The equation √x+4-√x-3+1=0 has:

(C) one real root and one imaginary root

For the equation (1+x)/(1-x)=(N+1)/(N) to be true where N is positive, x can have:

(A) any positive value less than 1

1956 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

The value of $x+x(x^x)$ when $x=2$ is:

$10$

$16$

$18$

$36$

$64$

Answer: A

Concepts:exponents order of operations substitution

Difficulty rating: 800

Small Hint:

Evaluate the exponent $x^x$ before doing the multiplication

Big Hint:

At $x=2,$ the second term is $2(2^2)$

Solution:

Substituting $x=2,$ and evaluating the exponent first, gives $\begin{aligned} x+x(x^x)&=2+2(2^2)\\ &=2+8=10. \end{aligned}$

Thus, the correct answer is A.

2.

Mr. Jones sold two pipes at $\$1.20$ each. Based on the cost his profit on one was $20\%$ and his loss on the other was $20\%.$ On the sale of the pipes, he:

broke even

lost $4$ ¢

gained $4$ ¢

lost $10$ ¢

gained $10$ ¢

Answer: D

Concepts:profit and loss reverse percentages cost price

Difficulty rating: 1340

Small Hint:

Recover each cost by dividing the sale price by $1.2$ or $0.8$

Big Hint:

Compare the combined cost with the combined selling price of $\$2.40$

Solution:

The pipe sold at a $20\%$ profit cost $\frac{\$1.20}{1.20}=\$1.00,$ while the pipe sold at a $20\%$ loss cost $\frac{\$1.20}{0.80}=\$1.50.$ Their combined cost was $\$2.50,$ but they sold for $\$2.40,$ so Mr. Jones lost $10$ cents.

Thus, the correct answer is D.

3.

The distance light travels in one year is approximately $5{,}870{,}000{,}000{,}000$ miles. The distance light travels in $100$ years is:

$587\times10^8$ miles

$587\times10^{10}$ miles

$587\times10^{-10}$ miles

$587\times10^{12}$ miles

$587\times10^{-12}$ miles

Answer: D

Concepts:scientific notation powers of ten multiplication

Difficulty rating: 950

Small Hint:

Multiplying by $100$ moves the decimal point two places to the right

Big Hint:

Rewrite the given distance as $587\times10^{10}$

Solution:

The given distance is $587\times10^{10}$ miles. Multiplying by $100=10^2$ gives $(587\times10^{10})10^2=587\times10^{12}$ miles.

Thus, the correct answer is D.

4.

A man has $\$10{,}000$ to invest. He invests $\$4{,}000$ at $5\%$ and $\$3{,}500$ at $4\%.$ In order to have a yearly income of $\$500,$ he must invest the remainder at:

$6\%$

$6.1\%$

$6.2\%$

$6.3\%$

$6.4\%$

Answer: E

Concepts:simple interest weighted investment percentages

Difficulty rating: 1410

Small Hint:

First find both the uninvested principal and the income still needed

Big Hint:

The first two investments earn $\$200$ and $\$140$ per year

Solution:

The amount left to invest is $10000-4000-3500=2500.$ The first two investments earn $0.05(4000)+0.04(3500)=340$ dollars, so the remaining investment must earn $500-340=160$ dollars. Its required rate is $\frac{160}{2500}=0.064=6.4\%.$

Thus, the correct answer is E.

5.

A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:

$4$

$5$

$6$

$8$

$12$

Answer: C

Concepts:circle packing central angles regular hexagon

Difficulty rating: 1070

Small Hint:

Join the center of the middle nickel to the centers of two neighboring nickels

Big Hint:

Those three mutually tangent equal circles make an equilateral triangle of centers

Solution:

The centers of the middle nickel and any two neighboring nickels are pairwise two radii apart, so they form an equilateral triangle. Therefore consecutive outer centers subtend $60^\circ$ at the center of the middle nickel. Exactly $\frac{360^\circ}{60^\circ}=6$ nickels fit around it.

Thus, the correct answer is C.

6.

In a group of cows and chickens, the number of legs was $14$ more than twice the number of heads. The number of cows was:

$5$

$7$

$10$

$12$

$14$

Answer: B

Concepts:linear equations heads and legs invariant

Difficulty rating: 1000

Small Hint:

Twice the number of heads already accounts for two legs per animal

Big Hint:

Each cow contributes two additional legs beyond that baseline

Solution:

Counting two legs for every head accounts for both legs of each chicken and two of each cow’s four legs. Thus every cow contributes exactly two of the $14$ extra legs. Hence the number of cows is $\frac{14}{2}=7.$

Thus, the correct answer is B.

