Standard Math Practice Test 1

A radical becomes a fractional exponent whose numerator is the power under the radical and whose denominator is the index of the root, so $\sqrt[3]{x^5} = x^{5/3}$. Reading the fraction upside down gives $x^{3/5}$, which is the cube root's fifth power turned inside out. Adding the index to the power instead of writing it beneath gives $x^{8/3}$, and multiplying the two gives $x^{15}$, which is the result of raising to the third power rather than taking a third root.

Consecutive odd integers rise by $2$, so calling the least of them $n$ gives $n + (n + 2) + (n + 4) + (n + 6) = 4n + 12 = 72$, and $n = 15$. The four are $15$, $17$, $19$, and $21$, so the greatest is $21$. Answering $15$ reports the least instead of the greatest, and $19$ the third. The average, $72 \div 4 = 18$, is not one of them at all, since four consecutive odd integers have an even number midway between the middle two.

Answer: B.

Because $\ell\parallel m$, the $52^\circ$ angle at $A$ and $\angle ABC$ are alternate interior angles, so $\angle ABC=52^\circ$. The $115^\circ$ angle at $C$ is an exterior angle of triangle $ABC$, so $\angle ACB=180^\circ-115^\circ=65^\circ$. The angles of triangle $ABC$ sum to $180^\circ$:

$x=180-52-65=63$

Check: The exterior angle at $C$ equals the sum of the two remote interior angles: $52^\circ+63^\circ=115^\circ$.

A: This is the measure of $\angle ABC$, which equals the given $52^\circ$ angle; it is not $\angle BAC$.

C: This is the measure of $\angle ACB$, the angle supplementary to the $115^\circ$ angle.

D: This treats $x$ as supplementary to the $52^\circ$ angle, but those two angles do not form a straight line.

Watch for: Label each angle of the triangle before adding; only the angle at $A$ between segments $AB$ and $AC$ is $x^\circ$.

Answer: A.

The original ordered values are $2,2,4,7,9$. Replacing 9 by 14 adds 5 to the sum, so the mean increases by $5/5=1$. The middle value remains 4.

Check: The means are $24/5=4.8$ and $29/5=5.8$; both medians are 4.

B: Only the mean changes.

C: Five is the change in the sum, not in the mean.

D: The mean changes; the median does not.

Watch for: A change to an extreme value affects the sum but need not change the middle position.

Answer: D.

Recover $a$ from the stated condition first: $2a+1=-3$ gives $2a=-4$, so $a=-2$. Then $f(-2)=(-2)^2-3(-2)=4+6=10$, and applying the outer function gives $g(10)=2(10)+1=21$.

Check: Composing symbolically instead, $g(f(x))=2(x^2-3x)+1=2x^2-6x+1$, and substituting $x=-2$ gives $8+12+1=21$.

A: $-2$ is the value of $a$ itself, the input the condition produces, not the value of $g(f(a))$.

B: 9 comes from solving $2a+1=-3$ as $a=\dfrac{-3+1}{2}=-1$, adding the 1 instead of subtracting it; that gives $f(-1)=4$ and $g(4)=9$.

C: 18 is $f(g(a))=f(-3)=9+9=18$, the same two functions applied in the wrong order.

Watch for: When the input to a composition is described by a condition instead of given outright, solve for it first, and check which of the two functions the condition is about.

Answer: D.

Let $b$ and $f$ be the masses, in kilograms, of one crate of bolts and one crate of fittings. The shipments give $4b+3f=87$ and $2b+5f=89$. Solving the system yields $b=12$ and $f=13$, so $b+f=25$ kilograms.

Check: $4(12)+3(13)=48+39=87$ and $2(12)+5(13)=24+65=89$, so both stated shipment masses are reproduced.

A: Subtracting the second equation from the first gives $2b-2f=-2$, so $f-b=1$. That is the difference of the two crate masses, not their sum.

B: $12$ is the mass of one crate of bolts by itself.

C: $13$ is the mass of one crate of fittings by itself.

Watch for: After solving a system, reread which combination of the variables the question asks for.

For a quadratic written as $ax^2 + bx + c = 0$, the two solutions add to $-\frac{b}{a}$.

Here $a$ is $2$ and $b$ is $-14$.

Substituting those values:

The two solutions add to $7$.

