Re: The weights of two wrestlers are in the ratio 7: 8. If each of the wre
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12 Jun 2025, 04:00
Let the weights of the two wrestlers be $7 x$ and $8 x$, respectively, where $x$ is a common multiplier.
If each wrestler wears additional fighting accessories weighing 2 pounds, their new weights would be:
- Wrestler 1's new weight: $\(7 x+2\)$
- Wrestler 2's new weight: $\(8 x+2\)$
The new ratio of their weights would be $\(\frac{7 x+2}{8 x+2}\)$.
The problem is that we don't know the value of $x$. Without knowing the actual weights of the wrestlers, or at least the value of $x$, we cannot determine the exact numerical ratio of their new weights.
For example:
- If $x=10$, then original weights are 70 lbs and 80 lbs . New weights are 72 lbs and 82 lbs . New ratio $\(=\frac{72}{82}=\frac{36}{41}\)$.
- If $x=20$, then original weights are 140 lbs and 160 lbs . New weights are 142 lbs and 162 lbs . New ratio $\(=\frac{142}{162}=\frac{71}{81}\)$.
The ratio changes depending on the value of $x$. Since $x$ is not given, the ratio cannot be uniquely determined.