1955 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Which one of the following is not equivalent to $0.000000375?$

$3.75\times10^{-7}$

$3\dfrac34\times10^{-7}$

$375\times10^{-9}$

$\dfrac38\times10^{-7}$

$\dfrac{3}{8000000}$

Concepts:scientific notation decimal conversion equivalent forms

Difficulty rating: 960

Small Hint:

Rewrite every choice with the same power of $10$

Big Hint:

The decimal is $3.75\times10^{-7}$

Solution:

The given decimal is $3.75\times10^{-7}.$ Choices A and B state this directly, while $375\times10^{-9}=3.75\times10^{-7}$ and $\frac38\times10^{-6}$ $=0.375\times10^{-6}$ $=3.75\times10^{-7}.$ But $\frac38\times10^{-7}=3.75\times10^{-8},$ which is ten times smaller.

Thus, the correct answer is D.

2.

The smaller angle between the hands of a clock at $12{:}25$ p.m. is:

$132^\circ30'$

$137^\circ30'$

$150^\circ$

$137^\circ32'$

$137^\circ$

Concepts:clock angle angular speed degrees and minutes

Difficulty rating: 1260

Small Hint:

In $25$ minutes, the minute hand moves $150^\circ$

Big Hint:

The hour hand moves $0.5^\circ$ per minute

Solution:

At $12{:}25,$ the minute hand is $25\cdot6^\circ=150^\circ$ past $12,$ while the hour hand is $25\cdot0.5^\circ=12.5^\circ.$ Their smaller separation is $150^\circ-12.5^\circ=137.5^\circ=137^\circ30'.$

Thus, the correct answer is B.

3.

If each number in a set of ten numbers is increased by $20,$ the arithmetic mean (average) of the original ten numbers:

remains the same

is increased by $20$

is increased by $200$

is increased by $10$

is increased by $2$

Concepts:arithmetic mean translation of data

Difficulty rating: 800

Small Hint:

Increasing all ten entries adds $10\cdot20$ to their sum

Big Hint:

The new total is still divided by $10$

Solution:

If the original sum is $S,$ the new sum is $S+10\cdot20.$ Hence the new mean is $\frac{S+200}{10}=\frac S{10}+20.$ The mean is increased by $20.$

Thus, the correct answer is B.

4.

The equality $\dfrac1{x-1}=\dfrac2{x-2}$ is satisfied by:

no real values of $x$

either $x=1$ or $x=2$

only $x=1$

only $x=2$

only $x=0$

Concepts:rational equation domain restrictions

Difficulty rating: 1060

Small Hint:

Cross-multiply, while remembering that $x\ne1,2$

Big Hint:

Solve $x-2=2x-2$

Solution:

For $x\ne1,2,$ cross-multiplication gives $x-2=2(x-1)=2x-2,$ so $x=0.$ This value makes both original denominators nonzero and satisfies the equality.

Thus, the correct answer is E.

5.

$y$ varies inversely as the square of $x.$ When $y=16,$ $x=1.$ When $x=8,$ $y$ equals:

$2$

$128$

$64$

$\dfrac14$

$1024$

Concepts:inverse variation squares

Difficulty rating: 920

Small Hint:

Write the variation as $y=\frac{k}{x^2}$

Big Hint:

Use the first pair of values to determine $k$

Solution:

Inverse-square variation gives $y=\frac{k}{x^2}.$ Since $16=\frac{k}{1^2},$ we have $k=16.$ At $x=8,$ $y=\frac{16}{8^2}=\frac14.$

Thus, the correct answer is D.

6.

A merchant buys a number of oranges at $3$ for $10$ ¢ and an equal number at $5$ for $20$ ¢. To “break even” he must sell all at:

$8$ for $30$ ¢

$3$ for $11$ ¢

$5$ for $18$ ¢

$11$ for $40$ ¢

$13$ for $50$ ¢

Concepts:unit price weighted average break-even

Difficulty rating: 1210

Small Hint:

Use $15$ oranges in each equal-sized purchase so that both quoted rates divide evenly

Big Hint:

Compare the total cost of $30$ oranges with each proposed selling rate

Solution:

Suppose he buys $15$ oranges at each rate. The first $15$ cost $50$ ¢ and the second $15$ cost $60$ ¢, for $110$ ¢ total. Thus $30$ oranges must sell for $110$ ¢, which is equivalent to $3$ for $11$ ¢.

Therefore, the correct answer is B.

7.

If a worker receives a $20$ percent cut in wages, he may regain his original pay exactly by obtaining a raise of:

$20$ percent

$25$ percent

$22\dfrac12$ percent

$\$20$

$\$25$

Concepts:percentage change reverse percentage

Difficulty rating: 920

Small Hint:

After the cut, the wage is $80\%$ of the original

Big Hint:

Find the percent of the reduced wage represented by the missing $20\%$ of the original

Solution:

If the original wage is $W,$ the reduced wage is $0.8W.$ The needed increase is $0.2W,$ which as a fraction of the reduced wage is $\frac{0.2W}{0.8W}=\frac14=25\%.$

Thus, the correct answer is B.

8.

The graph of $x^2-4y^2=0:$

is a hyperbola intersecting only the $x$ -axis

is a hyperbola intersecting only the $y$ -axis

is a hyperbola intersecting neither axis

is a pair of straight lines

does not exist

Concepts:factoring degenerate conic lines

Difficulty rating: 1260

Small Hint:

Factor the difference of squares

Big Hint:

A product is zero when at least one of its two linear factors is zero

Solution:

Factoring, $x^2-4y^2=(x-2y)(x+2y).$ Thus the graph is the union of the two straight lines $x=2y$ and $x=-2y.$

The correct answer is D.

