Standard Math Practice Test 2

The line passes through its $x$-intercept, $(6, 0)$, and its $y$-intercept, $(0, -4)$.

The slope is the change in $y$ divided by the change in $x$:

The point on the $y$-axis has $y$-coordinate $-4$, so $b$ is $-4$ in the form $y = mx + b$.

Filling in the slope and the intercept:

The input $-5$ is less than $1$, so the first rule applies.

The input $6$ is greater than $1$, so the second rule applies.

Adding the two outputs:

Each crate adds $45$ pounds, so $c$ crates weigh $45c$ pounds.

The empty pallet adds another $90$ pounds, so the loaded pallet weighs $45c + 90$ pounds.

The flat rate applies when that weight is at least $225$ pounds and at most $450$ pounds, which is written as

The ball is on the ground when its height is $0$.

A product is $0$ only when one of its factors is $0$. The factor $-4$ is never $0$, and the factor $t + 2$ is positive at every time after the throw.

So the height is $0$ when

Adding $8$ to both sides:

The ball reaches the ground $8$ seconds after it is thrown.

Answer: C.

Multiply both sides by $2$:

$3x-4=2x+10$

Subtracting $2x$ from both sides and adding $4$ gives $x=14$. Therefore $x-7=14-7=7$.

Check: With $x=14$, the left side is $\dfrac{42-4}{2}=19$ and the right side is $14+5=19$.

A: This comes from multiplying only the left side by $2$: $3x-4=x+5$ gives $x=4.5$, and $4.5-7=-2.5$.

B: This comes from doubling $x$ but not the $5$ on the right side: $3x-4=2x+5$ gives $x=9$, and $9-7=2$.

D: This is the value of $x$, not the value of $x-7$.

Watch for: When clearing a denominator, multiply every term on the other side, and answer the question asked ($x-7$), not $x$.

Multiplying both sides by $2$ gives $3a - b = 2c$. Adding $b$ gives $3a = 2c + b$, so $a = \dfrac{2c + b}{3}$. Subtracting $b$ instead of adding it gives $\dfrac{2c - b}{3}$. Forgetting to multiply by $2$ at all gives $\dfrac{c + b}{3}$. And multiplying the whole right side by $2$ after adding $b$, rather than before, gives $\dfrac{2(c + b)}{3}$.

Answer: C.

From the table, $f$ increases by $4$ for every $2$ units of $x$, so $f(x)=2x+3$. The function $g$ decreases by $2$ for every $2$ units of $x$, so $g(x)=-x+2$.

Solve $2x+3=(-x+2)+10$: $3x=9$, so $x=3$.

Check: $f(3)=9$ and $g(3)=-1$, and $9=-1+10$.

A: This solves $f(x)+10=g(x)$, adding $10$ to the wrong side.

B: This solves $f(x)=g(x)$ and ignores the $10$.

D: This uses $g(x)=x+2$, misreading the direction of change in the $g$ column.

Watch for: Build each rule from the table before combining them, and keep the $+10$ on the side of $g$.

Split the radicand into a perfect square times whatever is left. The largest perfect square factor of $50$ is $25$, and $x^7$ holds the perfect square $x^6$ with one $x$ left over.

The square root of $25x^6$ is $5x^3$, so the expression becomes

Answer: D.

The $500$ homes were selected at random from all $20{,}000$ homes the utility serves, so the results can be generalized to those homes. The showerheads were then randomly assigned, so the difference in water use can be attributed to the showerheads.

So each home is estimated to use about $400$ fewer gallons per month with a low-flow showerhead, and for all $20{,}000$ homes the decrease is about $400(20{,}000)=8{,}000{,}000$ gallons per month.

Check: Random selection supports applying the result to all $20{,}000$ homes, and random assignment supports a causal effect of $400$ gallons per home per month; $8{,}000{,}000\div20{,}000=400$.

A: $100{,}000$ is $400(250)$, the decrease for only the $250$ homes that received the showerhead in the study.

