1952 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

If the radius of a circle is a rational number, its area is given by a number which is:

Rational

Irrational

Integral

A perfect square

None of these

Answer: B

Concepts:circle circle area

Difficulty rating: 920

Small Hint:

Write the area in terms of the radius and $\pi$

Big Hint:

A circle’s radius is positive, so its rational square is nonzero

Solution:

If the positive rational radius is $r,$ then its area is $\pi r^2.$ The factor $r^2$ is a nonzero rational number, and a nonzero rational multiple of the irrational number $\pi$ is irrational.

Thus, the correct answer is B.

2.

Two high school classes took the same test. One class of $20$ students made an average grade of $80\%$ ; the other class of $30$ students made an average grade of $70\%.$ The average grade for all students in both classes is:

$75\%$

$74\%$

$72\%$

$77\%$

None of these

Answer: B

Concepts:weighted mean percentage

Difficulty rating: 1070

Small Hint:

Weight each class average by the number of students in that class

Big Hint:

The combined average is $\dfrac{20(80)+30(70)}{50}$

Solution:

The two classes earned a combined total of $20(80)+30(70)=3700$ percentage points among $50$ students. Their combined average is $\frac{3700}{50}=74\%.$

Thus, the correct answer is B.

3.

The expression $a^3-a^{-3}$ equals:

$\left(a-\dfrac1a\right)\left(a^2+1+\dfrac1{a^2}\right)$

$\left(\dfrac1a-a\right)\left(a^2-1+\dfrac1{a^2}\right)$

$\left(a-\dfrac1a\right)\left(a^2-2+\dfrac1{a^2}\right)$

$\left(\dfrac1a-a\right)\left(\dfrac1{a^2}+1+a^2\right)$

None of these

Answer: A

Concepts:sum and difference of cubes exponent factoring

Difficulty rating: 1430

Small Hint:

View the expression as a difference of two cubes

Big Hint:

Use $u^3-v^3=(u-v)(u^2+uv+v^2)$ with $v=\frac{1}{a}$

Solution:

Apply the difference-of-cubes identity with $u=a$ and $v=\frac{1}{a}.$ Since $uv=1,$ $\begin{aligned} a^3-\frac1{a^3} &=\left(a-\frac1a\right)\\ &\quad{}\cdot\left(a^2+1+\frac1{a^2}\right). \end{aligned}$

Thus, the correct answer is A.

4.

The cost $C$ of sending a parcel post package weighing $P$ pounds, $P$ an integer, is $10$ cents for the first pound and $3$ cents for each additional pound. The formula for the cost is:

$C=10+3P$

$C=10P+3$

$C=10+3(P-1)$

$C=9+3P$

$C=10P-7$

Answer: C

Concepts:money linear equation

Difficulty rating: 960

Small Hint:

Only the pounds after the first are charged at the additional-pound rate

Big Hint:

A $P$ -pound package has $P-1$ additional pounds

Solution:

The first pound costs $10$ cents. The remaining $P-1$ pounds cost $3(P-1)$ cents, so $C=10+3(P-1).$

Thus, the correct answer is C.

5.

The points $(6,12)$ and $(0,-6)$ are connected by a straight line. Another point on this line is:

$(3,3)$

$(2,1)$

$(7,16)$

$(-1,-4)$

$(-3,-8)$

Answer: A

Concepts:coordinate geometry slope

Difficulty rating: 1100

Small Hint:

Find the slope through the two given points

Big Hint:

The line has equation $y=3x-6$

Solution:

The slope is $\frac{12-(-6)}{6-0}=3,$ so the line is $y=3x-6.$ Substituting $x=3$ gives $y=3.$

Thus, the correct answer is A.

6.

The difference of the roots of $x^2-7x-9=0$ is:

$+7$

$+\dfrac72$

$+9$

$2\sqrt{85}$

$\sqrt{85}$

Answer: E

Concepts:quadratic

Difficulty rating: 1540

Small Hint:

Write the two roots using the quadratic formula

Big Hint:

Their separation is the square root of the discriminant because the leading coefficient is $1$

Solution:

The roots are $\dfrac{7+\sqrt{49+36}}2$ and $\dfrac{7-\sqrt{49+36}}2.$ Their positive difference is therefore $\sqrt{85}.$

Thus, the correct answer is E.

7.

When simplified, $(x^{-1}+y^{-1})^{-1}$ is equal to:

$x+y$

$\dfrac{xy}{x+y}$

$xy$

$\dfrac1{xy}$

$\dfrac{x+y}{xy}$

Answer: B

Concepts:exponent algebraic manipulation

Difficulty rating: 1290

Small Hint:

First combine $\frac{1}{x}+\frac{1}{y}$ into one fraction

Big Hint:

The exponent $-1$ then takes the reciprocal

Solution:

Combining the terms and then taking the reciprocal gives $\begin{aligned} \left(\frac1x+\frac1y\right)^{-1} &=\left(\frac{x+y}{xy}\right)^{-1}\\ &=\frac{xy}{x+y}. \end{aligned}$

Thus, the correct answer is B.

8.

Two equal circles in the same plane cannot have the following number of common tangents:

$1$

$2$

$3$

$4$

None of these

Answer: A

Concepts:circle tangent line casework

Difficulty rating: 1430

Small Hint:

Consider intersecting, externally tangent, and disjoint equal circles

Big Hint:

An internal tangency would require unequal radii unless the circles coincide

Solution:

Two distinct equal circles have $2$ common tangents when they intersect, $3$ when externally tangent, and $4$ when disjoint. They cannot be internally tangent, because internal tangency requires the center distance to equal the difference of the radii, which is $0.$ At center distance $0$ the equal circles coincide and have infinitely many common tangents.

Thus exactly $1$ common tangent is impossible, so the correct answer is A.

9.

