QUESTION

CSAT

Easy

Maths

Prelims 2017

There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there?

Select an option to attempt

Explanation

Step 1: Define the Number and Its Reverse: Let the number be 10x+y10x + y, where: xx = tens digit yy = ones digit The reverse is 10y+x10y + x.

Step 2: Set Up the Equation: The difference between the number and its reverse is given by: (10x+y)(10y+x)=27(10x + y) - (10y + x) = 27 Simplifying the equation: 9x9y=279x - 9y = 27 xy=3x - y = 3

Step 3: Solve for Possible Values of x and y: The equation xy=3x - y = 3 implies: Possible pairs of (x,y)(x, y) are: (4,1)(4, 1) (5,2)(5, 2) (6,3)(6, 3) (7,4)(7, 4) (8,5)(8, 5) (9,6)(9, 6)

Step 4: List the Corresponding Numbers: The corresponding two-digit numbers are: (4,1)41(4, 1) \rightarrow 41 (5,2)52(5, 2) \rightarrow 52 (6,3)63(6, 3) \rightarrow 63 (7,4)74(7, 4) \rightarrow 74 (8,5)85(8, 5) \rightarrow 85 (9,6)96(9, 6) \rightarrow 96

Step 5: Conclusion: There are 6 such numbers: 41, 52, 63, 74, 85, and 96. Therefore, the correct answer is D. None of the above.

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