9.

A circle is inscribed in a triangle with sides $8,$ $15,$ and $17.$ The radius of the circle is:

$6$

$2$

$5$

$3$

$7$

Concepts:right triangle inradius area and semiperimeter

Difficulty rating: 1210

Small Hint:

The side lengths form a Pythagorean triple

Big Hint:

For a right triangle, $r=\frac{a+b-c}{2}$

Solution:

Because $8^2+15^2=17^2,$ the triangle is right. Its inradius is $r=\frac{8+15-17}{2}=3.$

Thus, the correct answer is D.

10.

How many hours does it take a train traveling at an average rate of $40$ mph between stops to travel $a$ miles if it makes $n$ stops of $m$ minutes each?

$\dfrac{3a+2mn}{120}$

$3a+2mn$

$\dfrac{3a+2mn}{12}$

$\dfrac{a+mn}{40}$

$\dfrac{a+40mn}{40}$

Concepts:rate-time-distance unit conversion algebraic expression

Difficulty rating: 1410

Small Hint:

The moving time is $\frac{a}{40}$ hours

Big Hint:

Convert the total stopping time $mn$ from minutes to hours before adding

Solution:

The train moves for $\frac{a}{40}$ hours and is stopped for $\frac{mn}{60}$ hours. Hence the total time is $\frac a{40}+\frac{mn}{60} =\frac{3a+2mn}{120}.$

Thus, the correct answer is A.

11.

The negation of the statement “No slow learners attend this school,” is:

All slow learners attend this school.

All slow learners do not attend this school.

Some slow learners attend this school.

Some slow learners do not attend this school.

No slow learners do not attend this school.

Concepts:logic negation quantifiers

Difficulty rating: 1060

Small Hint:

“No” means that there does not exist even one example

Big Hint:

Negating a universal exclusion asserts the existence of a counterexample

Solution:

The original statement says that every slow learner is absent from the school. Its negation is that at least one slow learner attends the school.

Thus, “Some slow learners attend this school” is correct, so the answer is C.

12.

The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:

$x=2,$ $x=1$

$x=\dfrac23$

$x=2$

$x=1$

$x=0$

Concepts:radical equation domain extraneous roots

Difficulty rating: 1570

Small Hint:

The real-domain restriction $\sqrt{x-1}$ gives $x\ge1$

Big Hint:

Test the simplest endpoint before squaring the equation

Solution:

The domain requires $x\ge1.$ At $x=1,$ the left side is $2+0=2,$ so $x=1$ works. To see that there is no other solution, note that both radicals are nondecreasing for $x\ge1,$ and $\sqrt{5x-1}$ is strictly increasing. Therefore the sum exceeds $2$ for every $x>1.$

Thus, the correct answer is D.

13.

The fraction $\dfrac{a^{-4}-b^{-4}}{a^{-2}-b^{-2}}$ is equal to:

$a^{-6}-b^{-6}$

$a^{-2}-b^{-2}$

$a^{-2}+b^{-2}$

$a^2+b^2$

$a^2-b^2$

Concepts:difference of squares negative exponents rational expression

Difficulty rating: 1280

Small Hint:

Regard $a^{-4}-b^{-4}$ as a difference of squares

Big Hint:

Factor it using $a^{-2}$ and $b^{-2}$

Solution:

Factoring the numerator, $\begin{aligned} a^{-4}-b^{-4} &=(a^{-2}-b^{-2})\\ &\quad\cdot(a^{-2}+b^{-2}). \end{aligned}$ Where the original fraction is defined, canceling the common factor leaves $a^{-2}+b^{-2}.$

Thus, the correct answer is C.

14.

The length of rectangle $R$ is $10$ percent more than the side of square $S.$ The width of the rectangle is $10$ percent less than the side of the square. The ratio of the areas, $R:S,$ is:

$99:100$

$101:100$

$1:1$

$199:200$

$201:200$

Concepts:area ratio percentage change rectangle

Difficulty rating: 1180

Small Hint:

Let the square’s side length be $s$

Big Hint:

The rectangle’s area is $(1.1s)(0.9s)$

Solution:

If the square side is $s,$ then the rectangle has dimensions $1.1s$ and $0.9s.$ Therefore $\begin{aligned} \frac{[R]}{[S]} &=\frac{(1.1s)(0.9s)}{s^2}\\ &=0.99=\frac{99}{100}. \end{aligned}$

Thus, the correct answer is A.

15.

The ratio of the areas of two concentric circles is $1:3.$ If the radius of the smaller is $r,$ then the difference between the radii is best approximated by:

$0.41r$

$0.73$

$0.75$

$0.73r$

$0.75r$

Concepts:circle area ratio approximation

Difficulty rating: 1280

Small Hint:

Area ratios are the squares of radius ratios

Big Hint:

The larger radius is $r\sqrt3$ , so estimate $\sqrt3-1$

Solution:

If the larger radius is $R,$ then $\frac{\pi r^2}{\pi R^2}=\frac13,$ so $R=r\sqrt3.$ The difference is $R-r=(\sqrt3-1)r\approx0.732r.$

Thus, the best approximation is $0.73r,$ and the correct answer is D.

16.