B: $200{,}000$ is $400(500)$, the decrease if only the $500$ selected homes had the showerhead installed.

C: $4{,}000{,}000$ is $400(10{,}000)$, which applies the effect to only half of the homes, as if the study’s even split carried over to the whole utility.

Watch for: Random selection lets the result be applied to all $20{,}000$ homes, and random assignment lets the difference be attributed to the showerheads; multiply the per-home effect by all $20{,}000$ homes.

Answer: 270

The legs have ratio $5:12$, so the hypotenuse has ratio 13 by the Pythagorean theorem. Write the sides as $5k,12k,13k$. Then $30k=90$, so $k=3$. The area is $\tfrac12(15)(36)=270$.

Check: The side lengths 15, 36, and 39 sum to 90 and satisfy $15^2+36^2=39^2$.

Watch for: A trigonometric ratio supplies a shape, not the triangle’s scale.

Answer: 120

Let D be the route length. The time equation is $0.4D/3+0.6D/6=28$. The coefficient is $2/15+1/10=7/30$, so $(7/30)D=28$ and D = 120.

Check: The two distances are 48 and 72 meters, taking 16 and 12 minutes, respectively.

Watch for: Add times, not speeds; the percentages describe distances, not equal time intervals.

Dividing by a fraction is the same as multiplying by its reciprocal.

The numerator is a difference of two squares:

Cancel the shared factor $x + 3$, and reduce $\frac{8}{4}$ to $2$:

Let the original volume be $V$. A fall of $80\%$ leaves $20\%$ of it, so the volume after the drought is $0.20V$.

To return to $V$, the volume must grow by the amount it lost:

A percent rise is measured against the current amount, which is now $0.20V$, not against the original:

A rise equal to $4$ times the current volume is a rise of $400\%$.

An inscribed angle is half the central angle that opens onto the same arc. Angle $ABC$ and angle $AOC$ both open onto arc $AC$, so $x = \frac{1}{2}(104) = 52$. Halving a second time gives $26$. Taking the angle to equal the central angle gives $104$. Subtracting $52$ from $90$, as though the inscribed angle were the complement of the answer, gives $38$.

Answer: C. 3 The coefficient of x is zero when $a = 3$ or $a = -3$. At $a = 3$, the equation is $0 = 6$, so it has no solution. At $a = -3$, it is $0 = 0$, so every real x works. For all other a, the coefficient is nonzero and there is one solution.

Watch for: Check the right side at each value that makes the variable coefficient zero.

Answer: D.

The conditions give $36a+b=3$ and $9a+b=15$. Subtracting yields $27a=-12$, so $a=-4/9$. Then $b=19$. Since a is negative, the vertex height b is the maximum.

Check: The vertex is at x = 6; the fitted rule gives 3 and 15 at the specified inputs.

A: This is the modeled height at zero.

B: This is the horizontal vertex coordinate.

C: This is a given height away from the vertex.

Watch for: A vertex-form model does not give a maximum unless its squared-term coefficient is negative.

Answer: A.

Whenever g(x) = 0, $f(x-3)=-4$. Therefore the required original inputs are u = x - 3. From x = -1 and 5, obtain u = -4 and 2.

Check: Substitution gives g(-1) = f(-4) + 4 = 0 and g(5) = f(2) + 4 = 0.

B: This shifts the inputs in the wrong direction.

C: This leaves the horizontal input shift unapplied.

D: This uses the vertical shift as a horizontal shift.

Watch for: Translate from the new input to the old one using the inside expression, not the visual shift direction alone.

Answer: C.

Rearrange: $(a-2)x>15$. The solution set $x<-5$ has its inequality reversed, which happens only when both sides are divided by a negative number. So $a-2<0$, and the solution is $x<\dfrac{15}{a-2}$. Setting $\dfrac{15}{a-2}=-5$ gives $a-2=-3$, so $a=-1$.

Check: With $a=-1$: $-x-6>2x+9$, so $-3x>15$ and $x<-5$.

