SAT Math Full Practice Test, Intermediate Level

The answer is (A), $12$ minutes.

1. Let $t$ be the number of minutes after filling begins. The tank starts with $18$ liters and gains $4$ liters each minute, so the volume is $V = 18 + 4t$.
2. Set the volume equal to the target, $18 + 4t = 66$, so $4t = 48$.
3. Dividing by $4$ gives $t = \frac{48}{4} = 12$ minutes.
4. Check that $18 + 4(12) = 66$.

Choice (B), $16.5$, is $\frac{66}{4}$, which ignores the $18$ liters already in the tank and treats all the water as added during these minutes. Choice (C), $21$, is $\frac{66 + 18}{4}$, which adds the starting volume instead of subtracting it. Choice (D), $48$, is $66 - 18$, the number of liters added, which is a volume and not a time.

The answer is (B), $15$ labels per minute.

1. Machine A prints $\frac{540}{9} = 60$ labels per minute.
2. Machine B prints one label every $0.8$ second, so in the $60$ seconds of a minute it prints $\frac{60}{0.8} = 75$ labels.
3. The difference is $75 - 60 = 15$ labels per minute.
4. Check that $75(0.8) = 60$ seconds and $60(9) = 540$ labels.

Choice (A), $1.25$, is $\frac{1}{0.8}$, machine B’s rate in labels per second. That rate is valid, but it still needs converting to labels per minute, and the question asks for a difference. Choice (C), $60$, is machine A’s rate alone, and choice (D), $75$, is machine B’s rate alone.

The answer is (C), $34$.

1. The condition compares the outputs at the same input, so $5t - 5 = (3t + 7) + 6$.
2. Combine the constants on the right to get $5t - 5 = 3t + 13$.
3. Subtract $3t$ and add $5$ on both sides, which gives $2t = 18$, so $t = 9$.
4. The question asks for the output $f(t)$, so compute $f(9) = 3(9) + 7 = 34$.
5. Check that $g(9) = 5(9) - 5 = 40$ and $40 = 34 + 6$, so the six-unit difference holds.

Choice (A), $9$, is the input $t$, while the question asks for the output $f(t)$. Choice (B), $25$, comes from setting $g(t) = f(t)$, which gives $t = 6$ and $f(6) = 25$ and leaves out the stated difference of $6$. Choice (D), $40$, is $g(9)$, the output of the other function.

The answer is $-9$.

1. Both products share the factor $x - 2$, so the left side is $(x-2)\big[(3x+4) + (x-5)\big]$.
2. Combining inside the brackets gives $(x-2)(4x-1)$.
3. Expanding gives $4x^2 - x - 8x + 2 = 4x^2 - 9x + 2$, so $k = -9$.
4. As a check, expanding the original products gives $(3x^2 - 2x - 8) + (x^2 - 7x + 10) = 4x^2 - 9x + 2$.

A common factor of a sum can be factored out, but it cannot simply be erased. Keeping it is what produces both the $-8x$ term and the constant $2$. Enter $-9$.

The answer is (D), $60$ degrees.

1. Since $CD$ is parallel to $AB$ and $D$ and $B$ lie on opposite sides of line $AC$, angles $DCA$ and $CAB$ are alternate interior angles, so angle $CAB = 52^\circ$.
2. The angles of triangle $ABC$ add to $180^\circ$, so $52 + 68 + \angle ACB = 180$.
3. Hence angle $ACB = 180 - 120 = 60^\circ$.
4. Check that $52 + 60 + 68 = 180$.

The transfer of the $52^\circ$ angle rests on the stated parallel lines, not on how a drawing looks. Choice (A), $52$, repeats the transferred angle at $A$. Choice (B), $68$, repeats the given angle at $B$. Choice (C), $120$, is the sum $52 + 68$ of the two known angles, without subtracting it from $180$.

The answer is (B), $45$.

