Ratios, Rates, Proportional Relationships, and Units Flashcards

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?

Answer: 66.5 mL

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.

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?

Answer: About 25%

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%.

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?

Answer: 11.5 gallons

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.

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?

Answer: 90 minutes

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.

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?

Answer: 4.0 kg

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.)

An object's weight on a planet is directly proportional to the planet's mass and inversely proportional to the square of the planet's radius. Planet X has 3 times the mass and 1.5 times the radius of Planet Y. A probe weighs 540 N on Planet Y. What does it weigh on Planet X?

Answer: 720 N

Weight ∝ M/r². The ratio of weights is W_X / W_Y = (M_X / r_X²) ÷ (M_Y / r_Y²) = (3M_Y / (1.5 r_Y)²) ÷ (M_Y / r_Y²) = (3 / 2.25) = 4/3. Weight on Planet X = 540 × (4/3) = 720 N.