/25.

Answer: C


How do you know this uses inclusive counting?
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   14 Oct 2020, 19:49
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Asariol wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
Of a set of 25 consecutive integers beginning with 4, what is the probability that a number selected at random will be divisible by 3?

A. 13/25
B. 9/25
C. 8/25
D. 6/25
E. 3/8


We are given that there is a set of 25 consecutive integers, starting with the number 4. Thus, the set of integers is from 4 to 28, inclusive. We need to determine how many multiples of 3 fall within that range. To do so, we can use the following formula:

# of multiples of 3 = [(largest multiple of 3 in the set - smallest multiple of 3 in the set)/3] + 1

# of multiples of 3 = [(27 - 6)/3] +1 = 21/3 + 1 = 8

Thus, the probability of selecting a multiple of 3 is 8/25.

Answer: C


How do you know this uses inclusive counting?


The exact words in the question are "Of a set of 25 consecutive integers beginning with 4...". This means that we list the cosecutive integers starting with 4 until there are 25 integers in our list. The number of integers between a and b inclusive is b - a + 1. Working this formula backwards, the greatest integer in the list can be found by:

b - 4 + 1 = 25

b = 28

Indeed, if we list all the integers between 4 and 28 inclusive, we will obtain 25 consecutive integers (beginning with 4).
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink] New post   14 Oct 2020, 22:34
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Total numbers: 25

First number: 4

Last number: 25 = Last - 4 + 1 => Last = 28

From 4 till 28 , total numbers divisible by 3 are: 6, 9, 12, 15, 18, 21, 24 and 27 = 8

Probability: \(\frac{8}{25}\)

Answer C
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Re: Of a set of 25 consecutive integers beginning with 4, what is the prob [#permalink]
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