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

Question

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

vitaliyGMAT · Answer

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

rikson · Answer

Can any one explain these??..

rishit1080 · Answer

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.

sb0541 · Answer

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 .

rikson · Answer

Thanks to both of you(Rishit and Sb),,,actually I took these easy question in different format,,,now it's clear,,,thanks,,,,

ScottTargetTestPrep · Answer

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 =   + 1

# of multiples of 3 =   +1 = 21/3 + 1 = 8

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

Answer: C

habil · Answer

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

Asariol · Answer

How do you know this uses inclusive counting?

ScottTargetTestPrep · Answer

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).

MathRevolution · Answer

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

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

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

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