1. Let $u = F(3)$. Since $a$ and $b$ are positive, $u$ is positive.
2. Each step from input $2$ to $3$ and from $3$ to $4$ multiplies the output by the same $b$, so $\frac{u}{27} = \frac{75}{u}$.
3. Multiplying by the nonzero quantity $27u$ gives $u^2 = 27 \times 75 = 2025$.
4. The positive square root gives $u = 45$. The negative root is excluded because the exponential model is positive.
5. Check the successive ratios, $\frac{45}{27} = \frac{5}{3}$ and $\frac{75}{45} = \frac{5}{3}$.

Choice (A), $\frac{25}{9}$, is $\frac{F(4)}{F(2)}$, the multiplier across two input units rather than the middle output. Choice (C), $51$, is $\frac{27 + 75}{2}$, the arithmetic midpoint, which fits a constant additive change. Choice (D), $2025$, is $u^2$, before taking the positive square root.

The answer is (B), $15$ jugs.

1. With $n$ filled jugs, where $n$ is a nonnegative whole number, the mass condition is $6 + 2.4n \le 44$.
2. Subtracting $6$ gives $2.4n \le 38$, so $n \le \frac{95}{6}$, which is $15\frac{5}{6}$.
3. The greatest whole number that satisfies this is $n = 15$.
4. Check that $6 + 2.4(15) = 42 \le 44$, while $6 + 2.4(16) = 44.4 > 44$.

Choice (A), $14$, works but is not the greatest allowed count. Choice (C), $16$, comes from rounding $15\frac{5}{6}$ to the nearest whole number, which breaks the mass condition. Choice (D), $18$, rounds $\frac{44}{2.4}$ down and ignores the mass of the rack itself.

The answer is $-5$.

1. Substitute $x = -4$, which gives $3\big((-4) - k\big) = 2(-4) + 11$.
2. The right side is $3$, so $3(-4 - k) = 3$ and $-4 - k = 1$.
3. Then $-k = 5$, so $k = -5$.
4. Check that $3\big((-4) - (-5)\big) = 3 = 2(-4) + 11$.

In the check, the second minus sign comes from the value of $k$, while the first belongs to the original subtraction. Keeping both visible prevents a sign change from being lost. Enter $-5$.

The answer is (C), $-7$.

1. Factor $2$ from the $x$ terms, $h(x) = 2(x^2 - 6x) + 11$.
2. Complete the square, $h(x) = 2\big[(x-3)^2 - 9\big] + 11$.
3. Simplify to $h(x) = 2(x-3)^2 - 7$.
4. A square is never negative, so $h(x) \ge -7$, with equality at $x = 3$.
5. Check that $h(3) = 2(3)^2 - 12(3) + 11 = -7$.

Choice (A), $11$, is $h(0)$, the $y$-intercept. Choice (B), $3$, is the input at the vertex, not its output. Choice (D), $-18$, is the adjustment from completing the square, before adding $11$.

The answer is (D), $66$ grams.

1. A mean multiplied by the number of observations gives the total. The original total is $6 \times 52 = 312$ grams.
2. The new total is $8 \times 55 = 440$ grams.
3. The two added masses total $440 - 312 = 128$ grams.
4. Subtract the known added mass, $128 - 62 = 66$ grams.
5. Check that $\frac{312 + 62 + 66}{8} = 55$. No assumption about equal original masses is needed.

Choice (A), $48$, solves $\frac{62 + x}{2} = 55$, which wrongly gives the mean of all eight pieces to the two added pieces. Choice (B), $51$, solves $\frac{52 + 62 + x}{3} = 55$, which treats the mean of six pieces as one observation. Choice (C), $64$, is $\frac{128}{2}$, the mean of the two added pieces, not the mass of the other piece.

The answer is (D), $x + 3$ with $x \ne 4$.