If $m=\dfrac{cab}{a-b},$ then $b$ equals:

$\dfrac{m(a-b)}{ca}$

$\dfrac{cab-ma}{-m}$

$\dfrac1{1+c}$

$\dfrac{ma}{m+ca}$

$\dfrac{m+ca}{ma}$

Answer: D

Concepts:rational equation algebraic manipulation

Difficulty rating: 1580

Small Hint:

Clear the denominator before collecting the terms containing $b$

Big Hint:

The equation becomes $ma=b(m+ca)$

Solution:

Clearing the denominator gives $m(a-b)=cab.$ Hence $ma=b(m+ca)$ , so $b=\frac{ma}{m+ca}.$

Thus, the correct answer is D.

10.

An automobile went up a hill at a speed of $10$ miles an hour and down the same distance at a speed of $20$ miles an hour. The average speed for the round trip was:

$12\dfrac12$ mph

$13\dfrac13$ mph

$14\dfrac12$ mph

$15$ mph

None of these

Answer: B

Concepts:distance rate and time harmonic mean

Difficulty rating: 1240

Small Hint:

Use equal distances rather than averaging the two speeds directly

Big Hint:

For a one-mile trip each way, divide the total distance $2$ by the total time $\frac{1}{10}+\frac{1}{20}$

Solution:

For one mile in each direction, the total time is $\frac1{10}+\frac1{20}=\frac3{20}$ hour. Thus the average speed is $\frac{2}{\frac{3}{20}}=\frac{40}{3} =13\frac13\text{ mph}.$

Thus, the correct answer is B.

11.

If $y=f(x)=\dfrac{x+2}{x-1},$ then it is incorrect to say:

$x=\dfrac{y+2}{y-1}$

$f(0)=-2$

$f(1)=0$

$f(-2)=0$

$f(y)=x$

Answer: C

Concepts:function rational equation

Difficulty rating: 1580

Small Hint:

Check the domain of the rational function before evaluating it

Big Hint:

The denominator vanishes at one of the listed inputs

Solution:

The function is undefined at $x=1$ because its denominator is zero there, so $f(1)=0$ is false. The other statements hold: $f(0)=-2,$ $f(-2)=0,$ solving $y=\frac{x+2}{x-1}$ gives $x=\frac{y+2}{y-1},$ and this same formula shows that $f$ is its own inverse.

Thus, the correct answer is C.

12.

The sum to infinity of the terms of an infinite geometric progression is $6.$ The sum of the first two terms is $4\dfrac12.$ The first term of the progression is:

$3$ or $1\dfrac12$

$1$

$2\dfrac12$

$6$

$9$ or $3$

Answer: E

Concepts:geometric sequence quadratic

Difficulty rating: 1770

Small Hint:

Let the first term be $a$ and the common ratio be $r$

Big Hint:

Use $a=6(1-r)$ in the equation $a(1+r)=\frac{9}{2}$

Solution:

Let the first term be $a$ and the common ratio be $r.$ The infinite sum gives $a=6(1-r).$ The first two terms therefore satisfy $a(1+r)=6(1-r^2)=\frac92.$ Thus $r^2=\frac{1}{4},$ so $r=\frac{1}{2}$ gives $a=3,$ while $r=-\frac{1}{2}$ gives $a=9.$

Thus, the correct answer is E.

13.

The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:

$x=-p$

$x=\dfrac p2$

$x=-2p$

$x=\dfrac{p^2}{4q}$

$x=-\dfrac p2$

Answer: E

Concepts:quadratic completing the square

Difficulty rating: 1470

Small Hint:

Complete the square in the terms $x^2+px$

Big Hint:

The squared term is minimized when $x+\frac{p}{2}=0$

Solution:

Completing the square gives $\begin{aligned} x^2+px+q &=\left(x+\frac p2\right)^2\\ &\quad{}+q-\frac{p^2}{4}. \end{aligned}$ The square is least when $x=-\frac{p}{2}.$

Thus, the correct answer is E.

14.

A house and store were sold for $\$12000$ each. The house was sold at a loss of $20\%$ of the cost, and the store at a gain of $20\%$ of the cost. The entire transaction resulted in:

No loss or gain

Loss of $\$1000$

Gain of $\$1000$

Gain of $\$2000$

None of these

Answer: B

Concepts:percentage money

Difficulty rating: 1310

Small Hint:

Recover each original cost from its selling price separately

Big Hint:

The house sold for $80\%$ of cost, while the store sold for $120\%$ of cost

Solution:

The house cost $\frac{12000}{0.8}=15000$ dollars, and the store cost $\frac{12000}{1.2}=10000$ dollars. The combined cost was $25000$ dollars, while the combined selling price was $24000$ dollars, producing a loss of $1000$ dollars.

Thus, the correct answer is B.

15.

The sides of a triangle are in the ratio $6:8:9.$ Then:

The triangle is obtuse

The angles are in the ratio $6:8:9$

The triangle is acute

The angle opposite the largest side is double the angle opposite the smallest side

None of these

Answer: C

Concepts:Pythagorean Theorem triangle inequality

Difficulty rating: 1410

Small Hint:

Compare the square of the largest side with the sum of the squares of the other two

Big Hint:

Here $9^2$ is less than $6^2+8^2$

Solution:

For a triangle with largest side $9,$ $9^2=81\lt 6^2+8^2=100.$ By the converse of the Pythagorean inequality, the angle opposite side $9$ is acute. Since it is the largest angle, every angle is acute.

Thus, the correct answer is C.

16.

If the base of a rectangle is increased by $10\%$ and the area is unchanged, then the altitude is decreased by:

$9\%$

$10\%$

$11\%$

$11\dfrac19\%$

$9\dfrac1{11}\%$

Answer: E

Concepts:percentage area ratio and proportion

Difficulty rating: 1290

Small Hint:

The new altitude must be divided by the same factor by which the base was multiplied

Big Hint:

A $10\%$ base increase multiplies the base by $\frac{11}{10}$

Solution:

To keep the area fixed, the altitude is multiplied by $\frac{10}{11}.$ Its fractional decrease is $1-\frac{10}{11}=\frac1{11},$ which is $\frac{100}{11}\%=9\dfrac1{11}\%.$

Thus, the correct answer is E.

