Quantitative Reasoning & Test Mastery

Set up the mathematics before reaching for the calculator

DAT Quantitative Reasoning requires accurate setup as well as efficient calculation. This lesson covers algebra, percentages, rates, probability, statistics, data interpretation, data sufficiency, and quantitative comparison. You should be able to explain an answer's units, assumptions, and reasonable size before selecting it.

The section gives 40 questions in 45 minutes, or 67.5 seconds per question on average, and provides an on-screen calculator. The current ADA outline lists algebra and graphical analysis; data analysis, interpretation and sufficiency; quantitative comparison; probability/statistics; and applied word problems. Use that outline to plan preparation. Geometry may supply a context or useful background, but geometry and trigonometry should not be advertised as guaranteed standalone blocks. ADA candidate guide.

1. Make the equation match the question

Define the unknown, translate the relationship, and keep units consistent. In an equation, apply the same permissible operation to both sides. Check for excluded values before dividing by an expression.

Linear example: 3x + 7 = 22 gives 3x = 15 and x = 5. Substitution checks it: 3(5) + 7 = 22.

Inequality example: −3x > 12 gives x < −4, because dividing by a negative reverses the comparison. A check such as x = −5 gives 15 > 12; x = 0 fails.

Quadratic example: x² − 5x + 6 = 0 factors as (x−2)(x−3) = 0, so x = 2 or 3. If factoring is inconvenient, for ax² + bx + c = 0 with a ≠ 0, use:

x = (−b ± √(b² − 4ac)) / (2a)

The discriminant b²−4ac determines the number of distinct real roots: positive gives two, zero gives one repeated root, negative gives none. Squaring an equation can introduce extraneous solutions; check candidates in the original equation.

Exponents and absolute value

For positive bases, the usual real-exponent rules apply: xᵃxᵇ = xᵃ⁺ᵇ, xᵃ/xᵇ = xᵃ⁻ᵇ, and (xᵃ)ᵇ = xᵃᵇ. For integer exponents, negative bases can also be used where expressions are defined. A denominator cannot be zero; x⁰ = 1 requires x ≠ 0. Be careful extending fractional-exponent identities to negative bases over the reals.

√(x²) = |x|, not always x. Thus |x−2| = 3 has x−2 = 3 or x−2 = −3, giving x = 5 or −1. Absolute value is distance, so it cannot be negative.

2. Track the base of each percentage

For a positive original amount:

Percent change = (new − original) / original × 100%.

Successive changes multiply their factors. A 25% markup followed by a 20% discount gives 1.25 × 0.80 = 1.00, hence no net change. Starting with 100 makes the steps 100 → 125 → 100. Starting with P gives 0.80(1.25P) = P, not a product containing P twice.

By contrast, a 20% increase followed by a 20% decrease gives 1.20 × 0.80 = 0.96: a 4% decrease. The second percentage uses the changed base.

Choose 100 only when scaling does not change the question. It works for an unknown proportional starting price; it cannot erase a fixed $10 fee or determine an actual unknown dollar amount without enough information.

Try it: a rate rises from 20% to 25%. That is 5 percentage points, but a 25% relative increase: (25−20)/20 = 0.25. Label which quantity is requested.

3. Add rates, not completion times

If two workers independently complete the same divisible job in a and b hours, work at constant rates, and do not interfere, their combined rate is 1/a + 1/b jobs per hour. Their joint time is ab/(a+b).

Worked example: one worker takes 3 hours and the other 6. The rate is 1/3 + 1/6 = 1/2 job/hour; time = 2 hours. Under these assumptions, their joint time is shorter than either alone. Real work may have bottlenecks; the mathematical assumptions make the equation valid.

For travel, average speed = total distance / total time. A 60 km outward leg at 60 km/h takes 1 hour; the 60 km return at 30 km/h takes 2 hours. Average = 120/3 = 40 km/h, not 45. The slower speed lasts longer. An arithmetic mean of speeds works for equal-duration legs, not equal-distance legs in general.

Mixture example: mix 200 g of a 10% salt-by-mass solution with 300 g of a 20% salt-by-mass solution, without losses. Salt mass = 20 + 60 = 80 g; total mass = 500 g; concentration = 16% by mass. A plain average of 10% and 20% ignores the different quantities. Specify whether a concentration is by mass, by volume, or mass per volume before calculating.

For simple interest, I = Prt with r as a decimal annual rate and t in years. For nominal annual rate r compounded n times per year, A = P(1+r/n)ⁿᵗ under the stated model. The units and rate convention must match.

4. Distinguish probability rules

Counting favorable outcomes divided by total outcomes applies when the elementary outcomes are equally likely. In general, probabilities of disjoint elementary outcomes are added with their actual probabilities.