The value of $\dfrac3{a+b}$ when $a=4$ and $b=-4$ is:

$3$

$\dfrac38$

$0$

any finite number

meaningless

Concepts:division by zero rational expression substitution

Difficulty rating: 1060

Small Hint:

Substitute the two given values into the denominator first

Big Hint:

A fraction with denominator $0$ is undefined

Solution:

Substitution gives $a+b=4+(-4)=0.$ Hence the expression becomes $\frac{3}{0},$ which is undefined.

Thus, the expression is meaningless and the correct answer is E.

17.

If $\log x-5\log3=-2,$ then $x$ equals:

$1.25$

$0.81$

$2.43$

$0.8$

either $0.8$ or $1.25$

Concepts:common logarithm logarithm laws exponents

Difficulty rating: 1260

Small Hint:

Move $5\log3$ to the other side

Big Hint:

Use $-2=\log10^{-2}$ and combine logarithms

Solution:

Using common logarithms, $\begin{aligned} \log x&=5\log3-2\\ &=\log(3^5)-\log100\\ &=\log\left(\frac{243}{100}\right). \end{aligned}$ Therefore $x=2.43.$

The correct answer is C.

18.

The discriminant of the equation $x^2+2x\sqrt3+3=0$ is zero. Hence, its roots are:

real and equal

rational and equal

rational and unequal

irrational and unequal

imaginary

Concepts:quadratic discriminant repeated root real numbers

Difficulty rating: 1150

Small Hint:

Use the quadratic formula with the discriminant term equal to zero

Big Hint:

Then inspect the resulting value $-\frac{b}{2a}$ for the requested classifications

Solution:

The quadratic formula gives $x=\frac{-2\sqrt3}{2}=-\sqrt3$ twice because the discriminant is zero. The root is irrational, but the requested description that applies is “real and equal.”

Thus, the correct answer is A.

19.

Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:

$x^2-6x+7=0$

$x^2-6x-7=0$

$x^2+6x-8=0$

$x^2-6x+8=0$

$x^2+6x-7=0$

Concepts:sum and difference roots of a quadratic Vieta formulas

Difficulty rating: 1410

Small Hint:

Solve $u+v=6$ and $u-v=8$ after choosing an order

Big Hint:

A monic quadratic with roots $u,v$ is $x^2-(u+v)x+uv$

Solution:

Taking the larger number first, $u+v=6,\qquad u-v=8$ gives $u=7$ and $v=-1.$ Their product is $-7,$ so the monic equation with these roots is $x^2-6x-7=0.$

Thus, the correct answer is B.

20.

The expression $\sqrt{25-t^2}+5$ equals zero for:

no real or imaginary values of $t$

no real values of $t$ only

no imaginary values of $t$ only

$t=0$

$t=\pm5$

Concepts:principal square root radical equation complex numbers

Difficulty rating: 1340

Small Hint:

Isolate the radical before doing any algebra

Big Hint:

If you square, check every candidate in the original equation because the radical has a prescribed sign

Solution:

For the expression to vanish, one would need $\sqrt{25-t^2}=-5.$ The radical sign denotes the principal square root. It is nonnegative for a nonnegative real radicand and, over the complex numbers, is chosen with nonnegative real part; it therefore cannot equal $-5.$ Squaring would introduce the extraneous candidate $t=0,$ for which the original expression is $10.$

Thus, no real or imaginary value works, and the correct answer is A.

21.

Represent the hypotenuse of a right triangle by $c$ and the area by $A.$ The altitude on the hypotenuse is:

$\dfrac Ac$

$\dfrac{2A}{c}$

$\dfrac A{2c}$

$\dfrac{A^2}{c}$

$\dfrac A{c^2}$

Concepts:triangle area altitude algebra

Difficulty rating: 1060

Small Hint:

Use the hypotenuse as the base of the triangle

Big Hint:

If the corresponding altitude is $h,$ then $A=\frac{ch}{2}$

Solution:

Taking the hypotenuse as the base and writing its altitude as $h,$ $A=\frac12ch.$ Solving gives $h=\frac{2A}{c}.$

Thus, the correct answer is B.

22.

On a $\$10{,}000$ order a merchant has a choice between three successive discounts of $20\%,$ $20\%,$ and $10\%$ and three successive discounts of $40\%,$ $5\%,$ and $5\%.$ By choosing the better offer, he can save:

nothing at all

$\$440$

$\$330$

$\$345$

$\$360$

Concepts:successive discounts percentage multiplication comparison

Difficulty rating: 1570

Small Hint:

Successive discounts multiply the remaining-price factors

Big Hint:

Compare $0.8\cdot0.8\cdot0.9$ with $0.6\cdot0.95\cdot0.95$

Solution:

The first offer leaves a fraction $0.8\cdot0.8\cdot0.9=0.576$ of the price. The second leaves $0.6\cdot0.95^2=0.5415.$ The second is cheaper by $0.576-0.5415=0.0345$ of the order, or $0.0345(10000)=345$ dollars.

Thus, the correct answer is D.

23.

In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:

make no correction

subtract $11$ ¢

subtract $11x$ ¢

add $11x$ ¢

add $x$ ¢

Concepts:coin values accounting error linear expression

Difficulty rating: 1280

Small Hint:

Each nickel counted as a quarter makes the total too large by $20$ ¢

Big Hint:

Each dime counted as a cent makes the total too small by $9$ ¢

Solution:

The $x$ nickels counted as quarters overstate the total by $20x$ ¢. The $x$ dimes counted as cents understate it by $9x$ ¢. The net overstatement is $20x-9x=11x$ cents, so that amount must be subtracted.