A: $-5$ is the boundary value of $x$, not the value of $a$.

B: $-3$ is the value of $a-2$, not of $a$.

D: With $a=5$, the inequality becomes $3x>15$, whose solution set is $x>5$, not $x<-5$.

Watch for: A solution set of the form $x<\dots$ from a “$>$” inequality signals division by a negative number.

Answer: 8

The solutions of $\lvert x-4\rvert=k$ are $4+k$ and $4-k$. The solutions of $\lvert x+2\rvert=2k$ are $-2+2k$ and $-2-2k$. A common solution matches one from each list:

$4+k=-2+2k$ gives $k=6$.

$4-k=-2+2k$ gives $k=2$.

$4+k=-2-2k$ gives $k=-2$, and $4-k=-2-2k$ gives $k=-6$; neither is positive.

For $k=6$, the solutions are $10,-2$ and $10,-14$, so $10$ is the only common solution. For $k=2$, they are $6,2$ and $2,-6$, so $2$ is the only common solution. The sum of the possible values of $k$ is $6+2=8$.

Check: For $k=2$: $\lvert2-4\rvert=2$ and $\lvert2+2\rvert=4=2k$. For $k=6$: $\lvert10-4\rvert=6$ and $\lvert10+2\rvert=12=2k$.

Watch for: The most common wrong value is $6$, which matches only the two “plus” solutions; the solution $4-k$ of the first equation can also match $-2+2k$.

Answer: 16.25

The mean is $\dfrac{30+x}{5}$, and the median depends on where $x$ falls in the ordered list.

If $x\le4$, the median is $4$, so the mean is $5$: $30+x=25$ and $x=-5$, which satisfies $x\le4$.

If $4<x<8$, the median is $x$, so $\dfrac{30+x}{5}=x+1$: $30+x=5x+5$ and $x=6.25$, which satisfies $4<x<8$.

If $x\ge8$, the median is $8$, so the mean is $9$: $30+x=45$ and $x=15$, which satisfies $x\ge8$.

All three values work, and their sum is $-5+6.25+15=16.25$.

Check: For $x=-5$: median $4$, mean $\tfrac{25}{5}=5$. For $x=6.25$: median $6.25$, mean $\tfrac{36.25}{5}=7.25$. For $x=15$: median $8$, mean $\tfrac{45}{5}=9$.

Watch for: The most common wrong value is $15$, which assumes the median is $8$ and misses the cases in which $x$ itself is the median or is below it.

Answer: 5.76

By the Pythagorean theorem, $AC=\sqrt{9^2+12^2}=15$.

Triangles $BDC$ and $ABC$ are similar: each has a right angle, and they share angle $C$. So $\frac{DC}{BC}=\frac{BC}{AC}$, which gives $DC=\frac{12^2}{15}=\frac{48}{5}$.

Triangles $DEC$ and $ABC$ are also similar: each has a right angle, and they share angle $C$. So $\frac{DE}{DC}=\frac{AB}{AC}=\frac{9}{15}$, which gives

$DE=\frac{3}{5}\cdot\frac{48}{5}=\frac{144}{25}=5.76$

Check: With $B$ at $(0,0)$, $C$ at $(12,0)$, and $A$ at $(0,9)$, line $AC$ is $3x+4y=36$, and the foot of the perpendicular from the origin is $D=\left(\frac{108}{25},\frac{144}{25}\right)$. Point $E$ is directly below $D$ on the $x$-axis, so $DE=\frac{144}{25}=5.76$.

Watch for: The most common wrong values are $7.2$, the length of $\overline{BD}$, and $9.6$, the length of $\overline{DC}$; $\overline{DE}$ is a second perpendicular, inside triangle $BDC$.

Answer: A.

Complete the square: $f(x)=-(x-k)^2+k^2-k-6$, so the vertex is $(k,\,k^2-k-6)$.