7.

The roots of the equation $ax^2+bx+c=0$ will be reciprocal if:

$a=b$

$a=bc$

$c=a$

$c=b$

$c=ab$

Answer: C

Concepts:quadratic roots Vieta’s formulas reciprocal numbers

Difficulty rating: 1180

Small Hint:

If the roots are $r$ and $s,$ reciprocal roots satisfy $rs=1$

Big Hint:

Use Vieta’s formula $rs=\frac{c}{a}$

Solution:

By Vieta’s formulas, the product of the two roots is $\frac{c}{a}.$ Reciprocal roots have product $1,$ so $\frac ca=1,$ which requires $c=a.$

Thus, the correct answer is C.

8.

If $8\cdot2^x=5^{y+8},$ then, when $y=-8,$ $x=$

$-4$

$-3$

$0$

$4$

$8$

Answer: B

Concepts:exponential equations powers substitution

Difficulty rating: 1180

Small Hint:

Substitute $y=-8$ into the exponent on $5$

Big Hint:

Write $8$ as $2^3$

Solution:

When $y=-8,$ the right side is $5^0=1.$ Therefore $8\cdot2^x=2^{x+3}=1=2^0,$ so $x+3=0$ and $x=-3.$

Thus, the correct answer is B.

9.

Simplify $\left[\sqrt[3]{\sqrt[6]{a^9}}\right]^4\left[\sqrt[6]{\sqrt[3]{a^9}}\right]^4;$ the result is:

$a^{16}$

$a^{12}$

$a^8$

$a^4$

$a^2$

Answer: D

Concepts:rational exponents nested radicals exponent laws

Difficulty rating: 1280

Small Hint:

Replace each $n$ th root by an exponent of $\frac{1}{n}$

Big Hint:

Each bracket has exponent $9\cdot\frac16\cdot\frac13\cdot4$

Solution:

For each of the two factors, the nested roots and outer fourth power multiply the exponent of $a$ by $9\cdot\frac16\cdot\frac13\cdot4=2.$ Thus each factor is $a^2,$ and their product is $a^2a^2=a^4.$

Therefore, the correct answer is D.

10.

A circle of radius $10$ inches has its center at the vertex $C$ of an equilateral triangle $ABC$ and passes through the other two vertices. The side $AC$ extended through $C$ intersects the circle at $D.$ The number of degrees of angle $ADB$ is:

$15$

$30$

$60$

$90$

$120$

Answer: B

Concepts:inscribed angles equilateral triangle circle geometry

Difficulty rating: 1410

Small Hint:

The radii $CA$ and $CB$ form a $60^\circ$ central angle

Big Hint:

Angle $ADB$ is an inscribed angle intercepting the minor arc $AB$

Solution:

Since $ABC$ is equilateral, the central angle $\angle ACB$ is $60^\circ.$ The inscribed angle $\angle ADB$ intercepts the same minor arc $AB,$ so its measure is half the central angle: $\angle ADB=\frac{60^\circ}{2}=30^\circ.$

Thus, the correct answer is B.

11.

The expression $1-\dfrac1{1+\sqrt3}+\dfrac1{1-\sqrt3}$ equals:

$1-\sqrt3$

$1$

$-\sqrt3$

$\sqrt3$

$1+\sqrt3$

Answer: A

Concepts:rational expressions conjugates radicals

Difficulty rating: 1340

Small Hint:

Combine the two fractional terms over $(1+\sqrt3)(1-\sqrt3)$

Big Hint:

The product of the conjugate denominators is $1-3=-2$

Solution:

Combining the fractional terms, $\begin{gathered} -\frac1{1+\sqrt3}+\frac1{1-\sqrt3} \\ =\frac{(1+\sqrt3)-(1-\sqrt3)} {(1+\sqrt3)(1-\sqrt3)}\\ =\frac{2\sqrt3}{-2}\\ =-\sqrt3. \end{gathered}$ Adding the initial $1$ gives $1-\sqrt3.$

Thus, the correct answer is A.

12.

