Advanced Math Practice Test 3

Answer: D.

The coefficient of h is the model’s increase for one more minute, so it predicts 6.5 additional grooves inspected per minute.

Check: $R(h+1)-R(h)=6.5$ for any allowed h.

A: That value is 21.

B: That would be the reciprocal rate.

C: That value is 27.5.

Watch for: A model may have a noninteger average rate even when it counts whole objects.

Percent change is the change divided by the original value. Sales fell by $5{,}040 - 4{,}200 = 840$, and $840 \div 5{,}040 \approx 0.167$, so the decrease is about $16.7\%$. Dividing the change by the new value instead of the original gives $840 \div 4{,}200 = 20\%$. Dividing June by May gives $83.3\%$, the fraction remaining rather than the fraction lost. Dividing May by June gives $120\%$.

Answer: D.

Above the first line means $y\ge-x+6$ because the boundary is solid. Below the horizontal line means $y<8$ because that boundary is dashed.

Check: A point such as (2,6) satisfies both inequalities and lies inside the shaded region.

A: This places the region below the sloped line.

B: This places the region above the horizontal line.

C: This reverses which boundaries are included.

Watch for: Translate the side of each boundary and its inclusion style independently.

Answer: 32

Let m be missed checkpoints. Then 2m are skipped, so 50 - 3m are successful. The score equation is $6(50-3m)-4m=168$. Thus $300-22m=168$, m = 6, and the successful count is $50-18=32$.

Check: The counts are 32 successful, 6 missed, and 12 skipped. They sum to 50 and give score $192-24=168$.

Watch for: The skipped category consumes checkpoints even though it contributes zero points.

Answer: C.

Taking $12\%$ off saves more than $15 exactly when $0.12p>15$, that is, when the price $p$ is greater than $125.

If the $15 discount were used, the price would be $132+15=147$ dollars. But at $147, $12\%$ off saves $0.12\cdot147=17.64$ dollars, which is more than $15, so the coupon would have taken $12\%$ off instead. This case is impossible.

If $12\%$ off was used, then $0.88p=132$, so $p=150$. At $150, $12\%$ off saves $18, more than $15, so this case is consistent. The price was $150.00.

Check: For a $150 bicycle, the two discounts are $15 and $18; the coupon takes the larger, $18, and the customer pays $132.

A: $147.00 undoes a $15 discount, but at that price $12\%$ off saves $17.64, so the coupon would not have taken $15.

B: $147.84 adds $12\%$ of the amount paid, $1.12\cdot132$, instead of dividing by $0.88$.

D: $151.80 adds $15\%$ of the amount paid, $1.15\cdot132$, mixing the two offers.

Watch for: Check each possible discount against the rule “whichever saves more”; a price that fits one discount is valid only if that discount is actually the larger one.

Answer: 13.

Squaring both sides gives $2x+10=x^2-14x+49$, so $x^2-16x+39=0$, which factors as $(x-3)(x-13)=0$. Squaring can create roots that do not satisfy the original equation, so both candidates must be tested. At $x=13$, $\sqrt{36}=6$ and $13-7=6$, which agree; at $x=3$, $\sqrt{16}=4$ but $3-7=-4$, so $x=3$ must be discarded.

Check: The left side $\sqrt{2x+10}$ is never negative, so any solution needs $x-7\ge 0$, that is $x\ge 7$. The candidate $x=3$ fails that requirement outright, and $x=13$ meets it.

Watch for: After squaring a radical equation, substitute every candidate back into the original equation, because a principal square root can never equal a negative number.

Answer: C.

The correspondence is 6 model centimeters to 24 actual meters, or 4 actual meters per model centimeter. The actual long side is 9(4) = 36 meters. The area is 24(36) = 864 square meters.

Check: The actual dimensions retain the model’s 2:3 ratio.

A: This applies a length factor to the model area only once.

B: This uses an incorrect long-side conversion.

D: This treats the plaza as a 36-meter square.

