Advanced Math Practice Test 1

Squaring a binomial gives $(a - b)^2 = a^2 - 2ab + b^2$, so $(3x - 4)^2 = 9x^2 - 2(3x)(4) + 16 = 9x^2 - 24x + 16$. Squaring each term separately and dropping the middle term gives $9x^2 + 16$. Using $ab$ rather than $2ab$ for the middle term gives $-12x$. Doubling the coefficient instead of squaring it gives $6x^2$.

Answer: B.

The model predicts $200(0.8)^1=160$ at $x=1$ and $200(0.8)^3=200(0.512)=102.4$ at $x=3$. The decrease is $160-102.4=57.6$.

Check: The predictions at $x=1$, $2$, and $3$ are $160$, $128$, and $102.4$, so the decreases are $32$ and then $25.6$, which add to $57.6$.

A: $32$ is the decrease from $x=1$ to $x=2$ only.

C: $64$ assumes the model decreases by the same $32$ in each step, as a linear model would.

D: $97.6$ is the decrease from $x=0$ to $x=3$, starting at the initial value $200$.

Watch for: An exponential model decreases by $20\%$ of the current value each step, so the drops get smaller.

To factor $x^2 - 2x - 15$, look for two numbers that multiply to $-15$ and add to $-2$.

The numbers $-5$ and $3$ satisfy both conditions, since their product is $-15$ and their sum is $-2$.

So the equation is

This form displays the $x$-intercepts, $5$ and $-3$, as constants.

Answer: 5

The horizontal change is 3 and the vertical change is $(5a-2)-(2a+1)=3a-3$. Thus $(3a-3)/3=4$, giving a - 1 = 4 and a = 5.

Check: The two points become (5,11) and (8,23), whose slope is 12/3 = 4.

Watch for: Subtract the entire first y-coordinate, including its constant term.

Answer: C.

The line is $1{,}200$ centimeters long, so it has $\dfrac{1{,}200}{30}=40$ gaps between emitters and $41$ emitters. Twenty-five minutes is $\dfrac{25}{60}$ hour, so the total is

$41(1.8)\left(\dfrac{25}{60}\right)=30.75$

liters.

Check: Together the emitters release $41(1.8)=73.8$ liters per hour, and $\dfrac{25}{60}$ of $73.8$ is $30.75$.

A: This treats $25$ minutes as $0.25$ hour.

B: This counts $40$ emitters, the number of gaps, instead of $41$.

D: This multiplies the hourly rate by $25$ without converting minutes to hours.

Watch for: With an emitter at each end, the number of emitters is one more than the number of equal gaps.

Answer: B.

Triangle $ACD$ has a right angle at $C$. Since $\cos(\angle ADC)=\dfrac{7}{25}$, its sides are in the ratio $7:24:25$, with $DC$ adjacent to angle $ADC$, $AC$ opposite it, and $AD$ the hypotenuse. So

$\sin(\angle ADC)=\dfrac{AC}{AD}=\dfrac{24}{25}$

and $AD=48\cdot\dfrac{25}{24}=50$. Then $DC=\dfrac{7}{25}\cdot50=14$.

Since $BD=AD=50$, $BC=50+14=64$. In right triangle $ACB$, $\tan(\angle ABC)=\dfrac{AC}{BC}=\dfrac{48}{64}=\dfrac34$.

Check: $AD=\sqrt{14^2+48^2}=\sqrt{2{,}500}=50=BD$, and $\cos(\angle ADC)=\dfrac{14}{50}=\dfrac{7}{25}$.

A: $\dfrac{7}{24}$ is $\dfrac{DC}{AC}$, a ratio in triangle $ACD$, not in triangle $ACB$.

C: $\dfrac{24}{25}$ is $\dfrac{AC}{BD}$, which uses $BD=50$ as if it were all of $\overline{BC}$.

D: $\dfrac{24}{7}$ is $\tan(\angle ADC)$, the tangent of the angle at $D$ rather than at $B$.

Watch for: Find $AD$ from the side opposite angle $ADC$, then use $AD=BD$ to get all of $\overline{BC}$ before forming the tangent at $B$.

The polynomial $q(x)$ is the quotient when $p(x)$ is divided by $x - 2$, and $r$ is the remainder. Synthetic division with $2$ finds both. Bring down the leading coefficient $2$, then at each step multiply the running value by $2$ and add the next coefficient.

