Advanced Math Practice Test 2

Answer: A.

From n = 4 to n = 10, the coiled length decreases by 18 centimeters, so the slope is -3 centimeters per turn. The rule is $L=115-3n$. At n = 22, $L=49$.

Check: Starting at the second table entry, 12 more turns reduce 85 by $12(3)=36$, leaving 49.

B: This is the total decrease over 22 turns: $3(22)=66$. Subtract it from the initial length to find the length still coiled: $115-66=49$.

C: This is the length after 10 turns.

D: This is the initial coiled length.

Watch for: The input counts additional turns; the output is the amount still coiled, so the slope is negative.

The area of a trapezoid is $\frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(10 + 6)(4) = 32$. Forgetting the factor of $\frac{1}{2}$ gives $64$. Using only the shorter parallel side gives $6 \times 4 = 24$. Using only the longer one gives $10 \times 4 = 40$.

Answer: C.

The graph is a line through $(0,300)$ and $(40,0)$, so the hopper starts with $300$ kilograms and loses $300/40=7.5$ kilograms each minute, giving $M=300-7.5t$. Setting $300-7.5t=105$ gives $7.5t=195$, so $t=26$ minutes.

Check: In $26$ minutes the hopper discharges $26(7.5)=195$ kilograms, and $300-195=105$ kilograms are left.

A: $7.5$ is the discharge rate in kilograms per minute, not a number of minutes.

B: $105/7.5=14$ is how long it takes to discharge $105$ kilograms. The question asks when $105$ kilograms are still in the hopper.

D: $40$ minutes is when the hopper is empty, which the graph shows as its horizontal intercept.

Watch for: Decide whether a stated amount is what remains or what has already left before writing the equation.

Answer: B.

The unreimbursed charge for 140 centimeters is 136 dollars. Thus 60 extra centimeters add $136-94=42$ dollars, so the rate is 0.70 dollar per centimeter. The fixed charge is $94-0.70(80)=38$ dollars.

Check: The larger bill is $38+0.70(140)-9=127$, as given.

A: This leaves the reimbursement subtracted from the fixed charge.

C: This comes from an incorrect rate.

D: This is the variable cost of the 80-centimeter order.

Watch for: Undo a reimbursement or discount before comparing totals used to determine a unit rate.

Answer: D.

With the sampling method and the confidence level held fixed, a margin of error shrinks as the sample size grows, because a larger random sample pins down the population mean more precisely. Only choice D increases the sample size while keeping the selection random.

Check: A margin of error is inversely proportional to the square root of the sample size, so quadrupling a random sample from $100$ to $400$ households would cut the $12$-gallon margin roughly in half, to about $6$ gallons, while shrinking it to $40$ households would widen the margin to about $19$ gallons.

A: Raising the confidence level requires a wider interval in order to sustain that higher level of confidence, so the margin of error grows.

B: A smaller random sample carries less information about the population, so it would widen the margin of error rather than narrow it.

C: Choosing the households with the largest meters abandons random selection and biases the sample toward heavy users. A biased sample does not estimate the district’s mean more precisely.

Watch for: Only more data from the same random process narrows a margin of error; higher confidence widens it and a targeted sample biases it.

Answer: D.

Graph the function and locate its positive x-intercept. The quadratic formula also gives $x=(6.3+\sqrt{6.3^2+4(0.7)(18.4)})/1.4\approx11.3$.

Check: The product of the roots is negative, so exactly one root is positive.

A: This approximates the magnitude of the negative root.

B: This is the axis-of-symmetry input.

C: This is the sum of the roots.

Watch for: Select the x-intercept, not the vertex, and keep the positive root.

Answer: A.

Let $p=\dfrac{1}{m}$ and $q=\dfrac{1}{n}$, so the system becomes $3p+5q=19$ and $6p-2q=14$. Halving the second equation gives $3p-q=7$, and subtracting that from the first gives $6q=12$, so $q=2$ and then $p=3$. Therefore $m=\dfrac{1}{3}$ and $n=\dfrac{1}{2}$, so $mn=\dfrac{1}{6}$.