Watch for: Convert both corresponding lengths consistently before multiplying; a length scale is squared for area.

Answer: D.

At $x=-1$ the line gives $y=4(-1)+3=-1$, and the parabola must give the same value there: $2(-1)^2+k(-1)-5=-3-k$, so $-3-k=-1$ and $k=-2$. The graphs then meet where $2x^2-2x-5=4x+3$, that is, where $2x^2-6x-8=0$, or $2(x+1)(x-4)=0$. The intersections sit at $x=-1$ and $x=4$, and the line puts the second one at $y=4(4)+3=19$.

Check: The parabola reaches that same height at $x=4$: $2(4)^2-2(4)-5=32-8-5=19$, so $(4,19)$ lies on both graphs.

A: $-1$ is the $y$-coordinate of the point the stem already supplies, not of the second intersection.

B: 4 is the $x$-coordinate of the second intersection; the stem asks for its $y$-coordinate.

C: 11 reflects $x=-1$ across the parabola’s own axis of symmetry, $x=0.5$, landing at $x=2$. The two intersections are symmetric about the axis of the combined equation, $x=1.5$, not about the parabola’s axis.

Watch for: One stated intersection is enough to pin down an unknown constant; with the constant known the system collapses to a quadratic that factors, and the second root is read straight off.

Answer: C.

Using all $3{,}000$ milliliters of blue paint would need $\frac25(3{,}000)=1{,}200$ milliliters of white paint, but only $1{,}000$ milliliters are available. So the white paint limits the mixture.

With all $1{,}000$ milliliters of white paint, the blue paint is $\frac52(1{,}000)=2{,}500$ milliliters, and the total is $2{,}500+1{,}000=3{,}500$ milliliters.

Check: The ratio $2{,}500:1{,}000$ is $5:2$, and $2{,}500$ milliliters of blue paint is less than the $3{,}000$ milliliters available.

A: $1{,}400$ inverts the ratio: with $1{,}000$ milliliters of white paint it uses $\frac25(1{,}000)=400$ milliliters of blue paint.

B: $2{,}500$ is the volume of the blue paint only.

D: $4{,}200$ treats the blue paint as the paint that runs out first: all $3{,}000$ milliliters of blue paint would need $1{,}200$ milliliters of white paint, but only $1{,}000$ milliliters are available.

Watch for: Check which paint runs out first; convert liters to milliliters before comparing.

The equation is quadratic in $x^2$: substituting $u = x^2$ gives $u^2 - 13u + 36 = 0$, which factors as $(u - 4)(u - 9) = 0$, so $x^2 = 4$ or $x^2 = 9$. Each yields two solutions, $x = \pm 2$ and $x = \pm 3$, so the positive ones are $2$ and $3$ and their product is $6$. Multiplying all four solutions instead gives $36$, since the two negatives cancel. Answering $2$ or $3$ reports one of the pair rather than their product.

Answer: D.

Graph the two lines and read their intersection, $(4.5,8)$. The requested display is $D=3(4.5)+8=21.5$.

Check: $1.7(4.5)+2.3(8)=26.05$ and $2.6(4.5)-0.8(8)=5.3$. Both original equations are satisfied.

A: This is x, not the requested display.

B: This is y, not the requested display.

C: This gives x+y instead of 3x+y.

Watch for: An intersection gives two variables; calculate the expression actually requested.

Answer: B.

Silo A empties at $2.5(60)=150$ bushels per hour, so after $t$ hours it holds $6{,}400-150t$ bushels while silo B holds $2{,}800+90t$ bushels. Solving $2{,}800+90t>6{,}400-150t$ gives $240t>3{,}600$, so $t>15$, and the least whole number of hours is $16$.

Check: At $t=15$ each silo holds $4{,}150$ bushels, so B is not yet ahead. At $t=16$, silo A holds $4{,}000$ bushels and silo B holds $4{,}240$.

A: $15$ hours is when the two silos hold equal amounts. The question asks when B holds more, so $15$ is one hour too early.

