NIELIT 2017 DEC Scientist B - Section B: 53

Question

When the sum of all possible two digit numbers formed from three different one digit natural numbers are divided by sum of the original three numbers, the result is

• A. $26$
• B. $24$
• C. $20$
• D. $22$

Answer 1

let's take three diffrent natural no = 2, 3, 4

two digit no formed by these digits are :   23, 32 , 42, 24, 34, 43

sum of these two digits no =    23 + 32  + 42 + 24 + 34 + 43 = 198

sum of those three digits =    2+ 3 + 4 = 9

 

so, 198 / 9 = 22

So, "D is correct"

Answer 2

Consider $a,b$ and $c$ as the $3$ original natural numbers. Then the sum of all possible 2 digit numbers would give us: $$\frac{2(a+b+c)*10 + 2(a+b+c)}{a+b+c} = 22$$

Comments

Explain the numerator part plz...

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$a\color{blue}b+c\color{blue}b+a\color{blue}c+b\color{blue}c+b\color{blue}a+c\color{blue}a = (2(a+b+c)*10) + \color{blue}{(a+b+c)} $

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All possible Numerators are :
$ (10*a)+b \\(10*a)+c\\(10*b)+a\\(10*b)+c \\(10*c)+a\\(10*c)+b$

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Thanks both of you.

Answer 3

Answer D

Let 3 one digit numbers be a,b,c

From these possible 3 digits number of different 2 digits numbers that can be formed are: 6 (ab,ba, ac,ca ,bc ,cb)

Value of ab will be equal to 10a+b

ba =10b + a

ac = 10a + c

similarly all can be written

sum of all 2 digits number = 22(a+ b+c )

$\Rightarrow$$\frac{22(a+ b+c )}{(a+ b+ c)}$ = 22