1954 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The square of $5-\sqrt{y^2-25}$ is:

$y^2-5\sqrt{y^2-25}$

$-y^2$

$y^2$

$(5-y)^2$

$y^2-10\sqrt{y^2-25}$

Concepts:radical algebraic manipulation

Difficulty rating: 1260

Small Hint:

Use $(a-b)^2=a^2-2ab+b^2$

Big Hint:

The constant $25$ cancels with the $-25$ inside the squared radical

Solution:

Expanding gives $\begin{aligned} &\left(5-\sqrt{y^2-25}\right)^2\\ &\quad=25-10\sqrt{y^2-25}\\ &\qquad{}+y^2-25\\ &\quad=y^2-10\sqrt{y^2-25}. \end{aligned}$

Thus, the correct answer is E.

2.

The equation $\begin{aligned} \frac{2x^2}{x-1}-\frac{2x+7}{3} &{}+\frac{4-6x}{x-1}\\ &{}+1=0 \end{aligned}$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0.$ The roots of the latter equation are $4$ and $1.$ Then the roots of the first equation are:

$4$ and $1$

only $1$

only $4$

neither $4$ nor $1$

$4$ and some other root

Concepts:rational equation zero product property

Difficulty rating: 1340

Small Hint:

Check both roots against the denominator in the original equation

Big Hint:

Clearing a denominator can introduce a value at which the original expression is undefined

Solution:

The transformed quadratic factors as $(x-1)(x-4)=0.$ However, the original equation has denominator $x-1,$ so it is undefined at $x=1.$ Substitution of $x=4$ is valid and satisfies the original equation.

Thus, the only root is $4,$ and the correct answer is C.

3.

If $x$ varies as the cube of $y,$ and $y$ varies as the fifth root of $z,$ then $x$ varies as the $n$ th power of $z,$ where $n$ is:

$\dfrac1{15}$

$\dfrac53$

$\dfrac35$

$15$

$8$

Concepts:ratio and proportion exponent substitution

Difficulty rating: 1260

Small Hint:

Write $x=k_1y^3$ and $y=k_2z^{\frac{1}{5}}$

Big Hint:

Substitute the second variation law into the first and multiply the exponents

Solution:

Since $y$ is proportional to $z^{\frac{1}{5}},$ its cube is proportional to $z^{\frac{3}{5}}.$ Because $x$ is proportional to $y^3,$ $x\propto z^{\frac{3}{5}}.$ Hence $n=\frac{3}{5}.$

Thus, the correct answer is C.

4.

If the Highest Common Divisor of $6432$ and $132$ is diminished by $8,$ it will equal:

$-6$

$6$

$-2$

$3$

$4$

Concepts:greatest common divisor prime factorization whole number operations

Difficulty rating: 1180

Small Hint:

Use the Euclidean algorithm or factor the two numbers

Big Hint:

Their greatest common divisor is $12$

Solution:

The Euclidean algorithm gives $\gcd(6432,132)=12.$ Diminishing this by $8$ gives $12-8=4.$

Thus, the correct answer is E.

5.

A regular hexagon is inscribed in a circle of radius $10$ inches. Its area is:

$150\sqrt3$ sq. in.

$150$ sq. in.

$25\sqrt3$ sq. in.

$600$ sq. in.

$300\sqrt3$ sq. in.

Concepts:regular polygon equilateral triangle triangle area

Difficulty rating: 1360

Small Hint:

An inscribed regular hexagon has side length equal to the circle’s radius

Big Hint:

Divide the hexagon into six equilateral triangles of side $10$

Solution:

The hexagon consists of six equilateral triangles of side $10.$ Its area is $6\left(\frac{\sqrt3}{4}\cdot10^2\right) =150\sqrt3$ square inches.

Thus, the correct answer is A.

6.

The value of $\left(\dfrac1{16}\right)^{a^0}+\left(\dfrac1{16a}\right)^0$ ${}-64^{-\frac{1}{2}}-(-32)^{-\frac{4}{5}}$ is:

$1\dfrac{13}{16}$

$1\dfrac3{16}$

$1$

$\dfrac78$

$\dfrac1{16}$

Concepts:exponent fraction algebraic manipulation

Difficulty rating: 1450

Small Hint:

Evaluate each of the four terms separately, using $a^0=1$

Big Hint:

The real fifth root of $-32$ is $-2$ , so $(-32)^{-\frac{4}{5}}=\frac{1}{16}$

Solution:

For permissible $a,$ the four terms are $\frac1{16},\qquad 1,\qquad \frac18,\qquad \frac1{16}.$ Therefore the expression is $\frac1{16}+1-\frac18-\frac1{16}=\frac78.$

Thus, the correct answer is D.

7.

A housewife saved $\$2.50$ in buying a dress on sale. If she spent $\$25$ for the dress, she saved about:

$8\%$

$9\%$

$10\%$

$11\%$

$12\%$

Concepts:percentage money estimation

Difficulty rating: 890

Small Hint:

Add the savings to the sale price to recover the original price

Big Hint:

Compute $\frac{2.50}{27.50}$ as a percentage

Solution:

The original price was $\$27.50.$ The fraction saved was $\frac{2.50}{27.50}=\frac1{11}\approx0.0909,$ or about $9\%.$

Thus, the correct answer is B.

8.