Thus, the correct answer is C.

24.

The function $4x^2-12x-1:$

always increases as $x$ increases

always decreases as $x$ decreases to $1$

cannot equal $0$

has a maximum value when $x$ is negative

has a minimum value of $-10$

Concepts:quadratic function completing the square minimum

Difficulty rating: 1410

Small Hint:

Complete the square in $4x^2-12x-1$

Big Hint:

Factor $4$ from the quadratic terms before forming $(x-\frac{3}{2})^2$

Solution:

Completing the square, $\begin{aligned} 4x^2-12x-1 &=4\left(x-\frac32\right)^2\\ &\quad{}-10. \end{aligned}$ The squared term is nonnegative, so the minimum value is $-10,$ attained at $x=\frac{3}{2}.$

Thus, the correct answer is E.

25.

One of the factors of $x^4+2x^2+9$ is:

$x^2+3$

$x+1$

$x^2-3$

$x^2-2x-3$

none of these

Concepts:polynomial factorization difference of squares

Difficulty rating: 1400

Small Hint:

Rewrite the polynomial as $(x^2+3)^2-(2x)^2$

Big Hint:

Factor the resulting difference of squares and compare both factors with the choices

Solution:

We have $\begin{aligned} x^4+2x^2+9 &=(x^2+3)^2\\ &\quad{}-(2x)^2\\ &=(x^2-2x+3)\\ &\quad\cdot(x^2+2x+3). \end{aligned}$ Neither factor appears among choices A through D.

Thus, the correct answer is E.

26.

Mr. $A$ owns a house worth $\$10{,}000.$ He sells it to Mr. $B$ at $10\%$ profit. Mr. $B$ sells the house back to Mr. $A$ at a $10\%$ loss. Then:

Mr. $A$ comes out even

Mr. $A$ makes $\$100$

Mr. $A$ makes $\$1{,}000$

Mr. $B$ loses $\$100$

none of the above is correct

Concepts:percentage profit and loss successive transactions net change

Difficulty rating: 1280

Small Hint:

The first sale price is $1.10(10000)$

Big Hint:

Mr. $B$ ’s loss is $10\%$ of what he paid, not $10\%$ of the original house value

Solution:

Mr. $A$ first receives $1.10(10000)=11000$ dollars. Mr. $B$ then sells at a loss of $10\%$ of his $11000$ -dollar cost, so Mr. $A$ buys the house back for $0.90(11000)=9900$ dollars. Mr. $A$ again owns the house and has gained $11000-9900=1100$ dollars; Mr. $B$ has lost $1100$ dollars. None of the stated amounts is correct.

Thus, the correct answer is E.

27.

If $r$ and $s$ are the roots of $x^2-px+q=0,$ then $r^2+s^2$ equals:

$p^2+2q$

$p^2-2q$

$p^2+q^2$

$p^2-q^2$

$p^2$

Concepts:Vieta formulas symmetric expressions quadratic roots

Difficulty rating: 1260

Small Hint:

Use $r+s=p$ and $rs=q$

Big Hint:

Expand $(r+s)^2$ and isolate $r^2+s^2$

Solution:

By Vieta’s formulas, $r+s=p$ and $rs=q.$ Therefore $\begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=p^2-2q. \end{aligned}$

Thus, the correct answer is B.

28.

On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation. If $b\ne0$ and $c\ne0$ these two graphs intersect:

in two points, one on the $x$ -axis and one on the $y$ -axis

in one point located on neither axis

only at the origin

in one point on the $x$ -axis

in one point on the $y$ -axis

Concepts:parabola reflection simultaneous equations symmetry

Difficulty rating: 1340

Small Hint:

The reflected graph is $y=ax^2-bx+c$

Big Hint:

Set the two expressions for $y$ equal and use $b\ne0$

Solution:

Replacing $x$ by $-x$ gives $y=ax^2-bx+c.$ At an intersection, $ax^2+bx+c=ax^2-bx+c,$ so $2bx=0.$ Since $b\ne0,$ we must have $x=0,$ and then $y=c\ne0.$ There is exactly one intersection, $(0,c),$ on the $y$ -axis.

Thus, the correct answer is E.

29.

In the figure $\overline{PA}$ is tangent to semicircle $SAR;$ $\overline{PB}$ is tangent to semicircle $RBT;$ $SRT$ is a straight line; the arcs are indicated in the figure. Angle $APB$ is measured by:

$\dfrac12(a-b)$

$\dfrac12(a+b)$

$(c-a)-(d-b)$

$a-b$

$a+b$

Concepts:tangent-tangent angle circle arcs angle measure

Difficulty rating: 2310

Small Hint:

Draw $\overline{PR}$ , which is tangent to both semicircles at $R$

Big Hint:

Use the tangent-tangent angle theorem on each circle, then use $a+c=b+d=180^\circ$

Solution:

The line $PR$ is tangent to both semicircles at their common endpoint $R.$ For the larger circle, the tangent-tangent angle theorem gives $\angle APR=180^\circ-a=c.$ For the smaller circle, the same theorem gives $\angle RPB=180^\circ-b=d.$ Thus the reflex angle from $\overline{PA}$ to $\overline{PB}$ through $\overline{PR}$ has measure $c+d.$ Hence the other angle between the tangents is $\begin{aligned} 360^\circ-(c+d) &=(180^\circ-c)\\ &\quad{}+(180^\circ-d)\\ &=a+b, \end{aligned}$ because each upper semicircle has measure $180^\circ.$

Thus, the correct answer is E.