Check: Substituting back into the originals, $\dfrac{3}{1/3}+\dfrac{5}{1/2}=9+10=19$ and $\dfrac{6}{1/3}-\dfrac{2}{1/2}=18-4=14$, exactly as stated.

B: $\dfrac{2}{3}$ is $\dfrac{m}{n}$, the ratio of the two numbers rather than their product.

C: $\dfrac{5}{6}$ is $m+n$.

D: 6 is $pq$, the product of the substituted variables $\dfrac{1}{m}$ and $\dfrac{1}{n}$, with no conversion back to $m$ and $n$.

Watch for: After solving for a substituted variable, undo the substitution before answering, because the substituted value is usually sitting right there among the choices.

Answer: C.

Count the dots before comparing anything: variety R has $12$ plants and variety S has $18$. For variety R the $6$th and $7$th values are both $8$, so its median is $8$ tubers; for variety S the $9$th and $10$th values are both $7$, so its median is $7$ tubers. Variety R has the greater median.

Check: Only $4$ of variety R’s $12$ dots lie below $8$ and only $5$ lie above it, so $8$ has to be the middle value; for variety S, $10$ of its $18$ dots sit at $7$ or below and $12$ sit at $7$ or above, which pins its middle value at $7$.

A: Range runs from the lowest dot to the highest: variety R spans $5$ to $11$, a range of $6$, while variety S spans $4$ to $14$, a range of $10$. Variety S is the wider of the two, so this reverses the comparison; the lone dot at $14$ is the easiest one to miss.

B: Four variety S plants reached $10$ tubers against three variety R plants, so the raw count does favor variety S. But the plots hold different numbers of plants: $4$ out of $18$ is about $22\%$, below variety R’s $3$ out of $12$, or $25\%$. This choice compares counts where the wording calls for proportions.

D: Variety S’s dots at $8$ and above number $2+2+2+1+1=8$, fewer than half of its $18$ plants. Reading the tall stacks at $6$ and $7$ as though they sat at $8$ produces this choice.

Watch for: Two dot plots need not hold the same number of dots, so count them first, because every comparison that depends on position or on a share of the whole turns on that count.

A reflection across the $y$-axis sends each point $(a, b)$ to $(-a, b)$, so the reflected graph is $y = f(-x)$.

Substituting $-x$ for $x$:

Squaring $-x$ gives $x^2$, and $-6(-x)$ gives $6x$, so the reflected graph is

Answer: B.

Adding $5$ and dividing by $2$ gives $4<x<8$. Then

$12<3x<24$

$13<3x+1<25$

The number $x$ need not be an integer, so $3x+1$ can be any number between $13$ and $25$. The integers strictly between $13$ and $25$ are $14$ through $24$, which is $11$ values.

Check: Each of these values is reached: $3x+1=14$ when $x=\dfrac{13}{3}\approx4.33$, and $3x+1=24$ when $x=\dfrac{23}{3}\approx7.67$, both between $4$ and $8$.

A: $3$ counts the integer values of $x$ ($5$, $6$, and $7$); $x$ does not have to be an integer.

C: $12$ includes one endpoint, counting $13$ through $24$ or $14$ through $25$, but both inequalities are strict.

D: $13$ counts $13$ through $25$, treating both strict inequalities as inclusive.

Watch for: The question counts integer values of $3x+1$, not integer values of $x$; transform the inequality first, then count.

Answer: A.

The deck’s area is $60\cdot36=2{,}160$ square feet. One square yard is $3\cdot3=9$ square feet, so the area is $240$ square yards. Two coats cover $480$ square yards, which requires $\dfrac{480}{40}=12$ gallons, or $\dfrac{12}{3}=4$ pails.