In quadrant IV, the $x$-coordinate is positive, so $k>0$, and the $y$-coordinate is negative:

$k^2-k-6<0$

which is $(k-3)(k+2)<0$, so $-2<k<3$. Together with $k>0$, this gives $0<k<3$.

Check: For $k=1$ the vertex is $(1,-6)$, in quadrant IV. For $k=-1$ it is $(-1,-4)$, in quadrant III, and for $k=4$ it is $(4,6)$, in quadrant I.

B: This uses only the condition that the vertex is below the $x$-axis; for $-2<k\le0$ the vertex is not to the right of the $y$-axis.

C: This reverses the inequality, which gives $k<-2$ or $k>3$, and then keeps only the positive values; for $k>3$ the vertex is above the $x$-axis.

D: This reverses the inequality and ignores the condition on the $x$-coordinate.

Watch for: A point in quadrant IV needs a positive $x$-coordinate and a negative $y$-coordinate; use both conditions.

Answer: -6

The second equation must be twice the first. Doubling the coefficient of y gives $b=2(-3)=-6$.

Check: With b = -6, all three terms of the second equation are exactly doubled.

Watch for: For infinitely many solutions, the constants must share the same multiplier as both variable coefficients.

Answer: B.

The increase is $62-38=24$ stitches over 6 centimeters, so the rate is 4 stitches per centimeter. From the first row, the intercept is $38-4(6)=14$.

Check: The rule $S(d)=4d+14$ reproduces both rows.

A: Thirty-eight is the output after six centimeters.

C: Four is the rate per centimeter.

D: Twenty-four is the difference between the two rows.

Watch for: An inferred intercept is the model’s baseline, not its change per input unit.

The quadratic formula gives $x = \frac{8 \pm \sqrt{64 - 20}}{2} = \frac{8 \pm \sqrt{44}}{2}$. Since $\sqrt{44} = 2\sqrt{11}$, both terms divide by $2$, leaving $4 \pm \sqrt{11}$. The greater solution is $4 + \sqrt{11}$. Taking the lesser root gives $4 - \sqrt{11}$. Dividing only the $8$ by $2$ and leaving the radical untouched gives the two options containing $\sqrt{44}$, of which the greater is $4 + \sqrt{44}$.

The longest segment inside a rectangular box joins two opposite corners. It is the hypotenuse of a right triangle whose legs are one edge and the diagonal of the face the other two edges span.

The diagonal of the $6$ by $9$ face:

That diagonal and the $2$-inch edge meet at a right angle, so the corner-to-corner segment is

A rod $11$ inches long fits along that segment, and no longer rod fits.

Answer: B.

In order, the data have $8$ values of $5$ or less (positions $1$ through $8$) and then four $6$s (positions $9$ through $12$). With $n$ added values of $8$, there are $15+n$ values.

With $1$ added value there are $16$ values, and the median is the average of the $8$th and $9$th values, $\dfrac{5+6}{2}=5.5$. With $2$ added values there are $17$ values, and the median is the $9$th value, $6$. So the least number is $2$.

Check: Adding more $8$s moves the middle position later, but the median stays $6$ for anywhere from $2$ through $8$ added values, so $2$ is the least number.

A: One added value gives $16$ values, and the median averages $5$ and $6$ to get $5.5$.

C: Three added values also give a median of $6$, but $2$ is fewer.

D: Four added values give $19$ values with median the $10$th value, $6$, but fewer values suffice.

Watch for: With an even number of values, the median is the average of the two middle values, which can land between data values.

Answer: D.

The condition $|a-34|\le1.5$ means $32.5\le a\le35.5$. Multiplying through by 2 gives $65\le2a\le71$, and subtracting 11 gives $54\le r\le60$. That interval has center 57 and radius 3, so it is written $|r-57|\le3$.

Check: Test the endpoints directly: $a=32.5$ gives $r=2(32.5)-11=54$ and $a=35.5$ gives $r=2(35.5)-11=60$, and both are exactly 3 away from 57.

A: 23 is $34-11$, so the doubling in $r=2a-11$ was left out of the center.