30.

Each of the equations $3x^2-2=25,$ $(2x-1)^2=(x-1)^2,$ $\sqrt{x^2-7}=\sqrt{x-1}$ has:

two integral roots

no root greater than $3$

no root zero

only one root

one negative root and one positive root

Concepts:equation solving radical domain common property

Difficulty rating: 1670

Small Hint:

Solve each equation separately and compare the properties of their roots

Big Hint:

For the radical equation, reject candidates that make either radicand negative

Solution:

The first equation gives $x=\pm3.$ Factoring the difference of squares in the second gives $\begin{aligned} &[(2x-1)-(x-1)]\\ &\quad\cdot[(2x-1)+(x-1)]\\ &\qquad=x(3x-2)=0, \end{aligned}$ so $x=0$ or $\frac{2}{3}.$ Squaring the third gives $x^2-x-6=0,$ with candidates $3$ and $-2;$ only $3$ satisfies the real-domain restrictions. Every root obtained is at most $3.$

Thus, each equation has no root greater than $3,$ and the correct answer is B.

31.

An equilateral triangle whose side is $2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides. If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:

$\dfrac{\sqrt6}{2}$

$\sqrt2$

$2+\sqrt2$

$\dfrac{2+\sqrt2}{2}$

$\dfrac{2\sqrt3-\sqrt6}{2}$

Difficulty rating: 1630

Small Hint:

The small triangle has half the original area, so determine its side using similarity

Big Hint:

The trapezoid median is the average of its two parallel side lengths

Solution:

The small triangle also has half the original area. If its side parallel to the original base has length $b,$ similarity gives $\left(\frac b2\right)^2=\frac12,$ so $b=\sqrt2.$ The parallel sides of the trapezoid have lengths $2$ and $\sqrt2,$ so its median has length $\frac{2+\sqrt2}{2}.$

Thus, the correct answer is D.

32.

If the discriminant of $ax^2+2bx+c=0$ is zero, then another true statement about $a,$ $b,$ and $c$ is that:

they form an arithmetic progression

they form a geometric progression

they are unequal

they are all negative numbers

only $b$ is negative and $a$ and $c$ are positive

Concepts:quadratic discriminant geometric progression

Difficulty rating: 1260

Small Hint:

Set $(2b)^2-4ac$ equal to zero

Big Hint:

Compare the resulting relation with the defining tests for arithmetic and geometric progressions

Solution:

A zero discriminant gives $(2b)^2-4ac=0,$ hence $b^2=ac.$ Equivalently, $\frac{a}{b}=\frac{b}{c}$ where the ratios are defined, which is the defining relation for $a,b,c$ to form a geometric progression.

Thus, the correct answer is B.

33.

Henry starts a trip when the hands of the clock are together between $8$ a.m. and $9$ a.m. He arrives at his destination between $2$ p.m. and $3$ p.m. when the hands of the clock are exactly $180^\circ$ apart. The trip takes:

$6$ hr.

$6$ hr. $43\dfrac7{11}$ min.

$5$ hr. $16\dfrac4{11}$ min.

$6$ hr. $30$ min.

none of these

Concepts:clock hands angular speed elapsed time

Difficulty rating: 1880

Small Hint:

At the start, the minute hand must close a $240^\circ$ gap at $5.5^\circ$ per minute

Big Hint:

Compute the corresponding time after $2{:}00$ when the hands are opposite and compare the two offsets

Solution:

At $8{:}00,$ the hour hand is $240^\circ$ ahead. The minute hand gains at $5.5^\circ$ per minute, so the hands coincide $\frac{240}{5.5}=\frac{480}{11}$ minutes after $8.$ At $2{:}00,$ the hour hand is $60^\circ$ ahead. For the hands to be $180^\circ$ apart in the relevant direction, the minute hand must gain $240^\circ,$ again taking $\frac{480}{11}$ minutes. The start and finish therefore have the same minute offset within their hours, exactly six hours apart.

Thus, the correct answer is A.

34.

A $6$ -inch-diameter pole and an $18$ -inch-diameter pole are placed together and bound together with wire. The length of the shortest wire that will go around them is:

$12\sqrt3+16\pi$

$12\sqrt3+7\pi$

$12\sqrt3+14\pi$

$12+15\pi$

$24\pi$

Concepts:common external tangents circle arcs perimeter

Difficulty rating: 2150

Small Hint:

The wire consists of two common external tangent segments and one exposed arc on each circle

Big Hint:

Use radii $3$ and $9$ ; each tangent segment has length $\sqrt{12^2-6^2}$

Solution:

The centers are $12$ inches apart and their radii differ by $6.$ Thus each common external tangent segment has length $\sqrt{12^2-6^2}=6\sqrt3,$ contributing $12\sqrt3$ in all. The geometry of the tangent lines leaves a $120^\circ$ exposed arc on the radius- $3$ circle and a $240^\circ$ exposed arc on the radius- $9$ circle. Their lengths total $\begin{aligned} &\frac{120}{360}(2\pi\cdot3)\\ &\quad{}+\frac{240}{360}(2\pi\cdot9)\\ &\qquad=2\pi+12\pi\\ &\qquad=14\pi. \end{aligned}$ Hence the wire length is $12\sqrt3+14\pi.$

Thus, the correct answer is C.