17.

A merchant bought some goods at a discount of $20\%$ of the list price. He wants to mark them at such a price that he can give a discount of $20\%$ of the marked price and still make a profit of $20\%$ of the selling price. The percent of the list price at which he should mark them is:

$20$

$100$

$125$

$80$

$120$

Answer: C

Concepts:percentage money algebraic manipulation

Difficulty rating: 1740

Small Hint:

Normalize the list price to $1$ and distinguish cost, marked price, and selling price

Big Hint:

A profit equal to $20\%$ of selling price means the cost is $80\%$ of selling price

Solution:

Let the list price be $1.$ The merchant’s cost is $0.8.$ If the selling price is $S,$ then $S-0.8=0.2S,$ so $S=1.$ If the marked price is $M,$ the $20\%$ discount gives $0.8M=S=1,$ hence $M=1.25.$

The marked price must therefore be $125\%$ of the list price, so the correct answer is C.

18.

$\log p+\log q=\log(p+q)$ only if:

$p=q=0$

$p=\dfrac{q^2}{1-q}$

$p=q=1$

$p=\dfrac q{q-1}$

$p=\dfrac q{q+1}$

Answer: D

Concepts:logarithm rational equation

Difficulty rating: 1640

Small Hint:

Combine the logarithms on the left before equating their positive arguments

Big Hint:

Solve $pq=p+q$ for $p$

Solution:

The logarithm product rule changes the equation to $\log(pq)=\log(p+q),$ so the positive arguments must satisfy $pq=p+q.$ Therefore $p(q-1)=q,$ giving $p=\frac{q}{q-1}.$ Its admissible values have $q\gt1,$ which also keeps all logarithm arguments positive.

Thus, the correct answer is D.

19.

Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE,$ which meet $AC$ at $D$ and $E,$ respectively. Then:

$\dfrac{AD}{EC}=\dfrac{AE}{DC}$

$\dfrac{AD}{EC}=\dfrac{AB}{BC}$

$\dfrac{AD}{EC}=\dfrac{BD}{BE}$

$\dfrac{AD}{EC}=\dfrac{AB\cdot BD}{BE\cdot BC}$

$\dfrac{AD}{EC}=\dfrac{AE\cdot BD}{DC\cdot BE}$

Answer: D

Concepts:area ratio angle chasing

Difficulty rating: 2060

Small Hint:

Compare the areas of triangles $ABD$ and $EBC$ in two different ways

Big Hint:

They have bases on $AC$ , and their included angles at $B$ are equal trisection angles

Solution:

Triangles $ABD$ and $EBC$ have bases $AD$ and $EC$ on the same line $AC,$ so their common altitude from $B$ gives $\frac{[ABD]}{[EBC]}=\frac{AD}{EC}.$ Also, $\angle ABD=\angle EBC$ because both are one-third of angle $B.$ Using two sides and the included angle for the same area ratio gives $\frac{[ABD]}{[EBC]} =\frac{AB\cdot BD}{BE\cdot BC}.$

Thus, the correct answer is D.

20.

If $\dfrac xy=\dfrac34,$ then the incorrect expression in the following is:

$\dfrac{x+y}{y}=\dfrac74$

$\dfrac{y}{y-x}=\dfrac41$

$\dfrac{x+2y}{x}=\dfrac{11}3$

$\dfrac{x}{2y}=\dfrac38$

$\dfrac{x-y}{y}=\dfrac14$

Answer: E

Concepts:ratio and proportion algebraic manipulation

Difficulty rating: 1260

Small Hint:

Let $x=3k$ and $y=4k$

Big Hint:

Pay attention to the sign of $x-y$

Solution:

Set $x=3k$ and $y=4k.$ The first four expressions become $\frac{7}{4},$ $4,$ $\frac{11}{3},$ and $\frac{3}{8},$ respectively. But $\frac{x-y}{y}=\frac{3k-4k}{4k}=-\frac14,$ not $\frac{1}{4}.$

Thus, the correct answer is E.

21.

The sides of a regular polygon of $n$ sides, $n\gt4,$ are extended to form a star. The number of degrees at each point of the star is:

$\dfrac{360}{n}$

$\dfrac{(n-4)180}{n}$

$\dfrac{(n-2)180}{n}$

$180-\dfrac{90}{n}$

$\dfrac{180}{n}$

Answer: B

Concepts:regular polygon angle sum

Difficulty rating: 1500

Small Hint:

Relate a star point to the exterior angle at each of its two base vertices

Big Hint:

The angle at a star point and two exterior angles of the regular polygon form a straight-angle triangle

Solution:

Each exterior angle of a regular $n$ -gon is $\frac{360}{n}$ degrees. The two sides forming a point of the star, together with the intervening polygon side, make a triangle whose two base angles are those exterior angles. Hence its point angle is $180-2\left(\frac{360}{n}\right) =\frac{(n-4)180}{n}.$

Thus, the correct answer is B.

22.

On hypotenuse $AB$ of a right triangle $ABC$ a second right triangle $ABD$ is constructed with hypotenuse $AB.$ If $BC=1,$ $AC=b,$ and $AD=2,$ then $BD$ equals:

$\sqrt{b^2+1}$

$\sqrt{b^2-3}$

$\sqrt{b^2+1}+2$

$b^2+5$

$\sqrt{b^2+3}$

Answer: B

Concepts:right triangle Pythagorean Theorem

Difficulty rating: 1510

Small Hint:

Use triangle $ABC$ first to express the common hypotenuse $AB$

Big Hint:

Then apply the Pythagorean theorem to triangle $ABD$ , whose hypotenuse is $AB$

Solution:

In the first right triangle, $AB^2=AC^2+BC^2=b^2+1.$ In the second right triangle, $AB$ is the hypotenuse and $AD=2,$ so $\begin{aligned} BD^2&=AB^2-AD^2\\ &=b^2+1-4=b^2-3. \end{aligned}$ Therefore $BD=\sqrt{b^2-3}.$

Thus, the correct answer is B.