Second coefficient:

Third coefficient:

Last value, which is the remainder:

The first three values, $2$, $-1$ and $1$, are the coefficients of the quotient, one degree lower than $p(x)$:

Check by expanding:

Distributing the right side gives $ax + 11 = 10x - 15$, and collecting the variable terms gives $(a - 10)x = -26$. For any $a$ other than $10$ that has exactly one solution; only $a = 10$ makes the left side $0$ and leaves $0 = -26$, which no value of $x$ satisfies. Matching the coefficient visible inside the parentheses, without distributing the $5$, gives $2$. Taking the multiplier outside them to be the coefficient gives $5$. And $-15$ is the distributed constant, which decides whether the equation has no solution or infinitely many, not what $a$ must be.

Answer: 37

The total mass rises by $427-242=185$ grams when five samples are added. Each sample therefore has mass 185/5 = 37 grams.

Check: The carrier itself has mass $242-5(37)=57$ grams, consistent with the other row.

Watch for: Subtract the totals to remove the unchanged carrier mass.

Answer: C.

Let s and l be the counts. The equations are $4s+7l=98$ and $9s+18l=234$. Divide the second by 9 to get $s+2l=26$, or $s=26-2l$. Substitute into the time equation: $104-8l+7l=98$, so $l=6$ and $s=14$.

Check: Fourteen short and six long cycles use $56+42=98$ minutes and bind $126+108=234$ sheets.

A: This is the long-cycle count.

B: This does not satisfy both batch totals.

D: This is the total cycle count.

Watch for: The output per long cycle doubles, but the time per long cycle does not; keep separate equations.

Answer: C.

Let the playlist contain $j$ jazz songs and $n$ songs in all. Either addition makes $n+4$ songs:

$\dfrac{j+4}{n+4}=\dfrac12$, so $j+4=\dfrac{n+4}{2}$

$\dfrac{j}{n+4}=\dfrac13$, so $j=\dfrac{n+4}{3}$

Subtracting the second equation from the first gives $4=\dfrac{n+4}{2}-\dfrac{n+4}{3}=\dfrac{n+4}{6}$, so $n+4=24$ and $n=20$ (with $j=8$).

Check: With $8$ jazz and $12$ rock songs, adding $4$ jazz songs gives $\dfrac{12}{24}=\dfrac12$, and adding $4$ rock songs gives $\dfrac{8}{24}=\dfrac13$.

A: $8$ is the number of jazz songs, not the number of songs on the playlist.

B: $12$ is the number of rock songs.

D: $24$ is the number of songs after $4$ are added.

Watch for: Both scenarios add $4$ songs, so both probabilities have the same denominator, $n+4$; solve for $n$, not for $n+4$ or $j$.

Answer: D.

Divide both sides of $a<b$ by $b$. Because $b$ is negative, dividing reverses the inequality: $\frac{a}{b}>\frac{b}{b}=1$. So $\frac{a}{b}>1$ must be true.

The other statements are false for every such pair. For example, with $a=-4$ and $b=-2$: $a^2=16>4=b^2$, $\frac1a=-\frac14>-\frac12=\frac1b$, and $ab=8>0$.

Check: With $a=-4$ and $b=-2$, $\frac ab=2>1$; with $a=-3$ and $b=-1$, $\frac ab=3>1$.

A: $a$ is farther from $0$ than $b$, so $a^2>b^2$; the statement reverses the inequality.

B: For negative numbers, the one farther from $0$ has the reciprocal closer to $0$: with $a=-4$ and $b=-2$, $-\frac14>-\frac12$.

C: The product of two negative numbers is positive.

Watch for: Multiplying or dividing an inequality by a negative number reverses it, and squaring or taking reciprocals of negative numbers also reverses their order.

Answer: B.

Let $P$ be the price, in dollars.

Discount first, then coupon: $0.85P-4$.

Coupon first, then discount: $0.85(P-4)=0.85P-3.40$.

The difference is $(0.85P-3.40)-(0.85P-4)=0.60$, so applying the discount first costs $0.60 less, whatever the price.

Check: For a $100 item: discount first gives $85-4=81$ dollars; coupon first gives $0.85(96)=81.60$ dollars. The difference is $0.60.

A: $0.00 assumes that the order of a percent discount and a fixed coupon does not matter.

