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@DaniyalAlwani

To get integer which are divisible by 6 from 1 to 100; max would be 100/6: 16

["DaniyalAlwani"][url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Archit3110][b]Archit3110[/b][/url]

Terms divisible by 6 > (6 +96)/6 + 1

=17+1 = 18 terms.

What am I doing wrong here? Please explain[/quote]

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Bunuel
If an integer is to be chosen at random from the integer 1 to 100, inclusive, what is the probability that the integer chosen will be divisible by 6?

A. 3/20
B. 4/25
C. 17/100
D. 1/6
E. 1/3

First such number = 6
Last such number = 96 (figure out that 90 is divisible by 6 and then add 6)

Total number of numbers = (96-6)/6 = 90/6 = 15 + 1 = 16 (I added 1 because to be able to count all the extremes)

Now 16/100=4/25

Answer B
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