Question

Four machines A, B, C, D produce items in a ratio.
A produces 40 more than B.
C produces 20% more than A.
D produces half of B.
If total production is 860 items,
how many items did Machine C produce?

Hint:

When one value is defined as a percentage or offset of another, rewrite everything using a single variable. This keeps the total-sum equation simple and avoids mistakes.

Answer 1

Correct Answer: 269

Let production of Machine B be \(x\).
Step 1: Express all machines in terms of \(x\).
A produces 40 more than B: \[ A = x + 40 \] C produces 20% more than A: \[ C = 1.2A = 1.2(x + 40) \] D produces half of B: \[ D = \frac{x}{2} \] Step 2: Write the total production equation.
\[ B + A + C + D = 860 \] Substitute values: \[ x + (x + 40) + 1.2(x + 40) + \frac{x}{2} = 860 \] Step 3: Simplify and solve for \(x\).
Expand C: \[ 1.2(x+40) = 1.2x + 48 \] Combine all terms: \[ x + x + 40 + 1.2x + 48 + \frac{x}{2} = 860 \] Add like terms: \[ 3.7x + 88 + \frac{x}{2} = 860 \] Convert \( \frac{x}{2} \) to decimal: \[ 3.7x + 0.5x + 88 = 860 \] \[ 4.2x + 88 = 860 \] \[ 4.2x = 772 \] \[ x = 183.81 \] Step 4: Compute C’s production.
\[ C = 1.2(x + 40) = 1.2(183.81 + 40) \] \[ C = 1.2 \times 223.81 \] \[ C \approx 268.57 \approx 269 \] Final Answer: \(\boxed{269}\)

Answer 2

Approach: Express every machine's output as a linear function of B, substitute into the total, and solve the single resulting linear equation.

Given \(A=B+40\), \(C=1.2A\), \(D=0.5B\), write everything in terms of \(B\):
\(A=B+40\), \(C=1.2(B+40)=1.2B+48\), \(D=0.5B\).

Sum to the total: \[ (B+40)+B+(1.2B+48)+0.5B=860 \implies 3.7B+88=860 \implies B=\frac{772}{3.7}=\frac{7720}{37}. \]
Then \(A=B+40=\dfrac{9200}{37}\), and \[ C=1.2A=1.2\times\frac{9200}{37}=\frac{11040}{37}\approx\boxed{298.4 \text{ items}} \]

(Flag: this comes out to a non-integer number of items \(\left(\tfrac{11040}{37}\approx298.38\right)\), which is unusual for a "number of items produced" answer, worth checking the original numbers for a possible typo.)