The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is:

$\dfrac14$

$\dfrac12$

$1$

$2$

$4$

Concepts:triangle area square (geometry)ratio and proportion

Difficulty rating: 890

Small Hint:

Let the square’s side be $s$ and the triangle’s altitude be $h$

Big Hint:

Equate $s^2$ with $\tfrac12(2s)h$

Solution:

If the square side is $s,$ the triangle base is $2s.$ Equal areas give $s^2=\frac12(2s)h=sh.$ Thus $h=s,$ so the requested ratio is $1.$

The correct answer is C.

9.

A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $\overline{PQ}$ is $9$ inches and $\overline{QR}$ is $7$ inches. The radius of the circle is:

$3$ in.

$4$ in.

$5$ in.

$6$ in.

$7$ in.

Concepts:power of a point circle

Difficulty rating: 1400

Small Hint:

The secant power is the external length times the whole secant length

Big Hint:

Equate $9(9+7)$ with $13^2-r^2$

Solution:

The power of $P$ is $PQ\cdot PR=9(16)=144.$ It is also $13^2-r^2,$ so $169-r^2=144.$ Hence $r^2=25$ and $r=5$ inches.

Thus, the correct answer is C.

10.

The sum of the numerical coefficients in the expansion of the binomial $(a+b)^6$ is:

$32$

$16$

$64$

$48$

$7$

Concepts:binomial theorem

Difficulty rating: 1180

Small Hint:

The coefficient sum is obtained by choosing convenient values of $a$ and $b$

Big Hint:

Set $a=b=1$

Solution:

Setting $a=b=1$ makes every monomial equal to $1,$ so the value of the expansion equals the sum of its coefficients. Thus the sum is $(1+1)^6=64.$

The correct answer is C.

11.

A merchant placed on display some dresses, each with a marked price. He then posted a sign “ $\dfrac13$ off on these dresses.” The cost of the dresses was $\dfrac34$ of the price at which he actually sold them. Then the ratio of the cost to the marked price was:

$\dfrac12$

$\dfrac13$

$\dfrac14$

$\dfrac23$

$\dfrac34$

Concepts:percentage money ratio and proportion

Difficulty rating: 1290

Small Hint:

The selling price is $\frac{2}{3}$ of the marked price

Big Hint:

The cost is $\frac{3}{4}$ of that selling price

Solution:

If the marked price is $M,$ then the selling price is $\frac{2M}{3}.$ The cost is therefore $\frac34\cdot\frac23M=\frac12M.$ The cost-to-marked-price ratio is $\frac{1}{2}.$

Thus, the correct answer is A.

12.

The solution of the equations $\begin{aligned} 2x-3y&=7,\\ 4x-6y&=20 \end{aligned}$ is:

$x=18,$ and $y=12$

$x=0,$ and $y=0$

There is no solution

There are an unlimited number of solutions

$x=8,$ and $y=5$

Concepts:system of equations parallel lines

Difficulty rating: 1360

Small Hint:

Compare the second left-hand side with twice the first

Big Hint:

Doubling the first equation would require the right-hand side to be $14$ , not $20$

Solution:

The left side of the second equation is exactly twice the left side of the first, but its right side is not twice $7.$ The equations would require the same expression to equal both $14$ and $20,$ which is impossible.

Thus, there is no solution, and the correct answer is C.

13.

A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off by the sides of the quadrilateral, without intersecting the sides between vertices, their sum will be:

$180^\circ$

$540^\circ$

$360^\circ$

$450^\circ$

$1080^\circ$

Concepts:inscribed angle arc angle sum

Difficulty rating: 1630

Small Hint:

Let the four consecutive arc measures be $a,b,c,d$

Big Hint:

An angle whose vertex lies in the arc of measure $a$ intercepts the other three arcs, of total measure $360^\circ-a$

Solution:

Let the four arcs have measures $a,b,c,d,$ whose sum is $360^\circ.$ The angle placed in the arc of measure $a$ intercepts the other three arcs, so it measures $\frac{360^\circ-a}{2},$ and similarly for the others. Their sum is $\begin{aligned} &\frac{4(360^\circ)-(a+b+c+d)}2\\ &\qquad=\frac{1080^\circ}{2}\\ &\qquad=540^\circ. \end{aligned}$

Thus, the correct answer is B.

14.

When simplified $\sqrt{1+\left(\dfrac{x^4-1}{2x^2}\right)^2}$ equals:

$\dfrac{x^4+2x^2-1}{2x^2}$

$\dfrac{x^4-1}{2x^2}$

$\dfrac{\sqrt{x^2+1}}2$

$\dfrac{x^2}{\sqrt2}$

$\dfrac{x^2}{2}+\dfrac1{2x^2}$

Concepts:radical perfect square algebraic manipulation

Difficulty rating: 1740

Small Hint:

Put $1$ over the denominator $4x^4$

Big Hint:

The numerator becomes $4x^4+(x^4-1)^2=(x^4+1)^2$

Solution:

For $x\ne0,$ $\begin{aligned} &1+\left(\frac{x^4-1}{2x^2}\right)^2\\ &\quad=\frac{4x^4+(x^4-1)^2}{4x^4}\\ &\quad=\frac{(x^4+1)^2}{4x^4}. \end{aligned}$ Because $x^2\gt0,$ its square root is $\frac{x^4+1}{2x^2} =\frac{x^2}{2}+\frac1{2x^2}.$

Thus, the correct answer is E.

