Correct option is (C) 308
Let the required number be a0b where 100's digit is a & 10's digit is zero (0) and unit digit is b.
\(\therefore\) a0b = 100a + 10 \(\times\) 0 + b
\(\Rightarrow\) a0b = 100a + b __________(1)
Sum of other digits is a+b.
a+b = 11 __________(2) (Given)
If digits are reversed then formed number be b0a.
\(\therefore\) b0a = 100b + a __________(3)
According to given condition, we have
100b+a = 100a + b + 495
\(\Rightarrow\) 99b - 99a = 495
\(\Rightarrow\) b - a = \(\frac{495}{99}\) = 5
\(\Rightarrow\) b - a = 5 __________(4)
Adding equation (2) & (4), we get
(a+b) + (b - a) = 11+5
\(\Rightarrow\) 2b = 16
\(\Rightarrow\) b = \(\frac{16}2\) = 8
\(\therefore\) a = 11 - b
= 11 - 8 = 3 (From (2))
Hence, the required number is 100a+b = 300+8 = 308. (From (1))