C: $3.40 is how much the coupon lowers the final price when it is applied before the discount, $0.85(4)$, not the difference between the two orders.

D: $4.00 is the value of the coupon.

Watch for: When the coupon comes first, the discount is taken on $4 less, so the discount saves $0.15(4)=0.60$ dollars less; the price itself cancels.

Answer: 2

Let the inset dimensions be L and W, so $L+W=29$. The area added is $(L+2w)(W+2w)-LW=2w(L+W)+4w^2$. Thus $58w+4w^2=132$, or $2w^2+29w-66=0$. Factoring gives $(w-2)(2w+33)=0$. The positive width is 2.

Check: The four side strips contribute 116 square centimeters and the four corner squares contribute 16, totaling 132.

Watch for: Use perimeter to replace the unknown sum of dimensions; the inset’s individual sides are unnecessary.

Answer: B. $\frac{2592}{49}$ Let the vertical side be h and the horizontal side be $2h$. The hypotenuse joins (18, 0) and (0, 12), so a point on it satisfies $\frac{x}{18} + \frac{y}{12} = 1$. Thus $\frac{2h}{18} + \frac{h}{12} = 1$, giving $\frac{7h}{36} = 1$ and $h = \frac{36}{7}$. The area is $2h^{2} = \frac{2592}{49}$.

Watch for: The rectangle reaches the hypotenuse at one corner; its sides are not the full triangle legs.

Answer: 36 The second condition gives slope $\frac{48}{8} = 6$. Therefore $f(t + 2) = f(t) + 12$. Compare this with 3 f(t) to get $f(t) = 6$. The output five input units later is $f(t + 5) = 6 + 5(6) = 36$.

Watch for: The value of t itself is unnecessary; linear output changes depend only on input differences.

Answer: 19 The difference $g(x) - f(x) = x^{2} - (a + 4)x + (b - 12)$ has roots 1 and 5 and leading coefficient 1. It must equal $(x - 1)(x - 5) = x^{2} - 6x + 5$. Therefore $a + 4 = 6$ and $b - 12 = 5$, giving $a = 2$, $b = 17$, and sum 19.

Watch for: An intersection makes the difference of the functions zero, not either individual function zero.

Answer: C.

Let the radius be $r$ centimeters and the central angle $\theta$ radians. The arc has length $r\theta$, and the perimeter includes two radii:

$2r+r\theta=20$

The area is $\frac12r^2\theta=\frac12r(r\theta)=24$. From the first equation, $r\theta=20-2r$, so

$\frac12r(20-2r)=24$

which gives $r^2-10r+24=0$, so $r=4$ or $r=6$.

If $r=4$, then $\theta=\frac{20-8}{4}=3$. If $r=6$, then $\theta=\frac{20-12}{6}=\frac43$. Only $\theta=3$ is greater than $2$, so $r=4$.

Check: With $r=4$ and $\theta=3$: the perimeter is $4+4+4(3)=20$, and the area is $\frac12(16)(3)=24$.

A: $2.4$ leaves the two radii out of the perimeter, using $r\theta=20$, which gives $\frac12r(20)=24$.

B: $3$ is the central angle, in radians, not the radius.

D: $6$ is the other solution of the quadratic equation, but its central angle is $\frac43$ radian, which is less than $2$.

Watch for: The perimeter of a sector includes the two radii as well as the arc, and both solutions of the quadratic must be checked against the angle condition.

Answer: A.

At x = 1, $q(1)=a(4)(-6)=-24a=-48$, so a = 2. The roots -3 and 7 have midpoint 2. At that vertex, $q(2)=2(5)(-5)=-50$.

Check: Completing the square gives $q(x)=2(x-2)^2-50$, confirming the minimum.

B: This is the given value at x = 1, away from the vertex.

C: This is the unscaled product at x = 1.

D: This is the recovered leading coefficient, not the minimum.

Watch for: The midpoint of the zeros locates the vertex input; evaluate the function there to get the minimum output.

Answer: A.

The graph of $y=\lvert x-3\rvert$ has vertex $(3,0)$, a left branch $y=3-x$ for $x\le3$, and a right branch $y=x-3$ for $x\ge3$.

Left branch: $mx=3-x$ gives $x=\dfrac{3}{1+m}$. For $m>-1$, this is at most $3$ exactly when $m\ge0$. (For $m<-1$, it is negative, which is also on the left branch.)