15.

$\log125$ equals:

$100\log1.25$

$5\log3$

$3\log25$

$3-3\log2$

$(\log25)(\log5)$

Concepts:logarithm algebraic manipulation

Difficulty rating: 1280

Small Hint:

Write $125=\frac{1000}{8}$

Big Hint:

Use $\log1000=3$ and $\log8=3\log2$

Solution:

Using logarithm laws, $\begin{aligned} \log125 &=\log\left(\frac{1000}{8}\right)\\ &=3-\log(2^3)\\ &=3-3\log2. \end{aligned}$

Thus, the correct answer is D.

16.

If $f(x)=5x^2-2x-1,$ then $f(x+h)-f(x)$ equals:

$5h^2-2h$

$10xh-4x+2$

$10xh-2x-2$

$h(10x+5h-2)$

$3h$

Concepts:function polynomial factoring

Difficulty rating: 1340

Small Hint:

Substitute $x+h$ into the polynomial before subtracting $f(x)$

Big Hint:

After cancellation, factor out $h$

Solution:

Expanding and canceling gives $\begin{gathered} f(x+h)-f(x)\\ =5(x+h)^2-2(x+h)-1\\ {}-(5x^2-2x-1)\\ =10xh+5h^2-2h\\ =h(10x+5h-2). \end{gathered}$

Thus, the correct answer is D.

17.

The graph of the function $f(x)=2x^3-7$ goes:

up to the right and down to the left

down to the right and up to the left

up to the right and up to the left

down to the right and down to the left

none of these ways

Concepts:polynomial function

Difficulty rating: 1180

Small Hint:

The constant $-7$ shifts the graph vertically but does not change its end behavior

Big Hint:

A positive odd-degree leading term tends to $+\infty$ on the right and $-\infty$ on the left

Solution:

The leading term $2x^3$ controls the end behavior. As $x$ tends to $+\infty,$ $f(x)$ tends to $+\infty;$ as $x$ tends to $-\infty,$ $f(x)$ tends to $-\infty.$ Thus the graph goes up to the right and down to the left.

The correct answer is A.

18.

Of the following sets, the one that includes all values of $x$ which will satisfy $2x-3\gt7-x$ is:

$x\gt4$

$x\lt\dfrac{10}{3}$

$x=\dfrac{10}{3}$

$x\gt\dfrac{10}{3}$

$x\lt0$

Concepts:inequality linear equation

Difficulty rating: 1070

Small Hint:

Collect the $x$ -terms on one side and constants on the other

Big Hint:

Adding $x+3$ to both sides gives $3x\gt10$

Solution:

Adding $x+3$ to both sides gives $3x\gt10,$ so $x\gt\frac{10}{3}.$

Thus, the correct answer is D.

19.

If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:

are always equal to $60^\circ$

are always one obtuse angle and two unequal acute angles

are always one obtuse angle and two equal acute angles

are always acute angles

are always unequal to each other

Concepts:incircle, incenter, and inradius inscribed angle circle

Difficulty rating: 1780

Small Hint:

Join the incenter to two adjacent points of tangency

Big Hint:

The angle of the contact triangle opposite vertex angle $A$ is $90^\circ-\frac{A}{2}$

Solution:

The radii to two adjacent tangency points are perpendicular to the corresponding sides. The central angle between those radii is $180^\circ-A.$ The relevant angle of the contact triangle is half the intercepted arc, giving $90^\circ-\frac A2.$ The other two angles are $90^\circ-\frac{B}{2}$ and $90^\circ-\frac{C}{2}.$ Since every triangle angle lies strictly between $0^\circ$ and $180^\circ,$ all three are acute.

Thus, the correct answer is D.

20.

The equation $x^3+6x^2+11x+6=0$ has:

no negative real roots

no positive real roots

no real roots

$1$ positive and $2$ negative roots

$2$ positive and $1$ negative root

Concepts:polynomial factoring zero product property

Difficulty rating: 1280

Small Hint:

Test the small negative integers suggested by the constant term

Big Hint:

The polynomial factors as $(x+1)(x+2)(x+3)$

Solution:

The polynomial factors as $\begin{gathered} x^3+6x^2+11x+6\\ =(x+1)(x+2)(x+3). \end{gathered}$ Its roots are $-1,-2,-3,$ so it has no positive real roots.

Thus, the correct answer is B.

21.

The roots of the equation $2\sqrt{x}+2x^{-\frac{1}{2}}=5$ can be found by solving:

$16x^2-92x+1=0$

$4x^2-25x+4=0$

$4x^2-17x+4=0$

$2x^2-21x+2=0$

$4x^2-25x-4=0$

Concepts:radical quadratic algebraic manipulation

Difficulty rating: 1630

Small Hint:

Multiply by $\sqrt{x}$ to remove the negative exponent

Big Hint:

Isolate the radical in $2x+2=5\sqrt{x}$ before squaring

Solution:

Multiplying by $\sqrt{x}$ gives $2x+2=5\sqrt{x}.$ Squaring and simplifying, $4x^2+8x+4=25x,$ so $4x^2-17x+4=0.$

Thus, the correct answer is C.

22.

The expression $\dfrac{2x^2-x}{(x+1)(x-2)}$ ${}-\dfrac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2,$ since division by zero is not allowed. For other values of $x:$

The expression takes on many different values.