C: $24$ hours is when silo A alone falls to silo B’s starting level of $2{,}800$ bushels. It ignores the grain being added to silo B.

D: This never converts $2.5$ bushels per minute into an hourly rate, solving $2{,}800+90t>6{,}400-2.5t$ instead, which gives $t>38.9$.

Watch for: Put both rates in the same time unit first, and remember that the boundary value of a strict inequality is not itself a solution.

Answer: B.

At a depth of $12$ meters, the temperature is $18-0.4(12)=13.2^\circ\text{C}$, which is still above $12^\circ\text{C}$. The remaining drop of $1.2^\circ\text{C}$ happens at $0.1^\circ\text{C}$ per meter, which takes $\dfrac{1.2}{0.1}=12$ more meters. The depth is $12+12=24$ meters.

Check: At $24$ meters the temperature is $18-0.4(12)-0.1(12)$, which is $18-4.8-1.2=12$ degrees Celsius.

A: $15$ uses $0.4^\circ\text{C}$ per meter for the whole drop, $\dfrac{6}{0.4}$, ignoring the change in rate at $12$ meters.

C: $60$ uses $0.1^\circ\text{C}$ per meter for the whole drop, $\dfrac{6}{0.1}$.

D: $72$ applies $0.1^\circ\text{C}$ per meter to the whole $6$-degree drop and then adds the first $12$ meters.

Watch for: Find the temperature at $12$ meters first; only the drop below that depth uses the second rate.

Answer: A.

Reflecting across the $x$-axis changes $(x,y)$ to $(x,-y)$. The reflected line passes through $(5,2)$, so line $\ell$ passes through $(5,-2)$. Line $\ell$ also passes through $(2,7)$, so its slope is

$m=\dfrac{-2-7}{5-2}=-3$

Check: Line $\ell$ is $y=-3x+13$, which contains $(2,7)$. Its reflection is $y=3x-13$, which contains $(5,2)$.

B: $-\dfrac{5}{3}$ uses $(5,2)$ itself as a point on line $\ell$, ignoring the reflection.

C: $\dfrac{5}{7}$ reflects $(5,2)$ across the $y$-axis to $(-5,2)$ instead of across the $x$-axis.

D: $3$ is the slope of the reflected line, not of line $\ell$.

Watch for: The point on the reflected line corresponds to $(5,-2)$ on line $\ell$.

Answer: B.

The vertex of the graph is $(2,k)$. The graph meets the $x$-axis where $|2x-4|=k$, that is, where $x=2\pm\dfrac k2$. So the enclosed triangle has base $\dfrac k2-\left(-\dfrac k2\right)=k$ and height $k$, and its area is $\tfrac12k^2$. Then $\tfrac12k^2=25$, so $k^2=50$ and $k=5\sqrt2$.

Check: With $k=5\sqrt2\approx7.07$, the intercepts are about $-1.54$ and $5.54$, a base of about $7.07$, and $\tfrac12(7.07)(7.07)\approx25$.

A: $5$ solves $|x-2|=k$ instead of $|2x-4|=k$, which doubles the base to $2k$ and gives an area of $k^2$.

C: $10$ uses $\tfrac k2$, the distance from the vertex to one intercept, as the whole base, so the area is $\tfrac{k^2}{4}$.

D: $\tfrac{25}{2}$ uses the $4$ in $|2x-4|$ as the height: $\tfrac12\cdot k\cdot4=25$.

Watch for: The coefficient $2$ inside the absolute value halves the distance from the vertex to each intercept.

Answer: C.

Because $\overline{AB}\parallel\overline{DC}$, triangles $ABE$ and $CDE$ are similar. Their area ratio is $27:12=9:4$, so their similarity ratio is $3:2$, and $AE:EC=3:2$. Triangles $ABE$ and $CBE$ share the height from $B$ to line $AC$, so their areas are in the ratio $AE:EC=3:2$:

$\text{area}(CBE)=\tfrac23(27)=18$

Similarly, triangles $ADE$ and $CDE$ share the height from $D$, so $\text{area}(ADE)=\tfrac32(12)=18$. The trapezoid’s area is $27+12+18+18=75$.