23.

If $\dfrac{x^2-bx}{ax-c}=\dfrac{m-1}{m+1}$ has roots which are numerically equal but of opposite signs, the value of $m$ must be:

$\dfrac{a-b}{a+b}$

$\dfrac{a+b}{a-b}$

$c$

$\dfrac1c$

$1$

Answer: A

Concepts:quadratic Vieta’s Formulas rational equation

Difficulty rating: 1740

Small Hint:

Clear the denominator and collect the equation as a quadratic in $x$

Big Hint:

Opposite roots have sum zero, so the coefficient of $x$ must vanish

Solution:

Cross-multiplication gives $\begin{aligned} (m+1)x^2 &-(m+1)bx\\ &-(m-1)ax\\ &\quad{}+(m-1)c=0. \end{aligned}$ Roots that are equal in magnitude and opposite in sign have sum $0,$ so the coefficient of $x$ is $0.$ Thus $m(a+b)+(b-a)=0,$ and $m=\frac{a-b}{a+b}.$

Thus, the correct answer is A.

24.

In the figure, it is given that angle $C=90^\circ,$ $AD=DB,$ $DE\perp AB,$ $AB=20,$ and $AC=12.$ The area of quadrilateral $ADEC$ is:

$75$

$58\dfrac12$

$48$

$37\dfrac12$

None of these

Answer: B

Concepts:coordinate geometry right triangle area

Difficulty rating: 1760

Small Hint:

Place $AB$ on the $x$ -axis and find the coordinates of $C$ from the two side lengths

Big Hint:

The perpendicular through midpoint $D$ meets $CB$ at $E$ ; then use the shoelace formula on $A,D,E,C$

Solution:

Since $AB=20,$ $AC=12,$ and angle $C$ is right, $BC=16.$ Put $A=(0,0)$ and $B=(20,0).$ Solving $\begin{aligned} x_C^2+y_C^2&=144,\\ (x_C-20)^2+y_C^2&=256 \end{aligned}$ gives $C=(\frac{36}{5},\frac{48}{5}).$ Also $D=(10,0).$

The line $CB$ meets $x=10$ at $E=(10,\frac{15}{2}).$ The shoelace formula for $A,D,E,C$ now gives $\begin{aligned} [ADEC] &=\frac12\left( 10\cdot\frac{15}{2} +10\cdot\frac{48}{5}\right.\\ &\qquad\left.{} -\frac{15}{2}\cdot\frac{36}{5}\right)\\ &=\frac{117}{2}=58\frac12. \end{aligned}$

Thus, the correct answer is B.

25.

A powderman set a fuse for a blast to take place in $30$ seconds. He ran away at a rate of $8$ yards per second. Sound travels at the rate of $1080$ feet per second. When the powderman heard the blast, he had run approximately:

$200$ yd.

$352$ yd.

$300$ yd.

$245$ yd.

$512$ yd.

Answer: D

Concepts:distance rate and time unit conversion

Difficulty rating: 1560

Small Hint:

Let $T$ be the total number of seconds from lighting the fuse until the sound is heard

Big Hint:

At time $T$ , the sound has traveled for $T-30$ seconds and must cover the runner’s distance

Solution:

Let $T$ be the total time in seconds. The powderman is $8T$ yards, or $24T$ feet, from the blast. Because the sound starts after $30$ seconds, $1080(T-30)=24T.$ Thus $T=\frac{1350}{44}$ seconds, and the distance he ran is $8T=\frac{2700}{11}\approx245.45\text{ yards}.$

Approximately $245$ yards is the listed value, so the correct answer is D.

26.

If $\left(r+\dfrac1r\right)^2=3,$ then $r^3+\dfrac1{r^3}$ equals

$1$

$2$

$0$

$3$

$6$

Answer: C

Concepts:algebraic manipulation exponent

Difficulty rating: 1420

Small Hint:

Let $u=r+\frac{1}{r}$ and express the requested sum in terms of $u$

Big Hint:

Use $r^3+r^{-3}=u^3-3u$

Solution:

Let $u=r+\frac{1}{r}.$ Expanding $u^3$ gives $r^3+\frac1{r^3}=u^3-3u.$ Since $u^2=3,$ this becomes $u(u^2-3)=0.$

Thus the value is $0,$ so the correct answer is C.

27.

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:

$1:2$

$1:3$

$1:\sqrt3$

$\sqrt3:2$

$2:3$

Answer: E

Concepts:equilateral triangle circle ratio and proportion

Difficulty rating: 1560

Small Hint:

Express each equilateral triangle’s side length in terms of the same circle radius $R$

Big Hint:

An equilateral triangle of altitude $R$ has side $\frac{2R}{\sqrt3}$ , while an inscribed one has side $\sqrt3R$

Solution:

An equilateral triangle with side $s$ has altitude $\frac{s\sqrt3}{2}.$ Thus the first triangle has side $\frac{2R}{\sqrt3}$ and perimeter $2\sqrt3R.$ An equilateral triangle inscribed in a circle of radius $R$ has side $\sqrt3R$ and perimeter $3\sqrt3R.$

The perimeter ratio is therefore $2:3,$ so the correct answer is E.

28.

In the table shown, the formula relating $x$ and $y$ is:

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$3$	$7$	$13$	$21$	$31$

$y=4x-1$

$y=x^3-x^2+x+2$

$y=x^2+x+1$

$y=(x^2+x+1)(x-1)$

None of these

Answer: C

Concepts:data and graph interpretation quadratic pattern recognition

Difficulty rating: 1290

Small Hint:

Look at the first and second differences of the $y$ -values

Big Hint:

The constant second difference $2$ suggests a monic quadratic; test the listed quadratic

Solution:

The successive differences of the $y$ -values are $4,6,8,10,$ whose successive differences are all $2.$ Thus the data fit a monic quadratic. Substitution shows $y=x^2+x+1$ gives $3,7,13,21,31$ at $x=1,2,3,4,5,$ respectively.