If $x^{-1}-1$ is divided by $x-1$ the quotient is:

$1$

$\dfrac1{x-1}$

$-\dfrac1{x-1}$

$\dfrac1x$

$-\dfrac1x$

Answer: E

Concepts:negative exponents rational expressions factoring

Difficulty rating: 1180

Small Hint:

Rewrite $x^{-1}-1$ as $\frac{1}{x}-1$

Big Hint:

Factor $1-x=-(x-1)$

Solution:

Where the quotient is defined, $\begin{aligned} \frac{x^{-1}-1}{x-1} &=\frac{\frac{1-x}{x}}{x-1}\\ &=\frac{-(x-1)}{x(x-1)}\\ &=-\frac1x. \end{aligned}$

Thus, the correct answer is E.

13.

Given two positive integers $x$ and $y$ with $x\lt y.$ The percent that $x$ is less than $y$ is:

$\dfrac{100(y-x)}x$

$\dfrac{100(x-y)}x$

$\dfrac{100(y-x)}y$

$100(y-x)$

$100(x-y)$

Answer: C

Concepts:percent decrease reference quantity algebraic expression

Difficulty rating: 1390

Small Hint:

The amount by which $x$ is smaller is $y-x$

Big Hint:

Because the comparison is to $y,$ use $y$ as the denominator

Solution:

The difference is $y-x.$ Measured as a fraction of the reference value $y,$ this is $\frac{y-x}{y}.$ Multiplying by $100$ converts the fraction to a percent: $\frac{100(y-x)}y.$

Thus, the correct answer is C.

14.

The points $A,$ $B,$ and $C$ are on a circle $O.$ The tangent line at $A$ and the secant $BC$ intersect at $P,$ $B$ lying between $C$ and $P.$ If $BC=20$ and $PA=10\sqrt3,$ then $PB$ equals:

$5$

$10$

$10\sqrt3$

$20$

$30$

Answer: B

Concepts:power of a point tangent-secant theorem quadratic equation

Difficulty rating: 1570

Small Hint:

Use $PA^2=PB\cdot PC$

Big Hint:

If $PB=t,$ then $PC=t+20$

Solution:

Let $PB=t.$ Since $B$ lies between $P$ and $C,$ we have $PC=t+20.$ The tangent-secant theorem gives $(10\sqrt3)^2=t(t+20).$ Thus $t^2+20t-300=0,$ or $(t-10)(t+30)=0.$ A length is positive, so $t=10.$

Therefore, the correct answer is B.

15.

The root(s) of $\dfrac{15}{x^2-4}-\dfrac2{x-2}=1$ is (are):

$-5$ and $3$

$\pm2$

$2$ only

$-3$ and $5$

$3$ only

Answer: A

Concepts:rational equation factoring domain restrictions

Difficulty rating: 1630

Small Hint:

The original equation requires $x\ne\pm2$

Big Hint:

Multiply through by $(x-2)(x+2)$ and collect terms

Solution:

For $x\ne\pm2,$ multiplying by $x^2-4$ gives $15-2(x+2)=x^2-4.$ Hence $x^2+2x-15=0,$ so $(x+5)(x-3)=0.$ Both $x=-5$ and $x=3$ satisfy the domain restriction.

Thus, the correct answer is A.

16.

The sum of three numbers is $98.$ The ratio of the first to the second is $\dfrac23,$ and the ratio of the second to the third is $\dfrac58.$ The second number is:

$15$

$20$

$30$

$32$

$33$

Answer: C

Concepts:ratios proportional quantities linear equation

Difficulty rating: 1380

Small Hint:

Choose a common scaling so the three numbers have ratio $10:15:24$

Big Hint:

Those ratio parts add to $49$

Solution:

The first-to-second ratio $2:3$ and second-to-third ratio $5:8$ combine to give the three-number ratio $10:15:24.$ The $49$ total parts equal $98,$ so each part is $2.$ The second number is $15\cdot2=30.$

Thus, the correct answer is C.

17.

The fraction $\dfrac{5x-11}{2x^2+x-6}$ was obtained by adding the two fractions $\dfrac A{x+2}$ and $\dfrac B{2x-3}.$ The values of $A$ and $B$ must be, respectively:

$5x,$ $-11$

$-11,$ $5x$

$-1,$ $3$

$3,$ $-1$

$5,$ $-11$

Answer: D

Concepts:partial fractions polynomial identity linear system

Difficulty rating: 1550

Small Hint:

Factor $2x^2+x-6=(x+2)(2x-3)$

Big Hint:

After combining the two fractions, match coefficients in $A(2x-3)+B(x+2)=5x-11$

Solution:

Combining the proposed partial fractions gives $\frac{A(2x-3)+B(x+2)}{(x+2)(2x-3)}.$ Matching its numerator with $5x-11$ yields $\begin{aligned} 2A+B&=5,\\ -3A+2B&=-11. \end{aligned}$ Solving gives $A=3$ and $B=-1.$

Thus, the correct answer is D.

