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A rocket plus its fuel weighs 5200kg. After a quarter of the fuel is used, the rocket and the remaining fuel weigh 4600kg. Find the weight of the rocket in kg

Respuesta :

Let's denote the weight of the rocket as \( R \) kg and the weight of the fuel as \( F \) kg.

Given:
1. \( R + F = 5200 \) (initial weight of the rocket plus fuel)
2. \( R + \frac{3}{4}F = 4600 \) (weight of the rocket plus 3/4 of the fuel after using a quarter of the fuel)

From equation 2, we can express \( F \) in terms of \( R \):
\[ R + \frac{3}{4}F = 4600 \]
\[ \frac{3}{4}F = 4600 - R \]
\[ F = \frac{4}{3}(4600 - R) \]

Now, substitute the expression for \( F \) into equation 1:
\[ R + \frac{4}{3}(4600 - R) = 5200 \]

Solve for \( R \):
\[ R + \frac{4}{3}(4600 - R) = 5200 \]
\[ R + \frac{4}{3}(4600) - \frac{4}{3}R = 5200 \]
\[ R + \frac{4}{3}(4600) - \frac{4R}{3} = 5200 \]
\[ \frac{3R}{3} + \frac{4(4600)}{3} - \frac{4R}{3} = 5200 \]
\[ \frac{3R - 4R}{3} = 5200 - \frac{4(4600)}{3} \]
\[ -\frac{R}{3} = 5200 - \frac{4(4600)}{3} \]
\[ R = -3\left(5200 - \frac{4(4600)}{3}\right) \]

Now, calculate \( R \):
\[ R = -3\left(5200 - \frac{4(4600)}{3}\right) \]
\[ R = -3\left(5200 - \frac{18400}{3}\right) \]
\[ R = -3\left(5200 - 6133.33\right) \]
\[ R = -3(-933.33) \]
\[ R = 2800 \]

So, the weight of the rocket is \( 2800 \) kg.

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