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Free SSAT Math Practice Test — Grade 10

Math only, deliberately — the real SSAT (published by the Enrollment Management Association) also has verbal and reading sections, which we don't cover.

20 real practice questions at SSAT Upper Level difficulty, drawn from the same calibrated bank our adaptive practice uses. Work through them right here — every question has the answer and a step-by-step explanation one tap away.

Question 1 · MathIntro
Model Cell Dimensions
Volume (cu cm)Side Length (cm)
11
82
273
18\frac{1}{8}?
A science class is modeling cells using cubes. The table below shows the volume of each model cell and the side length students need to compute. For the cell with a volume of 18\frac{1}{8} cubic centimeter, what is the side length in centimeters?
A2
B12\frac{1}{2}
C124\frac{1}{24}
D1512\frac{1}{512}
E38\frac{3}{8}
Show answer & explanation
Let's work it out
1.The side length of a cube equals the cube root of its volume.
2.To find the cube root of 18\frac{1}{8}, take the cube root of the numerator and the cube root of the denominator separately: the cube root of 1 is 1, and the cube root of 8 is 2, giving a side length of 12\frac{1}{2}.
3.This is confirmed by the pattern in the table, where the side length cubed always equals the volume.
Answer: B) 12\frac{1}{2}
Question 2 · Coordinate GraphsIntro
A biology student records the height (in centimeters) of five plant seedlings at two different times and models each seedling's growth as a linear function of days elapsed. The student wants the seedling whose growth rate is exactly 1.5 centimeters per day -- that is, whose line has a slope of 1.5. The two recorded (day, height) measurements for each seedling are listed as the five options. Which seedling has a growth rate of exactly 1.5 cm per day?
ADay 0: 2.0 cm; Day 4: 7.0 cm
BDay 0: 0.6 cm; Day 3: 2.6 cm
CDay 0: 0.4 cm; Day 4: 8.4 cm
DDay 0: 1.8 cm; Day 2: 4.8 cm
EDay 0: 3.0 cm; Day 4: 7.0 cm
Show answer & explanation
Let's work it out
1.Slope equals the change in height divided by the change in days elapsed.
2.For seedling D, measured at Day 0 with height 1.8 cm and Day 2 with height 4.8 cm: slope = 4.8−1.82−0\frac{4.8 − 1.8}{2 − 0} = 3.0 / 2.0 = 1.5 cm per day, which matches the required growth rate exactly.
Answer: D) Day 0: 1.8 cm; Day 2: 4.8 cm
Question 3 · Data & StatisticsIntro
School Supply Shop Sales (in dollars)
MonthOnlineIn-Store
September1,200800
October1,5001,000
November2,100900
December1,8001,500
The table shows monthly sales figures (in dollars) for a school supply shop, broken down by online and in-store purchases over four months. What fraction of the shop's total four-month sales came from online purchases?
A29\frac{2}{9}
B718\frac{7}{18}
C1625\frac{16}{25}
D710\frac{7}{10}
E1118\frac{11}{18}
Show answer & explanation
Let's work it out
1.Add the online sales across all four months: 1,200 + 1,500 + 2,100 + 1,800 = 6,600 dollars.
2.Then add every month's total to get the grand total: (1,200 + 800) + (1,500 + 1,000) + (2,100 + 900) + (1,800 + 1,500) = 2,000 + 2,500 + 3,000 + 3,300 = 10,800 dollars.
3.The fraction of total sales that came from online purchases is 6,60010\frac{600}{10},800, which simplifies by dividing both numerator and denominator by 600 to give 1118\frac{11}{18}.
Answer: E) 1118\frac{11}{18}
Question 4 · Arithmetic & Number SenseEasy
Cone Distances in the Six-Round Drill
RoundCone Distance in Meters
212.8
36.4
51.6
A basketball player completes a six-round conditioning drill. In each round, the player sprints from the starting line to a cone and then returns to the starting line. The cone distances form a geometric sequence, as shown in the table. How many meters farther does the player sprint during the first four rounds combined than during the final two rounds combined?
