Once we get to this point...
Quantity A: 10x
Quantity B: 20

...we can divide both quantities by 10 to get:
Quantity A: x
Quantity B: 2

We're told that x is an integer greater than 1. So, x COULD equal 2 or it COULD be greater.



Cheers,
Brent

Can we simplify it as 8x+6x?

x is an integer and x>1
Column A
\(\sqrt{64x^2+36x^2}\) = \(\sqrt{x^2(64+36)}\)
= \(\sqrt{100x^2}\) \(\sqrt{(10x)^2}\)
= |10x| [As square root of a number is always positive]
= 10 * |x|
Now, we know that x > 1 => |x| = x
=> 10|x| = 10x
Now, x is an integer greater than 1, so x can be 2, 3,...
=> 10x can we be 10*2, 10*3,...
=> 10x can we be 20,30,....

Column B
20

Now, Column A can be 20 or greater than 20 too.
So, column A can be = Column B or can be greater than Column B too.
So, Answer will be Option D "Answer Cannot be determined"
Hope it helps!

Reetika1990 : We cannot split \(\sqrt{64x^2+36x^2}\) as 8x+6x Directly.
Example: we cannot split \(\sqrt{x+y}\) as \(\sqrt{x}\) + \(\sqrt{y}\)
Numerical example
we cannot split \(\sqrt{2+3}\) as \(\sqrt{2}\) + \(\sqrt{3}\)
\(\sqrt{2+3}\) = \(\sqrt{5}\) = 2.23
Where as, \(\sqrt{2}\) + \(\sqrt{3}\) = 1.414 + 1.732 = 3.146

Hope it helps!

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