35.

Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:

is none of the following

cannot be determined from the given data

is $20$ or $26$

is $14$ or $32$

is $8$ or $38$

Concepts:fractions of a remainder integer constraints underdetermined problem

Difficulty rating: 1490

Small Hint:

Let the original number be $n$ and express the remainder after the first boy

Big Hint:

Translate the last two boys’ share condition into an equation and see whether it determines $n$ uniquely

Solution:

The first boy takes $\frac{n}{2}+1,$ leaving $\frac{n}{2}-1.$ The second takes one third of that remainder, and the third receives the other two thirds, automatically twice the second boy’s share. Thus the share condition imposes no unique value of $n.$ It only requires $\frac{n}{2}-1$ to be a nonnegative multiple of $3,$ so many values such as $8,14,20,26,\ldots$ work.

Therefore the original number cannot be determined, and the correct answer is B.

36.

A cylindrical oil tank, lying horizontally, has an interior length of $10$ feet and an interior diameter of $6$ feet. If the rectangular surface of the oil has an area of $40$ square feet, the depth of the oil is:

$\sqrt5$

$2\sqrt5$

$3-\sqrt5$

$3+\sqrt5$

either $3-\sqrt5$ or $3+\sqrt5$

Concepts:circle chord horizontal cylinder Pythagorean theorem

Difficulty rating: 1740

Small Hint:

The oil surface is a rectangle of length $10$ , so its width is $4$

Big Hint:

In the circular end, a chord of length $4$ lies $\sqrt{3^2-2^2}$ from the center

Solution:

The rectangular surface has length $10,$ so its chord width in the circular cross-section is $\frac{40}{10}=4.$ Half the chord is $2,$ and the radius is $3,$ so the distance from the center to the chord is $\sqrt{3^2-2^2}=\sqrt5.$ A chord of this length can lie either below or above the center. Measured from the bottom of the tank, the corresponding depths are $3-\sqrt5$ and $3+\sqrt5.$

Thus, the correct answer is E.

37.

A three-digit number has, from left to right, the digits $h,$ $t,$ and $u$ with $h>u.$ When the number with the digits reversed is subtracted from the original number, the units’ digit in the difference is $4.$ The next two digits, from right to left, are:

$5$ and $9$

$9$ and $5$

impossible to tell

$5$ and $4$

$4$ and $5$

Concepts:digit reversal modular arithmetic subtraction

Difficulty rating: 1570

Small Hint:

Subtract algebraically: $(100h+10t+u)$ ${}-(100u+10t+h)=99(h-u)$

Big Hint:

Find the digit $h-u$ for which $99(h-u)$ ends in $4$

Solution:

The difference is $99(h-u).$ Since $h-u$ is an integer from $1$ through $9,$ and the units digit is $4,$ we need $9(h-u)\equiv4\pmod{10}.$ This gives $h-u=6,$ so the difference is $99\cdot6=594.$ Moving from right to left after the units digit $4,$ the next digits are $9$ and $5.$

Thus, the correct answer is B.

38.

Four positive integers are given. Select any three of these integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers $29,$ $23,$ $21$ and $17$ are obtained. One of the original integers is:

$19$

$21$

$23$

$29$

$17$

Concepts:linear system averages sum invariant

Difficulty rating: 1840

Small Hint:

If the original sum is $S$ and the singled-out integer is $x,$ the result is $\frac{S+2x}{3}$

Big Hint:

Sum all four reported results to determine $S$

Solution:

If the original integers sum to $S,$ the result associated with singled-out integer $x$ is $x+\frac{S-x}{3}=\frac{S+2x}{3}.$ Summing all four reported results counts the total as $2S,$ so $S=\frac{29+23+21+17}{2}=45.$ The result $29$ therefore comes from $x=\dfrac{3(29)-45}{2}=21.$

Thus, one original integer is $21,$ and the correct answer is B.

39.

If $y=x^2+px+q,$ then if the least possible value of $y$ is zero, $q$ is equal to:

$0$

$\dfrac{p^2}{4}$

$\dfrac p2$

$-\dfrac p2$

$\dfrac{p^2}{4}-q$

Concepts:quadratic minimum completing the square vertex

Difficulty rating: 1260

Small Hint:

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

Big Hint:

The minimum occurs when $x=-\frac{p}{2}$

Solution:

Completing the square, $y=\left(x+\frac p2\right)^2+q-\frac{p^2}{4}.$ Its least value is $q-\frac{p^2}{4}.$ Setting this equal to zero gives $q=\frac{p^2}{4}.$

Thus, the correct answer is B.

40.

If $b\ne d,$ the fractions $\dfrac{ax+b}{cx+d}$ and $\dfrac bd$ are unequal if:

$a=c=1$ and $x\ne0$

$a=b=0$

$a=c=0$

$x=0$

$ad=bc$

Concepts:rational expressions cross multiplication inequalities

Difficulty rating: 1590

Small Hint:

Cross-multiply the equality $\frac{ax+b}{cx+d}=\frac{b}{d}$

Big Hint:

After cancellation, equality is governed by $x(ad-bc)=0$

Solution:

Where the fractions are defined, equality would require $d(ax+b)=b(cx+d),$ which simplifies to $x(ad-bc)=0.$ Under choice A, $a=c=1$ gives $ad-bc=d-b\ne0,$ and $x\ne0,$ so equality is impossible. Each of B through E instead forces equality directly.