A92.8
B91.2
C45.6
D78.4
E100.8
Show answer & explanation
Let's work it out
1.The common ratio is 6.4 ÷ 12.8 = 0.5, so the six cone distances are 25.6, 12.8, 6.4, 3.2, 1.6, and 0.8 meters.
2.Doubling for each return trip gives 96 meters during the first four rounds and 4.8 meters during the final two rounds.
3.The player therefore sprints 96 - 4.8 = 91.2 meters farther during the first four rounds.
Answer: B) 91.2
Question 5 · PercentagesEasy
Changes in Snow Depth
StageSnow depth information
Before the storm16.00 centimeters
New snowfallIncrease of 37.5%
Snow settlesDecrease of 10%
Afternoon meltingDecrease of 20%
The table shows how the snow depth beside a school weather station changes during a storm and the following afternoon. After all the stages shown, another snowfall begins. How many centimeters of new snow must accumulate to bring the depth to exactly 18.00 centimeters?
A17.56
B6.21
C1.80
D2.16
E0.80
Show answer & explanation
Let's work it out
1.The new snowfall changes the depth to 16.00 × 1.375 = 22.00 centimeters.
2.Successive decreases then give 22.00 × 0.90 × 0.80 = 15.84 centimeters.
3.Therefore, 18.00 − 15.84 = 2.16 centimeters of additional snow must accumulate.
Answer: D) 2.16
Question 6 · Data & StatisticsEasy
Set P contains half as many numbers as Set Q, and the mean of Set P is 34\frac{3}{4}. One-fourth of the numbers in Set Q, with a mean of 12\frac{1}{2}, are removed. The mean of the numbers in Set P together with the numbers remaining in Set Q is 710\frac{7}{10}. What was the original mean of Set Q?
A710\frac{7}{10}
B58\frac{5}{8}
C12\frac{1}{2}
D23\frac{2}{3}
E712\frac{7}{12}
Show answer & explanation
Let's work it out
1.Let Set Q contain 8 numbers, so Set P contains 4 and 2 are removed from Set Q.
2.The combined sum after removal is 10 times 710\frac{7}{10}, or 7; subtracting the sum of Set P leaves 4 for the remaining part of Set Q.
3.Restoring the removed sum gives an original Set Q total of 5, so its mean is 58\frac{5}{8}.
Answer: B) 58\frac{5}{8}
Question 7 · Arithmetic & Number SenseEasy
Classroom Reading Challenge
WeekWords Read
214000
526000
During a classroom reading challenge, the number of words read by the class each week forms an arithmetic sequence. According to the table, how many words does the class read altogether during the first 8 weeks?
A352000
B38000
C208000
D192000
E160000
Show answer & explanation
Let's work it out
1.From week 2 to week 5, the total increases by 12000 words over 3 equal steps, so the weekly increase is 4000 words.
2.Thus, the first and eighth weekly totals are 10000 and 38000, respectively.
3.Their average is 24000, so the total for 8 weeks is 8 × 24000 = 192000.
Answer: D) 192000
Question 8 · Data & StatisticsEasy
Data Sets
PositionSet PSet Q
17.83.9
24.68.4
39.36.2
45.75.1
How much greater is the sum of the two largest entries in Set P than the sum of the two smallest entries in Set Q?
A1.3
B26.1
C0.3
D13.2
E8.1
Show answer & explanation
Let's work it out
1.The two largest entries in Set P sum to 9.3 + 7.8 = 17.1.
2.The two smallest entries in Set Q sum to 3.9 + 5.1 = 9.0.
3.The difference is 17.1 − 9.0 = 8.1.
Answer: E) 8.1
Question 9 · Equations & InequalitiesStandard
If (34\frac{3}{4})[x - 2y−z3\frac{2y - z}{3}] - x+z5\frac{x + z}{5} = (23\frac{2}{3})(y - x2\frac{x}{2}), which relation must also hold?