Answer: 18

The pass total is 54 + 36 = 90. If this is 75% of all hinges, the total is 90/0.75 = 120. The three known cells sum to 102, so n = 18.

Check: The completed failure total is 30, one quarter of 120.

Watch for: An overall proportion uses the grand total, not either orientation’s row total.

Answer: C.

Substituting the reported depth gives $0.4+0.15p=3.25$, so $0.15p=2.85$ and $p=19$ passes. Twice as many passes means $p=38$, and the model then gives $d=0.4+0.15(38)=0.4+5.7=6.10$ millimeters.

Check: Each pass adds $0.15$ millimeter, so the $19$ extra passes add $19(0.15)=2.85$ millimeters to the original $3.25$ millimeters, which is $6.10$.

A: This adds the starting depth $0.4$ to the given $3.25$ a second time instead of doubling the number of passes.

B: This computes $0.15(38)=5.70$ but drops the constant term $0.4$, which is the depth the groove already has before any pass is made.

D: This doubles the depth itself, $2(3.25)=6.50$. Doubling the input doubles the output only for a proportional relationship, and here the $0.4$ does not double.

Watch for: Doubling the input of a linear model does not double the output unless the constant term is zero.

Answer: 2

Write each difference in terms of $a$ and $b$:

$f(2)-f(1)=ab^2-ab=ab(b-1)=12$

$f(3)-f(2)=ab^3-ab^2=ab^2(b-1)=36$

Dividing the second equation by the first gives $b=3$. Then $a(3)(3-1)=12$, so $6a=12$ and $a=2$. Because $f(0)=a\cdot b^0=a$, the value of $f(0)$ is $2$.

Check: With $f(x)=2\cdot3^x$, the values are $f(1)=6$, $f(2)=18$, and $f(3)=54$. The differences are $18-6=12$ and $54-18=36$.

Watch for: The most common wrong value is $6$, from writing $f(2)-f(1)$ as $a(b-1)$ and leaving out the factor $b$. The value $6$ is also $f(1)$, not $f(0)$.

Answer: 57

The sum of the $20$ numbers is $20(50)=1{,}000$.

Without the greatest number, the sum is $19(48)=912$, so the greatest number is $1{,}000-912=88$.

Without the least number, the sum is $19(51)=969$, so the least number is $1{,}000-969=31$.

The range is $88-31=57$.

Check: Removing $88$ from a sum of $1{,}000$ leaves $912$, and $\frac{912}{19}=48$; removing $31$ leaves $969$, and $\frac{969}{19}=51$.

Watch for: The most common wrong value is $3$, the difference between the two new means; find each removed number from the change in the sum.

Answer: 20/3

The midpoint of the segment is $(4,4)$, and the segment’s slope is $\dfrac{6-2}{7-1}=\dfrac23$. Line $\ell$ passes through $(4,4)$ with slope $-\dfrac32$, so $y-4=-\dfrac32(x-4)$. Setting $y=0$ gives $x-4=\dfrac83$, so $x=\dfrac{20}{3}$.

Check: The point $\left(\tfrac{20}{3},0\right)$ is equidistant from both endpoints: $\left(\tfrac{17}{3}\right)^2+2^2=\tfrac{325}{9}$ and $\left(\tfrac{1}{3}\right)^2+6^2=\tfrac{325}{9}$.

Watch for: The most common wrong value is $\tfrac43$, from using slope $+\tfrac32$ (the reciprocal without the sign change); another is $10$, the $y$-intercept.

The vertex is at $(h, k) = (2, -3)$. The graph also passes through $(5, 3)$, so $3 = a|5 - 2| - 3$, giving $3a = 6$ and $a = 2$. Then $a + h + k = 2 + 2 - 3 = 1$. Reading the vertex as $(2, 3)$, with the sign of $k$ reversed, gives $7$. Reporting $k$ alone gives $-3$, and reporting $a$ alone gives $2$.

Answer: B.

In the undiluted bath every $3$ liters of concentrate is matched by $22$ liters of water, so $54$ liters of concentrate calls for $54\times\dfrac{22}{3}=396$ liters of water. The extra $90$ liters raise the water to $486$ liters while the concentrate stays at $54$, and $54:486$ reduces to $1:9$.

Check: The reduction is exact because $54\times 9=486$, and the diluted bath totals $540$ liters, of which the concentrate is one part in ten, consistent with one part concentrate to nine parts water.