Check: The deck is $20$ yards by $12$ yards. Each coat uses $\dfrac{240}{40}=6$ gallons, so two coats use $12$ gallons, exactly $4$ pails.

B: $6$ is the number of gallons needed for one coat, not the number of pails for two coats.

C: $12$ is the number of gallons, not the number of $3$-gallon pails. (Converting square feet to square yards by dividing by $3$ also leads to $12$ pails.)

D: $36$ pails treats $40$ square yards per gallon as $40$ square feet per gallon: each coat then needs $\tfrac{2{,}160}{40}=54$ gallons, so two coats need $108$ gallons, or $36$ pails.

Watch for: Convert area units by the square of the length factor, and answer in the unit asked for (pails, not gallons).

Answer: B.

Substituting $(2,1)$ into the first system gives $2a+b=7$ and $a+2b=5$, so $a=3$ and $b=1$.

The second system is then $3x+y=5$ and $x+3y=7$. Subtracting $3$ times the second equation from the first gives $-8y=-16$, so $y=2$ and $x=1$. The solution is $(1,2)$.

(Swapping $x$ and $y$ turns $ax+by$ into $bx+ay$, so swapping the right sides swaps the solution.)

Check: $3(1)+1(2)=5$ and $1(1)+3(2)=7$.

A: $(-2,-1)$ negates the original solution, as if changing the constants reversed the signs.

C: $\left(\dfrac{10}{7},\dfrac{5}{7}\right)$ scales the original solution by $\dfrac57$, which fits the first new equation but not the second.

D: $(2,1)$ is the solution to the original system; the constants on the right have changed.

Watch for: Find $a$ and $b$ first, or notice that swapping $x$ and $y$ exchanges the two left sides.

Rewriting the given line as $y = \dfrac{3}{4}x - 2$ shows its slope is $\dfrac{3}{4}$, so a perpendicular line has slope $-\dfrac{4}{3}$. Then $5 = -\dfrac{4}{3}(3) + b$ gives $b = 9$. Using the original slope $\dfrac{3}{4}$ gives $2.75$, and using its opposite $-\dfrac{3}{4}$ gives $7.25$. Taking the reciprocal without changing the sign, as $\dfrac{4}{3}$, gives $1$.

Answer: 5

Group the terms: $x^2(x-a)-4(x-a)=(x-a)(x^2-4)$, so

$p(x)=(x-a)(x-2)(x+2)$

The zeros are $a$, $2$ and $-2$. Because $a>2$, the greatest zero is $a$ and the least is $-2$, so $a-(-2)=7$ and $a=5$.

Check: With $a=5$, $p(x)=(x-5)(x-2)(x+2)$ has zeros $5$, $2$ and $-2$, and $5-(-2)=7$.

Watch for: The most common wrong value is $9$, from taking the least zero to be $2$ and missing the negative zero from $x^2-4$.

Answer: C. $\frac{21}{2}$ Let $f(x) = mx + b$. The left side becomes $m(2x + 1) + b + m(3x - 2) + b = 5mx - m + 2b$. Matching coefficients gives $m = 4$ and $-4 + 2b = 17$, so $b = \frac{21}{2}$. This intercept is f(0).

Watch for: Each function evaluation contributes its own copy of the intercept.

Answer: 7

The graphs intersect where $x^2+3=kx-1$, or

$x^2-kx+4=0$

They do not intersect when this equation has no real solutions, which happens when its discriminant is negative:

$k^2-4(1)(4)<0$

So $k^2<16$, which means $-4<k<4$. The integers in this interval are $-3$, $-2$, $-1$, $0$, $1$, $2$, and $3$, so there are $7$ possible values.

Check: At $k=4$, the equation is $(x-2)^2=0$, so the line touches the parabola at $(2,7)$; values $k=\pm4$ must be excluded. At $k=0$, the line $y=-1$ lies below the parabola, which is never lower than $3$.