Ax = 70y−27z53\frac{70y - 27z}{53}
Bx = 70y−3z63\frac{70y - 3z}{63}
Cx = 70y−3z53\frac{70y - 3z}{53}
Dx = 70y−3z13\frac{70y - 3z}{13}
Ex = 70y+27z53\frac{70y + 27z}{53}
Show answer & explanation
Let's work it out
1.Multiply the relation by 60 and distribute to obtain 45x - 30y + 15z - 12x - 12z = 40y - 20x.
2.Combining like terms and collecting the x-terms gives 53x = 70y - 3z.
3.Therefore, x = 70y−3z53\frac{70y - 3z}{53}.
Answer: C) x = 70y−3z53\frac{70y - 3z}{53}
Question 10 · MathStandard
In a music app, the sound-processing rule q is applied before the rule p, where p(x) = √20−xx−12\frac{20 - x}{x - 12} and q(n) = 129−n−2\displaystyle \frac{12}{\sqrt{9 - n} - 2}. Which whole-number setting n can be entered into q but produces a value that cannot then be entered into p?
A1
B9
C8
D0
E5
Show answer & explanation
Let's work it out
1.The rule q accepts 0 because its radicand is nonnegative and its denominator is not zero.
2.Substitution gives q(0) = 12.
Answer: D) 0
Why the other choices are tempting: The value 12 cannot be entered into p because it makes the denominator of p equal zero, so the required setting is 0.
Question 11 · Equations & InequalitiesStandard
If -32a−b4\frac{2a - b}{4} + 5a+2b6\frac{a + 2b}{6} = c - d, which expression is equal to b?
A14a+12c−12d32\frac{14a + 12c - 12d}{32}
B−8a+12c−12d29\frac{-8a + 12c - 12d}{29}
C8a+12c−12d29\frac{8a + 12c - 12d}{29}
D8a+12c−12d11\frac{8a + 12c - 12d}{11}
E8a+12c−d29\frac{8a + 12c - d}{29}
Show answer & explanation
Let's work it out
1.Multiply the equation by 12 to obtain -9(2a - b) + 10(a + 2b) = 12c - 12d.
2.Distributing and combining like terms gives -8a + 29b = 12c - 12d.
3.Isolating b yields b = 8a+12c−12d29\frac{8a + 12c - 12d}{29}.
Answer: C) 8a+12c−12d29\frac{8a + 12c - 12d}{29}
Question 12 · MathStandard
During a classroom activity, students compare the exponential function E(x) = (12\frac{1}{2})rx with the linear function L(x) = 12\frac{1}{2} + 5x/16. Which candidate value of r makes both E(2) = L(2) and E(3) > L(3) true?
A516\frac{5}{16}
B94\frac{9}{4}
C32\frac{3}{2}
D98\frac{9}{8}
E-32\frac{3}{2}
Show answer & explanation
Let's work it out
1.Evaluating the linear function gives L(2) = 98\frac{9}{8}, so (12\frac{1}{2})r2 = 98\frac{9}{8} and r2 = 94\frac{9}{4}.
2.Testing the resulting candidates at x = 3 shows that r = 32\frac{3}{2} gives E(3) = 2716\frac{27}{16}, which is greater than L(3) = 2316\frac{23}{16}.
Answer: C) 32\frac{3}{2}
Question 13 · MathStandard
Values of P, Q, and R
xP(x)Q(x)R(x)
018\frac{1}{8}18\frac{1}{8}18\frac{1}{8}
138\frac{3}{8}28\frac{2}{8}48\frac{4}{8}
258\frac{5}{8}48\frac{4}{8}98\frac{9}{8}
378\frac{7}{8}88\frac{8}{8}168\frac{16}{8}
498\frac{9}{8}168\frac{16}{8}258\frac{25}{8}
The table gives values of three functions. If each displayed pattern continues, which lists the functions in order from slowest eventual growth to fastest eventual growth?