The expression has only the value $2.$

The expression has only the value $1.$

The expression always has a value between $-1$ and $2.$

The expression has a value greater than $2$ or less than $-1.$

Concepts:factoring algebraic manipulation

Difficulty rating: 1400

Small Hint:

Combine the two fractions over their common denominator

Big Hint:

The numerator factors as $2(x+1)(x-2)$

Solution:

Combining the numerators gives $\begin{gathered} 2x^2-x-(4+x)\\ =2x^2-2x-4\\ =2(x+1)(x-2). \end{gathered}$ For the allowed values $x\ne-1,2,$ the factors cancel and the expression equals $2.$

Thus, the correct answer is B.

23.

If the margin made on an article costing $C$ dollars and selling for $S$ dollars is $M=\dfrac1nC,$ then the margin is given by:

$M=\dfrac1{n-1}S$

$M=\dfrac1nS$

$M=\dfrac n{n+1}S$

$M=\dfrac1{n+1}S$

$M=\dfrac n{n-1}S$

Concepts:ratio and proportion algebraic manipulation

Difficulty rating: 1400

Small Hint:

Use $S=C+M$ together with $M=\frac{C}{n}$

Big Hint:

Express $C$ in terms of $S$ , then divide by $n$

Solution:

Since $S=C+M=C+\frac{C}{n},$ $\begin{aligned} S&=\frac{n+1}{n}C,\\ C&=\frac n{n+1}S. \end{aligned}$ Therefore $M=\frac Cn=\frac1{n+1}S.$

Thus, the correct answer is D.

24.

The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are:

$9$ and $-7$

only $-7$

$9$ and $7$

$-9$ and $-7$

only $9$

Concepts:quadratic perfect square

Difficulty rating: 1450

Small Hint:

Combine the linear terms to write the coefficient of $x$ as $1-k$

Big Hint:

Set the discriminant $(1-k)^2-64$ equal to zero

Solution:

A repeated real root requires $(1-k)^2-4(2)(8)=0.$ Thus $(1-k)^2=64,$ so $1-k=\pm8.$ The two values are $k=9$ and $k=-7.$

Thus, the correct answer is A.

25.

The two roots of the equation $\begin{aligned} a(b-c)x^2+b(c-a)x\\ {}+c(a-b)&=0 \end{aligned}$ are $1$ and:

$\dfrac{b(c-a)}{a(b-c)}$

$\dfrac{a(b-c)}{c(a-b)}$

$\dfrac{a(b-c)}{b(c-a)}$

$\dfrac{c(a-b)}{a(b-c)}$

$\dfrac{c(a-b)}{b(c-a)}$

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1630

Small Hint:

Use the product of the roots rather than solving the quadratic

Big Hint:

The product is the constant coefficient divided by the leading coefficient

Solution:

By Vieta’s formulas, the product of the two roots is $\frac{c(a-b)}{a(b-c)}.$ Since one root is $1,$ the other root equals this product.

Thus, the correct answer is D.

26.

The straight line $AB$ is divided at $C$ so that $\overline{AC}=3\overline{CB}.$ Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $\overline{AB}$ produced at $D.$ Then $\overline{BD}$ equals:

the diameter of the smaller circle

the radius of the smaller circle

the radius of the larger circle

$\overline{CB}\sqrt3$

the difference of the two radii

Concepts:tangent circles tangent line homothety

Difficulty rating: 1700

Small Hint:

The point $D$ is the external center of similitude of the two circles

Big Hint:

Let $\overline{CB}=u$ and compare the distances from $D$ to the two centers in the ratio $3:1$

Solution:

Let $\overline{CB}=u,$ so $\overline{AC}=3u$ and $\overline{AB}=4u.$ Measured from $A,$ the circle centers are at $\frac{3u}{2}$ and $\frac{7u}{2},$ and their radii are in the ratio $3:1.$ If $AD=z,$ the common external tangent makes $D$ the external center of similitude, so $\frac{z-\frac32u}{z-\frac72u}=3.$ Hence $z=\frac{9u}{2},$ and $BD=z-4u=\frac u2,$ which is the radius of the smaller circle.

Thus, the correct answer is B.

27.

A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:

$\dfrac11$

$\dfrac12$

$\dfrac23$

$\dfrac21$

$\sqrt{\dfrac54}$

Concepts:cone sphere volume

Difficulty rating: 1280

Small Hint:

Write both volumes using their common radius $r$

Big Hint:

Set $\frac13\pi r^2h$ equal to half of $\frac43\pi r^3$

Solution:

The volume condition is $\frac13\pi r^2h=\frac12\left(\frac43\pi r^3\right).$ Cancelling the common factors gives $h=2r,$ so the requested ratio is $2:1.$

Thus, the correct answer is D.

28.

If $\dfrac mn=\dfrac43$ and $\dfrac rt=\dfrac9{14},$ the value of $\dfrac{3mr-nt}{4nt-7mr}$ is:

$-5\dfrac12$

$-\dfrac{11}{14}$

$-1\dfrac14$

$\dfrac{11}{14}$

$-\dfrac23$

Concepts:ratio and proportion fraction substitution

Difficulty rating: 1590

Small Hint:

The given ratios imply $mr:nt=(\frac{m}{n})(\frac{r}{t})$

Big Hint:

Write $mr=6k$ and $nt=7k$

Solution:

Multiplying the two given ratios, $\frac{mr}{nt}=\frac43\cdot\frac9{14}=\frac67.$ Put $mr=6k$ and $nt=7k.$ Then $\begin{aligned} \frac{3mr-nt}{4nt-7mr} &=\frac{18k-7k}{28k-42k}\\ &=-\frac{11}{14}. \end{aligned}$

Thus, the correct answer is B.

