Latest Elitmus Aptitude Question SOLUTION: Asked on 17th JAN, 2016 A number is of form abbbbbba , where a is not equal to b, and neither a nor b equals 0(zero). eg. 12222221 . What is the probability that such possible n

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(#M40155308) ELITMUS Numerical Ability question Probability Keep an EYE

Asked on 17th JAN, 2016

A number is of form abbbbbba , where a is not equal to b, and neither a nor b equals 0(zero). eg. 12222221 . What is the probability that such possible numbers are divisible by 3.

Asked In Elitmus Amit Sankhyan (8 years ago)

Unsolved Read Solution (10)

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kundan prasad  (8 years ago)

i think
(2a+6b)/3
2a/3+6b/3 (6b alwayes divisible by 3, so we have chek only for 3a part)
2a/3 divisible by 3 only when
a is multiple of 3 like a=3,6,9
so we have only three condition at which (2a+6b) or abbbbbba will divisible 3

Is this solution Helpfull? Yes (38) | No   

dytort  (8 years ago)

total of such numbers = 72
no. of such numbers = 24
probablity = 1/3

Is this solution Helpfull? Yes (30) | No (7)  

akhilnaarajkn  (8 years ago)

Ans: 1/3

a can be any of the 9 numbers from (1,2,3,...,9).
b can be any of the remaining 8 numbers.
total: 9×8 = 72

i.e., there can be 72 possible numbers.

If the number is divisible by 3,
a+b+b+b+b+b+b+a is divisible by 3.
=> 2a + 6b is divisible by 3.
=> 2a is divisible by 3 (because 6b is always divisible by 3).
=> a can be (3,6,9)

Thus, if the number is divisible by 3,
a can be any of the 3 numbers from (3,6,9).
b can be any of the 8 remaining numbers.
total: 3×8 = 24

i.e., there can be 24 possible numbers which are divisible by 3

Required probability =24/72=1/3

Is this solution Helpfull? Yes (29) | No (1)  

ashish thomas  (8 years ago)

answer 8/27
i dont know whether this is correct or not but here is what according to me can be the way to solve it.
a number to be divisible by 3 sum of its unit digit must be divisible by 3 therefore
6b+2a=3z
dividing above equation by 3 we get
2b+2/3a=z
2/3a should be integer therefore a can only be 3,6,9.
now
if a=3
probablity of selecting 3 is 1/9
probality of selecting any number other than 3 for b is 8/9
so 1/9*8/9=8/81
similiarly we will do for a= 6 and a=9
and we will get 8/81 for all of them.
now we will add all the three
3(8/81)
hence answer is 8/27

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Amit Sankhyan  (8 years ago)

@Dytort: The total number of such numbers are 72 ? how ? It should be 36 instead !

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Deepak Sehgal  (8 years ago)

~The total no. Of such numbers should be- 9×8×1×1×1×1×1×1= 72
~The number of such numbers divisible by 3 will be-
1) By keeping 3 at 1st and last place- 9numbers
2) By keeping 6 at 1st and last place- 9 numbers
3) By keeping 9 at 1st and last place- 9numbers
so total of such numbers= 9+9+9= 27
~ Probability= 72/27 = 8/3

Is this solution Helpfull? Yes (1) | No (5)  

Sneha Singh  (8 years ago)

total such nos=72
no. of such numbers will be 24
hence p=24/72

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The Trojan  (8 years ago)

total of such numbers = 36
no. of such numbers = 24
probability = 2/3

(3c1*8c1)/9c2

Is this solution Helpfull? Yes | No (2)  

Saquib Akhter  (8 years ago)

@Deepak how 9. If u have already chose a number and ques says that a & b cant be 0

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SAUMRIT  (6 years ago)

Here b can be any value between 0-9 and a can be only multiple of 3 i.e 3,6,9.
So probability depends upon what value a take i.e 3/10
Hence ans is 3/10.

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