Right branch: $mx=x-3$ gives $x=\dfrac{3}{1-m}$, which is at least $3$ exactly when $0\le m<1$.

When $m=0$, both branches give the same point, the vertex $(3,0)$, so there is only one intersection point. Two different intersection points require an intersection with each branch away from the vertex, which happens exactly when $0<m<1$.

Check: For $m=\dfrac12$, the points are $(2,1)$ on the left branch and $(6,3)$ on the right branch. For $m=1$, the line is parallel to the right branch and meets only the left branch, at $\left(\dfrac32,\dfrac32\right)$.

B: For slopes between $-1$ and $0$, the line misses the graph entirely: to the right of the origin the line is below the $x$-axis, and to the left it rises more slowly than the left branch, which starts $3$ units higher.

C: A slope of $1$ or more is at least as steep as the right branch, so the line meets only the left branch, giving one point.

D: This includes $m\le0$: for $-1\le m<0$ there are no intersection points, and for $m=0$ or $m<-1$ there is only one.

Watch for: Test the boundary slopes: at $m=0$ the line touches only the vertex, and at $m=1$ it is parallel to the right branch.

Answer: 18

The original values are $1$, $2$, $3$, $3$, $4$, $6$, $7$, $7$, $8$, $9$. Their sum is $50$, so the mean is $5$, and the median is $\dfrac{4+6}{2}=5$.

The new data set has $12$ values and mean $6$, so its sum is $72$, and the two added values, $x\le y$, satisfy $x+y=22$.

The new median is the average of the $6$th and $7$th values. If $x\le4$ and $y\ge6$, these are $4$ and $6$, and the median stays $5$. If $4<x<6$, the $6$th and $7$th values are $x$ and $6$, whose average is more than $5$. If $x\ge6$, both are at least $6$. So the median is unchanged only when $x\le4$, which makes $y=22-x\ge18$. The least possible value of the greater added value is $18$, with $x=4$.

Check: Adding $4$ and $18$ gives $1,2,3,3,4,4,6,7,7,8,9,18$. The sum is $72$, so the mean is $6$, and the $6$th and $7$th values are $4$ and $6$, so the median is $5$.

Watch for: The most common wrong value is $11$, from splitting the sum $22$ equally; adding $11$ and $11$ makes the median $6.5$.

Answer: A.

Two distinct real solutions require the equation to be quadratic, so $k-3\neq0$, and require a positive discriminant. The discriminant is $(2k+2)^2-4(k-3)k=4k^2+8k+4-4k^2+12k=20k+4$, which is positive exactly when $k>-\dfrac{1}{5}$. Combining the two requirements gives $k>-\dfrac{1}{5}$ with $k\neq3$.

Check: Test the boundary cases. At $k=3$ the equation reduces to $8x+3=0$, which has one solution; at $k=-\dfrac{1}{5}$ it becomes $-\dfrac{16}{5}x^2+\dfrac{8}{5}x-\dfrac{1}{5}=0$, whose only solution is $x=\dfrac{1}{4}$; and at $k=5$ the equation $2x^2+12x+5=0$ does have two distinct real solutions.

B: B excludes $k=0$ needlessly; at $k=0$ the equation $-3x^2+2x=0$ has the two distinct solutions $0$ and $\dfrac{2}{3}$.

C: This applies the discriminant condition but keeps $k=3$, where the leading coefficient vanishes and the equation is linear.

D: D allows $k=-\dfrac{1}{5}$, where the discriminant is zero and the two solutions coincide.

Watch for: Before reaching for the discriminant, confirm the equation is actually quadratic, since a parameter in the leading coefficient can make it linear.

The vertex form of a quadratic is written using the vertex $(h, k)$.

With vertex $(2, -3)$, this becomes

Since the graph passes through $(4, 5)$, substituting those coordinates gives

which simplifies to

Adding $3$ to both sides:

Dividing both sides by $4$:

Substituting $2$ for $a$ gives

Both terms holding $m$ sit on the right side, so start by clearing the constant. Adding $7$ to both sides:

Factoring $m$ out of the right side:

Dividing both sides by $x + 3$, which the given restriction $x \neq -3$ keeps away from zero:

Answer: B.

The bakeries chose their own fermentation step rather than being randomly assigned one, so the bakeries that volunteered may differ from the others in flour, ovens, or handling. The design therefore supports an association within this particular group of 48, and nothing stronger.

