natural number

[maihuna]

the sum of all possible two digits number formed from three different one digit natural number when divided by the sum of original three numbers is equal to;


1. 18
2. 22
3. 36
4. 54
5. none

[Musiq]

This is a hard question.... 700+, in my opinion...maybe even more.

Anyways here goes:

Let the 3 single digits be A , B and C.

The different 2 digit numbers that can be created are:
AB ------> The numerical value is 10*A + 1*B
BA ------> The numerical value is 10*B + 1*A
AC ------> The numerical value is 10*A + 1*C
CA ------> The numerical value is 10*C + 1*A
BC ------> The numerical value is 10*B + 1*C
CB ------> The numerical value is 10*C + 1*B

Add all these numbers to get (22*A ) + (22*B) + (22*C) as the value of the sum of all possible 2 digit numbers. This can be written as {22} {A +B+C)

We want to determine outcome when above is divided by {A+B+C}.

Answer is 22...or option B.

[dumb.doofus]

I think the simplest way to do this question is plug in values..
just take 1,2 & 3

quickly write the numbers.. 12, 13, 21,23,31,32.. sum is 132 and so divide it by 6 i.e. 1+2+3

doesnt take much time.. answer 22

[Musiq]

I think the simplest way to do this question is plug in values..
just take 1,2 & 3

quickly write the numbers.. 12, 13, 21,23,31,32.. sum is 132 and so divide it by 6 i.e. 1+2+3

doesnt take much time.. answer 22

Picking numbers as a strategy immediately loses a LOT OF WIND as soon as one of the options reads " None of the above".

You can never be sure...unless of course, the question says "Which of the following COULD".

This is not a "COULD" but a "MUST" situation. I would highly recommend NOT picking numbers and instead learning the simple rules of digits.

Of course, this advise is only for those of us who are planning to take the test.

For the afficionados and the connoisseurs, my rant doesnt mean much.

[m&m]

you can actually do this question in your head, here's how:

digit 1= x
digit 2=y
digit 3=z

dividing by the sum you have:

x' = x/(x+y+z)
y' = x/(x+y+z)
z' = z/(x+y+z)


ignoring that we have a 2 digit number for a moment, we know that if we form groups of 2, each x', y', z' will get picked twice. so sum of all that is:

(2x + 2y + 2z)/(x+y+z) = 2

now we want to form a 2 digit number and the unit will be 2 and the tens will be 2. So 2*10 + 2*1 = 22