Thus, the fractions are unequal under choice A.

41.

A train traveling from Aytown to Beetown meets with an accident after $1$ hr. It is stopped for $\dfrac12$ hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown $2$ hr. late. If the train had covered $80$ miles more before the accident, it would have been just $1$ hr. late. The usual rate of the train is:

$20$ mph

$30$ mph

$40$ mph

$50$ mph

$60$ mph

Concepts:rate-time-distance delay simultaneous conditions

Difficulty rating: 2310

Small Hint:

Traveling at $\frac{4}{5}$ speed adds one-fourth of the normal time for the affected distance

Big Hint:

Compare the two delays; moving the accident point $80$ miles changes the delay by $1$ hour

Solution:

Let the usual rate be $R$ mph and the total distance be $D.$ After the first hour, the normal time for the remaining distance is $\frac{D}{R}-1.$ Traveling it at $\frac{4R}{5}$ adds one-fourth of that time, so $\frac12+\frac14\left(\frac DR-1\right)=2.$ If the accident occurs $80$ miles later, $\frac12+\frac14\left(\frac DR-1-\frac{80}{R}\right)=1.$ Subtracting the second equation from the first gives $\frac{20}{R}=1,$ hence $R=20$ mph.

Thus, the correct answer is A.

42.

If $a,$ $b,$ and $c$ are positive integers, the radicals $\sqrt{a+\dfrac bc}$ and $a\sqrt{\dfrac bc}$ are equal when and only when:

$a=b=c=1$

$a=b$ and $c=a=1$

$c=\dfrac{b(a^2-1)}a$

$a=b$ and $c$ is any value

$a=b$ and $c=a-1$

Concepts:radical equation algebraic equivalence positive quantities

Difficulty rating: 1550

Small Hint:

Both sides are positive, so square the equality

Big Hint:

Multiply the resulting equation by $c$ and isolate $c$

Solution:

Because all quantities are positive, squaring is reversible: $a+\frac bc=a^2\frac bc.$ Multiplying by $c$ gives $ac+b=a^2b,$ so $c=\frac{b(a^2-1)}a.$

Thus, the correct answer is C.

43.

The pairs of values of $x$ and $y$ that are the common solutions of the equations $y=(x+1)^2$ and $xy+y=1$ are:

$3$ real pairs

$4$ real pairs

$4$ imaginary pairs

$2$ real and $2$ imaginary pairs

$1$ real and $2$ imaginary pairs

Concepts:simultaneous equations cube roots of unity complex solutions

Difficulty rating: 1720

Small Hint:

Rewrite the second equation as $y(x+1)=1$

Big Hint:

Substitute $y=(x+1)^2$ to obtain a cubic in $x+1$

Solution:

Substitution into $y(x+1)=1$ gives $(x+1)^3=1.$ The three cube roots of $1$ consist of one real root and two nonreal roots. Each determines exactly one corresponding value $y=(x+1)^2.$ Therefore there is one real pair and two imaginary pairs.

Thus, the correct answer is E.

44.

In circle $O$ chord $\overline{AB}$ is produced so that $\overline{BC}$ equals a radius of the circle. $\overline{CO}$ is drawn and extended to $D.$ $\overline{AO}$ is drawn. Which of the following expresses the relationship between angles $x$ and $y?$

$x=3y$

$x=2y$

$x=60^\circ$

there is no special relationship between $x$ and $y$

$x=2y$ or $x=3y,$ depending upon the length of $\overline{AB}$

Concepts:isosceles triangles exterior angle circle geometry

Difficulty rating: 1740

Small Hint:

Since $OB=BC,$ triangle $OBC$ is isosceles

Big Hint:

Use the exterior angle at $B$ , then use $OA=OB$

Solution:

Because $OB=BC,$ triangle $OBC$ is isosceles, so $\angle BOC=\angle BCO=y.$ Its exterior angle at $B$ is therefore $\angle ABO=2y.$ Since $OA=OB,$ triangle $AOB$ is also isosceles and $\angle OAB=2y.$ In triangle $AOC,$ the angles at $A$ and $C$ are $2y$ and $y,$ so $\angle AOC=180^\circ-3y.$ Angle $x=\angle AOD$ is supplementary to $\angle AOC,$ hence $x=3y.$

Thus, the correct answer is A.

45.

Given a geometric sequence with the first term $\ne0$ and $r\ne0$ and an arithmetic sequence with the first term $=0.$ A third sequence $1,$ $1,$ $2,$ $\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:

$978$

$557$

$467$

$1068$

not possible to determine from the information given

Concepts:geometric sequence arithmetic sequence systems of equations

Difficulty rating: 1930

Small Hint:

Write the two sequences as $a,ar,ar^2,\ldots$ and $0,d,2d,\ldots$

Big Hint:

Use the first three sums and the condition $r\ne0$ to determine $a,r,d$

Solution:

The first three termwise sums give $\begin{aligned} a&=1,\\ r+d&=1,\\ r^2+2d&=2. \end{aligned}$ Substituting $d=1-r$ yields $r(r-2)=0.$ Since $r\ne0,$ we have $r=2$ and $d=-1.$ The first ten geometric terms sum to $2^{10}-1=1023,$ while the first ten arithmetic terms sum to $0-1-\cdots-9=-45.$ Their combined sum is $1023-45=978.$

Thus, the correct answer is A.

46.