Watch for: The most common wrong value is $9$, which includes $k=-4$ and $k=4$; at those values the line is tangent to the parabola, so the graphs do intersect.

Answer: 20

Multiply the two outputs:

$f(2)\cdot f(8)=ab^2\cdot ab^8=a^2b^{10}=\left(ab^5\right)^2$

and $ab^5=f(5)$. So $\left(f(5)\right)^2=400$. Since $a$ and $b$ are positive, $f(5)$ is positive, and $f(5)=20$.

The individual values of $a$ and $b$ are not needed; $5$ is the average of $2$ and $8$, so $f(5)$ is the positive square root of the product.

Check: With $a=\dfrac58$ and $b=2$, $f(2)=\dfrac52$ and $f(8)=160$, whose product is $400$, and $f(5)=\dfrac58\cdot32=20$.

Watch for: The most common wrong value is $200$, half of the product; because the inputs $2$ and $8$ average to $5$, the product is the square of $f(5)$.

Answer: A.

Suppose there are $1{,}000$ vehicles in stock. Then $400$ are trucks and $600$ are cars.

There are $0.30(400)=120$ electric trucks. Because half of the electric vehicles are trucks, there are $240$ electric vehicles, and $240-120=120$ of them are cars.

The probability that a randomly selected car is electric is $\frac{120}{600}=\frac15$.

Check: Overall, $30\%$ of the trucks and $20\%$ of the cars are electric. Weighted by $40\%$ trucks and $60\%$ cars: $0.4(0.3)+0.6(0.2)=0.24$, or $240$ of $1{,}000$ vehicles, half of them trucks.

B: $\frac{6}{25}$ is the probability that a randomly selected vehicle of either kind is electric, $\frac{240}{1{,}000}$.

C: $\frac{3}{10}$ assumes that cars are electric at the same rate as trucks.

D: $\frac12$ is the probability that a randomly selected electric vehicle is a car, the reverse conditional probability.

Watch for: Trucks and cars have the same number of electric vehicles but different totals, so divide the electric cars by the number of cars.

Answer: A.

Let $\angle BAC=x^\circ$. Because $AD=BD$, triangle $ABD$ is isosceles, so $\angle ABD=x^\circ$. Angle $BDC$ is an exterior angle of triangle $ABD$, so $\angle BDC=2x^\circ$. Because $BD=BC$, triangle $BDC$ is isosceles, so $\angle BCD=\angle BDC=2x^\circ$. Because $AB=AC$, $\angle ABC=\angle ACB=2x^\circ$. The angles of triangle $ABC$ sum to $180^\circ$:

$x+2x+2x=180$

so $x=36$.

Check: The angles are $36^\circ$, $72^\circ$, and $72^\circ$. Then $\angle ABD=36^\circ$ and $\angle DBC=72^\circ-36^\circ=36^\circ$, so triangle $BDC$ has angles $36^\circ$, $72^\circ$, and $72^\circ$, which matches $BD=BC$.

B: This takes $\angle BDC$ to equal $x^\circ$ instead of $2x^\circ$, so all three angles of triangle $ABC$ come out equal.

C: This is the measure of each base angle, $\angle ABC$ or $\angle ACB$.

D: This is the measure of $\angle ADB$, which is supplementary to $\angle BDC$.

Watch for: An exterior angle of a triangle equals the sum of the two remote interior angles; use it to carry $x$ from one isosceles triangle to the next.

Answer: C.

Rewriting the first equation as $cx+6y=15$, the system has infinitely many solutions only when the second equation is a constant multiple of the first, so $\dfrac{2}{c}=\dfrac{c+1}{6}=\dfrac{d}{15}$. The first two ratios give $c(c+1)=12$, so $c=3$ or $c=-4$, and $c>0$ forces $c=3$. The multiplier is then $\dfrac{2}{3}$, so $d=\dfrac{2}{3}(15)=10$.

