If a and b are two-digit positive numbers and c is a three digit posit

Solution

Step 1: Analyse Question Stem

• a and b are positive two-digit numbers.

o a < 100 and b < 100
• c is a three-digit positive number
• c = a + b ……………Eq.(i)

o So, c < 100 + 100 ⟹ c < 200
o So, c must be in the form of 1xy, where x and y are single digit integers.
• We need to find if the value of y is equal to the unit’s digit of a or not.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

• According to this statement: x = 1 and y = 1

o Or c = 111 …….Eq.(ii)
• Also, since all the digits of a are same, so possible value of a are { 11, 22, 33, 44, 55, 66, 77, 88, 99}

o Now, unit’s digit of c and a will be equal only if a = 11, else they won’t be. So let’s check if a can be 11 or not.

 If a = 11, so from Eq.(i) and Eq.(ii), we have, b = c – a = 111-11 = 100

• Which is not possible as b < 100
• Hence, unit’s digit of c ≠ unit’s digit of a.

• Let us assume that a = 10p + q, where p is a single-digit positive integer and q is a single digit non-negative integer.
• Thus, b must be in the form of b = 10p + r, where r is a non-negative single digit integer.
• Therefore, c = a + b = 20p + q + r.

o Now, unit’s digit of c = ( 0 + q + r) = (q + r)
o However, we don’t know the value of r.

 If r = 0 then unit’s digit of c = unit’s digit of a
 And if r ≠ 0, then unit’s digit of c ≠ unit’s digit of a
• We are getting contradictory results.