AQ, P, R
BQ, R, P
CR, P, Q
DP, Q, R
EP, R, Q
Show answer & explanation
Let's work it out
1.To determine eventual growth rates, identify what type of function each one is.
2.For P: compute consecutive differences of the outputs.
3.If the table shows outputs increasing by a constant amount each step, P has a constant first difference — making it a **linear** function.
4.For R: check whether the outputs follow a pattern where the numerators are perfect squares (1, 4, 9, 16, …).
5.If so, R grows as a **quadratic** function.
6.For Q: check whether consecutive outputs have a constant ratio (each output is a fixed multiple of the previous one).
7.A constant multiplicative factor means Q is an **exponential** function.
8.Once the function types are identified, apply the fundamental growth-rate hierarchy: linear functions are eventually outpaced by quadratic functions, which are eventually outpaced by exponential functions — regardless of how the values compare in the early rows of the table.
9.Therefore, from slowest to fastest eventual growth: **P (linear), R (quadratic), Q (exponential)**, which corresponds to answer choice **E**.
Answer: E) P, R, Q
Question 14 · Algebraic ExpressionsHard
If x is negative and x2+5x+6x+2\frac{x² + 5x + 6}{x + 2} = -4, what is the value of x2−49x+7\frac{x² - 49}{x + 7}?
A-7
B14
Cundefined
D-14
E0
Show answer & explanation
Let's work it out
1.Factoring the first numerator gives (x + 2)(x + 3), so the relation simplifies to x + 3 = -4 and therefore x = -7.
2.Substituting -7 into the denominator x + 7 gives 0, so the requested expression is undefined.
Answer: C) undefined
Question 15 · GeometryHard
A bakery prices each wedge of a uniformly thick circular tart according to the fraction of the tart contained in the wedge. A whole tart has a marked price of 26 14\frac{1}{4} dollars. After a discount of 15\frac{1}{5} of its marked price, one wedge costs 7 dollars. A point P is chosen on the major arc between the endpoints of the wedge, and straight lines are drawn from P to those endpoints. What is the measure of the angle formed at P?
A48°
B30°
C60°
D36°
E120°
Show answer & explanation
Let's work it out
1.Because the customer paid 45\frac{4}{5} of the marked price, the wedge was marked at 7 ÷ 45\frac{4}{5} = 354\frac{35}{4} dollars.
2.Compared with the whole-tart price of 1054\frac{105}{4} dollars, the wedge is 13\frac{1}{3} of the tart, so its arc measures 120 degrees.
3.The angle at P is an inscribed angle intercepting that arc, so it measures 120 ÷ 2 = 60 degrees.
Answer: C) 60°
Question 16 · Coordinate GraphsHard
A microscope image uses a coordinate grid in which each coordinate unit represents 0.4 millimeter. One endpoint of a needle-shaped organism is A=(1.5,2.0), and its center is M=(3.5,3.5), the midpoint of its two endpoints. A food particle is located at F=(7.5,6.5). How many millimeters separate the food particle from the other endpoint of the organism?
A2.5
B1
C1.4
D6.25
E2
Show answer & explanation
Let's work it out
1.Reflecting A across center M gives the other endpoint B=(5.5,5.0).
2.Its coordinate-grid distance from F is (7.5−5.5)2+(6.5−5.0)2\sqrt{(7.5-5.5)^2+(6.5-5.0)^2}=2.5 units, which represents 2.5*0.4=1 millimeter.
Answer: B) 1
Question 17 · Algebraic ExpressionsHard
If x = 30,000, evaluate x2−100,000,000x2−20,000x+100,000,000\frac{x^2 - 100,000,000}{x^2 - 20,000x + 100,000,000} - x2−400,000,000x2−40,000x+400,000,000\frac{x^2 - 400,000,000}{x^2 - 40,000x + 400,000,000}.