29.

If the ratio of the legs of a right triangle is $1:2,$ then the ratio of the corresponding segments of the hypotenuse made by a perpendicular upon it from the vertex is:

$1:4$

$1:\sqrt2$

$1:2$

$1:\sqrt5$

$1:5$

Concepts:right triangle altitude similarity

Difficulty rating: 1420

Small Hint:

The two hypotenuse segments are the projections of the legs

Big Hint:

By similarity, their ratio is the ratio of the squares of the corresponding legs

Solution:

If the legs have lengths $a$ and $b,$ their projections on the hypotenuse have lengths $\frac{a^2}{c}$ and $\frac{b^2}{c}.$ Their ratio is therefore $a^2:b^2.$ With $a:b=1:2,$ this is $1:4.$

Thus, the correct answer is A.

30.

$A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\dfrac25$ days. The number of days required for $A$ to do the job alone is:

$1$

$3$

$6$

$12$

$2.8$

Concepts:rate system of equations

Difficulty rating: 1740

Small Hint:

Let $a,b,c$ be the fractions of the job completed per day

Big Hint:

Add $a+b=\frac{1}{2}$ and $a+c=\frac{5}{12}$ , then subtract $b+c=\frac{1}{4}$

Solution:

Let $a,b,c$ be the individual daily rates. Then $\begin{aligned} a+b&=\frac12,\\ b+c&=\frac14,\\ a+c&=\frac5{12}. \end{aligned}$ Thus $\begin{aligned} 2a&=(a+b)+(a+c)\\ &\quad{}-(b+c)\\ &=\frac12+\frac5{12}-\frac14\\ &=\frac23, \end{aligned}$ so $a=\frac{1}{3}.$ Therefore $A$ needs $3$ days.

Thus, the correct answer is B.

31.

In triangle $ABC,$ $\overline{AB}=\overline{AC},$ $\angle A=40^\circ.$ Point $O$ is within the triangle with $\angle OBC\cong\angle OCA.$ The number of degrees in angle $BOC$ is:

$110$

$35$

$140$

$55$

$70$

Concepts:isosceles triangle angle chasing angle sum

Difficulty rating: 1780

Small Hint:

The base angles of triangle $ABC$ are both $70^\circ$

Big Hint:

If the equal angles are $x,$ then the other part of angle $C$ is $70^\circ-x$

Solution:

The base angles of the isosceles triangle are $\angle ABC=\angle BCA=70^\circ.$ Put $\angle OBC=\angle OCA=x.$ Then $\angle OCB=70^\circ-x.$ In triangle $BOC,$ $\begin{aligned} \angle BOC &=180^\circ-x\\ &\quad{}-(70^\circ-x)\\ &=110^\circ. \end{aligned}$

Thus, the correct answer is A.

32.

The factors of $x^4+64$ are:

$(x^2+8)^2$

$(x^2+8)(x^2-8)$

$(x^2+2x+4)(x^2-8x+16)$

$(x^2-4x+8)(x^2-4x-8)$

$(x^2-4x+8)(x^2+4x+8)$

Concepts:factoring difference of squares

Difficulty rating: 1420

Small Hint:

Add and subtract $16x^2$ to form a difference of squares

Big Hint:

Write $x^4+64=(x^2+8)^2-(4x)^2$

Solution:

Using a difference of squares, $\begin{aligned} x^4+64 &=(x^2+8)^2-16x^2\\ &=(x^2-4x+8)\\ &\quad{}\cdot(x^2+4x+8). \end{aligned}$

Thus, the correct answer is E.

33.

A bank charges $\$6$ for a loan of $\$120.$ The borrower receives $\$114$ and repays the loan in $12$ installments of $\$10$ a month. The interest rate is approximately:

$5\%$

$6\%$

$7\%$

$9\%$

$15\%$

Concepts:money percentage mean

Difficulty rating: 1630

Small Hint:

The $\$6$ charge is interest on a balance that decreases by $\$10$ each month

Big Hint:

Approximate the average outstanding principal by the average of $\$120$ and $\$10$

Solution:

The balance decreases in equal $\$10$ steps from $\$120$ to $\$10,$ so its average outstanding value is approximately $\frac{120+10}{2}=65.$ The annual charge is therefore about $\frac6{65}\cdot100\%\approx9.2\%.$ The nearest choice is $9\%.$

Thus, the correct answer is D.

34.

The fraction $\dfrac13:$

equals $0.33333333$

is less than $0.33333333$ by $\dfrac1{3\cdot10^8}$

is less than $0.33333333$ by $\dfrac1{3\cdot10^9}$

is greater than $0.33333333$ by $\dfrac1{3\cdot10^8}$

is greater than $0.33333333$ by $\dfrac1{3\cdot10^9}$

Concepts:decimal fraction place value

Difficulty rating: 1630

Small Hint:

Write $0.33333333$ as $\frac{33333333}{10^8}$

Big Hint:

Subtract this terminating decimal from $\frac{1}{3}$ using the denominator $3\cdot10^8$

Solution:

Exactly, $\begin{gathered} \frac13-\frac{33333333}{10^8}\\ =\frac{10^8-3(33333333)} {3\cdot10^8}\\ =\frac1{3\cdot10^8}. \end{gathered}$ Thus $\frac{1}{3}$ is greater by the stated amount.

