SAT Math Practice — Texas CBE™

Question 1 of 10

Advanced MathMedium

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

A10

C14

Explanation

First compute g(1) = 3(1) + (2) = 5. Then f(5) = 3(5) + (-1) = 14.

Question 2 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?

A46

B45

C48

D44

Explanation

By the Pythagorean theorem: b² = c² − a² = 52² − 20² = 2704 − 400 = 2304. So b = √2304 = 48.

Question 3 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?

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 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

B36,000 liters

C0.6 liters

D9 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

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?

A27/8π

B27/4π

C243/8π

D18π

Explanation

Arc length = (angle/360°) × circumference = (135/360) × 2π(9) = (135/360) × 18π = 27/4π.

Question 6 of 10

Advanced MathMedium

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

A46

B49

C21

D16

Explanation

First compute g(6) = 4(6) + (-3) = 21. Then f(21) = 2(21) + (4) = 46.

Question 7 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 8 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

B7/2

C−45/4

D45/4

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 9 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?

B−9

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 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

B83.5

C84.4

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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