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CAT 2021 Quant Question [Slot 1] with Solution 11

Question
How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?
Option: 70
Solution:
Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.
Then (100c+10b+a)-(100a+10b+c)=198
99c-99a=198
c-a = 2.
Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.
Thus, there can be 7 cases.
B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be $7 \times \;10 = 70$.
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