Question

One column of numbers consists of 33, 58, and 17. When the digits of the

Answer 1

Answer:

Only (B) is correct.

Step-by-step explanation:

Let, if the numbers in 2nd column are added  , the sum is  same as the sum of numbers in the 1st column and obviously all are integers and the sum is,

   (33 + 58 +17) =108---------(1)

Now , let the  3 numbers in 2nd column are,

 1. $a \times 100 + b \times 10 + c$

2.  $d \times 100 + e \times 10 + f$

and

3. $g \times 100 + h \times 10 + i$

So, the sum of the digits of the numbers  in the 2nd colum  are,

 ( a + b +c +d +e +f +g +h + i)  -------------(2)

Now, c +f +i =  8 or 18 (since the last digit is 8 and  0 ≤ c + f +i ≤ 27)

Now,

 sum of the numbers in 2nd column,

= $(a + d + g) \times 100 + (b +e +h) \times 10 + (c + f + i) =108$

  = $1 \times 100 + 0 \times 10 + 8$  -----------(3)

or = $0 \times 100 + 10 \times 10 + 8$ -------------(4)

or = $0 \times 100 + 9 \times 10 + 18$ -----------(5)

it must fall in one of above 3 forms i.e., (3) or (4) or (5)

in case (3),

(a + b + c + d + e + f + g + h + i) =  (1 +8) = 9 ---- (6)

in case (4)

(a + b + c + d + e + f + g + h + i) =  (10+8) = 18 and (1 + 8) =9 ---------(7)

in case (5)

(a + b + c + d + e + f + g + h + i) = (9 + 18) = 27 and 2 + 7 = 9 -------(8)

So, when the sum of numbers in the 2nd column is equal to the sum of numbers in the 1st column, the sum of digits of the numbers in  the 2nd column will be 9.

So, when the sum of digits of the numbers in  the 2nd column is not  9, the sum of those numbers won't be equal to the sum of numbers in the 1st column.

So, only (B) is correct.

[Note , that (C) is false. e.g. when the numbers in 2nd column is 6 , 1 and 2,  then (6 +1 +2) = 9 ≠ 108 .]