Independent is not the same as mutually exclusive. Two disjoint events with nonzero probabilities cannot be independent: knowing one happened rules out the other.

Worked complement: three appointments each independently have a 10% cancellation probability. P(none cancel) = 0.9³ = 0.729. Therefore P(at least one) = 0.271 = 27.1%. Adding three 10% values double-counts overlapping cancellation events.

Without replacement: a bag has 3 red and 2 blue counters. P(two red on two draws) = (3/5)(2/4) = 3/10. The second denominator and numerator change. With replacement and independent draws, it would be (3/5)² = 9/25.

5. Summarize and interpret data

Mean = total/count. Median = middle sorted value, or the mean of the middle pair. Mode = most frequent value, possibly tied or absent under common conventions. Range = maximum − minimum. Standard deviation describes spread around the mean; distinguish population from sample formulas when calculation is required.

Worked data: for 2, 3, 3, 4, 18, the mean is 30/5 = 6, median 3, mode 3, and range 16. The high outlier pulls the mean upward more than the median.

Weighted mean: ten observations average 70 and thirty average 90. Combined mean = (10×70 + 30×90)/40 = 85, not 80. Weight by the number of observations.

Read a table before forming a conclusion

Group B has more successes in count, but A has the higher success rate: 60% versus 40%. The difference is 20 percentage points. A's rate is 50% higher relative to B's rate: (0.60−0.40)/0.40 = 0.50.

On a graph, inspect axis labels, units, and scale. A truncated vertical axis can make a modest difference look dramatic. For a straight line through (1,3) and (4,9), slope = (9−3)/(4−1) = 2, and the equation is y = 2x + 1. Slope is change in y per unit x, not y/x unless the line passes through the origin.

6. Data sufficiency: can you determine a unique answer?

A statement can help without being sufficient. Test each statement separately before combining them, and use the answer-choice scheme printed on the actual item.

Question: what is the value of real x?

• Statement 1: x² = 9.
• Statement 2: x > 0.

Statement 1 permits x = −3 or 3: insufficient. Statement 2 permits infinitely many positive values: insufficient. Together they force x = 3, so the two statements together are sufficient, while neither alone is.

Important distinction: if the question had asked for x², Statement 1 alone would be sufficient. Sufficiency depends on what must be determined, not merely whether x is known.

Try it: for a rectangle, perimeter 20 alone does not determine area; length 6 alone does not either. Together, 2(6+w) = 20 gives w = 4 and area = 24. The prompt's rectangle assumption matters.

7. Quantitative comparison: seek a proof or a counterexample

Suppose a comparison task asks whether Quantity A is larger, Quantity B is larger, they are equal, or the relationship cannot be determined.

Quantity A = x²; Quantity B = x; condition: x > 0.

• x = 2 gives x² > x.
• x = 1/2 gives x² < x.
• x = 1 gives equality.

Therefore the relationship cannot be determined from x > 0 alone. Trying only positive integers would miss the smaller-than-one case. If instead x > 1, multiplying x > 1 by positive x proves x² > x.

Check zero, negative values, fractions, and boundary values when the conditions allow them. One counterexample can refute a universal claim; several supporting examples do not prove it.

8. Estimate, then check the required precision

47×23 is near 50×20 = 1,000; its exact value is 1,081. Likewise, (8.97×10²)(3.12×10⁻⁴) is near (9×3)×10⁻² = 0.27, but its exact value is 0.279864. The estimate verifies magnitude, not that 0.27 is a correctly rounded exact result.

Use estimation to detect impossible signs, units, or decimal places. It only selects a unique option if the remaining choices are far enough apart. Practice the calculator interface and check entries when needed; repeating every calculation automatically wastes time.

Useful background example: a rectangle 24 mm by 32 mm has diagonal √(24²+32²) = 40 mm. The 3–4–5 ratio scaled by eight provides a second check. This is a geometry refresher, not a promise about the frequency of geometry questions on the current exam.

9. Plan a first pass and a return pass

A possible plan is 30 minutes first pass + 13 minutes return pass + 2 minutes final check = 45 minutes. That first pass averages 45 seconds per item if you reach all 40; difficult items need an early provisional answer and flag. A 90-second allowance for every first-pass question would break the plan.

There is no penalty for guessing. For a hypothetical five-choice item, eliminating two wrong choices leaves three; guessing uniformly gives 1/3 ≈ 33.3%, provided the correct choice remains. Use the actual option count and avoid claiming that all scored questions contribute identical scaled-score increments.

After practice, label each miss as setup, concept, arithmetic, units, interpretation, or timing. Rework it untimed before repeating timed practice. Speed practice helps only when the underlying method is sound.