Thus, the correct answer is C.

29.

In a circle of radius $5$ units, $CD$ and $AB$ are perpendicular diameters. A chord $CH$ cutting $AB$ at $K$ is $8$ units long. The diameter $AB$ is divided into two segments whose dimensions are:

$1.25,$ $8.75$

$2.75,$ $7.25$

$2,$ $8$

$4,$ $6$

None of these

Answer: A

Concepts:circle chord coordinate geometry

Difficulty rating: 1760

Small Hint:

Put the center at the origin, take $C=(0,5)$ , and represent the other endpoint $H$ of the chord by coordinates

Big Hint:

Use $CH=8$ and $OH=5$ , then find where line $CH$ crosses the horizontal diameter

Solution:

Put the circle at the origin with $C=(0,5)$ and let $H=(x,y).$ The equations $\begin{aligned} x^2+y^2&=25,\\ x^2+(y-5)^2&=64 \end{aligned}$ give $y=-\frac{7}{5}$ and $x=\pm\frac{24}{5}.$ Taking $x=\frac{24}{5}$ does not change the segment lengths.

The line from $C=(0,5)$ to $H=(\frac{24}{5},-\frac{7}{5})$ crosses $y=0$ at $x=\frac{15}{4}.$ Its distances to the endpoints $(-5,0)$ and $(5,0)$ of diameter $AB$ are therefore $\begin{aligned} 5+\frac{15}{4}&=\frac{35}{4}=8.75,\\ 5-\frac{15}{4}&=\frac54=1.25. \end{aligned}$

Thus, the correct answer is A.

30.

When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is:

$1:2$

$2:1$

$1:4$

$4:1$

$1:1$

Answer: A

Concepts:arithmetic sequence summation ratio and proportion

Difficulty rating: 1590

Small Hint:

Write both partial sums in terms of the first term $a$ and common difference $d$

Big Hint:

Use $S_n=\dfrac n2(2a+(n-1)d)$ and simplify $S_{10}=4S_5$

Solution:

The sum formula gives $\begin{aligned} S_{10}&=5(2a+9d),\\ S_5&=\frac52(2a+4d). \end{aligned}$ The equation $S_{10}=4S_5$ becomes $2a+9d=2(2a+4d),$ so $d=2a.$ Hence $a:d=1:2.$

Thus, the correct answer is A.

31.

Given $12$ points in a plane no three of which are collinear, the number of lines they determine is:

$24$

$54$

$120$

$66$

None of these

Answer: D

Concepts:combinations basic counting

Difficulty rating: 1150

Small Hint:

Each line is determined by choosing two of the points

Big Hint:

The condition that no three are collinear ensures that different pairs determine different lines

Solution:

Every pair of points determines one line, and no line is counted by more than one pair because no three points are collinear. Therefore the number of lines is $\binom{12}{2}=\frac{12\cdot11}{2}=66.$

Thus, the correct answer is D.

32.

$K$ takes $30$ minutes less time than $M$ to travel a distance of $30$ miles. $K$ travels $\dfrac13$ mile per hour faster than $M.$ If $x$ is $K$ ’s rate of speed in miles per hour, then $K$ ’s time for the distance is:

$\dfrac{x+\frac{1}{3}}{30}$

$\dfrac{x-\frac{1}{3}}{30}$

$\dfrac{30}{x+\frac{1}{3}}$

$\dfrac{30}{x}$

$\dfrac{x}{30}$

Answer: D

Concepts:distance rate and time rate

Difficulty rating: 920

Small Hint:

Use the basic relation $\text{time}=\frac{\text{distance}}{\text{rate}}$

Big Hint:

The question asks only for $K$ ’s time, and both $K$ ’s distance and rate are already given

Solution:

For $K,$ the distance is $30$ miles and the rate is $x$ miles per hour. Thus $\text{time}=\frac{\text{distance}}{\text{rate}} =\frac{30}{x}\text{ hours}.$ The comparison with $M$ is not needed for this expression.

Thus, the correct answer is D.

33.

A circle and a square have the same perimeter. Then:

Their areas are equal

The area of the circle is the greater

The area of the square is the greater

The area of the circle is $\pi$ times the area of the square

None of these

Answer: B

Concepts:circle square (geometry)area perimeter

Difficulty rating: 1070

Small Hint:

Call the common perimeter $P$ and express each area in terms of $P$

Big Hint:

The circle has area $\frac{P^2}{4\pi}$ , while the square has area $\frac{P^2}{16}$

Solution:

With common perimeter $P,$ the circle’s radius is $\frac{P}{2\pi}$ , so its area is $\frac{P^2}{4\pi}.$ The square’s side is $\frac{P}{4},$ so its area is $\frac{P^2}{16}.$ Since $4\pi\lt16,$ $\frac{P^2}{4\pi}\gt\frac{P^2}{16}.$

The circle has the greater area, so the correct answer is B.

34.

The price of an article was increased $p\%.$ Later the new price was decreased $p\%.$ If the last price was one dollar, the original price was:

$\dfrac{1-p^2}{200}$

$\dfrac{\sqrt{1-p^2}}{100}$

One dollar

$1-\dfrac{p^2}{10{,}000-p^2}$

$\dfrac{10{,}000}{10{,}000-p^2}$

Answer: E

Concepts:percentage algebraic manipulation

Difficulty rating: 1560

Small Hint:

Represent the increase and decrease by multiplicative factors

Big Hint:

If the original price is $P,$ then $P(1+\frac{p}{100})(1-\frac{p}{100})=1$

Solution:

If the original price is $P$ dollars, then after both changes its price is $\begin{aligned} &P\left(1+\frac p{100}\right) \left(1-\frac p{100}\right)\\ &\qquad=P\left(1-\frac{p^2}{10{,}000}\right). \end{aligned}$ Setting this equal to $1$ gives $P=\frac{10{,}000}{10{,}000-p^2}.$

Thus, the correct answer is E.

35.