Check: With $c=3$ and $d=10$ the equations are $3x+6y=15$ and $2x+4y=10$. Dividing the first by 3 and the second by 2 turns both into $x+2y=5$, so every point on that line solves the system, including $(1,2)$ and $(5,0)$.

A: $-7.5$ is the value of $d$ that goes with the root $c=-4$, which the condition $c>0$ rules out.

B: 3 is the value of $c$, not of $d$.

D: 22.5 uses $\dfrac{d}{15}=\dfrac{c}{2}$, inverting the multiplier that carries the first equation to the second.

Watch for: Infinitely many solutions means one equation is a multiple of the other, so every corresponding pair of coefficients shares the same ratio, constant terms included.

Answer: -2

Isolate one radical: $\sqrt{2x+7}=1+\sqrt{x+3}$. Squaring both sides gives

$2x+7=1+2\sqrt{x+3}+x+3$

which simplifies to $x+3=2\sqrt{x+3}$. Let $u=\sqrt{x+3}$, so $u\ge0$ and $u^2=2u$. Then $u=0$ or $u=2$, which gives $x=-3$ or $x=1$. Both satisfy the original equation, so the sum is $-3+1=-2$.

Check: At $x=-3$: $\sqrt{1}-\sqrt{0}=1$. At $x=1$: $\sqrt{9}-\sqrt{4}=1$.

Watch for: The most common wrong value is $1$, which comes from dividing $x+3=2\sqrt{x+3}$ by $\sqrt{x+3}$ and losing the solution $x=-3$.

Answer: 26

The top side of the rectangle is horizontal, so its two vertices on the parabola have the same $y$-value and are $(a,12-a^2)$ and $(-a,12-a^2)$ for some $a$ with $0<a<\sqrt{12}$. The rectangle’s width is $2a$ and its height is $12-a^2$, so its perimeter is

$P=2(2a)+2(12-a^2)$

$P=-2a^2+4a+24$

This quadratic in $a$ has a negative leading coefficient, so its greatest value is at $a=-\frac{4}{2(-2)}=1$. The greatest perimeter is $-2+4+24=26$.

Check: At $a=1$ the rectangle is $2$ units wide and $11$ units tall, with perimeter $2(2)+2(11)=26$. Nearby rectangles have smaller perimeters: at $a=0.5$ and at $a=1.5$ the perimeter is $25.5$.

Watch for: The width of the rectangle is $2a$, not $a$; using $a$ as the width gives $2a+2(12-a^2)$, whose greatest value is $24.5$. Also maximize the whole perimeter, not half of it, which gives $13$.

Answer: C.

No money remains when $P=0$:

$1{,}250-45w=0$

so $w=\dfrac{1{,}250}{45}$. This value is the number of weeks after the card was loaded when the amount remaining reaches $0$.

Check: $\dfrac{1{,}250}{45}\approx27.8$, and $P(27.8)\approx1{,}250-1{,}251\approx0$.

A: The amount spent each week is $45$, the rate of decrease.

B: The amount loaded onto the card is $1{,}250$, the value of $P$ when $w=0$.

D: The amount remaining after $45$ weeks is $1{,}250-45(45)$, which is negative; it is not $\dfrac{1{,}250}{45}$.

Watch for: Dividing the starting amount by the weekly decrease gives the number of weeks until the amount reaches $0$.

Answer: A.

Combine the fractions in the numerator:

$\frac1x-\frac13=\frac{3-x}{3x}$

Then divide by $x-3$. Because $3-x=-(x-3)$,

$\frac{3-x}{3x}\cdot\frac{1}{x-3}=\frac{-(x-3)}{3x(x-3)}=-\frac{1}{3x}$

Check: At $x=1$, the original expression is $\frac{1-\frac13}{1-3}=\frac{\frac23}{-2}=-\frac13$, and $-\frac{1}{3(1)}=-\frac13$.