Check: Triangles $ABD$ and $ABC$ have the same base $\overline{AB}$ and the same height, so they have equal areas: $27+18=45$ each. Also, $18^2=27\cdot12$, as the two side triangles require.

A: This adds only the two given triangles and leaves out triangles $ADE$ and $BCE$.

B: This assumes each of triangles $ADE$ and $BCE$ has the same area as triangle $CDE$, $12$: $27+12+12+12=63$.

D: This uses the area ratio $9:4$ as the ratio $AE:EC$, which makes the side triangles $\tfrac49(27)=12$ and $\tfrac94(12)=27$, for a total of $78$.

Watch for: Similar triangles have areas in the ratio of the squares of their sides; take the square root before using the ratio for lengths.

Answer: 23 Let x and y count the 9- and 14-kilogram containers. Then 9x + 14y = 235, with x, y nonnegative integers. To maximize x, minimize y. Testing y = 0, 1, 2 gives remainders 235, 221, 207 for the 9-kilogram containers. The first divisible by 9 is 207 = 9(23), so y = 2 and x = 23. Larger y only decreases x.

Watch for: The continuous bound 235/9 is not attainable with whole containers of the given sizes.

Answer: -3

The zeros of $p$ are $0$, $3$, and $-a$. The graph touches the $x$-axis without crossing it at a zero that is repeated an even number of times, so one of the three zeros must repeat.

If $-a=0$, then $a=0$ and $p(x)=x^2(x-3)$, which touches the axis at $x=0$ and crosses it at $x=3$.

If $-a=3$, then $a=-3$ and $p(x)=x(x-3)^2$, which touches the axis at $x=3$ and crosses it at $x=0$.

For any other value of $a$, the three zeros are different and the graph crosses the axis at each of them. The possible values are $0$ and $-3$, so the least is $-3$.

Check: For $a=-3$, $p(2)=2(-1)^2=2$ and $p(4)=4(1)^2=4$ are both positive, so the graph stays above the axis on both sides of $x=3$.

Watch for: The most common wrong value is $0$, which makes $x=0$ the repeated zero; the zero $x=3$ can also repeat, when $a=-3$.

Answer: 3

Let the intersection point be $(t,0)$. The slope of $\ell$ is $\dfrac{2-0}{4-t}$, and the slope of $m$ is $\dfrac{3-0}{-1-t}$. Perpendicular lines have slopes whose product is $-1$:

$\dfrac{2}{4-t}\cdot\dfrac{3}{-1-t}=-1$

$6=(4-t)(1+t)$

$t^2-3t+2=0$

So $(t-1)(t-2)=0$, and $t=1$ or $t=2$. Both work, and the sum of the possible $x$-coordinates is $3$.

Check: For $t=1$, the slopes are $\dfrac23$ and $-\dfrac32$; for $t=2$, they are $1$ and $-1$. Each pair has product $-1$.

Watch for: The most common wrong values are $1$ or $2$ alone; the condition gives a quadratic equation, and both of its solutions work.

Answer: B.

Draw the altitude from $A$ to line $BC$, meeting it at $D$. Let $BD=x$ and $AD=h$, so $DC=21-x$. The two right triangles give

$x^2+h^2=100$

$(21-x)^2+h^2=289$

Subtracting the first equation from the second gives $441-42x=189$, so $x=6$. Then $h^2=100-36=64$, so $h=8$, and $DC=21-6=15$. Because $0<x<21$, point $D$ lies between $B$ and $C$.

In right triangle $ADC$, $\tan C=\frac{AD}{DC}=\frac{8}{15}$.

Check: Right triangle $ADC$ has legs $8$ and $15$ and hypotenuse $\sqrt{64+225}=17=AC$, and right triangle $ADB$ has legs $6$ and $8$ and hypotenuse $10=AB$.