A: $1:10$ compares concentrate with the whole diluted bath, $54:540$, rather than with the water alone.

C: $3:22$ is the ratio before the extra water goes in; the $90$ liters were never counted.

D: $4:11$ adds the $90$ liters to the concentrate instead of to the water, giving $144:396$.

Watch for: Adding one ingredient changes a part-to-part ratio but leaves the other part untouched, so settle which part the addition joins before reducing.

Answer: C.

Each hour the count is multiplied by $0.6$, so over two hours it is multiplied by $(0.6)^2=0.36$. A multiplier of $0.36$ leaves $36\%$ of the parcels, so the decrease is $100\%-36\%=64\%$.

Check: Starting from $1{,}200$ parcels, the model gives $1{,}200(0.6)=720$ after one hour and $720(0.6)=432$ after two. The drop is $1{,}200-432=768$, and $768/1{,}200=0.64$.

A: $36\%$ is the fraction of parcels still unscanned after two hours, not the percent by which the count fell.

B: $40\%$ is the decrease over a single hour, not over two.

D: This doubles the hourly $40\%$ decrease. Repeated percent changes multiply rather than add, so two hours leave $0.6(0.6)=36\%$, not $20\%$.

Watch for: Percent decreases compound by multiplying the remaining fractions; they never simply add.

Answer: A. −4 The first inequality gives $2a + 2 > 9$, or $a > 3.5$. The second gives $4a - 2 < 17$, or $a < 4.75$. The only integer is $a = 4$. Then $f(x) = 2x + 8$, whose zero is $x = -4$.

Watch for: Find the allowed integer parameter before using the line's intercept formula.

Answer: A.

Tangency to the x-axis makes the radius 4. The point condition gives $(5-a)^2+(6-4)^2=16$, so $(5-a)^2=12$. Since a < 5, $5-a=2\sqrt3$, giving $a=5-2\sqrt3$.

Check: The horizontal and vertical offsets to the point have squares 12 and 4, totaling the radius squared 16.

B: This violates a < 5.

C: This has both the wrong displacement and the wrong side.

D: This doubles the required horizontal displacement.

Watch for: Use the side condition to select the sign after taking a square root.

Answer: B.

Because $f(-2)=f(6)$, the axis of symmetry of the parabola is halfway between $-2$ and $6$, at $x=2$.

Both $0$ and $1$ are to the left of the axis, and $0$ is farther from it. Since $f(0)>f(1)$, the values increase with distance from the axis, so the parabola opens upward and $f$ has its minimum at $x=2$.

For an upward-opening parabola, the input farther from the axis has the greater output. The input $5$ is $3$ units from the axis and $0$ is $2$ units from it, so $f(5)>f(0)$ must be true.

Check: The function $f(x)=(x-2)^2$ satisfies both conditions: $f(-2)=16=f(6)$ and $f(0)=4>1=f(1)$. It gives $f(5)=9>4=f(0)$, while the statements in choices A, C, and D are false for it.

A: This would be true if the parabola opened downward; but $f(0)>f(1)$ means the parabola opens upward, so $f(2)$ is the minimum and $f(2)<f(1)$.

C: $4$ and $0$ are both $2$ units from the axis $x=2$, so $f(4)=f(0)$.

D: $3$ is only $1$ unit from the axis and $0$ is $2$ units away, so $f(3)<f(0)$ for an upward-opening parabola.

Watch for: Use the equal outputs to find the axis, then use the inequality to decide which way the parabola opens before comparing distances from the axis.

Let the $y$-intercept be $(0, b)$. Then the $x$-intercept is $(2b, 0)$.

The slope between the two intercepts is the change in $y$ over the change in $x$:

The slope is $-\dfrac{1}{2}$ no matter what $b$ is, so the line has the form

Substitute the point $(4, 3)$:

Simplify the right side and solve:

So the line is $y = -\dfrac{1}{2}x + 5$. Its $y$-intercept is $5$ and its $x$-intercept is $10$, which is twice $5$.

Answer: B.