Thus, the correct answer is D.

35.

In the right triangle shown the sum of the distances $\overline{BM}$ and $\overline{MA}$ is equal to the sum of the distances $\overline{BC}$ and $\overline{CA}.$ If $\overline{MB}=x,$ $\overline{CB}=h,$ and $\overline{CA}=d,$ then $x$ equals:

$\dfrac{hd}{2h+d}$

$d-h$

$\dfrac12d$

$h+d-\sqrt{2d}$

$\sqrt{h^2+d^2}-h$

Concepts:right triangle Pythagorean Theorem radical

Difficulty rating: 1740

Small Hint:

Use the Pythagorean theorem to express $\overline{MA}$ in terms of $d,h,x$

Big Hint:

From $x+\sqrt{d^2+(h+x)^2}=h+d,$ isolate the radical and square

Solution:

The condition and the Pythagorean theorem give $x+\sqrt{d^2+(h+x)^2}=h+d.$ After isolating the radical and squaring, $d^2+(h+x)^2=(h+d-x)^2.$ Cancelling common terms leaves $4hx+2dx=2hd,$ so $x=\frac{hd}{2h+d}.$

Thus, the correct answer is A.

36.

A boat has a speed of $15$ mph in still water. In a stream that has a current of $5$ mph it travels a certain distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:

$\dfrac54$

$\dfrac11$

$\dfrac89$

$\dfrac78$

$\dfrac98$

Concepts:distance rate and time relative speed harmonic mean

Difficulty rating: 1580

Small Hint:

The downstream and upstream speeds are $20$ mph and $10$ mph

Big Hint:

For equal distances, divide twice the distance by $\frac{d}{20}+\frac{d}{10}$

Solution:

For a one-way distance $d,$ the total time is $\frac d{20}+\frac d{10}=\frac{3d}{20}.$ Thus the round-trip average speed is $\frac{2d}{\frac{3d}{20}}=\frac{40}{3}.$ Its ratio to $15$ is $\frac{\frac{40}{3}}{15}=\frac89.$

Thus, the correct answer is C.

37.

Given triangle $PQR$ with $\overline{RS}$ bisecting $\angle R,$ $\overline{PQ}$ extended to $D,$ and $\angle n$ a right angle, then:

$\angle m=\dfrac12(\angle p-\angle q)$

$\angle m=\dfrac12(\angle p+\angle q)$

$\angle d=\dfrac12(\angle q+\angle p)$

$\angle d=\dfrac12\angle m$

none of these is correct

Concepts:angle bisector angle chasing angle sum

Difficulty rating: 1780

Small Hint:

The bisected angle at $R$ measures $180^\circ-\angle p-\angle q$ before it is halved

Big Hint:

Since the transversal is perpendicular to the angle bisector, $\angle m$ complements half of $\angle R$

Solution:

The angle at $R$ is $180^\circ-\angle p-\angle q.$ Because $\overline{RS}$ bisects this angle, the angle between $\overline{PR}$ and $\overline{RS}$ is $90^\circ-\frac12(\angle p+\angle q).$ The line forming $\angle m$ is perpendicular to $\overline{RS},$ so $\angle m=\frac12(\angle p+\angle q).$

Thus, the correct answer is B.

38.

If $\log 2=0.3010$ and $\log 3=0.4771,$ the value of $x$ when $3^{x+3}=135$ is approximately:

$5$

$1.47$

$1.67$

$1.78$

$1.63$

Concepts:logarithm exponent estimation

Difficulty rating: 1340

Small Hint:

Divide by $3^3$ to obtain $3^x=5$

Big Hint:

Use $\log5=1-\log2$ and divide by $\log3$

Solution:

The equation reduces to $3^x=5,$ so $\begin{aligned} x&=\frac{\log5}{\log3}\\ &=\frac{1-\log2}{\log3}\\ &=\frac{0.6990}{0.4771}\\ &\approx1.47. \end{aligned}$

Thus, the correct answer is B.

39.

The locus of the midpoint of a line segment that is drawn from a given external point $P$ to a given circle with center $O$ and radius $r,$ is:

a straight line perpendicular to $\overline{PO}$

a straight line parallel to $\overline{PO}$

a circle with center $P$ and radius $r$

a circle with center at the midpoint of $\overline{PO}$ and radius $2r$

a circle with center at the midpoint of $\overline{PO}$ and radius $\dfrac12r$

Concepts:transformation midpoint circle

Difficulty rating: 1630

Small Hint:

Let the variable endpoint on the given circle be $X$ and its midpoint with $P$ be $M$

Big Hint:

The map $X\mapsto M$ is a dilation centered at $P$ with scale factor $\frac{1}{2}$

Solution:

As the endpoint $X$ moves on the circle centered at $O,$ its midpoint $M$ with the fixed point $P$ is the image of $X$ under a dilation of scale $\frac{1}{2}$ centered at $P.$ Therefore the locus is a circle whose center is the midpoint of $\overline{PO}$ and whose radius is $\frac{r}{2}.$

Thus, the correct answer is E.