18.

If $10^{2y}=25,$ then $10^{-y}$ equals:

$-\dfrac15$

$\dfrac1{625}$

$\dfrac1{50}$

$\dfrac1{25}$

$\dfrac15$

Answer: E

Concepts:exponential equations positive square root reciprocal

Difficulty rating: 1210

Small Hint:

Write $10^{2y}$ as $(10^y)^2$

Big Hint:

The quantity $10^y$ is positive

Solution:

We have $(10^y)^2=25.$ Since $10^y$ is positive, $10^y=5.$ Taking the reciprocal gives $10^{-y}=\frac{1}{5}.$

Thus, the correct answer is E.

19.

Two candles of the same height are lighted at the same time. The first is consumed in $4$ hours and the second in $3$ hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the height of the second?

$\dfrac34$ hr.

$1\dfrac12$ hr.

$2$ hr.

$2\dfrac25$ hr.

$2\dfrac12$ hr.

Answer: D

Concepts:linear rates proportional height equation modeling

Difficulty rating: 1540

Small Hint:

Scale the common initial height to $1$

Big Hint:

After $t$ hours the remaining fractions are $1-\frac{t}{4}$ and $1-\frac{t}{3}$

Solution:

Let each initial height be $1.$ After $t$ hours the remaining heights are $1-\frac{t}{4}$ and $1-\frac{t}{3}.$ The required condition is $1-\frac t4=2\left(1-\frac t3\right).$ Multiplying by $12$ gives $12-3t=24-8t,$ so $5t=12$ and $t=\frac{12}{5}=2\dfrac25.$

Thus, the correct answer is D.

20.

If $(0.2)^x=2$ and $\log 2=0.3010,$ then the value of $x$ to the nearest tenth is:

$-10.0$

$-0.5$

$-0.4$

$-0.2$

$10.0$

Answer: C

Concepts:logarithms exponential equation approximation

Difficulty rating: 1360

Small Hint:

Take common logarithms of both sides

Big Hint:

Use $\log(0.2)=\log2-1=-0.6990$

Solution:

Taking common logarithms gives $x\log(0.2)=\log2.$ Since $\log(0.2)=\log2-1=-0.6990,$ $x=\frac{0.3010}{-0.6990}\approx-0.4306.$ To the nearest tenth, this is $-0.4.$

Thus, the correct answer is C.

21.

If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is:

$2$

$2$ or $3$

$2$ or $4$

$3$ or $4$

$2,$ $3,$ or $4$

Answer: E

Concepts:hyperbola line-conic intersections counting distinct points

Difficulty rating: 1770

Small Hint:

A non-tangent line that intersects a hyperbola can meet it in either one or two points

Big Hint:

The two lines may share one hyperbola point, since the lines themselves intersect

Solution:

A line can meet a hyperbola in either one finite point, when its direction is parallel to an asymptote, or two finite points. By choosing the two lines so that their finite intersection sets on the hyperbola are disjoint, they can therefore contribute $1+1=2,$ $1+2=3,$ or $2+2=4$ distinct intersections. If their own intersection lies on the hyperbola, one point is shared instead.

Thus, $2,$ $3,$ or $4$ are possible, and the correct answer is E.

22.

Jones covered a distance of $50$ miles on his first trip. On a later trip he traveled $300$ miles while going three times as fast. His new time compared with the old time was:

three times as much

twice as much

the same

half as much

a third as much

Answer: B

Concepts:rate-time-distance proportional reasoning

Difficulty rating: 1150

Small Hint:

Write each travel time as distance divided by speed

Big Hint:

If the old speed is $v,$ compare $\frac{50}{v}$ with $\frac{300}{3v}$

Solution:

If the first speed was $v,$ the old time was $\frac{50}{v}.$ The later time was $\frac{300}{3v}=\frac{100}{v}=2\left(\frac{50}{v}\right).$ Thus the new time was twice the old time.

