Bluebook SAT Test Ratios, Rates, Proportional Relationships, and Units 5

Question 1

A farmer wants to blend two fertilizers: Brand A contains 40% nitrogen by weight, and Brand B contains 10% nitrogen by weight. If the farmer needs a final blend that is exactly 22% nitrogen, what is the required ratio of Brand A to Brand B by weight?

Accepted Answer

2 : 3

Answer

Use the alligation method: the number of parts of each component equals how far the OTHER component's concentration is from the target. Parts of A = 22 − 10 = 12. Parts of B = 40 − 22 = 18. So A : B = 12 : 18 = 2 : 3. Verify: (2 × 40 + 3 × 10) ÷ 5 = (80 + 30) ÷ 5 = 110 ÷ 5 = 22%. The key insight is that the ratio is the INVERSE of the distance each concentration is from the target, not the direct distance.

Question 2

Cyclist 1 rides from Town P to Town Q at 15 mph, then immediately returns along the same route at 10 mph. Cyclist 2 covers the identical round trip at a constant speed of 12 mph throughout. Which cyclist completes the round trip first?

Accepted Answer

They finish at exactly the same time; both average 12 mph over the round trip

Answer

When the same distance is traveled at two different speeds, the correct average speed for the whole trip is the harmonic mean — NOT the arithmetic mean. Harmonic mean = 2(15)(10) ÷ (15 + 10) = 300 ÷ 25 = 12 mph. Since Cyclist 1's effective average speed is exactly 12 mph (same as Cyclist 2's constant speed), they tie. The common trap is averaging arithmetically: (15 + 10) ÷ 2 = 12.5 mph, which overstates Cyclist 1's speed because they spend more time at the slower pace.

Question 3

The variable p varies directly as the square of q and inversely as r. When q = 3 and r = 2, the value of p is 18. What is the value of p when q = 5 and r = 10?

Accepted Answer

10

Answer

Write the combined variation equation: p = k·q²/r. Substitute the known values to find k: 18 = k·(3²)/2 = k·9/2, so k = 18·2/9 = 4. Now apply the formula with the new values: p = 4·(5²)/10 = 4·25/10 = 100/10 = 10. A common error is forgetting to square q first before substituting, or misidentifying whether the variation is direct or inverse for each variable.

Question 4

In a chemical reaction, the rate at which reactant R is consumed is proportional to the square of its current concentration. Initially, the concentration is 0.80 mol/L and the consumption rate is 0.192 mol/(L·s). What is the consumption rate when the concentration falls to 0.50 mol/L?

Accepted Answer

0.075 mol/(L·s)

Answer

Since rate = k·[R]², first find k: 0.192 = k·(0.80)² = k·0.64, so k = 0.192 ÷ 0.64 = 0.300. Then at [R] = 0.50: rate = 0.300·(0.50)² = 0.300·0.25 = 0.075 mol/(L·s). Alternatively, use the ratio shortcut: rate₂/rate₁ = ([R]₂/[R]₁)² = (0.50/0.80)² = (0.625)² = 0.390625; rate₂ = 0.192 × 0.390625 = 0.075. Choice B (0.120) is the trap answer for students who use a simple linear proportion instead of squaring.

Question 5

A European car's fuel efficiency is rated at 6.5 liters per 100 kilometers. A comparable American car achieves 38 miles per gallon. Which car is more fuel-efficient, and by approximately what percentage? (Use: 1 mile = 1.609 km; 1 gallon = 3.785 liters.)

Accepted Answer

The American car is more efficient, by approximately 5%

Answer

Convert the European rating to mpg: 100 km ÷ 1.609 km/mi ≈ 62.15 miles per 6.5 liters; 6.5 L ÷ 3.785 L/gal ≈ 1.717 gallons. So the European car gets 62.15 ÷ 1.717 ≈ 36.2 mpg. The American car gets 38 mpg. The American car is more efficient by (38 − 36.2) ÷ 36.2 × 100 ≈ 5.0%. The trap answers exploit unit confusion — dividing in the wrong order or skipping the two-step conversion gives wildly different results.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Ratios, Rates, Proportional Relationships, and Units 5