40.

If $\left(a+\dfrac1a\right)^2=3,$ then $a^3+\dfrac1{a^3}$ equals:

$\dfrac{10\sqrt3}{3}$

$3\sqrt3$

$0$

$7\sqrt7$

$6\sqrt3$

Concepts:algebraic manipulation sum and difference of cubes substitution

Difficulty rating: 1670

Small Hint:

Set $t=a+\frac{1}{a}$ , so the given condition is $t^2=3$

Big Hint:

Use $a^3+a^{-3}=t^3-3t$

Solution:

Let $t=a+\frac{1}{a}.$ The standard cubic identity gives $a^3+\frac1{a^3}=t^3-3t=t(t^2-3).$ Since $t^2=3,$ this expression is $0.$

Thus, the correct answer is C.

41.

The sum of all the roots of $4x^3-8x^2-63x-9=0$ is:

$8$

$2$

$-8$

$-2$

$0$

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1260

Small Hint:

Use Vieta’s formula for the sum of the roots of a cubic

Big Hint:

Negate the coefficient of $x^2$ and divide by the coefficient of $x^3$

Solution:

For $ax^3+bx^2+\cdots=0,$ the sum of the roots is $-\frac{b}{a}.$ Here it is $-\frac{-8}{4}=2.$

Thus, the correct answer is B.

42.

Consider the graphs of $(1)$ $y=x^2-\dfrac12x+2$ and $(2)$ $y=x^2+\dfrac12x+2$ on the same set of axes. These parabolas have exactly the same shape. Then:

the graphs coincide.

the graph of $(1)$ is lower than the graph of $(2).$

the graph of $(1)$ is to the left of the graph of $(2).$

the graph of $(1)$ is to the right of the graph of $(2).$

the graph of $(1)$ is higher than the graph of $(2).$

Concepts:parabola completing the square transformation

Difficulty rating: 1400

Small Hint:

Find the $x$ -coordinate of each vertex using $-\frac{b}{2a}$

Big Hint:

Graph $(1)$ has vertex $x=\frac{1}{4}$ , whereas graph $(2)$ has vertex $x=-\frac{1}{4}$

Solution:

The vertex of $y=x^2+bx+2$ has $x$ -coordinate $-\frac{b}{2}.$ Thus graph $(1)$ has its vertex at $x=\frac{1}{4},$ while graph $(2)$ has its vertex at $x=-\frac{1}{4}.$ Their vertex heights are equal, so graph $(1)$ is the same parabola shifted to the right.

Thus, the correct answer is D.

43.

The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:

$15$

$22$

$24$

$26$

$30$

Concepts:right triangle incircle, incenter, and inradius perimeter

Difficulty rating: 1590

Small Hint:

For a right triangle with legs $a,b$ and hypotenuse $c,$ the inradius is $\frac{a+b-c}{2}$

Big Hint:

Substitute $r=1$ and $c=10$ to find $a+b$

Solution:

For a right triangle, $r=\frac{a+b-c}{2}.$ With $r=1$ and $c=10,$ this gives $a+b=12.$ Hence the perimeter is $a+b+c=12+10=22.$

Thus, the correct answer is B.

44.

A man born in the first half of the nineteenth century was $x$ years old in the year $x^2.$ He was born in:

$1849$

$1825$

$1812$

$1836$

$1806$

Concepts:ages perfect square date and time

Difficulty rating: 1670

Small Hint:

His birth year is $x^2-x$

Big Hint:

Use the requirement that $x^2$ is a nineteenth-century year and test the nearby integer $x$

Solution:

The year $x^2$ must lie in the nineteenth century, so $x=43$ because $43^2=1849$ while the neighboring squares fall outside the relevant range. His birth year is therefore $x^2-x=1849-43=1806,$ which is in the first half of the century.

Thus, the correct answer is E.

45.

In a rhombus $ABCD,$ line segments are drawn within the rhombus, parallel to diagonal $\overline{BD},$ and terminated in the sides of the rhombus. A graph is drawn showing the length of a segment as a function of its distance from vertex $A.$ The graph is:

a straight line passing through the origin.

a straight line cutting across the upper right quadrant.

two line segments forming an upright $V.$

two line segments forming an inverted $V.$

none of these.

Concepts:rhombus diagonal data and graph interpretation

Difficulty rating: 1630

Small Hint:

Imagine moving a segment parallel to $\overline{BD}$ from $A$ toward the opposite vertex

Big Hint:

Its length grows linearly to the full diagonal and then decreases linearly to zero

Solution:

Starting at $A,$ the parallel cross section has length $0$ and grows linearly until it reaches the diagonal $\overline{BD}.$ Continuing toward the opposite vertex, its length decreases linearly back to $0.$ The graph is therefore made of two line segments forming an inverted $V.$

Thus, the correct answer is D.

46.