The correct answer is B.

23.

About the equation $ax^2-2x\sqrt2+c=0,$ with $a$ and $c$ real constants, we are told that the discriminant is zero. The roots are necessarily:

equal and integral

equal and rational

equal and real

equal and irrational

equal and imaginary

Answer: C

Concepts:quadratic discriminant repeated root real coefficients

Difficulty rating: 1280

Small Hint:

A zero discriminant makes the two quadratic-formula values coincide

Big Hint:

The repeated root is $-\frac{b}{2a}=\frac{\sqrt2}{a}$

Solution:

A quadratic with real coefficients and discriminant zero has the repeated root $x=\frac{-(-2\sqrt2)}{2a}=\frac{\sqrt2}{a}.$ Since $a$ is a nonzero real number, this root is real. It need not always be rational or always be irrational.

Thus, the roots are necessarily equal and real, so the correct answer is C.

24.

In the figure $AB=AC,$ angle $BAD=30^\circ,$ and $AE=AD.$ Then angle $CDE$ equals:

$7\dfrac12^\circ$

$10^\circ$

$12\dfrac12^\circ$

$15^\circ$

$20^\circ$

Answer: D

Concepts:isosceles triangles angle chasing supplementary angles

Difficulty rating: 2070

Small Hint:

Let $\angle DAE=\alpha$ and express the base angle at $C$ using isosceles triangle $ABC$

Big Hint:

Use $AE=AD$ to express $\angle ADE,$ then split the straight angle at $D$

Solution:

Let $\angle DAE=\alpha.$ Since $AB=AC$ and $AE=AD,$ $\begin{aligned} \angle ACB&=75^\circ-\frac\alpha2,\\ \angle ADE&=90^\circ-\frac\alpha2. \end{aligned}$ Triangle $ACD$ gives $\angle ADC=180^\circ-\alpha-\angle ACB,$ so $\angle ADC=105^\circ-\frac\alpha2.$ Since $\angle ADC=\angle ADE+x,$ $\begin{aligned} x&=\left(105^\circ-\frac\alpha2\right) -\left(90^\circ-\frac\alpha2\right)\\ &=15^\circ. \end{aligned}$

Thus, the correct answer is D.

25.

The sum of all numbers of the form $2k+1,$ where $k$ takes on integral values from $1$ to $n$ is:

$n^2$

$n(n+1)$

$n(n+2)$

$(n+1)^2$

$(n+1)(n+2)$

Answer: C

Concepts:arithmetic series summation algebraic simplification

Difficulty rating: 1490

Small Hint:

Separate $\sum_{k=1}^n(2k+1)$ into two sums

Big Hint:

Use $1+2+\cdots+n=\frac{n(n+1)}{2}$

Solution:

We compute $\begin{aligned} \sum_{k=1}^n(2k+1) &=2\sum_{k=1}^n k+\sum_{k=1}^n1\\ &=n(n+1)+n\\ &=n(n+2). \end{aligned}$

Thus, the correct answer is C.

26.

Which one of the following combinations of given parts does not determine the indicated triangle?

base angle and vertex angle; isosceles triangle

vertex angle and the base; isosceles triangle

the radius of the circumscribed circle; equilateral triangle

one arm and the radius of the inscribed circle; right triangle

two angles and a side opposite one of them; scalene triangle

Answer: A

Concepts:triangle determination similarity congruence data

Difficulty rating: 1570

Small Hint:

Ask whether each set of information fixes scale as well as shape

Big Hint:

Two angles determine only a similarity class unless some length is also supplied

Solution:

In an isosceles triangle, a base angle and the vertex angle determine all three angles, but no side length is specified. Therefore triangles of every scale have the same given data, so the triangle is not determined. Each other choice includes enough length information, directly or through a radius, to fix the scale as well as the shape.

Thus, the correct answer is A.

27.

If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:

$2$

$3$

$4$

$6$

more than $6$

Answer: C

Concepts:triangle area included angle scaling

Difficulty rating: 1180

Small Hint:

Use the area formula $\frac12ab\sin C$ for two sides and their included angle

Big Hint:

Doubling both side factors multiplies their product by $2\cdot2$

Solution:

If the included sides are $a$ and $b,$ and their unchanged angle is $C,$ the original area is $\frac12ab\sin C.$ After both sides are doubled, the area is $\begin{gathered} \frac12(2a)(2b)\sin C \\ =2ab\sin C\\ =4\left(\frac12ab\sin C\right). \end{gathered}$

Thus, the correct answer is C.