B: $\frac{1}{3x}$ cancels $3-x$ with $x-3$ as if they were equal; they are opposites, so the result is negative.

C: $\frac{3-x}{3x}$ is only the simplified numerator; it has not been divided by $x-3$.

D: $\frac{1}{(x-3)^2}$ rewrites $\frac1x-\frac13$ as $\frac{1}{x-3}$, which is not a valid way to subtract fractions.

Watch for: The factors $3-x$ and $x-3$ are opposites, so they cancel to $-1$, not $1$.

Adding water changes the volume but not the salt, and the salt is $0.15 \times 240 = 36$ milliliters. If $V$ is the final volume, then $36 = 0.09V$, so $V = 400$ and the water added is $400 - 240 = 160$. Answering with the final volume gives $400$. Applying the $40\%$ drop in concentration to the volume, as $0.4 \times 240$, gives $96$, and subtracting that from $240$ gives $144$.

Answer: D.

Because $S$ is positive, the inequality $\lvert T-S\rvert\le0.04S$ is equivalent to

$-0.04S\le T-S\le0.04S$

so $0.96S\le T\le1.04S$. The greatest allowed temperature is $1.04S$:

$1.04S=72.8$

so $S=70$.

Check: With $S=70$, $0.04S=2.8$, so the allowed temperatures are from $67.2$ to $72.8$ degrees Fahrenheit, and the greatest is $72.8$.

A: $67.2$ is the least temperature the thermostat allows when $S=70$, $0.96(70)$, not the set temperature.

B: $68.8$ treats $4\%$ as $4$ degrees and subtracts $4$ from $72.8$.

C: $69.888$ subtracts $4\%$ of $72.8$; the tolerance is $4\%$ of $S$, not of the greatest temperature.

Watch for: The $4\%$ is a percent of the unknown set temperature $S$, so undo the multiplication by $1.04$ instead of taking $4\%$ of $72.8$.

Answer: -36

Expand: $(x-a)^3=x^3-3ax^2+3a^2x-a^3$. Matching the $x$ terms gives $3a^2=27$, so $a^2=9$ and, since $a>0$, $a=3$. Then $b=-3a=-9$ and $c=-a^3=-27$, so $b+c=-36$.

Check: At $x=1$: $(1-3)^3=-8$ and $1-9+27-27=-8$.

Watch for: The most common wrong values are $36$, from taking both $b=9$ and $c=27$, and $18$, from taking $c=27$; with $a>0$, both $-3a$ and $-a^3$ are negative.

Answer: C.

Multiply all parts by 4 to obtain $-12\le5-2x<8$. Subtract 5: $-17\le-2x<3$. Division by -2 reverses both comparisons, giving $-3/2<x\le17/2$.

Check: The lower endpoint makes the expression 2 and is excluded. The upper endpoint makes it -3 and is included.

A: This fails to reverse and reorder correctly after division by -2.

B: The endpoint inclusions are reversed.

D: The lower endpoint has the wrong sign.

Watch for: Reverse inequality signs when dividing by a negative number, and carry each endpoint’s inclusion correctly.

Answer: D.

The line through $(1,3)$ and $(4,9)$ has slope $\dfrac{9-3}{4-1}=2$ and equation $y=2x+1$.

The point $(k+2,2k+1)$ lies on this line only if $2k+1=2(k+2)+1$, which simplifies to $1=5$. That is false for every $k$.

Every point $(k+2,2k+1)$ lies on the line $y=2x-3$, which is parallel to $y=2x+1$ and never meets it. So there are $0$ such values of $k$.

Check: For example, $k=0$ gives $(2,1)$, and the slope from $(1,3)$ to $(2,1)$ is $-2$, not $2$. For any $k\ne-1$, the slope from $(1,3)$ to $(k+2,2k+1)$ is $\dfrac{2k-2}{k+1}$, which equals $2$ only if $-2=2$.