1. The original expression is defined only when $x \ne 4$, since the denominator is zero at $x = 4$.
2. Factor the numerator, $x^2 - x - 12 = (x-4)(x+3)$.
3. For $x \ne 4$, the factor cancels and $\frac{(x-4)(x+3)}{x-4} = x + 3$.
4. The graph is the line $y = x + 3$ with a hole at $(4, 7)$.
5. Check at $x = 2$, where the original gives $\frac{-10}{-2} = 5 = 2 + 3$. At $x = -3$ the original gives $\frac{0}{-7} = 0$, so $-3$ stays in the domain.

Choice (A) extends the domain to $x = 4$, where the original denominator is zero. Choice (B) has the wrong sign in the remaining factor. Choice (C) excludes a zero of the numerator instead of the zero of the denominator, and wrongly includes $4$.

The answer is $4$.

1. The slope of the line is $\frac{-3 - 9}{4 - (-2)} = \frac{-12}{6} = -2$.
2. Write $y = -2x + c$ and substitute $(-2, 9)$, which gives $9 = -2(-2) + c$, so $c = 5$.
3. Because $(a, b)$ lies on the line, $b = -2a + 5$. Combine this with $b = 3a$.
4. Setting $3a = -2a + 5$ gives $5a = 5$ and $a = 1$. Then $b = 3$.
5. The requested sum is $a + b = 1 + 3 = 4$. The point $(1, 3)$ satisfies both equations, since $3 = -2(1) + 5$ and $3 = 3(1)$.

The answer is (A), $83$ square meters.

1. The enclosing rectangle has area $12 \times 9 = 108$ square meters.
2. Removing the square of area $25$ leaves $108 - 25 = 83$ square meters.
3. As a check, split the region into a lower rectangle $12$ by $4$ and an upper-left rectangle $7$ by $5$.
4. Their areas add to $12(4) + 7(5) = 48 + 35 = 83$.

Choice (B), $108$, is the whole rectangle and still includes the missing corner. Choice (C), $73$, is $108 - 35$, which removes the upper-left piece instead of the stated square. Choice (D), $35$, counts only the upper-left piece.

The answer is (B), $3$.

1. Infinitely many solutions means both equations describe the same line, so the first equation is a multiple of the second. The $y$ coefficients give the scale $\frac{6}{8} = \frac{3}{4}$.
2. The constants give the same scale, $\frac{15}{20} = \frac{3}{4}$, so $k = \frac{3}{4}(4) = 3$.
3. Check that multiplying $3x + 6y = 15$ by $\frac{4}{3}$ gives $4x + 8y = 20$. Both equations contain $(0, \frac{5}{2})$ and $(5, 0)$, since $3(0) + 6(\frac{5}{2}) = 15$ and $3(5) + 6(0) = 15$.
4. For any other value, subtracting $\frac{3}{4}$ of the second equation from the first gives $(k - 3)x = 0$, so $x = 0$ and the only solution is $(0, \frac{5}{2})$.

Choice (A), $0$, does not keep the coefficient ratio $\frac{3}{4}$ and gives one intersection point. Choice (C), $4$, copies the $x$ coefficient of the second equation without scaling it. Choice (D), $6$, copies the $y$ coefficient of the first equation instead of applying the common scale.

The answer is (C), $3$.

1. A square root is never negative, so any solution needs $x \ge 0$.
2. Squaring both sides gives $x + 6 = x^2$, or $x^2 - x - 6 = 0$.
3. Factoring gives $(x - 3)(x + 2) = 0$, so the candidates are $3$ and $-2$.
4. Check both in the original equation. At $x = 3$, $\sqrt{3 + 6} = 3$, which works. At $x = -2$, $\sqrt{-2 + 6} = 2 \ne -2$, so $-2$ is extraneous.

Squaring makes $2$ and $-2$ look the same, because both squares are $4$. That loss of sign is why the original equation must decide between the candidates. Choice (A), $-2$, is the extraneous candidate introduced by squaring. Choice (B), $2$, confuses the value $\sqrt{4} = 2$ from the check with a solution, and $2$ itself fails, since $\sqrt{8} \ne 2$. Choice (D) accepts both roots of the squared equation without the sign condition of the original.