With a rational denominator, the expression $\dfrac{\sqrt2}{\sqrt2+\sqrt3-\sqrt5}$ is equivalent to:

$\dfrac{3+\sqrt6+\sqrt{15}}6$

$\dfrac{\sqrt6-2+\sqrt{10}}6$

$\dfrac{2+\sqrt6+\sqrt{10}}{10}$

$\dfrac{2+\sqrt6-\sqrt{10}}6$

None of these

Answer: A

Concepts:radical rationalizing denominator algebraic manipulation

Difficulty rating: 2210

Small Hint:

First treat the denominator as $(\sqrt2+\sqrt3)-\sqrt5$ and multiply by its conjugate

Big Hint:

After the first rationalization, a denominator containing $\sqrt6$ remains; use another conjugate

Solution:

Multiply first by the conjugate $\sqrt2+\sqrt3+\sqrt5.$ This gives $\begin{aligned} &\frac{\sqrt2(\sqrt2+\sqrt3+\sqrt5)} {(\sqrt2+\sqrt3)^2-5}\\ &\qquad=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt6}. \end{aligned}$ Multiplying numerator and denominator by $\sqrt6$ yields $\begin{gathered} \frac{2\sqrt6+6+\sqrt{60}}{12} \\ =\frac{3+\sqrt6+\sqrt{15}}6. \end{gathered}$

Thus, the correct answer is A.

36.

To be continuous at $x=-1,$ the value of $\dfrac{x^3+1}{x^2-1}$ is taken to be:

$-2$

$0$

$\dfrac32$

$\infty$

$-\dfrac32$

Answer: E

Concepts:calculus factoring

Difficulty rating: 1450

Small Hint:

Factor both numerator and denominator before substituting $x=-1$

Big Hint:

Cancel the common factor $x+1$ and evaluate the remaining expression

Solution:

For $x\ne-1,$ $\begin{aligned} \frac{x^3+1}{x^2-1} &=\frac{(x+1)(x^2-x+1)} {(x+1)(x-1)}\\ &=\frac{x^2-x+1}{x-1}. \end{aligned}$ Its limit at $x=-1$ is $\frac{1+1+1}{-2}=-\frac32.$ Assigning this value removes the discontinuity.

Thus, the correct answer is E.

37.

Two equal parallel chords are drawn $8$ inches apart in a circle of radius $8$ inches. The area of that part of the circle that lies between the chords is:

$21\dfrac13\pi-32\sqrt3$

$32\sqrt3+21\dfrac13\pi$

$32\sqrt3+42\dfrac23\pi$

$16\sqrt3+42\dfrac23\pi$

$42\dfrac23\pi$

Answer: B

Concepts:circle chord sector area

Difficulty rating: 2270

Small Hint:

Equal parallel chords lie the same distance from the center, so each is $4$ inches from it

Big Hint:

Subtract the two congruent outer circular segments from the whole circle

Solution:

The equal chords are symmetrically $4$ inches from the center. For either chord, the half-angle at the center satisfies $\cos\theta=\frac{4}{8}=\frac{1}{2},$ so $\theta=60^\circ.$ Each outer segment is a $120^\circ$ sector minus the isosceles triangle: $\begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3. \end{aligned}$ The area between the chords is therefore $\begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3. \end{aligned}$

Thus, the correct answer is B.

38.

The area of a trapezoidal field is $1400$ square yards. Its altitude is $50$ yards. Find the two bases, if the number of yards in each base is an integer divisible by $8.$ The number of solutions to this problem is:

None

One

Two

Three

More than three

Answer: D

Concepts:trapezoid divisibility basic counting

Difficulty rating: 1450

Small Hint:

Use the trapezoid area formula to find the sum of the two bases

Big Hint:

List the unordered pairs of positive multiples of $8$ with that sum

Solution:

If the bases are $b_1,b_2,$ then $1400=\frac12(b_1+b_2)(50),$ so $b_1+b_2=56.$ The unordered positive pairs of multiples of $8$ are $(8,48),\quad(16,40),\quad(24,32).$ There are three solutions.

Thus, the correct answer is D.

39.

If the perimeter of a rectangle is $p$ and its diagonal is $d,$ the difference between the length and width of the rectangle is:

$\dfrac{\sqrt{8d^2-p^2}}2$

$\dfrac{\sqrt{8d^2+p^2}}2$

$\dfrac{\sqrt{6d^2-p^2}}2$

$\dfrac{\sqrt{6d^2+p^2}}2$

$\dfrac{\sqrt{8d^2-p^2}}4$

Answer: A

Concepts:rectangle difference of squares diagonal

Difficulty rating: 1640

Small Hint:

Let the side lengths be $L$ and $W$ , and write equations for $L+W$ and $L^2+W^2$

Big Hint:

Write $2(L^2+W^2)$ , then subtract $(L+W)^2$ to obtain $(L-W)^2$

Solution:

The perimeter and diagonal give $\begin{aligned} L+W&=\frac p2,\\ L^2+W^2&=d^2. \end{aligned}$ Hence $\begin{aligned} (L-W)^2 &=2(L^2+W^2)\\ &\quad{}-(L+W)^2\\ &=2d^2-\frac{p^2}{4}\\ &=\frac{8d^2-p^2}{4}. \end{aligned}$ Taking the nonnegative square root gives $L-W=\frac{\sqrt{8d^2-p^2}}2.$

Thus, the correct answer is A.

40.

In order to draw a graph of $f(x)=ax^2+bx+c,$ a table of values was constructed. These values of the function for a set of equally spaced increasing values of $x$ were $3844,$ $3969,$ $4096,$ $4227,$ $4356,$ $4489,$ $4624,$ and $4761.$ The one which is incorrect is:

$4096$

$4356$

$4489$

$4761$

None of these

Answer: E

Concepts:quadratic pattern recognition perfect square

Difficulty rating: 1410

Small Hint:

A quadratic sampled at equally spaced inputs has constant second differences

Big Hint:

The surrounding values also suggest consecutive perfect squares; check every displayed table value before checking which values appear among the choices

Solution:

The values are intended to be the consecutive squares $\begin{gathered} 62^2,63^2,64^2,65^2,\\ 66^2,67^2,68^2,69^2. \end{gathered}$ These are $3844,3969,4096,4225,$ $4356,4489,4624,4761.$ Thus the incorrect table entry is $4227,$ but it is not one of choices A-D.