The graphs of $2x+3y-6=0,$ $4x-3y-6=0,$ $x=2,$ and $y=\dfrac23$ intersect in:

$6$ points

$1$ point

$2$ points

no points

an unlimited number of points

Concepts:linear equations common intersection coordinate geometry

Difficulty rating: 1340

Small Hint:

Solve the first two linear equations simultaneously

Big Hint:

Check whether that solution also satisfies each of the last two displayed equations

Solution:

Adding the first two equations gives $6x-12=0,$ so $x=2.$ Substitution gives $3y=2,$ hence $y=\frac{2}{3}.$ This point also lies on the last two given lines. Therefore all four graphs have the single common point $(2,\frac{2}{3}).$

Thus, the correct answer is B.

47.

The expressions $a+bc$ and $(a+b)(a+c)$ are:

always equal

never equal

equal when $a+b+c=1$

equal when $a+b+c=0$

equal only when $a=b=c=0$

Concepts:algebraic expansion sufficient condition equality

Difficulty rating: 1280

Small Hint:

Expand $(a+b)(a+c)$

Big Hint:

Subtract $a+bc$ and factor the difference

Solution:

The difference is $\begin{aligned} &(a+b)(a+c)-(a+bc)\\ &\qquad=a^2+ab+ac-a\\ &\qquad=a(a+b+c-1). \end{aligned}$ In particular, whenever $a+b+c=1,$ this difference is zero and the expressions are equal.

Thus, the correct answer is C.

48.

Given triangle $ABC$ with medians $\overline{AE},$ $\overline{BF},$ $\overline{CD};$ $\overline{FH}$ parallel and equal in length to $\overline{AE};$ $\overline{BH}$ and $\overline{HE}$ are drawn; $\overline{FE}$ extended meets $\overline{BH}$ in $G.$ Which one of the following statements is not necessarily correct?

$AEHF$ is a parallelogram

$\overline{HE}=\overline{HG}$

$\overline{BH}=\overline{DC}$

$\overline{FG}=\dfrac34\overline{AB}$

$\overline{FG}$ is a median of triangle $BFH$

Concepts:medians vectors parallelogram affine geometry

Difficulty rating: 2310

Small Hint:

Place $A=(0,0),$ $B=(2,0),$ $C=(u,v)$ and compute the midpoints $D,E,F$

Big Hint:

Use $\overrightarrow{FH}=\overrightarrow{AE}$ , then locate $G$ where $BH$ reaches height $\frac{v}{2}$

Solution:

Set $A=(0,0),$ $B=(2,0),$ $C=(u,v).$ Then $\begin{aligned} D&=(1,0),\\ E&=\left(\frac{u+2}{2},\frac v2\right),\\ F&=\left(\frac u2,\frac v2\right). \end{aligned}$ Since $\overrightarrow{FH}=\overrightarrow{AE},$ we get $H=(u+1,v).$ The horizontal line $FE$ meets $BH$ halfway from $B$ to $H,$ at $G=(\frac{u+3}{2},\frac{v}{2}).$ These coordinates verify that $AEHF$ is a parallelogram, $\overrightarrow{BH}=\overrightarrow{DC},$ $FG=\frac{3AB}{4},$ and $G$ is the midpoint of $BH,$ making $FG$ a median. But $\begin{aligned} HE&=\frac12\sqrt{u^2+v^2},\\ HG&=\frac12\sqrt{(u-1)^2+v^2}, \end{aligned}$ which are not generally equal.

Thus, statement B is not necessarily correct.

49.

The graphs of $y=\dfrac{x^2-4}{x-2}$ and $y=2x$ intersect in:

one point whose abscissa is $2$

one point whose abscissa is $0$

no points

two distinct points

two identical points

Concepts:rational function removable discontinuity graph intersection

Difficulty rating: 1400

Small Hint:

Factor $x^2-4$ , but keep the original restriction $x\ne2$

Big Hint:

Compare the algebraic intersection of the simplified lines with that domain restriction

Solution:

For $x\ne2,$ $\frac{x^2-4}{x-2}=x+2.$ The lines $y=x+2$ and $y=2x$ would meet at $x=2,\ y=4.$ But $x=2$ is excluded from the original rational function, so that point is a hole and there is no intersection.

Thus, the correct answer is C.

50.

In order to pass $B$ going $40$ mph on a two-lane highway $A,$ going $50$ mph, must gain $30$ feet. Meantime, $C,$ $210$ feet from $A,$ is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, then, in order to pass safely, $A$ must increase his speed by:

$30$ mph

$10$ mph

$5$ mph

$15$ mph

$3$ mph

Concepts:relative speed rate-time-distance passing

Difficulty rating: 1870

Small Hint:

Let $v$ be $A$ ’s new speed and equate the passing time to the time before $A$ and $C$ meet

Big Hint:

Use relative distances and speeds: $\frac{30}{v-40}=\frac{210}{v+50}$

Solution:

Let $v$ be $A$ ’s new speed. Relative to $B,$ $A$ gains at $v-40$ mph, so the passing time is proportional to $\frac{30}{v-40}.$ $A$ and $C$ close their $210$ -foot separation at $v+50$ mph, so their meeting time is proportional to $\frac{210}{v+50}.$ At the limiting safe time, $\frac{30}{v-40}=\frac{210}{v+50}.$ Thus $30v+1500=210v-8400,$ giving $v=55$ mph. $A$ must increase his original speed by $5$ mph.

Therefore, the correct answer is C.