Let's solve the blending ratio for Aggregates A and B step by step. ## **Step 1: Let the blend be \( x \) of Agg. A and \( (1-x) \) of Agg. B** Let: - \( x \) = fraction of Agg. A - \( 1-x \) = fraction of Agg. B The **blended percent passing** for each sieve size is: \[ \text{Blended} = x \times (\text{A}) + (1-x) \times (\text{B}) \] ## **Step 2: Write equations for critical sieve sizes** We will focus on the tightest (most restrictive) ranges, usually where aggregate curves are closest to the limits. Let's write **blended percent passing** formulas for each sieve where both aggregates are significantly different, and the range is not –100. | Sieve (mm) | Req. Range | A (%) | B (%) | Equation | |------------|------------|-------|-------|----------| | 12.5 | 80–100 | 90 | 100 | \( 90x + 100(1-x) \) | | 9.5 | 70–90 | 59 | 100 | \( 59x + 100(1-x) \) | | 4.75 | 50–70 | 16 | 96 | \( 16x + 96(1-x) \) | | 2.36 | 35–50 | 3 | 75 | \( 3x + 75(1-x) \) | | .6 | 18–29 | | 51 | \( x + 51(1-x) \) | | .3 | 13–23 | | 30 | \( x + 30(1-x) \) | | .15 | 8–16 | | 18 | \( x + 18(1-x) \) | | .075 | 4–10 | | 9 | \( x + 9(1-x) \) | **For each size, blended value must be within the required range.** --- ## **Step 3: Set up inequalities** Let's do this for each critical sieve: ### **12.5 mm:** \[ 80 \leq 90x + 100(1-x) \leq 100 \] \[ 80 \leq 90x + 100 - 100x \leq 100 \] \[ 80 \leq 100 - 10x \leq 100 \] Solve for \( x \): Lower limit: \[ 80 \leq 100 - 10x \implies 10x \leq 20 \implies x \leq 2 \] (automatically satisfied, since \( x \leq 1 \)) Upper limit: \[ 100 - 10x \leq 100 \implies -10x \leq \implies x \geq \] (always true) So, **no restriction** from this sieve. --- ### **9.5 mm:** \[ 70 \leq 59x + 100(1-x) \leq 90 \] \[ 70 \leq 100 - 41x \leq 90 \] Lower limit: \[ 70 \leq 100 - 41x \implies 41x \leq 30 \implies x \leq \frac{30}{41} \approx .73 \] Upper limit: \[ 100 - 41x \leq 90 \implies -41x \leq -10 \implies x \geq \frac{10}{41} \approx .244 \] --- ### **4.75 mm:** \[ 50 \leq 16x + 96(1-x) \leq 70 \] \[ 50 \leq 96 - 80x \leq 70 \] Lower limit: \[ 50 \leq 96 - 80x \implies 80x \leq 46 \implies x \leq \frac{46}{80} = .575 \] Upper limit: \[ 96 - 80x \leq 70 \implies -80x \leq -26 \implies x \geq \frac{26}{80} = .325 \] --- ### **2.36 mm:** \[ 35 \leq 3x + 75(1-x) \leq 50 \] \[ 35 \leq 75 - 72x \leq 50 \] Lower limit: \[ 35 \leq 75 - 72x \implies 72x \leq 40 \implies x \leq \frac{40}{72} \approx .556 \] Upper limit: \[ 75 - 72x \leq 50 \implies -72x \leq -25 \implies x \geq \frac{25}{72} \approx .347 \] --- ### **.6 mm:** \[ 18 \leq x + 51(1-x) \leq 29 \] \[ 18 \leq 51 - 51x \leq 29 \] Lower limit: \[ 18 \leq 51 - 51x \implies 51x \leq 33 \implies x \leq \frac{33}{51} \approx .647 \] Upper limit: \[ 51 - 51x \leq 29 \implies -51x \leq -22 \implies x \geq \frac{22}{51} \approx .431 \] --- ### **.3 mm:** \[ 13 \leq x + 30(1-x) \leq 23 \] \[ 13 \leq 30 - 30x \leq 23 \] Lower limit: \[ 13 \leq 30 - 30x \implies 30x \leq 17 \implies x \leq \frac{17}{30} \approx .567 \] Upper limit: \[ 30 - 30x \leq 23 \implies -30x \leq -7 \implies x \geq \frac{7}{30} \approx .233 \] --- ### **.15 mm:** \[ 8 \leq x + 18(1-x) \leq 16 \] \[ 8 \leq 18 - 18x \leq 16 \] Lower limit: \[ 8 \leq 18 - 18x \implies 18x \leq 10 \implies x \leq \frac{10}{18} \approx .556 \] Upper limit: \[ 18 - 18x \leq 16 \implies -18x \leq -2 \implies x \geq \frac{2}{18} \approx .111 \] --- ### **.075 mm:** \[ 4 \leq x + 9(1-x) \leq 10 \] \[ 4 \leq 9 - 9x \leq 10 \] Lower limit: \[ 4 \leq 9 - 9x \implies 9x \leq 5 \implies x \leq \frac{5}{9} \approx .556 \] Upper limit: \[ 9 - 9x \leq 10 \implies -9x \leq 1 \implies x \geq -\frac{1}{9} \approx -.111 \] (But \( x \geq \), so ignore negative bound) --- ## **Step 4: Find the feasible range for x** Collect all lower and upper bounds for \( x \): - **Upper bounds:** .73, .575, .556, .567, .556, .556, .647 - **Lower bounds:** .244, .325, .347, .431, .233, .111, **Maximum of lower bounds:** \[ \max(.244, .325, .347, .431, .233, .111, ) = .431 \] **Minimum of upper bounds:** \[ \min(.73, .575, .556, .567, .556, .556, .647) = .556 \] So, **feasible range** for \( x \) is: \[ .431 \leq x \leq .556 \] --- ## **Step 5: Final Answer** ### **Blending ratio for A and B:** - **A:** 43.1% to 55.6% - **B:** 44.4% to 56.9% **Any value of \( x \) in this range will satisfy all requirements.** ### **Typical choice:** Use the midpoint for a balanced blend: \[ x = \frac{.431 + .556}{2} = .494 \] **So, blend:** - **A:** 49.4% - **B:** 50.6% --- ## **Summary Table** | Aggregate | Ratio (%) | |-----------|-----------| | A | 43.1 – 55.6 | | B | 44.4 – 56.9 | **Typical value:** - A = 49.4% - B = 50.6% --- ### **Explanation** We formed an equation for the percent passing for each sieve size, set up inequalities for the required range, and found the intersection of all possible values for \( x \) (portion of A). The blend ratio ensures all blended sieve passing values fit within the specified requirements. --- **If you need the detailed blended percent passing table for a specific blend, let me know!**