Consequently none of the four listed numbers is incorrect, so the correct answer is E.

41.

Increasing the radius of a cylinder by $6$ units increases the volume by $y$ cubic units. Increasing the altitude of the cylinder by $6$ units also increases the volume by $y$ cubic units. If the original altitude is $2,$ then the original radius is:

$2$

$4$

$6$

$6\pi$

$8$

Answer: C

Concepts:cylinder volume quadratic

Difficulty rating: 1770

Small Hint:

Let the original radius be $r$ and write each of the two volume increases

Big Hint:

Equate $2\pi((r+6)^2-r^2)$ and $6\pi r^2$

Solution:

With original height $2$ and radius $r,$ increasing the radius adds $2\pi\bigl((r+6)^2-r^2\bigr)=24\pi r+72\pi$ cubic units. Increasing the height by $6$ adds $6\pi r^2$ cubic units. Equating these gives $6r^2=24r+72,$ or $r^2-4r-12=0.$ Its positive root is $r=6.$

Thus, the correct answer is C.

42.

Let $D$ represent a repeating decimal. If $P$ denotes the $r$ figures of $D$ which do not repeat themselves, and $Q$ denotes the $s$ figures which do repeat themselves, then the incorrect expression is:

$D=0.PQQQ\ldots$

$10^rD=P.QQQ\ldots$

$10^{r+s}D=PQ.QQQ\ldots$

$10^r(10^s-1)D=Q(P-1)$

$10^r\cdot10^{2s}D=$ $PQQ.QQQ\ldots$

Answer: D

Concepts:repeating decimal place value algebraic manipulation

Difficulty rating: 2060

Small Hint:

Shift the decimal point first past the $r$ nonrepeating digits and then past one block of $s$ repeating digits

Big Hint:

Subtract $10^rD$ from $10^{r+s}D$ and compare the result with choice D

Solution:

After shifting past the nonrepeating block, $10^rD=P.QQQ\ldots,$ and after shifting one repeating block farther, $10^{r+s}D=PQ.QQQ\ldots.$ Subtracting yields the standard relation $\begin{gathered} 10^r(10^s-1)D \\ =P(10^s-1)+Q. \end{gathered}$ This is not $Q(P-1),$ as stated in choice D.

Thus, the incorrect expression, and hence the correct answer, is D.

43.

The diameter of a circle is divided into $n$ equal parts. On each part a semicircle is constructed. As $n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:

Equal to the semi-circumference of the original circle

Equal to the diameter of the original circle

Greater than the diameter but less than the semi-circumference of the original circle

That is infinite

Greater than the semi-circumference but finite

Answer: A

Concepts:circle arc circumference

Difficulty rating: 1360

Small Hint:

Let the original diameter be $d$ , so each small semicircle has diameter $\frac{d}{n}$

Big Hint:

Multiply the arc length of one small semicircle by the number $n$ of parts

Solution:

Each small semicircle has diameter $\frac{d}{n},$ so its arc length is $\frac{\pi d}{2n}.$ The sum of all $n$ arc lengths is $n\cdot\frac{\pi d}{2n}=\frac{\pi d}{2},$ exactly the semi-circumference of the original circle. This equality holds for every positive $n,$ not merely in the limit.

Thus, the correct answer is A.

44.

If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by:

$(9-k)$

$(10-k)$

$(11-k)$

$(k-1)$

$(k+1)$

Answer: C

Concepts:digits algebraic manipulation

Difficulty rating: 1420

Small Hint:

Let the two digits be $a$ and $b$ , and express both the original and reversed numbers

Big Hint:

The two numbers add to $11(a+b)$

Solution:

Let the original number be $10a+b=k(a+b).$ The reversed number is $10b+a.$ Since $\begin{gathered} (10a+b)+(10b+a) \\ =11(a+b), \end{gathered}$ the reversed number equals $\begin{aligned} &11(a+b)-k(a+b)\\ &\qquad=(11-k)(a+b). \end{aligned}$

Thus, the correct answer is C.

45.

If $a$ and $b$ are two unequal positive numbers, then:

$\dfrac{2ab}{a+b}\gt\sqrt{ab}\gt\dfrac{a+b}{2}$

$\sqrt{ab}\gt\dfrac{2ab}{a+b}\gt\dfrac{a+b}{2}$

$\dfrac{2ab}{a+b}\gt\dfrac{a+b}{2}\gt\sqrt{ab}$

$\dfrac{a+b}{2}\gt\dfrac{2ab}{a+b}\gt\sqrt{ab}$

$\dfrac{a+b}{2}\gt\sqrt{ab}\gt\dfrac{2ab}{a+b}$

Answer: E

Concepts:AM-GM Inequality harmonic mean inequality

Difficulty rating: 1450

Small Hint:

Recognize the arithmetic, geometric, and harmonic means of $a$ and $b$

Big Hint:

For unequal positive numbers, the mean inequalities are strict

Solution:

The arithmetic-geometric mean inequality gives $\frac{a+b}{2}\gt\sqrt{ab}$ for unequal positive $a,b.$ Applying the same inequality to $\frac{1}{a}$ and $\frac{1}{b}$ gives $\sqrt{ab}\gt\frac{2ab}{a+b}.$ Therefore the decreasing order is arithmetic mean, geometric mean, harmonic mean.

Thus, the correct answer is E.

46.