28.

Mr. J left his entire estate to his wife, his daughter, his son, and the cook. His daughter and son got half the estate, sharing in the ratio of $4$ to $3.$ His wife got twice as much as the son. If the cook received a bequest of $\$500,$ then the entire estate was:

$\$3500$

$\$5500$

$\$6500$

$\$7000$

$\$7500$

Answer: D

Concepts:ratios estate distribution linear equation

Difficulty rating: 1660

Small Hint:

Let the daughter’s and son’s shares be $4x$ and $3x$

Big Hint:

Their $7x$ is half the estate, while the other half is the wife’s $6x$ plus $\$500$

Solution:

Let the daughter receive $4x$ and the son $3x.$ Together they receive $7x,$ which is half the estate. The wife receives $6x,$ so the other half gives $7x=6x+500.$ Thus $x=500,$ and the whole estate is $2(7x)=14(500)=7000$ dollars.

Therefore, the correct answer is D.

29.

The points of intersection of $xy=12$ and $x^2+y^2=25$ are joined in succession. The resulting figure is:

a straight line

an equilateral triangle

a parallelogram

a rectangle

a square

Answer: D

Concepts:systems of equations coordinate geometry quadrilaterals

Difficulty rating: 1840

Small Hint:

Use $(x+y)^2=x^2+y^2+2xy$ to find the possible sums $x+y$

Big Hint:

The four intersection points are permutations and negatives of $(3,4)$

Solution:

At an intersection, $\begin{aligned} (x+y)^2&=x^2+y^2+2xy\\ &=25+24=49, \end{aligned}$ so $x+y=\pm7.$ Together with $xy=12,$ this gives the four points $\begin{aligned} &(3,4),\ (4,3),\\ &(-3,-4),\ (-4,-3). \end{aligned}$ In their cyclic order around the circle, adjacent side vectors are perpendicular, so the quadrilateral is a rectangle. Its adjacent side lengths are unequal, so it is not a square.

Thus, the correct answer is D.

30.

If the altitude of an equilateral triangle is $\sqrt6,$ then the area is:

$2\sqrt2$

$2\sqrt3$

$3\sqrt3$

$6\sqrt2$

$12$

Answer: B

Concepts:equilateral triangle altitude area

Difficulty rating: 1490

Small Hint:

For side length $s,$ an equilateral triangle has altitude $\frac{s\sqrt3}{2}$

Big Hint:

Once $s$ is known, use area $\frac{sh}{2}$

Solution:

Let the side length be $s.$ From $\frac{s\sqrt3}{2}=\sqrt6,$ we get $s=2\sqrt2.$ Hence the area is $\frac12s\sqrt6 =\frac12(2\sqrt2)(\sqrt6)=2\sqrt3.$

Thus, the correct answer is B.

31.

In our number system the base is ten. If the base were changed to four you would count as follows: $1,$ $2,$ $3,$ $10,$ $11,$ $12,$ $13,$ $20,$ $21,$ $22,$ $23,$ $30,$ $\ldots$ The twentieth number would be:

$20$

$38$

$44$

$104$

$110$

Answer: E

Concepts:base four place value number representation

Difficulty rating: 1470

Small Hint:

The twentieth positive number represents the ordinary value $20$

Big Hint:

Divide $20$ successively by $4,$ or express it using the place values $16,$ $4,$ and $1$

Solution:

The twentieth positive integer has ordinary value $20.$ Since $20=1\cdot4^2+1\cdot4+0,$ its base-four representation is $110.$

Thus, the correct answer is E.

32.

George and Henry started a race from opposite ends of the pool. After a minute and a half, they passed each other in the center of the pool. If they lost no time in turning and maintained their respective speeds, how many minutes after starting did they pass each other the second time?

$3$

$4\dfrac12$

$6$

$7\dfrac12$

$9$

Answer: B

Concepts:constant speed reflected motion periodic motion

Difficulty rating: 1550

Small Hint:

Meeting at the center means the two swimmers have equal speeds

Big Hint:

They reach the opposite ends at $3$ minutes, turn, and need another half-pool each

Solution:

Because they started simultaneously from opposite ends and first met at the center, their speeds are equal. Each reaches the opposite end after $3$ minutes. They turn immediately, and each then needs another $1.5$ minutes to return to the center, where they meet again. The second meeting occurs after $3+1.5=4.5=4\frac12$ minutes.

