SAT Math — Quick Drill
Quick Drill · 10 Questions · 15 min
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Question 1 of 10
Geometry & TrigMedium Diagram

In the circle shown, the radius is 9 and an arc subtends a central angle of 135°. What is the arc length, in terms of π?

r = 9135°arc
A243/8π
B18π
C27/8π
D27/4π
Explanation
Arc length = (angle/360°) × circumference = (135/360) × 2π(9) = (135/360) × 18π = 27/4π.
Question 2 of 10
AlgebraMedium

A line has a slope of 4 and passes through the point (-1, -2). Which equation represents the line?

Ay = 4x + 2
By = -4x + 2
Cy = 4x − 2
Dy = 5x + 2
Explanation
Use point-slope form: y − -2 = 4(x − (-1)). Solve for y: y = 4x + 2.
Question 3 of 10
Advanced MathMedium

If f(x) = 2x + 4 and g(x) = 4x − 3, what is f(g(6))?

A16
B21
C49
D46
Explanation
First compute g(6) = 4(6) + (-3) = 21. Then f(21) = 2(21) + (4) = 46.
Question 4 of 10
Problem-Solving & DataMedium Calc Word

A laboratory pump moves fluid at a steady rate of 2.5 milliliters per second. At this rate, how many liters of fluid does the pump move in 4 hours? (1 liter = 1,000 milliliters)

A36 liters
B9 liters
C36,000 liters
D0.6 liters
Explanation
This is a unit-rate conversion done by dimensional analysis. One hour contains 60 × 60 = 3,600 seconds, so 4 hours contains 4 × 3,600 = 14,400 seconds. At 2.5 milliliters per second, the pump moves 2.5 × 14,400 = 36,000 milliliters. Dividing by 1,000 milliliters per liter gives 36,000 ÷ 1,000 = 36 liters. A student who stops after multiplying and never converts milliliters to liters gets 36,000. A student who treats an hour as 60 seconds computes 2.5 × 60 × 4 = 600 mL = 0.6 liters. A student who converts correctly but forgets the 4-hour multiplier finds only one hour's output, 2.5 × 3,600 ÷ 1,000 = 9 liters.
Question 5 of 10
Advanced MathMedium

If f(x) = 3x − 1 and g(x) = 3x + 2, what is f(g(1))?

A10
B5
C14
D2
Explanation
First compute g(1) = 3(1) + (2) = 5. Then f(5) = 3(5) + (-1) = 14.
Question 6 of 10
Geometry & TrigMedium Diagram

Triangle ABC is similar to triangle DEF, shown below. Side AB has length 8 and corresponds to side DE of length 40. If side BC has length 7, what is the length of side EF?

AB = 8△ABCDE = 40△DEFFigure not drawn to scale
A35
B15
C12
D42
Explanation
Similar triangles have proportional sides. Scale factor = DE/AB = 40/8 = 5. So EF = BC × 5 = 7 × 5 = 35.
Question 7 of 10
Geometry & TrigMedium Diagram

In the right triangle shown, the hypotenuse has length 52 and one leg has length 20. What is the length of the other leg?

20?52
A45
B46
C48
D44
Explanation
By the Pythagorean theorem: b² = c² − a² = 52² − 20² = 2704 − 400 = 2304. So b = √2304 = 48.
Question 8 of 10
Advanced MathHard

The quadratic equation x² − kx + 2k = 0, where k is a constant, has two distinct real solutions r and s such that r² + s² = 45. What is one possible value of k?

A5
B9
C−9
D8
Explanation
By Vieta's formulas, for x² − kx + 2k = 0 the solutions satisfy r + s = k and rs = 2k. Using the identity r² + s² = (r + s)² − 2rs gives k² − 2(2k) = 45, so k² − 4k − 45 = 0, which factors as (k − 9)(k + 5) = 0, giving k = 9 or k = −5. Testing k = 9: the equation becomes x² − 9x + 18 = 0, whose roots are 3 and 6 — distinct and real — and 3² + 6² = 9 + 36 = 45, as required. Writing the identity with the wrong sign, r² + s² = (r + s)² + 2rs, leads to k² + 4k = 45 and the values 5 and −9; the value 5 in fact gives discriminant 25 − 40 < 0, so that equation has no real solutions at all, and −9 gives roots whose squares sum to 117, not 45. Setting the discriminant k² − 8k equal to zero — confusing "two distinct real solutions" with "exactly one solution" — produces 8, but then both roots equal 4 and the sum of squares is only 32.
Question 9 of 10
AlgebraHard

In the system of equations below, k is a constant.

kx + 10y = 45 2x + 5y = 7

The system has no solution. What is the x-intercept of the graph of kx + 10y = 45 in the xy-plane?

A9/2
B45/4
C−45/4
D7/2
Explanation
A system of two linear equations has no solution exactly when the two lines are parallel: the x- and y-coefficients are proportional, but the constants break that proportion. Comparing coefficients, k/2 must equal 10/5 = 2, so k = 4. Checking the constants, 45/7 ≠ 2, so with k = 4 the lines really are parallel and never meet — the no-solution condition holds. The line in question is therefore 4x + 10y = 45. Its x-intercept comes from setting y = 0: 4x = 45, so x = 45/4. A common slip is to set x = 0 instead, which produces the y-intercept 45/10 = 9/2 rather than the x-intercept. Confusing "no solution" with "infinitely many solutions" leads a student to scale the entire second equation by 45/7 so that the constants match too; that forces k = 90/7 and gives an x-intercept of 45 ÷ (90/7) = 7/2. Finally, a sign error while isolating x — writing 4x = −45 as if 45 sat on the left side — produces −45/4.
Question 10 of 10
Problem-Solving & DataHard Calc Word

A teacher computes the mean of 30 quiz scores as 82. She then discovers two data-entry errors: a student who earned 58 was recorded as 85, and a student who earned 91 was recorded as 19. After both errors are corrected, what is the mean of the 30 scores?

A81.1
B84.4
C83.5
D80.5
Explanation
The mean is the sum divided by the count, so the recorded sum is 30 × 82 = 2,460. Correcting the data means removing the two wrong entries and inserting the two true scores: the sum changes by (−85 − 19) + (58 + 91) = −104 + 149 = +45. The corrected sum is 2,460 + 45 = 2,505, and the corrected mean is 2,505 ÷ 30 = 83.5. Note the shortcut: only the net change matters, and it spreads over all 30 scores, so the mean rises by 45/30 = 1.5. A student who mixes up which values were recorded and which were earned reverses the correction and gets 82 − 1.5 = 80.5. A student who fixes only the 19-that-should-be-91 error adds 72/30 and gets 84.4; one who fixes only the 85-that-should-be-58 error subtracts 27/30 and gets 81.1.

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