A: $\frac{8}{17}$ is $\sin C$, the altitude divided by the hypotenuse $AC$.

C: $\frac43$ is $\tan B$, from right triangle $ADB$.

D: $\frac{15}{8}$ is $\frac{DC}{AD}$, the reciprocal of $\tan C$.

Watch for: Angle $C$ is not in a right triangle until you draw the altitude from $A$; then use the two right triangles that share it.

Answer: B.

Fit $d(x)=0.6x^2-4.4x+10.2$ to the table. The vertex input is $-b/(2a)=4.4/1.2=11/3\approx3.7$. Since a > 0, the vertex gives the minimum.

Check: The table’s second differences are 1.2, giving a = 0.6. This independently recovers the same vertex.

A: This approximates the vertex’s output rather than input.

C: This is the magnitude of the linear coefficient.

D: This is the modeled output at zero.

Watch for: The input of a minimum is the vertex’s horizontal coordinate, not its height.

Answer: 12

Multiply both sides by $(x-2)(x+1)$, remembering that $x\ne2$ and $x\ne-1$:

$x(x+1)-4(x-2)=k$

so $x^2-3x+8-k=0$. The original equation has exactly one real solution in two kinds of cases.

The quadratic has a repeated root: $9-4(8-k)=0$ gives $k=\tfrac{23}{4}$, with root $x=\tfrac32$, which is allowed.

The quadratic has two roots and one of them is excluded. If $x=2$ is a root, $4-6+8-k=0$ gives $k=6$, and the roots are $1$ and $2$, leaving only $x=1$. If $x=-1$ is a root, $1+3+8-k=0$ gives $k=12$, and the roots are $-1$ and $4$, leaving only $x=4$.

So $k$ can be $\tfrac{23}{4}$, $6$, or $12$, and the greatest possible value is $12$.

Check: With $k=12$ and $x=4$: $\dfrac42-\dfrac45=\dfrac65$, and $\dfrac{12}{(2)(5)}=\dfrac65$. The other root of $x^2-3x-4=0$ is $-1$, which makes a denominator zero.

Watch for: The most common wrong value is $\tfrac{23}{4}$, which considers only a repeated root and misses the values of $k$ for which one root is excluded from the domain.

The numerator is a difference of squares, $(x - 4)(x + 4)$, and the denominator factors as $(x - 4)(x + 3)$, since $-4$ and $3$ multiply to $-12$ and add to $-1$. Cancelling the common factor $x - 4$ leaves $\frac{x + 4}{x + 3}$. Cancelling the wrong factor leaves $x - 4$ on top. Reversing the signs when factoring the denominator produces $x - 3$ below.

The crates already aboard weigh $12 \times 95 = 1{,}140$ pounds, leaving $1{,}800 - 1{,}140 = 660$ pounds. Since $660 \div 65 \approx 10.2$, and a partial crate cannot be loaded, the answer is $10$. Rounding up rather than down gives $11$, which exceeds the limit. Dividing the load already aboard by $65$ gives about $17$. Ignoring the crates already aboard and dividing $1{,}800$ by $65$ gives about $27$.

Answer: C.

If the survey reports $p\%$, the plausible values run from $(p-4.5)\%$ to $(p+4.5)\%$. For all of them to be greater than $50\%$, the lowest one must be:

$p-4.5>50$

so $p>54.5$. The least whole-number percent is $55\%$.

Check: A report of $55\%$ gives plausible values from $50.5\%$ to $59.5\%$, all above $50\%$. A report of $54\%$ gives values from $49.5\%$ to $58.5\%$, which include values that are not above $50\%$.

A: $51\%$ puts the estimate above $50\%$ but ignores the margin of error, so the interval extends down to $46.5\%$.

B: $54\%$ rounds $54.5$ down instead of up; its interval extends down to $49.5\%$.

D: $59\%$ adds the margin of error twice, $50+2(4.5)$, as if the whole width of the interval had to lie above $50\%$; only the lower end, $p-4.5$, must be greater than $50$, so $59\%$ works but is not the least.