The answer is $\frac{9}{16}$, which can be entered as $9/16$, $.5625$ or $0.5625$.

1. The selection is made from club members only, and there are $18 + 14 = 32$ of them.
2. Of these, $18$ travel by bus, so $P(\text{bus} \mid \text{club}) = \frac{18}{32}$.
3. Simplify, $\frac{18}{32} = \frac{9}{16} = 0.5625$.
4. Check with the walkers, $\frac{18}{32} + \frac{14}{32} = 1$.

The whole group has $70$ students, but the selection was restricted to club members. Using $70$ as the denominator would answer a different question, about a student selected from the whole group.

The answer is (D), $(-4, 2)$.

1. The only known value of $f$ is $f(-1) = 4$, so choose $x$ with $x + 3 = -1$.
2. That gives $x = -4$, and indeed $(-4) + 3 = -1$.
3. Then $g(-4) = f(-1) - 2 = 4 - 2 = 2$.
4. The point $(-1, 4)$ on the graph of $f$ corresponds to $(-4, 2)$ on the graph of $g$.

Moving three units right would use $x = 2$, which makes the inside input $5$, and the problem gives no value for $f(5)$. Adding $2$ to the output would reverse the stated subtraction. Choice (A) moves right instead of solving $x + 3 = -1$. Choice (B) shifts the output up by $2$ instead of down. Choice (C) reverses both the horizontal and the vertical shift.

The answer is (A), $60$ square centimeters.

1. A perpendicular from the center of a circle to a chord bisects the chord, so $AM = MB = 12$ centimeters.
2. In right triangle $OMA$, the radius $OA = 13$ is the hypotenuse. Let $OM = h$.
3. Then $h^2 + 12^2 = 13^2$, so $h^2 = 25$ and $h = 5$, because a length is positive.
4. For triangle $OAB$, the full base is $AB = 24$ and the perpendicular height is $OM = 5$.
5. The area is $\frac{1}{2}(24)(5) = 60$ square centimeters.

Choice (B), $30$, uses $AM = 12$ as if it were the whole base of triangle $OAB$, which gives the area of one half-triangle. Choice (C), $120$, multiplies base and height without the factor $\frac{1}{2}$. Choice (D), $156$, uses the radius $13$ as the height, $\frac{1}{2}(24)(13)$.

The answer is (A), $15$.

1. Subtract the first equation from the second, $(5x + 4y) - (2x + 3y) = 32 - 17$.
2. The left side simplifies to $3x + y$, so $3x + y = 15$, which is exactly the target.
3. As a check, solve for each variable. Multiplying gives $4(2x + 3y) = 68$ and $3(5x + 4y) = 96$, and subtracting these gives $7x = 28$.
4. So $x = 4$ and $y = 3$, and $3(4) + 3 = 15$.

Choice (B), $3$, is the value of $y$, and choice (C), $4$, is the value of $x$, not $3x + y$. Choice (D), $49$, adds the right sides instead of using the difference that matches the target.

The answer is $180$.

1. Eight hours contain $\frac{8}{4} = 2$ intervals of $4$ hours.
2. If $N$ is the count at time $0$, then $N(1.5)^2 = 405$.
3. Hence $N = \frac{405}{9/4} = 405\left(\frac{4}{9}\right) = 180$.
4. Check forward, $180\left(\frac{3}{2}\right) = 270$ and $270\left(\frac{3}{2}\right) = 405$.

Dividing by $1.5$ only once would give the count at $4$ hours. Using eight multiplications would treat the four-hour multiplier as an hourly one. The length of the interval decides the exponent.

The answer is (B), $3$ items.

1. At $6$ hours the model predicts $y = 1.5(6) + 8 = 9 + 8 = 17$ items.
2. The residual is $20 - 17 = 3$ items.
3. The observed value is above the prediction, so the residual is positive.
4. Check that the prediction plus the residual gives the observation, $17 + 3 = 20$.