Check: Ask what would have to change to license a stronger claim. Randomly assigning the step to bakeries would support a causal conclusion, and randomly sampling from all rye bakeries would support generalization; this study did neither.

A: A generalizes to all rye bakeries, but these 48 were not a random sample of that larger population.

C: C treats equal group sizes as a substitute for random assignment, but matching the counts does nothing about whatever led some bakeries to volunteer.

D: This contradicts the stated analysis, which already found the observed difference unlikely to arise by chance.

Watch for: Random assignment is what licenses a causal claim and random sampling is what licenses generalization; a study can have neither, one, or both.

Answer: D.

Rewrite the negative exponents as reciprocals and add the fractions:

$x^{-1}+y^{-1}=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{y+x}{xy}$

The outer exponent $-1$ takes the reciprocal of this sum, which is $\dfrac{xy}{x+y}$.

Check: For $x=2$ and $y=3$: $\left(\dfrac{1}{2}+\dfrac{1}{3}\right)^{-1}=\left(\dfrac{5}{6}\right)^{-1}=\dfrac{6}{5}$, and $\dfrac{2\cdot3}{2+3}=\dfrac{6}{5}$.

A: $x+y$ applies the outer exponent to each term separately, as if $(a+b)^{-1}=a^{-1}+b^{-1}$.

B: $\dfrac{1}{x+y}$ adds the reciprocals by adding denominators, $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x+y}$, and then leaves out the outer exponent $-1$.

C: $\dfrac{x+y}{xy}$ is $x^{-1}+y^{-1}$ itself; it leaves out the outer exponent $-1$.

Watch for: An exponent of $-1$ on a sum takes the reciprocal of the whole sum, not of each term.

A fraction is undefined where its denominator is zero, so set the denominator equal to zero:

Two numbers with product $2 \cdot (-15) = -30$ and sum $7$ are $10$ and $-3$, so split the middle term and factor by grouping:

The first factor is zero when

The second factor is zero when

Neither value makes the numerator zero, so the expression is undefined at exactly those two values, $-5$ and $\dfrac{3}{2}$.

Answer: 700

For every $x$,

$g(x)=5\cdot2^{x+3}=5\cdot2^x\cdot2^3=8f(x)$

So $g(x)$ is $8$ times $f(x)$. A value that is $8$ times another is greater than it by $7$ times the other, which is $700\%$ greater. So $p=700$.

Check: At $x=0$: $f(0)=5$ and $g(0)=f(3)=40$. The increase is $35$, and $\frac{35}{5}=7$, or $700\%$.

Watch for: The most common wrong value is $800$, which treats “$8$ times as large” as “$800\%$ greater”; another is $300$, which treats the shift of $3$ units as a $300\%$ change.

Answer: B.

Let $R$ and $r$ be the radii and $d$ the distance from the center to the chord. The perpendicular from the center bisects both the chord of the larger circle and the chord of the smaller circle. So $R^2=d^2+12^2$ and $r^2=d^2+7^2$. Subtracting, $R^2-r^2=144-49=95$, and the region’s area is $\pi(R^2-r^2)=95\pi$.

Check: For example, $d=5$ gives $R=13$ and $r=\sqrt{74}$, and $169-74=95$.

A: $25\pi$ squares half the difference of the two chord lengths, $\left(\tfrac{24-14}{2}\right)^2$.

C: $100\pi$ squares the difference of the two chord lengths, $(24-14)^2$.

D: $380\pi$ subtracts the squares of the whole chord lengths, $24^2-14^2$, instead of their halves.

Watch for: Use half of each chord: the perpendicular from the center bisects every chord it meets, and the unknown distance $d$ cancels.

Answer: B.

The solutions of $\lvert x-3\rvert<2$ are $1<x<5$, and the solutions of $\lvert x-a\rvert<4$ are $a-4<x<a+4$.

Every $x$ with $1<x<5$ must satisfy $a-4<x<a+4$. This happens exactly when $a-4\le1$ and $a+4\ge5$; equality is allowed because neither interval includes its endpoints. So $a\le5$ and $a\ge1$, which gives $1\le a\le5$.

Check: For $a=1$, the second inequality gives $-3<x<5$, which contains every $x$ with $1<x<5$. For $a=5.1$, it gives $1.1<x<9.1$, which misses $x=1.05$.