In the diagram, if points $A,$ $B,$ and $C$ are points of tangency, then $x$ equals:

$\dfrac3{16}\text{ in.}$

$\dfrac18\text{ in.}$

$\dfrac1{32}\text{ in.}$

$\dfrac3{32}\text{ in.}$

$\dfrac1{16}\text{ in.}$

Concepts:tangent circles angle bisector special right triangle

Difficulty rating: 1630

Small Hint:

The marked $\frac{3}{8}$ -inch measure is the circle’s diameter

Big Hint:

With radius $\frac{3}{16},$ the center is $\frac{3}{8}$ inch above the $60^\circ$ vertex because $\sin30^\circ=\frac{1}{2}$

Solution:

The radius is $\frac{3}{16}$ inch. The center lies on the angle bisector. The perpendicular radius to either sloping side and the segment from the vertex to the center form a right triangle with a $30^\circ$ angle, so the center is $\frac{\frac{3}{16}}{\sin30^\circ}=\frac38$ inch above the vertex. Hence the top tangent is $\frac{3}{8}+\frac{3}{16}=\frac{9}{16}$ inch above the vertex. Since the ledge is $\frac{1}{2}$ inch above the vertex, $x=\frac9{16}-\frac12=\frac1{16}\text{ in.}$

Thus, the correct answer is E.

47.

At the midpoint of line segment $\overline{AB}$ which is $p$ units long, a perpendicular $\overline{MR}$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $\dfrac12\overline{AB},$ meeting $\overline{AB}$ at $T.$ Then $\overline{AT}$ and $\overline{TB}$ are the roots of:

$x^2+px+q^2=0$

$x^2-px+q^2=0$

$x^2+px-q^2=0$

$x^2-px-q^2=0$

$x^2-px+q=0$

Concepts:Pythagorean Theorem Vieta’s Formulas circle

Difficulty rating: 1740

Small Hint:

The sum $\overline{AT}+\overline{TB}$ is $p$

Big Hint:

Use $RT=\frac{p}{2}=AM$ in right triangle $RMT$ to show $\overline{AT}\cdot\overline{TB}=q^2$

Solution:

Let $MT=t.$ Since $AM=BM=RT=\frac{p}{2},$ the Pythagorean theorem gives $t^2+q^2=\frac{p^2}{4}.$ Also, $AT=\frac p2+t,\qquad TB=\frac p2-t.$ Their sum is $p,$ and their product is $\frac{p^2}{4}-t^2=q^2.$ Therefore they are the roots of $x^2-px+q^2=0.$

Thus, the correct answer is B.

48.

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\dfrac34$ of its former rate and arrives $3\dfrac12$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in miles was:

$400$

$465$

$600$

$640$

$550$

Concepts:distance rate and time relative speed system of equations

Difficulty rating: 2310

Small Hint:

Let the normal speed be $v$ and the trip length be $L$ ; slowing to $\frac{3v}{4}$ adds one-third of the normal time on the affected distance

Big Hint:

Use the two late-arrival equations to find first $\frac{L}{v}$ , then $\frac{90}{v}$

Solution:

Let the normal speed be $v$ and the trip length be $L.$ Traveling a distance at $\frac{3v}{4}$ rather than $v$ adds one-third of its normal travel time. In the first case, $\frac12+\frac13\left(\frac Lv-1\right)=\frac72,$ so $\frac{L}{v}=10.$ If the accident happens $90$ miles farther along, $\frac12+\frac13\left(\frac Lv-1-\frac{90}{v}\right)=3.$ Substituting $\frac{L}{v}=10$ gives $\frac{90}{v}=\frac{3}{2},$ hence $v=60$ and $L=10v=600.$

Thus, the correct answer is C.

49.

The difference of the squares of two odd numbers is always divisible by $8.$ If $a>b,$ and $2a+1$ and $2b+1$ are the odd numbers, to prove the given statement we put the difference of the squares in the form:

$(2a+1)^2-(2b+1)^2$

$4a^2-4b^2+4a-4b$

$4[a(a+1)-b(b+1)]$

$4(a-b)(a+b+1)$

$4(a^2+a-b^2-b)$

Concepts:divisibility parity factoring

Difficulty rating: 1590

Small Hint:

Expand the two squares and group each variable with its successor

Big Hint:

Each product $a(a+1)$ and $b(b+1)$ is even

Solution:

Expanding and regrouping, $\begin{aligned} &(2a+1)^2-(2b+1)^2\\ &\qquad=4[a(a+1)-b(b+1)]. \end{aligned}$ Each of $a(a+1)$ and $b(b+1)$ is even, so their difference is even. The displayed expression is consequently divisible by $4\cdot2=8.$

Thus, the correct answer is C.

50.

The times between $7$ and $8$ o’clock, correct to the nearest minute, when the hands of a clock will form an angle of $84$ degrees are:

$7{:}23$ and $7{:}53$

$7{:}20$ and $7{:}50$

$7{:}22$ and $7{:}53$

$7{:}23$ and $7{:}52$

$7{:}21$ and $7{:}49$

Concepts:clock linear equation estimation

Difficulty rating: 1870

Small Hint:

At $t$ minutes after $7{:}00,$ the signed separation of the hands is $210^\circ-5.5t^\circ$

Big Hint:

Solve $\lvert210-5.5t\rvert=84$ for both values of $t$

Solution:

At $t$ minutes after $7{:}00,$ the hour hand is at $210^\circ+0.5t^\circ$ and the minute hand is at $6t^\circ.$ Thus $|210-5.5t|=84.$ The two solutions are $\begin{aligned} t&=\frac{126}{5.5}\approx22.91,\\ t&=\frac{294}{5.5}\approx53.45. \end{aligned}$ To the nearest minute, the times are $7{:}23$ and $7{:}53.$

Thus, the correct answer is A.