The base of a new rectangle equals the sum of the diagonal and the greater side of a given rectangle, while the altitude of the new rectangle equals the difference of the diagonal and the greater side of the given rectangle. The area of the new rectangle is:

Greater than the area of the given rectangle

Equal to the area of the given rectangle

Equal to the area of a square with its side equal to the smaller side of the given rectangle

Equal to the area of a square with its side equal to the greater side of the given rectangle

Equal to the area of a rectangle whose dimensions are the diagonal and shorter side of the given rectangle

Answer: C

Concepts:rectangle Pythagorean Theorem difference of squares

Difficulty rating: 1500

Small Hint:

Let $d$ be the diagonal and $L,W$ the greater and smaller sides

Big Hint:

The new area is $(d+L)(d-L)=d^2-L^2$

Solution:

Let the greater and smaller side lengths be $L$ and $W,$ and let the diagonal be $d.$ The new rectangle’s area is $(d+L)(d-L)=d^2-L^2.$ The Pythagorean theorem gives $d^2=L^2+W^2,$ so the new area is $W^2.$ This is the area of a square whose side equals the smaller side of the original rectangle.

Thus, the correct answer is C.

47.

In the set of equations $z^x=y^{2x},$ $2^z=2\cdot4^x,$ $x+y+z=16,$ the integral roots in the order $x,$ $y,$ $z$ are:

$3,$ $4,$ $9$

$9,$ $-5,$ $12$

$12,$ $-5,$ $9$

$4,$ $3,$ $9$

$4,$ $9,$ $3$

Answer: D

Concepts:exponent Diophantine Equation system of equations

Difficulty rating: 1810

Small Hint:

Compare exponents in $2^z=2\cdot4^x$ to express $z$ in terms of $x$

Big Hint:

For the intended positive solution, $z^x=y^{2x}$ gives $z=y^2$ ; substitute both relations into the sum

Solution:

For the intended positive integral solution, the first equation gives $z=y^2.$ The second equation is $2^z=2^{2x+1},$ so $z=2x+1$ and $x=\frac{y^2-1}{2}.$ Substituting into $x+y+z=16$ gives $\frac{y^2-1}{2}+y+y^2=16.$ The positive integral solution is $y=3,$ which gives $x=4$ and $z=9.$ Direct substitution verifies all three displayed equations.

Thus, the intended listed answer is D.

48.

Two cyclists, $k$ miles apart, and starting at the same time, would be together in $r$ hours if they traveled in the same direction, but would pass each other in $t$ hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:

$\dfrac{r+t}{r-t}$

$\dfrac r{r-t}$

$\dfrac{r+t}{r}$

$\dfrac rt$

$\dfrac{r+k}{t-k}$

Answer: A

Concepts:relative speed rate ratio and proportion

Difficulty rating: 1770

Small Hint:

Let the faster and slower speeds be $u$ and $v$ , and write the same-direction and opposite-direction relative-speed equations

Big Hint:

Use $(u-v)r=k$ and $(u+v)t=k$ , then solve for $\frac{u}{v}$

Solution:

Let the speeds be $u\gt v.$ The two meeting conditions give $u-v=\frac kr,\qquad u+v=\frac kt.$ Adding and subtracting these equations yields $\begin{aligned} u&=\frac k2\left(\frac1r+\frac1t\right),\\ v&=\frac k2\left(\frac1t-\frac1r\right). \end{aligned}$ Therefore $\frac uv=\frac{r+t}{r-t}.$

Thus, the correct answer is A.

49.

In the figure, $CD,$ $AE,$ and $BF$ are one-third of their respective sides. It follows that $AN_2:N_2N_1:N_1D=3:3:1,$ and similarly for lines $BE$ and $CF.$ Then the area of triangle $N_1N_2N_3$ is:

$\dfrac1{10}\triangle ABC$

$\dfrac19\triangle ABC$

$\dfrac17\triangle ABC$

$\dfrac16\triangle ABC$

None of these

Answer: C

Concepts:coordinate geometry area ratio

Difficulty rating: 2380

Small Hint:

Area ratios are affine-invariant, so choose convenient coordinates for triangle $ABC$

Big Hint:

Locate $D,E,F$ by the one-third conditions, intersect the three cevians, and compare the two areas with determinants

Solution:

Use $A=(0,1),$ $B=(0,0),$ $C=(1,0).$ Then $\begin{aligned} D&=\left(\frac23,0\right),\\ E&=\left(\frac13,\frac23\right),\\ F&=\left(0,\frac13\right). \end{aligned}$ Intersecting $AD,BE,CF$ in pairs gives $\begin{aligned} N_1&=\left(\frac47,\frac17\right),\\ N_2&=\left(\frac27,\frac47\right),\\ N_3&=\left(\frac17,\frac27\right). \end{aligned}$ The determinant area formula gives $\begin{aligned} [N_1N_2N_3]&=\frac1{14},\\ [ABC]&=\frac12. \end{aligned}$ Hence $\frac{[N_1N_2N_3]}{[ABC]}=\frac{1}{7}.$

Thus, the correct answer is C.

50.

A line initially $1$ inch long grows according to the following law, where the first term is the initial length. $\begin{gathered} 1+\frac14\sqrt2+\frac14+\frac1{16}\sqrt2\\ {}+\frac1{16}+\frac1{64}\sqrt2+\frac1{64}+\cdots. \end{gathered}$ If the growth process continues forever, the limit of the length of the line is:

$\infty$

$\dfrac43$

$\dfrac38$

$\dfrac13(4+\sqrt2)$

$\dfrac23(4+\sqrt2)$

Answer: D

Concepts:geometric sequence summation radical

Difficulty rating: 1640

Small Hint:

Group each pair having the same power of $\frac{1}{4}$

Big Hint:

The terms after the initial $1$ equal $(1+\sqrt2)\sum_{k=1}^{\infty}4^{-k}$

Solution:

After the initial term, each power $4^{-k}$ appears once by itself and once multiplied by $\sqrt2.$ Since $\sum_{k=1}^{\infty}4^{-k}=\frac{1}{3},$ the limit is $1+\frac{1+\sqrt2}{3}=\frac{4+\sqrt2}{3}.$

Thus, the correct answer is D.