OE



Given $\frac{(a-b)(b-c)}{(a-c)}= 1$ (equation 1)

Then $(a-b)(b-c)= (a-c)$

c(2-a)=1 (equation 2)

Since, the answer choices represent the values of a so let us put them in the eqn.2
and check which of them cannot be the value of a.
A. 0
If 𝑎 = 0, the eqn.2 reduces to
𝑐(2 − 0) = 1
i.e. 𝑐 =1/2

which is satisfying eqn.1

Hence, this can be the value of a.

B. 1

If 𝑎 = 1, the eqn.2 reduces to
𝑐 (2 − 1) = 1
i.e. 𝑐 = 1
If we put 𝑎 = 1 and 𝑐 = 1 in eqn.1 then we are getting denominator 0 which is possible.’

Hence, 𝑎 ≠ 1

Since this question has only one answer, so we do not need to check rest of
the answer choices.
Ans. (B)