Thus, the correct answer is B.

33.

The number $\sqrt2$ is equal to:

a rational fraction

a finite decimal

$1.41421$

an infinite repeating decimal

an infinite non-repeating decimal

Answer: E

Concepts:irrational numbers decimal expansions square roots

Difficulty rating: 1340

Small Hint:

Recall what kind of decimal expansion every rational number has

Big Hint:

The number $\sqrt2$ is irrational, so its decimal neither terminates nor repeats

Solution:

The number $\sqrt2$ is irrational. A rational number has a decimal expansion that either terminates or eventually repeats, whereas an irrational number has an infinite non-repeating decimal expansion. The finite decimal $1.41421$ is only an approximation.

Thus, the correct answer is E.

34.

If $n$ is any whole number, $n^2(n^2-1)$ is always divisible by:

$12$

$24$

any multiple of $12$

$12-n$

$12$ and $24$

Answer: A

Concepts:divisibility consecutive integers parity

Difficulty rating: 1770

Small Hint:

Factor the expression as $n^2(n-1)(n+1)$

Big Hint:

Among three consecutive integers there is a multiple of $3,$ and the factors supply at least two powers of $2$

Solution:

We have $n^2(n^2-1)=n^2(n-1)(n+1).$ Among $n-1,$ $n,$ and $n+1,$ one is divisible by $3.$ If $n$ is even, $n^2$ is divisible by $4;$ if $n$ is odd, both $n-1$ and $n+1$ are even, so their product is divisible by $4.$ Thus the expression is always divisible by $12.$ It is not always divisible by $24,$ since $n=2$ gives $12.$

Therefore, the correct answer is A.

35.

A rhombus is formed by two radii and two chords of a circle whose radius is $16$ feet. The area of the rhombus in square feet is:

$128$

$128\sqrt3$

$256$

$512$

$512\sqrt3$

Answer: B

Concepts:rhombus chord length central angle area

Difficulty rating: 1880

Small Hint:

Every side of the rhombus has length $16,$ so each chord equals the radius

Big Hint:

A chord equal to the radius subtends a $60^\circ$ central angle; use $s^2\sin\theta$

Solution:

All four sides of the rhombus equal the circle’s radius, $16.$ A chord of length equal to the radius forms an equilateral triangle with the two radii to its endpoints, so the included central angle is $60^\circ.$ Therefore the rhombus has area $\begin{aligned} 16^2\sin60^\circ &=256\cdot\frac{\sqrt3}{2}\\ &=128\sqrt3. \end{aligned}$

Thus, the correct answer is B.

36.

If the sum $1+2+3+\cdots+K$ is a perfect square $N^2$ and if $N$ is less than $100,$ then the possible values for $K$ are:

only $1$

$1$ and $8$

only $8$

$8$ and $49$

$1,$ $8,$ and $49$

Answer: E

Concepts:triangular square numbers Pell equation recurrence

Difficulty rating: 2450

Small Hint:

Rewrite $\frac{K(K+1)}{2}=N^2$ as $(2K+1)^2-8N^2=1$

Big Hint:

Generate successive positive solutions by multiplying $(2K+1)+N\sqrt8$ by $3+\sqrt8,$ stopping when $N\ge100$

Solution:

The condition is $\frac{K(K+1)}2=N^2,$ or $(2K+1)^2-8N^2=1.$ The positive solutions of this Pell equation are generated from the fundamental solution $3+\sqrt8.$ Their first pairs $(2K+1,N)$ are $\begin{aligned} &(3,1),\ (17,6),\\ &(99,35),\ (577,204),\ldots \end{aligned}$ Thus the values with $N\lt100$ are $K=1,8,49.$

Therefore, the correct answer is E.

37.

