IAS 2026 — CSAT Maths

The correct option is C - Both I and II.

Solution

Let the two distinct numbers be x and y.
According to the fundamental property of numbers:
$HCF(x, y) \times LCM(x, y) = x \times y$HCF(x,y)×LCM(x,y)=x×y

The question states that the product of HCF and LCM is the cube of one of the numbers. Let that number be x.
So, $HCF \times LCM = x^{3}$HCF×LCM=x3

Equating the two expressions:
$x \times y = x^{3}$x×y=x3
Dividing both sides by x (since x is a number and cannot be zero):
$y = x^{2}$y=x2

This tells us that one number (y) is the square of the other number (x). Since the numbers are distinct, x cannot be 1 (because if x=1, y=1^2=1, making them identical).

Testing Statement II: One of the numbers is a perfect square.
Since $y = x^{2}$y=x2, y is definitely a perfect square. Thus, Statement II is correct.

Testing Statement I: The difference of the numbers is an even number.
The difference is $(y - x)$(y−x) or $(x^{2} - x)$(x2−x).
$x^{2} - x$x2−x can be written as $x(x - 1)$x(x−1).
In mathematics, the product of any two consecutive integers (x and x-1) is always even, because one of them must be even.
Example 1: If x=2, y=4. Difference = $4-2 = 2$4−2=2 (Even).
Example 2: If x=3, y=9. Difference = $9-3 = 6$9−3=6 (Even).
Example 3: If x=4, y=16. Difference = $16-4 = 12$16−4=12 (Even).
Thus, Statement I is correct.

Why the other options are incorrect

• Option (a) - I only: This is incorrect because it ignores the fact that Statement II is also mathematically proven to be true ($y = x^{2}$y=x2).
• Option (b) - II only: This is incorrect because it overlooks the algebraic property that the difference between a square and its root $(x^{2} - x)$(x2−x) is always an even number.
• Option (d) - Neither I nor II: This is incorrect because both statements are logically and mathematically derived from the given condition.

Key Concept

The relationship between HCF, LCM, and the product of two numbers ($HCF \times LCM = Product$HCF×LCM=Product), combined with the property that the product of consecutive integers is always even.