Choice (A), $-3$, reverses the subtraction and computes predicted minus observed. Choice (C), $17$, is the predicted output, and choice (D), $20$, is the observed output. Neither alone is the difference the question asks for.

The answer is (C), $Tv - a$.

1. Multiply both sides by $v$, which is allowed since $v > 0$, to get $vT = a + b$.
2. Subtract $a$ to get $b = Tv - a$.
3. The units agree, since a time multiplied by a distance per unit time is a distance.
4. A numerical check with $T = 3$, $v = 20$ and $a = 18$ gives $b = 3(20) - 18 = 42$, and $\frac{18 + 42}{20} = 3$.

The subtraction happens after finding the full distance. Choice (A) divides by $v$ a second time instead of undoing the original division. Choice (B) puts $T - a$ inside one bracket, which subtracts a distance from a time. Choice (D) reverses the final subtraction and gives the negative of the second length.

The answer is (D), $-19$.

1. The zeros $-2$ and $7$ give the factors, so $p(x) = (x + 2)(x - 7)$.
2. Expanding gives $p(x) = x^2 - 5x - 14$.
3. So $b = -5$ and $c = -14$, and $b + c = -19$.
4. Check that $p(-2) = 4 + 10 - 14 = 0$ and $p(7) = 49 - 35 - 14 = 0$.

The sum of the zeros is $5$, but the linear coefficient is its negative, $-5$. The constant is the product of the zeros, $-14$. Expanding the factors gives these signs directly, so a memorized sign rule cannot be reversed by mistake. Choice (A), $9$, uses $b = -5$ but changes $c$ to $14$. Choice (B), $19$, changes both signs to $5$ and $14$. Choice (C), $-9$, uses the sum of the zeros, $5$, for $b$ while keeping $c = -14$.

The answer is $-4$.

1. The slope is $\frac{6 - 14}{1 - (-3)} = \frac{-8}{4} = -2$.
2. From input $1$ to input $6$ the output changes by $L(6) - L(1) = (-2)(6 - 1) = -10$.
3. So $L(6) = 6 - 10 = -4$.
4. As a check, the function is $L(x) = -2x + 8$, which gives $L(-3) = 14$, $L(1) = 6$ and $L(6) = -4$.

Subtracting $8$ again for the next step would overlook that the input gaps, $4$ and $5$, are unequal. A linear function keeps the rate per unit constant, so each output change must use its own input change.

The answer is (A), $15$.

1. Since $DE$ is parallel to $BC$, triangle $ADE$ is similar to triangle $ABC$.
2. The full side is $AB = 6 + 4 = 10$, so the scale factor is $\frac{AD}{AB} = \frac{6}{10} = \frac{3}{5}$.
3. Corresponding sides give $\frac{9}{BC} = \frac{3}{5}$, so $BC = 15$.
4. Check that $\frac{3}{5}(15) = 9$ and $\frac{3}{5}(10) = 6$.

The length $4$ is only the leftover piece $DB$. Pairing $AD$ with that leftover compares a whole side of the small triangle with only part of a side of the large one, and similarity ratios must use complete corresponding sides. Choice (B), $13.5$, is $9\left(\frac{6}{4}\right)$, which compares $AD$ with $DB$ instead of with $AB$. Choice (C), $9$, assumes parallel segments have equal lengths. Choice (D), $6$, is $9\left(\frac{4}{6}\right)$, built from mismatched sides.

The answer is (B), $7$.

1. The absolute value equals $7$ when $2x - 5 = 7$ or $2x - 5 = -7$.
2. The two branches give $x = 6$ or $x = -1$.
3. The greater minus the lesser is $6 - (-1) = 7$.
4. Check that $|2(6) - 5| = 7$ and $|2(-1) - 5| = 7$.

Choice (A), $1$, is the size of the lesser solution, not the difference. Choice (C), $6$, is only the greater solution, while the question asks for a difference between two solutions. Choice (D), $12$, assumes the solutions are opposites and doubles $6$, but the lesser solution is $-1$, not $-6$.

