Let's break down the problem:

1. We know that a load of bricks is twice as heavy as a load of sticks. Let's denote the weight of a load of sticks as 's' and the weight of a load of bricks as 'b'. So, we can write this relationship as: b = 2s.

2. We also know that the total weight of 4 loads of bricks and 4 loads of sticks is 771 kilograms. We can write this as: 4b + 4s = 771.

3. We can substitute the first equation into the second to solve for 's'. This gives us: 4(2s) + 4s = 771, which simplifies to 8s + 4s = 771, and further simplifies to 12s = 771. Solving for 's', we get s = 771 / 12 = 64.25 kilograms.

4. Now that we know 's', we can substitute it back into the first equation to find 'b'. This gives us: b = 2(64.25) = 128.5 kilograms.

5. Finally, we can find the total weight of 1 load of bricks and 3 loads of sticks by adding the weights together: 128.5 + 3(64.25) = 321.25 kilograms.

Question

Let's break down the problem:

1. We know that a load of bricks is twice as heavy as a load of sticks. Let's denote the weight of a load of sticks as 's' and the weight of a load of bricks as 'b'. So, we can write this relationship as: b = 2s.

2. We also know that the total weight of 4 loads of bricks and 4 loads of sticks is 771 kilograms. We can write this as: 4b + 4s = 771.

3. We can substitute the first equation into the second to solve for 's'. This gives us: 4(2s) + 4s = 771, which simplifies to 8s + 4s = 771, and further simplifies to 12s = 771. Solving for 's', we get s = 771 / 12 = 64.25 kilograms.

4. Now that we know 's', we can substitute it back into the first equation to find 'b'. This gives us: b = 2(64.25) = 128.5 kilograms.

5. Finally, we can find the total weight of 1 load of bricks and 3 loads of sticks by adding the weights together: 128.5 + 3(64.25) = 321.25 kilograms.

Knowee AI · Accepted Answer

Let's break down the problem:

1. We know that a load of bricks is twice as heavy as a load of sticks. Let's denote the weight of a load of sticks as 's' and the weight of a load of bricks as 'b'. So, we can write this relationship as: b = 2s.

2. We also know that the total weight of 4 loads of bricks and 4 loads of sticks is 771 kilograms. We can write this as: 4b + 4s = 771.

3. We can substitute the first equation into the second to solve for 's'. This gives us: 4(2s) + 4s = 771, which simplifies to 8s + 4s = 771, and further simplifies to 12s = 771. Solving for 's', we get s = 771 / 12 = 64.25 kilograms.

4. Now that we know 's', we can substitute it back into the first equation to find 'b'. This gives us: b = 2(64.25) = 128.5 kilograms.

5. Finally, we can find the total weight of 1 load of bricks and 3 loads of sticks by adding the weights together: 128.5 + 3(64.25) = 321.25 kilograms.

Read and imagine what is happening in this problem.A builder discovers that a load of bricks is twice as heavy as a load of sticks. The total weight of 4 loads of bricks and 4 loads of sticks is 771 kilograms.What is the total weight of 1 load of bricks and 3 loads of sticks?

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