A1
B7
C-3
D0
E3
Show answer & explanation
Let's work it out
1.The first fraction factors and simplifies to x+10,000x−10,000\frac{x + 10,000}{x - 10,000}, which has value 2 at x = 30,000.
2.The second simplifies to x+20,000x−20,000\frac{x + 20,000}{x - 20,000}, which has value 5.
3.Their difference is 2 - 5 = -3.
Answer: C) -3
Question 18 · GeometryHard
A circle has radius 12. A sector has area 48π. An inscribed angle intercepting the sector's arc measures (2x + 100)°. What is x?
A-20
B10
C-852\frac{85}{2}
D80
E-35
Show answer & explanation
Let's work it out
1.The entire circle has area 144π, so the sector with area 48π occupies 13\frac{1}{3} of the circle and intercepts a 120° arc.
2.An inscribed angle intercepting that arc measures 60°.
3.Thus, 2x + 100 = 60, so x = -20.
Answer: A) -20
Question 19 · Coordinate GraphsHard
Coordinate gridCoordinate grid, x-axis from 0 to 3, y-axis from 0 to 3. Points: A (1/2, 1/2) at (0.5, 0.5), B (5/2, 3/2) at (2.5, 1.5), P (1/2, 5/2) at (0.5, 2.5), Q (9/4, 3/2) at (2.25, 1.5). Also shown: a line labeled AB. x y 123123A (1/2, 1/2)B (5/2, 3/2)P (1/2, 5/2)Q (9/4, 3/2)
Mira is laying out a quilt design on a coordinate grid. The figure shows the fabric grain guide AB and two marked points, P and Q. She draws one chalk line through P parallel to AB and a second chalk line through Q perpendicular to AB. A decorative square will be centered where the two chalk lines meet. What is the x-coordinate of that center?
A1310\frac{13}{10}
B920\frac{9}{20}
C38\frac{3}{8}
D72\frac{7}{2}
E32\frac{3}{2}
Show answer & explanation
Let's work it out
1.Identify the slope of grain guide AB from the figure.
2.Using the two marked points on AB, the slope is 12\frac{1}{2}.
3.A line through P parallel to AB shares this slope; using P's coordinates from the figure, that chalk line has equation y = (12\frac{1}{2})x + 94\frac{9}{4}.
4.A line through Q perpendicular to AB has slope equal to the negative reciprocal of 12\frac{1}{2}, which is −2; using Q's coordinates, that chalk line has equation y = −2x + 6.
5.The center of the decorative square is where these two lines meet, so set the expressions equal: (12\frac{1}{2})x + 94\frac{9}{4} = −2x + 6.
6.Add 2x to both sides: (52\frac{5}{2})x + 94\frac{9}{4} = 6.
7.Subtract 94\frac{9}{4}: (52\frac{5}{2})x = 6 − 94\frac{9}{4} = 154\frac{15}{4}.
8.Multiply both sides by 25\frac{2}{5}: x = (154\frac{15}{4})(25\frac{2}{5}) = 3020\frac{30}{20} = 32\frac{3}{2}.
9.The x-coordinate of the center is 32\frac{3}{2}.
Answer: E) 32\frac{3}{2}
Question 20 · Algebraic ExpressionsHard
If x is a real number and x2−0.36x2+1.6x+0.6\frac{x^2 - 0.36}{x^2 + 1.6x + 0.6} = 0.2, which value of x satisfies the relation?
A0.8
B-0.6
C2.0
D1.0
E-0.5
Show answer & explanation
Let's work it out
1.Factor the rational expression and cancel the common factor x + 0.6, giving x−0.6x+1\frac{x - 0.6}{x + 1} = 0.2.
2.Cross-multiplication gives x - 0.6 = 0.2x + 0.2, so 0.8x = 0.8 and x = 1.0.
3.This value does not make the original denominator zero.
Answer: D) 1.0

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