B: The center 57 is correct, but the tolerance was not doubled along with the angle.

C: 68 is $2(34)$, so the subtraction of 11 was left out of the center.

Watch for: Under a rule $r=ma+b$, the center of an interval moves by the whole rule but the radius is scaled by $|m|$ only.

Answer: A.

The factor $(x-1)^2$ is positive for every $x$ except $x=1$, where it is $0$.

The product $(x+2)(x-4)$ is negative exactly when $-2<x<4$.

So $p(x)<0$ exactly when $-2<x<4$ and $x\ne1$. The integers that qualify are $-1$, $0$, $2$, and $3$, which is $4$ values.

Check: $p(0)=(1)(2)(-4)=-8$ and $p(3)=(4)(5)(-1)=-20$ are negative, while $p(1)=0$, $p(-2)=0$, and $p(4)=0$ are not.

B: $5$ counts $-1$, $0$, $1$, $2$, and $3$, but $p(1)=0$, which is not less than $0$.

C: $6$ solves $(x+2)(x-4)\le0$ as $-2\le x\le4$, which contains $7$ integers, and then removes only the double root $x=1$; the zeros $-2$ and $4$ make the product $0$, not negative.

D: $7$ counts every integer from $-2$ through $4$, which solves $p(x)\le0$.

Watch for: A squared factor does not change sign, but it makes the function $0$ at its zero, so that $x$-value must be excluded from a strict inequality.

Answer: B.

Each interior angle of a regular hexagon measures $120^\circ$.

Triangle $ABC$ is isosceles with $AB=BC$ and $\angle ABC=120^\circ$, so each base angle is $\dfrac{180-120}{2}=30$ degrees. Thus $\angle BAC=30^\circ$.

Similarly, in isosceles triangle $BCD$, $\angle CBD=30^\circ$, so $\angle ABD=120^\circ-30^\circ=90^\circ$.

In triangle $ABP$, $\angle APB=180^\circ-30^\circ-90^\circ=60^\circ$.

Check: In triangle $BPC$, $\angle PBC=30^\circ$ and $\angle PCB=\angle ACB=30^\circ$, so $\angle BPC=120^\circ$. Angles $APB$ and $BPC$ form a straight angle, and $60+120=180$.

A: $30^\circ$ is the measure of $\angle BAC$ (and of $\angle CBD$), one of the base angles, not the angle at $P$.

C: $90^\circ$ is the measure of $\angle ABD$, the angle of triangle $ABP$ at $B$.

D: $120^\circ$ is an interior angle of the hexagon, and also $\angle BPC$, the angle next to $\angle APB$.

Watch for: Break the hexagon into isosceles triangles and find the angles at the vertices before finding the angle at the intersection.

Answer: A.

Let $n$ and $p$ be last month’s prices, in dollars, of a folder and a marker. Then

$4n+3p=14.50$

$4(1.1n)+3(0.8p)=14.60$

The second equation is $4.4n+2.4p=14.60$. Multiplying the first equation by $0.8$ gives $3.2n+2.4p=11.60$. Subtracting, $1.2n=3$, so $n=2.50$. Then $3p=14.50-10=4.50$, so $p=1.50$.

This month’s price of a marker is $0.8(1.50)=1.20$ dollars.

Check: This month a folder costs $1.1(2.50)=2.75$ dollars, and $4(2.75)+3(1.20)=11.00+3.60=14.60$ dollars.

B: $1.50 is last month’s price of a marker; the question asks for this month’s price.

C: $1.80 increases last month’s marker price by $20\%$ instead of decreasing it.

D: $2.75 is this month’s price of a folder.

Watch for: Solve for last month’s prices first, and then apply the $20\%$ decrease to the marker price.

Answer: D.

Combine the first two fractions:

$\dfrac{1}{x-1}-\dfrac{1}{x+1}=\dfrac{(x+1)-(x-1)}{x^2-1}=\dfrac{2}{x^2-1}$