Let $H$ be the tree’s height at the beginning of 2010. Because the height increased linearly, it increased by the same amount in every $4$-year period. From 2010 to 2014 that amount was $0.3H$, so the height was $1.3H$ in 2014 and

$1.3H+2(0.3H)=1.9H$

in 2022. From 2014 to 2022, the height increased by $0.6H$, and as a percent of the 2014 height, that increase is

$\dfrac{0.6H}{1.3H}\approx0.462$

or about $46\%$.

Check: If the tree was $10$ meters tall in 2010, it grew $3$ meters every $4$ years: $13$ meters in 2014 and $19$ meters in 2022. The increase from 2014 to 2022 is $\dfrac{6}{13}\approx46\%$ of the 2014 height.

A: $23\%$ is the increase for only the first $4$ years after 2014: $\dfrac{0.3H}{1.3H}\approx23\%$.

C: $60\%$ adds a $30\%$ increase for each of the two $4$-year periods, measuring both against the 2010 height instead of the 2014 height.

D: $69\%$ applies a $30\%$ increase twice, $1.3^2=1.69$. That is exponential growth, not linear growth.

Watch for: In linear growth, the amount of increase stays the same for each equal time period, but the percent increase gets smaller, because the height it is measured against keeps growing.

Answer: C.

Because $p$ has leading coefficient $1$ and both $x+3$ and $x-4$ are factors, $p(x)=(x+3)(x-4)(x-r)$ for some constant $r$. By the remainder theorem the division statement says $p(1)=60$, so $(4)(-3)(1-r)=60$, which gives $1-r=-5$ and $r=6$. The constant term of a polynomial is its value at $x=0$, so $d=p(0)=(3)(-4)(-6)=72$.

Check: Expanding $(x+3)(x-4)(x-6)$ gives $x^3-7x^2-6x+72$, so $d=72$; dividing that polynomial by $x-1$ gives the quotient $x^2-6x-12$ with remainder $60$, and dividing it by $x+3$ or by $x-4$ leaves remainder $0$.

A: This multiplies the three zeros $-3$, $4$, and $6$ together. The constant term is $p(0)=(0+3)(0-4)(0-6)$, where each zero enters with the opposite sign, so that product is $-72$ while $d$ is $+72$.

B: $6$ is the third zero, the value the work produces on the way to the answer; the question asks for the constant $d$, not for that zero.

D: This reverses each factor when reading off zeros, using $x=3$ and $x=-4$ and so $p(x)=(x-3)(x+4)(x-r)$. That version satisfies $p(1)=60$ at $r=7$ and has constant term $84$, but a factor of $x+3$ corresponds to the zero $x=-3$.

Watch for: Two known factors of a monic cubic determine it up to a single unknown zero, and once that zero is found the constant term is read off as $p(0)$, not as the product of the zeros.

Answer: 4

A point $3$ units from the $y$-axis has $x=3$ or $x=-3$. On the line $y=2x+1$, these are the points $(3,7)$ and $(-3,-5)$.

If the lines meet at $(3,7)$, then $7=3m+5$, so $m=\frac23$.

If they meet at $(-3,-5)$, then $-5=-3m+5$, so $m=\frac{10}{3}$.

Neither value is $2$, so in each case the lines are not parallel and meet only at that point. The sum is $\frac23+\frac{10}{3}=4$.

Check: For $m=\frac23$: $2x+1=\frac23x+5$ gives $x=3$. For $m=\frac{10}{3}$: $2x+1=\frac{10}{3}x+5$ gives $x=-3$.

Watch for: The most common wrong value is $\frac23$ (entered as $0.666$ or $0.667$), which uses only $x=3$; a point $3$ units from the $y$-axis can also have $x=-3$.

Answer: B.

Right triangle JNM has hypotenuse JM = 10 and leg JN = 6. Its height is $MN=\sqrt{100-36}=8$. The parallelogram’s area is base times perpendicular height, $18(8)=144$.

Check: The triangle is a 6-8-10 right triangle; the height is shorter than the slanted side.

A: This uses the horizontal projection as the height.

C: This uses the slanted side as the height.

D: This uses an incorrect height of 12.

Watch for: A parallelogram’s adjacent side is not its perpendicular height unless there is a right angle.

Answer: D.

Among baskets with a lid, 18 have loop handles and 12 do not, for a total of 30. The probability is $12/30=2/5$.

Check: The complement of the loop-handle share is $1-18/30=2/5$.

A: This divides the 12 favorable baskets by all 90 baskets.