The answer is (C), $6$ hours.

1. The bill is more than the flat $6, so the time is beyond $2$ hours. For $t > 2$ the charge in dollars is $C(t) = 6 + 4(t - 2)$.
2. Set $6 + 4(t - 2) = 22$, so $4(t - 2) = 16$.
3. Then $t - 2 = 4$ and $t = 6$.
4. Check that $C(6) = 6 + 4(4) = 22$.

Choice (A), $4$, is only the added time, without the first $2$ hours. Choice (B), $5.5$, is $\frac{22}{4}$, which treats every dollar of the bill as an hourly charge, including the flat fee. Choice (D), $8$, doubles the four extra hours instead of adding the first two.

The answer is $12$.

1. Substitute the given input and output into the rule, $2^2 + 2c + 36 = 10$.
2. This gives $40 + 2c = 10$, so $2c = -30$ and $c = -15$.
3. The completed rule is $q(x) = x^2 - 15x + 36$.
4. To factor, find two numbers with product $36$ and sum $-15$, which are $-3$ and $-12$.
5. Thus $q(x) = (x - 3)(x - 12)$, with real zeros $3$ and $12$. The larger is $12$.
6. Check that $q(2) = 4 - 30 + 36 = 10$ and $q(12) = 144 - 180 + 36 = 0$.

The answer is (B), $7080$ to $8280$ registered members.

1. The interval for the proportion runs from $0.64 - 0.05 = 0.59$ to $0.64 + 0.05 = 0.69$.
2. Multiply each endpoint by the number of registered members, $0.59(12{,}000)$ and $0.69(12{,}000)$.
3. This gives $7080$ to $8280$ registered members.
4. As a check, the center is $0.64(12{,}000) = 7680$ and the margin is $0.05(12{,}000) = 600$, and $7680 - 600 = 7080$ and $7680 + 600 = 8280$.

This is a reported estimation interval, so it does not give an exact count of supporters. The confidence level describes how often this interval method covers the true value across repeated random samples. The random sample supports inference to registered library members, not automatically to everyone in the town.

Choice (A) multiplies by the sample size $400$ instead of the $12{,}000$ members, which gives counts on the scale of the sample. Choice (C) keeps the point estimate as the lower endpoint instead of subtracting the margin. Choice (D) takes $5$ percent of $64$ percent, which turns a margin of $5$ percentage points into a different quantity.

The answer is (D), $4$.

1. At an intersection the $y$ values agree, so $x^2 - 4x + 1 = 2x - 4$.
2. Rearranging gives $x^2 - 6x + 5 = 0$, and factoring gives $(x - 1)(x - 5) = 0$.
3. The line gives the $y$ values, $2(1) - 4 = -2$ at $x = 1$ and $2(5) - 4 = 6$ at $x = 5$.
4. The sum of the $y$-coordinates is $-2 + 6 = 4$.
5. Check in the parabola, which gives $1 - 4 + 1 = -2$ and $25 - 20 + 1 = 6$.

Choice (A), $6$, adds the $x$-coordinates $1 + 5$, or gives only the larger $y$ value. Choice (B), $5$, is the larger $x$-coordinate. Choice (C), $2$, counts the intersection points instead of adding their $y$-coordinates.

The answer is (A), $\frac{21}{29}$.

1. The right angle is at $Q$, so $PR$ is the hypotenuse and $PR^2 = PQ^2 + QR^2$.
2. This gives $PR^2 = 400 + 441 = 841$, so $PR = 29$.
3. At angle $R$ the adjacent leg is $QR$, so $\cos R = \frac{QR}{PR} = \frac{21}{29}$.
4. The orientation of a picture does not change these roles. A rotation keeps the angle at $R$ and every side length, so adjacent and opposite are decided by the chosen vertex, not by horizontal or vertical position.
5. Check that $29^2 = 841 = 20^2 + 21^2$. Also $\sin P = \cos R = \frac{21}{29}$, as expected for the two acute angles.