Question

1. a construction company is building a brick wall. crew a provides 1/3 of the required bricks, and crew b provides 1/2 of the required bricks. if the wall needed 2,000 more bricks, the combined crews would have had exactly half of the bricks necessary to build the wall. how many bricks are needed for the original wall?

Explanation:

Step1: Define the variable

Let \( x \) be the number of bricks needed for the original wall.

Step2: Calculate the number of bricks provided by the crews

Crew A provides \( \frac{1}{3}x \) bricks and Crew B provides \( \frac{1}{2}x \) bricks. So the combined number of bricks provided by the crews is \( \frac{1}{3}x+\frac{1}{2}x \).
First, find a common denominator for the fractions: \( \frac{1}{3}x+\frac{1}{2}x=\frac{2}{6}x + \frac{3}{6}x=\frac{2x + 3x}{6}=\frac{5x}{6} \).

Step3: Set up the equation based on the problem statement

If the wall needed 2000 more bricks, the combined crews would have had exactly half of the bricks necessary to build the wall. So we can write the equation:
\( \frac{5x}{6}+ 2000=\frac{1}{2}(x + 2000) \)

Step4: Solve the equation

First, expand the right - hand side: \( \frac{5x}{6}+2000=\frac{1}{2}x + 1000 \)
Subtract \( \frac{1}{2}x \) from both sides: \( \frac{5x}{6}-\frac{1}{2}x+2000 = 1000 \)
Find a common denominator for the \( x \) terms: \( \frac{5x}{6}-\frac{3x}{6}+2000 = 1000 \)
Simplify the \( x \) terms: \( \frac{5x - 3x}{6}+2000=1000\Rightarrow\frac{2x}{6}+2000 = 1000\Rightarrow\frac{x}{3}+2000 = 1000 \)
Subtract 2000 from both sides: \( \frac{x}{3}=1000 - 2000=- 1000 \)
Multiply both sides by 3: \( x=-3000 \)
Wait, this gives a negative number, which doesn't make sense. We must have made a mistake in setting up the equation. Let's re - examine the problem.

The correct equation should be: The number of bricks the crews have plus 2000 is equal to half of the total bricks needed for the wall (original + 2000? No, the problem says "if the wall needed 2,000 more bricks, the combined crews would have had exactly half of the bricks necessary to build the wall". The "bricks necessary to build the wall" in this case is the original number of bricks (because we are talking about the wall that needed 2000 more bricks? Wait, no. Let's re - interpret.

Let \( x \) be the original number of bricks needed. The combined crews have \( \frac{1}{3}x+\frac{1}{2}x=\frac{5x}{6} \) bricks. If we add 2000 bricks to the combined crews' bricks, this sum is equal to half of the original number of bricks \( x \). Wait, no, the problem says "the combined crews would have had exactly half of the bricks necessary to build the wall" when the wall needed 2000 more bricks. So the "bricks necessary to build the wall" is still \( x \)? No, maybe the problem means that if the total number of bricks needed for the wall was \( x + 2000 \) (because the wall needed 2000 more bricks), then the combined crews' bricks plus 2000 is half of \( x+2000 \). Wait, no, let's re - read: "If the wall needed 2,000 more bricks, the combined crews would have had exactly half of the bricks necessary to build the wall."

Let's start over. Let \( x \) be the original number of bricks needed.

Crew A: \( \frac{1}{3}x \), Crew B: \( \frac{1}{2}x \), combined: \( \frac{1}{3}x+\frac{1}{2}x=\frac{2x + 3x}{6}=\frac{5x}{6} \)

The situation is: combined bricks (\( \frac{5x}{6} \)) plus 2000 bricks equals half of the bricks needed for the wall (which is still \( x \)? No, that can't be. Wait, the problem says "the combined crews would have had exactly half of the bricks necessary to build the wall" when the wall needed 2000 more bricks. So the "bricks necessary to build the wall" is \( x \), and the combined crews' bricks plus 2000 is \( \frac{1}{2}x \)? No, that also doesn't make sense.

Wait, maybe the correct interpretation is: The combined number of bricks the crews have is equal to half of the bricks needed for the wall minus 2000.

So \( \frac{5x}{6}=\frac{1}{2}x-2000 \)

Subtract \( \frac{1}{2}x \) from both sides: \( \frac{5x}{6}-\frac{3x}{6}=- 2000\Rightarrow\frac{2x}{6}=-2000\Rightarrow\frac{x}{3}=-2000\Rightarrow x=-6000 \). Still negative.

Wait, I think I misinterpret the problem. Let's re - read: "If the wall needed 2,000 more bricks, the combined crews would have had exactly half of the bricks necessary to build the wall."

Let me define \( x \) as the original number of bricks needed. The combined crews have \( C=\frac{1}{3}x+\frac{1}{2}x=\frac{5x}{6} \)

The "bricks necessary to build the wall" when it needs 2000 more bricks is \( x + 2000 \)? No, the wall is the same, just that we are considering a situation where we need 2000 more bricks. Wait, no, the problem is: The combined crews' bricks (which is \( \frac{5x}{6} \)) plus 2000 bricks is equal to half of the bricks needed for the wall (which is \( x \))? No, that gives a negative answer.

Wait, maybe the problem is: The combined crews' bricks is equal to half of the bricks needed for the wall minus 2000.

So \( \frac{5x}{6}=\frac{1}{2}x-2000 \)

Multiply both sides by 6 to eliminate denominators: \( 5x = 3x-12000 \)

Subtract \( 3x \) from both sides: \( 2x=-12000\Rightarrow x = - 6000 \). Still wrong.

Wait, maybe the problem is stated as: If the wall needed 2000 more bricks (i.e., the total bricks needed for the wall is \( x+2000 \)), then the combined crews' bricks is half of \( x + 2000 \). And the combined crews' bricks is \( \frac{1}{3}x+\frac{1}{2}x \).

So the correct equation is \( \frac{1}{3}x+\frac{1}{2}x=\frac{1}{2}(x + 2000) \)

Let's solve this equation again:

Left - hand side: \( \frac{1}{3}x+\frac{1}{2}x=\frac{2x+3x}{6}=\frac{5x}{6} \)

Right - hand side: \( \frac{1}{2}x + 1000 \)

So \( \frac{5x}{6}=\frac{1}{2}x + 1000 \)

Subtract \( \frac{1}{2}x \) from both sides: \( \frac{5x}{6}-\frac{3x}{6}=1000\Rightarrow\frac{2x}{6}=1000\Rightarrow\frac{x}{3}=1000\Rightarrow x = 3000 \)

Answer:

The number of bricks needed for the original wall is 3000.