B: This could result from using 12/60, with the baskets without a lid as the denominator, instead of the 30 baskets with a lid.

C: This is 30/90, the probability that a randomly chosen basket has a lid, rather than the probability of no loop handle among the 30 lidded baskets.

Watch for: Apply the complement inside the specified group, not across all baskets.

Answer: A.

The midpoint of the diameter is $(1,1)$. The squared diameter length is $8^2+6^2=100$, so the radius squared is $100/4=25$. Hence the equation is $(x-1)^2+(y-1)^2=25$.

Check: Both endpoints are 5 units from $(1,1)$: their coordinate offsets are $(-4,-3)$ and $(4,3)$.

B: This uses the squared diameter instead of the squared radius.

C: Both the center and the radius are incorrect.

D: The signs put the center at (-1,-1).

Watch for: Use the midpoint for the center and half the diameter for the radius.

Let $w$ be the number of kilometers walked and $r$ the number ridden.

The two legs cover $36$ kilometers in all:

Time is distance divided by speed, and the two times add to $4$ hours:

Multiplying that second equation by $12$:

Subtracting the first equation from this one:

So $w$ is $6$, and the ridden distance is

Answer: B.

The fitted line goes through (0,10) and (5,70), so the slope is 12 minutes per hundred rivets. Fifty rivets is 0.5 hundred, giving $12(0.5)=6$ additional minutes.

Check: The line’s vertical change over a half-unit horizontal interval is 6 minutes.

A: This is the rate per individual rivet, not the time for 50.

C: This is the time for 100 additional rivets.

D: This treats the per-hundred slope as a per-rivet slope.

Watch for: The horizontal-axis unit is hundreds of rivets, not individual rivets.

Answer: B.

Angle ADC intercepts the arc from A through B to C, which measures $72+118=190$ degrees. An inscribed angle measures half its intercepted arc, so $\angle ADC=95^\circ$.

Check: The intercepted arc does not contain vertex D; it is the 190-degree arc, not the other 170-degree arc.

A: This halves the other arc, which contains D.

C: This repeats one of the given arc measures.

D: This forgets to halve the intercepted arc.

Watch for: The intercepted arc can exceed 180 degrees; choose the arc that does not contain the angle’s vertex.

Answer: A.

For $h$ hours, garage A charges $5+3(h-1)=3h+2$ dollars. Garage B charges $4h$ dollars when $4h\le26$, that is, for $h\le6$, and $26 for $h\ge7$.

For $h\le6$: $3h+2<4h$ when $h>2$, which gives $h=3$, $4$, $5$, and $6$.

For $h\ge7$: $3h+2<26$ when $h<8$, which gives $h=7$.

So garage A costs less for $h=3$ through $h=7$, which is $5$ whole numbers.

Check: At $h=2$ both garages charge $8, and at $h=8$ both charge $26, so neither is counted; at $h=7$, garage A charges $23 and garage B charges $26.

B: $6$ counts one of the two ties, at $2$ hours or at $8$ hours, as a time when garage A costs less.

C: $7$ counts both ties, $h=2$ and $h=8$.

D: $8$ ignores garage B’s daily maximum and counts every $h$ from $3$ to $10$.

Watch for: Garage B’s charge stops at $26, so garage A is cheaper only through $7$ hours; the costs tie at $2$ hours and at $8$ hours.

The line of sight, the horizontal line from the surveyor's eyes, and the mast form a right triangle. Its vertical leg runs from eye level to the top of the mast, not from the ground:

The tangent of the angle of elevation is the vertical leg over the horizontal leg. With $d$ as the horizontal distance:

Solving for $d$:

So the surveyor stands $210$ meters from the mast.

Apply $f$ to $f(x)$ by substituting $ax + b$ for $x$:

This equals $4x + 9$ for every $x$, so the coefficients of $x$ match:

Since $a$ is positive:

The constant terms match too:

Substitute $2$ for $a$:

So $b$ is $3$, and the sum is

Answer: A.

Multiply both sides by $12$:

$8x-3(x-c)=5x+2$

The left side is $8x-3x+3c=5x+3c$, so the equation becomes $5x+3c=5x+2$, which is equivalent to $3c=2$. The $x$-terms cancel for every $c$. If $c=\dfrac23$, the equation is true for every $x$; otherwise it is never true.