CSAT 2024

Q. Let X be a two-digit number and Y be another two-digit number formed by interchanging the digits of X.

If (X+Y) is the greatest two-digit number, then what is the number of possible values of X?

a) 2
b) 4
c) 6
d) 8
Correct answer: d) 8

Question from UPSC Prelims 2024 CSAT

Explanation : 

Two-Digit Number Sum by Interchanging Digits of X

Let’s analyze the problem step by step.

Given:
– Let X = 10a + b be a two-digit number where a is the tens digit (1-9) and b is the units digit (0-9).
– Let Y = 10b + a be the number formed by interchanging the digits of X.

Sum of X and Y:
X + Y = (10a + b) + (10b + a) = 11(a + b)

Objective:
– To maximize X + Y such that it remains a two-digit number.

Maximum Two-Digit Number:
– The greatest two-digit number is 99.
– So, 11(a + b) ≤ 99 which implies a + b ≤ 9.

Maximizing the Sum:
– To achieve the largest possible two-digit sum, set a + b = 9.
– Thus, X + Y = 99.

Possible Combinations:
We need to find pairs (a, b) where a + b = 9 and a ranges from 1 to 9 while b ranges from 0 to 9.

Here are the possible pairs:
(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)

Excluding Invalid Cases:
– When a = 9, b = 0 leads to Y = 09, which is not a valid two-digit number.
– Therefore, exclude (9,0).

Valid Pairs:
(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)

Number of Possible Values of X:
– There are 8 valid pairs.

Answer: d) 8

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