The weight of Jacob’s backpack is made up of the weight of the contents of the backpack as well as the weight of the backpack itself. Seventy percent of the total weight is textbooks. His notebooks weigh a total of 4 pounds, and the backpack itself weighs 2 pounds. If the backpack contains only textbooks and notebooks, which equation can be used to determine t, the weight of the textbooks?

Question

0.7(t) = t – 4 – 2
0.7(t) = t + 4 + 2
0.7t(4 + 2) = t
0.7(t + 4 + 2) = t

Answer 1

To determine the correct equation that can be used to solve for \( t \), the weight of the textbooks, we need to analyze the situation based on the information provided.

1. Weight of the Backpack: The weight of the backpack itself is 2 pounds.

2. Weight of the Notebooks: The notebooks weigh 4 pounds.

3. Total Weight of the Backpack: The total weight of the backpack is the weight of the textbooks plus the weight of the notebooks plus the weight of the backpack itself: \[ \text{Total Weight} = t + 4 + 2 = t + 6 \]

4. Weight of Textbooks: It is given that 70% (or 0.7) of the total weight is the weight of the textbooks. Therefore: \[ 0.7 \times \text{Total Weight} = t \] Substituting the expression for total weight: \[ 0.7(t + 6) = t \]

However, this option does not directly appear in the choices you provided. Instead, we can manipulate the choices to find the closest one that should be correct.

Let's analyze each option:

1. Option A: \( 0.7(t) = t - 4 - 2 \)

   • Rearranging gives \( 0.7t = t - 6 \), which does not correctly relate the weights.

2. Option B: \( 0.7(t) = t + 4 + 2 \)

   • Rearranging gives \( 0.7t = t + 6 \), which is incorrect.

3. Option C: \( 0.7t(4 + 2) = t \)

   • This simplifies to \( 0.7t \cdot 6 = t \), which does not hold because \( 0.7t \cdot 6 \neq t \).

4. Option D: \( 0.7(t + 4 + 2) = t \)

   • This simplifies to \( 0.7(t + 6) = t \), which matches our derived equation.

Therefore, the correct equation is: \[ \boxed{0.7(t + 4 + 2) = t} \]