Watch for: The whole interval of plausible values, not just the estimate, must lie above $50\%$, so the estimate must exceed $50\%$ by more than the margin of error.

Factoring the left side gives

The product of two factors is negative only when they have opposite signs, which happens between the zeros $4$ and $-2$.

So the solution set is

The rate of change is the change in level divided by the change in time.

So the level drops $4$ centimeters per hour. Monitoring began $3$ hours before the first reading, when the level was higher by $3$ times that rate.

The water level was $52$ centimeters when monitoring began.

Answer: A. −3 The ratio over two input units is $\frac{3}{12} = \frac{1}{4}$, so $b^{2} = \frac{1}{4}$ and $b = \frac{1}{2}$. Since $f(1) = \frac{a}{2} = 3$, $a = 6$. Thus $6(\frac{1}{2})^{k} = 48$, or $(\frac{1}{2})^{k} = 8 = 2^{3}$, giving $k = -3$.

Watch for: For a decreasing exponential, larger outputs occur at smaller inputs.

Answer: D.

In the model, each additional square meter adds $0.25$ hour to the time. The question asks the reverse: how much area corresponds to $1$ additional hour.

If $t$ increases by $1$, then $0.25a$ must increase by $1$, so $a$ increases by $\dfrac{1}{0.25}=4$.

The crew can paint $4$ additional square meters for each additional hour.

Check: For $a=20$, $t=0.25(20)+3=8$. For $a=24$, $t=0.25(24)+3=9$. An increase of $4$ square meters matches an increase of $1$ hour.

A: $0.25$ is the number of additional hours for each additional square meter, which is the reverse of what is asked.

B: $3$ is the estimated time, in hours, before any area is painted (the fixed part of the model).

C: $3.25$ is the estimated time, in hours, to paint $1$ square meter, found by substituting $a=1$.

Watch for: The slope $0.25$ is measured in hours per square meter; the question asks for square meters per hour, which is its reciprocal.

Setting the expressions equal gives $x^2 - 4x + 7 = 2x + c$, so $x^2 - 6x + (7 - c) = 0$. A tangent line meets the parabola exactly once, so the discriminant is zero: $36 - 4(7 - c) = 0$, giving $8 + 4c = 0$ and $c = -2$. Dropping the sign inside the parentheses, as $36 - 4(7 + c) = 0$, gives $2$. Reading $c$ off the parabola's constant term gives $7$. And $9$ is $\left(\dfrac{6}{2}\right)^2$, the constant that completes the square, not $c$.

Answer: B.

Each radius to a point of tangency is perpendicular to the tangent, so $\angle OAP=\angle OBP=90^\circ$. By symmetry, $\overline{OP}$ bisects $\angle APB$, so triangle $OAP$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with $OA=6$ opposite the $30^\circ$ angle. Then $PA=6\sqrt{3}$, and quadrilateral $OAPB$ has area

$2\cdot\dfrac{1}{2}(6)(6\sqrt{3})=36\sqrt{3}$

The angles of quadrilateral $OAPB$ sum to $360^\circ$, so $\angle AOB=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ$. The sector $AOB$ inside the quadrilateral has area $\dfrac{120}{360}\pi(6)^2=12\pi$. The shaded region is the quadrilateral minus the sector: $36\sqrt{3}-12\pi$.

Check: Numerically, $36\sqrt{3}-12\pi\approx62.35-37.70=24.65$, which is positive and less than the area of the quadrilateral, $62.35$, as it must be.

A: $18\sqrt{3}-6\pi$ is the area of only the half of the region on one side of $\overline{OP}$.

C: $36\sqrt{3}-6\pi$ subtracts a $60^\circ$ sector, using $\angle APB$ as the central angle instead of $\angle AOB=120^\circ$.

D: $36\sqrt{3}$ is the area of quadrilateral $OAPB$ without removing the sector.

Watch for: The central angle $\angle AOB$ is $180^\circ-60^\circ=120^\circ$, not $60^\circ$.