Latest Elitmus Aptitude Question SOLUTION: If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place

When you convert a decimal number into octal form, think which numbers end with zero. The multiples of 8.
For example (8)10=(10)8, (16)10=(20)8, (24)10=(30)8, (32)10=(40)8, (40)10=(50)8, . . .
So out of five numbers end with zero in octal form, only one (50)8=(40)10 has zero in decimal format. So the required probability is 1/5

My initial thinking:
Let the number be "ab0" in octal form. Remember here a, b < 8. So total possibilities are 64.
If we convert the given number into decimal format, then (ab0)8 = 64a + 8b.
We will find for how many values of 64a + 8b ends in zero.
64a + 8b = 60a + (4a + 8b)
For (0,0), (1, 2), (1, 7), (2, 4), (3, 1), (3, 6), (4, 3), (5, 5), (6, 2), (6, 7), (7, 4), (0, 0) the above expression gives unit digit 0.
So required probability = 11/64

never 0 .hoga..

1/5. For the no.s like (50)base8 = 5x8=(40)base10 , then (100)base8 = 10x8=(80)base10, so on ; octal no.s like 50, 100, 150, ..... every 5th no. ending with zero.

answer will be 1/10 because only we have 10 decimal numbers so last digit will be any thing from these numbers .
The possibility of zero at unit place is 1/10

if we convert any octal no. with zero as unit digit into decimal value then only two cases arise between 8to 80 where we get 0 as unit digit that pattern will repeat for next pattern.total outcomes=10,desired outcomes=2 so required probability= 2/10=1/5

For simplification, initially imagine the octal no. x has 2 digits ( _ _ ). Given, last digit is 0, so our no. becomes ( _ 0 ).
Now, what can be the values that can fill our blank?
( 0 0), ( 1 0), ( 2 0), ( 3 0), ( 4 0), ( 5 0), ( 6 0), ( 7 0) [remember, all are in hexadecimal]
Which when changed to decimal are respectively:
0, 8, 16, 24, 32, 40, 48, 56
Out of which, how many end with 0 at unit's place? 2
so, for a 2 digit octal no. having 0 in unit's place, 2 out of 8 numbers have 0 in its units place of its respective decimal no.
So, required probability for a 2-digit no. = 2/8 = 1/4

Now, let's imagine the hexadecimal no. is 3 digit. ( _ _ 0 )
The two blanks will have 64 combinations, right?
Out of those 64 combinations, how many combinations will have 0 in its unit's place of its respective decimal representation? (Do some work taking help from the 2 digit part)
(Hint: Write all multiples of 8 ending with 0 until 512 in decimal, then write its respective hexadecimal)
After working, you will get the combinations as:
000, 050, 120, 170, 240, 310, 360, 430, 500, 550, 620, 670, 740 [All nos. are in hexadecimal]
Which are 13 nos. among a combination of 64
So, required probability for a 3 digit no.: 13/64

As the probability depends on the no. of digits in the hexadecimal no., the information in the question is not enough to solve the problem.

(50)base8 also gives (40)base10....then how will be (NEVER) ans@jha

@shivam,how many digits does X have? because the logic behind answer 1/5 is not compatible with three digit octal number or 4 digit octal number.

in 10th place 1 to 9 any no. can be there so total possib will be 9, when only 5 iss on that place then it will give 0 at unit place.

Its not possible to answer this without knowing the number of digits in "number x" (octal form)