Bluebook SAT Test Ratios, Rates, Proportional Relationships, and Units 6

Question 1

A chemist mixes solution A (12% acid) with solution B (30% acid) to produce 360 mL of a 23% acid solution. She then adds pure acid until the concentration reaches 35%. How many additional milliliters of pure acid must she add?

Accepted Answer

66.5 mL

Answer

Step 1 — find the initial acid content: let x = mL of solution A. 0.12x + 0.30(360 − x) = 0.23(360) → 0.12x + 108 − 0.30x = 82.8 → −0.18x = −25.2 → x = 140 mL. Acid in the 360 mL mixture = 0.23 × 360 = 82.8 mL. Step 2 — add y mL pure acid to reach 35%: (82.8 + y)/(360 + y) = 0.35 → 82.8 + y = 126 + 0.35y → 0.65y = 43.2 → y ≈ 66.5 mL.

Question 2

A map has a scale of 1 cm : 4 km. A rectangular lake on the map measures 4.8 cm × 3.5 cm. A government report claims the lake's actual area is 336 km². By approximately what percentage does the report's figure exceed the area implied by the map?

Accepted Answer

About 25%

Answer

Linear scale: 1 cm = 4 km, so 1 cm² = 4² = 16 km². Map area = 4.8 × 3.5 = 16.8 cm². Actual area from map = 16.8 × 16 = 268.8 km². Report says 336 km². Percent excess = (336 − 268.8) / 268.8 × 100 = 67.2 / 268.8 ≈ 25%.

Question 3

A car rated at 35 mpg under standard conditions consumes 18% more fuel on the highway, and consumes an additional 3% more fuel for every 1,000 feet of elevation gain. A 300-mile highway trip has a net elevation gain of 4,500 feet. Approximately how many gallons does the trip require?

Accepted Answer

11.5 gallons

Answer

Each penalty multiplies the consumption rate. Highway factor: ×1.18. Elevation: 4,500 ÷ 1,000 = 4.5 increments × 3% = 13.5% more consumption → factor of 1.135. Combined factor: 1.18 × 1.135 = 1.3393. Adjusted efficiency = 35 ÷ 1.3393 ≈ 26.13 mpg. Fuel needed = 300 ÷ 26.13 ≈ 11.5 gallons.

Question 4

A tank drains through pipe A alone in 60 minutes and simultaneously fills through pipe B alone in 2 hours. Both pipes are open when the tank is 75% full. How many minutes until the tank is completely empty?

Accepted Answer

90 minutes

Answer

Pipe A drains at 1/60 tank per minute. Pipe B fills at 1/120 tank per minute (2 hours = 120 minutes). Net drain rate = 1/60 − 1/120 = 2/120 − 1/120 = 1/120 tank per minute. Starting volume = 0.75 tank. Time to empty = 0.75 ÷ (1/120) = 0.75 × 120 = 90 minutes.

Question 5

A recipe calls for flour, sugar, and butter in the ratio 8 : 3 : 2 by weight. A baker has exactly 2.4 kg of butter and wants to use all of it while maintaining the ratio. However, sugar is limited to a maximum of 1.5 kg. What is the maximum amount of flour (in kg) the baker can actually use?

Accepted Answer

4.0 kg

Answer

If the baker uses all 2.4 kg of butter (ratio part = 2), the scale factor = 2.4 ÷ 2 = 1.2. This would require sugar = 3 × 1.2 = 3.6 kg — but sugar is capped at 1.5 kg. Sugar is therefore the binding constraint. Sugar scale factor = 1.5 ÷ 3 = 0.5. Maximum flour = 8 × 0.5 = 4.0 kg. (The baker cannot use all the butter; the sugar cap is the limiting factor.)

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Ratios, Rates, Proportional Relationships, and Units 6