Solved to be 7 after arithmetic

Question

I recently made a blunder while trying to explain a question asked to me in an interview, The question was

• Think of $X$
• Add $X$ to itself ($X+X = y$)
• Times the result by $3$ ($y\times 3 = z$)
• Divide that result by the original answer ($z/x = q$)
• Add 1 to that result ($q + 1 = w$)
• The result is $7$

I was unable to articulate the reason for this and I hope someone could just break it down for me.

Answer 1

The outcomes of the calculations are, in order

$X$

$2X$

$6X$

$6X/X=6$

$6+1=7$

Answer 2

Add X to itself: $$x+x=2x$$ Times the result by 3: $$3\times2x = 6x$$ Divide that result by the original answer: $$6x\div x = 6 \quad with \quad(x\neq 0)\\$$ As you see x is cancelled out leaving 6 alone $$ $$Add 1 to that result: $$ 6+1 = 7$$

Answer 3

You can write the whole operation as one expression: $3\times (x + x)\div x + 1$. $x+x$ is just $2\times x$, so you have $3\times 2\times x\div x + 1$. The only thing you do with $x$ is multiple and divide the same value by it, which are inverse operations, so the result is independent of the value of $x$.

Answer 4

I think it is easier to intuit the answer if you see the whole computation at once:

$$ \frac{3(x+x)}{x}+1 $$ The numerator is easily factored:

$$ 3(x+x)=3(2x)=6x $$ We then have $\frac{6x}{x}+1$ which, if you're familiar with the rules of division in algebra, simplifies to $6+1$, or $7$.