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26 Feb 2024, 02:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:20) correct
30%
(02:28)
wrong
based on 263
sessions
History
Date
Time
Result
Not Attempted Yet
What is the probability that a two digit positive integer selected at random will be divisible by either 3 or 7?
A. 2/9
B. 3/9
C. 42/90
D. 46/90
E. 39/90
A. 2/9
B. 3/9
C. 42/90
D. 46/90
E. 39/90
26 Feb 2024, 04:06
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numbers divisible by 3 = 30
numbers divisibe by 7 =13
numbers common (both divisible by 3 )=4
And total number of 2 digit numbers =90
P=39/90
E
numbers divisibe by 7 =13
numbers common (both divisible by 3 )=4
And total number of 2 digit numbers =90
P=39/90
E
jack5397
Joined: 13 Sep 2020
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Location: India
Concentration: General Management, Strategy
Schools: ISB '27 IIMA '26 INSEAD '26
GMAT Focus 1: 575 Q79 V79 DI77 
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26 Feb 2024, 05:05
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IMO - E
Numbers divisible by 3 and 7 can be found using the AP.
For numbers divisible by 3 : 99 = 12+(n-1)*3 ; n=30
For numbers divisible by 3 : 98 = 14+(n-1)*7 ; n=13
Now eliminate common terms multiples of 3&7 which are multiple of 21 : 21,42,63,84
Total numbers divisible by 3 or 7 = 30+13-4= 39 ; total 2 digit numbers = 90
39/90.
Numbers divisible by 3 and 7 can be found using the AP.
For numbers divisible by 3 : 99 = 12+(n-1)*3 ; n=30
For numbers divisible by 3 : 98 = 14+(n-1)*7 ; n=13
Now eliminate common terms multiples of 3&7 which are multiple of 21 : 21,42,63,84
Total numbers divisible by 3 or 7 = 30+13-4= 39 ; total 2 digit numbers = 90
39/90.