On a map whose scale is $400$ miles to an inch and a half, a certain estate is represented by a rhombus having a $60^\circ$ angle. The diagonal opposite $60^\circ$ is $\dfrac3{16}$ in. The area of the estate in square miles is:

$\dfrac{2500}{\sqrt3}$

$\dfrac{1250}{\sqrt3}$

$1250$

$\dfrac{5625\sqrt3}{2}$

$1250\sqrt3$

Answer: E

Concepts:map scale rhombus equilateral triangles area

Difficulty rating: 2210

Small Hint:

In a $60^\circ$ rhombus, the shorter diagonal equals the side length

Big Hint:

The scale is $\frac{800}{3}$ miles per inch, so convert the $\frac{3}{16}$ -inch side before finding area

Solution:

The diagonal opposite the $60^\circ$ angle divides the rhombus into two equilateral triangles, so its length equals the rhombus side. The map scale is $\frac{800}{3}$ miles per inch, hence the actual side length is $\frac3{16}\cdot\frac{800}{3}=50$ miles. Therefore the rhombus area is $\begin{aligned} 50^2\sin60^\circ &=2500\cdot\frac{\sqrt3}{2}\\ &=1250\sqrt3. \end{aligned}$

Thus, the correct answer is E.

38.

In a right triangle with sides $a$ and $b,$ and hypotenuse $c,$ the altitude drawn on the hypotenuse is $x.$ Then:

$ab=x^2$

$\dfrac1a+\dfrac1b=\dfrac1x$

$a^2+b^2=2x^2$

$\dfrac1{x^2}=\dfrac1{a^2}+\dfrac1{b^2}$

$\dfrac1x=\dfrac ba$

Answer: D

Concepts:right triangle altitude to hypotenuse area reciprocal identity

Difficulty rating: 1810

Small Hint:

Compute the triangle’s area using either the legs or the hypotenuse and its altitude

Big Hint:

From $ab=cx,$ substitute $c^2=a^2+b^2$ and divide by $a^2b^2x^2$

Solution:

Equating two area formulas gives $\frac12ab=\frac12cx,$ so $ab=cx.$ Squaring and using $c^2=a^2+b^2$ yields $a^2b^2=x^2(a^2+b^2).$ Dividing by $a^2b^2x^2$ gives $\frac1{x^2}=\frac1{a^2}+\frac1{b^2}.$

Thus, the correct answer is D.

39.

The hypotenuse $c$ and one arm $a$ of a right triangle are consecutive integers. The square of the second arm is:

$ca$

$\dfrac ca$

$c+a$

$c-a$

none of these

Answer: C

Concepts:Pythagorean theorem difference of squares consecutive integers

Difficulty rating: 1470

Small Hint:

If the second arm is $b,$ then $b^2=c^2-a^2$

Big Hint:

Factor the difference of squares and use $c-a=1$

Solution:

By the Pythagorean theorem, $b^2=c^2-a^2=(c-a)(c+a).$ Since $c$ and $a$ are consecutive and $c\gt a,$ we have $c-a=1.$ Hence $b^2=c+a.$

Thus, the correct answer is C.

40.

If $V=gt+V_0$ and $S=\dfrac12gt^2+V_0t,$ then $t$ equals:

$\dfrac{2S}{V+V_0}$

$\dfrac{2S}{V-V_0}$

$\dfrac{2S}{V_0-V}$

$\dfrac{2S}{V}$

$2S-V$

Answer: A

Concepts:kinematics formulas algebraic elimination average velocity

Difficulty rating: 1530

Small Hint:

Use the first equation to replace $gt$ in the second

Big Hint:

Factor $t$ after writing $gt=V-V_0$

Solution:

From the first equation, $gt=V-V_0.$ Therefore $\begin{aligned} S&=\frac12(gt)t+V_0t\\ &=\frac12(V-V_0)t+V_0t\\ &=\frac12(V+V_0)t. \end{aligned}$ Solving gives $t=\frac{2S}{V+V_0}.$

Thus, the correct answer is A.

41.

The equation $3y^2+y+4=2(6x^2+y+2)$ where $y=2x$ is satisfied by:

no value of $x$

all values of $x$

$x=0$ only

all integral values of $x$ only

all rational values of $x$ only

Answer: C

Concepts:substitution quadratic cancellation linear equation

Difficulty rating: 1280

Small Hint:

Substitute $y=2x$ on both sides before expanding

Big Hint:

The $12x^2$ and constant terms cancel

Solution:

Substituting $y=2x$ gives $12x^2+2x+4=12x^2+4x+4.$ Cancelling the common quadratic and constant terms leaves $2x=4x,$ so $x=0.$ This value satisfies both equations.

Thus, the correct answer is C.

42.

The equation $\sqrt{x+4}-\sqrt{x-3}+1=0$ has:

no root

one real root

one real root and one imaginary root

two imaginary roots

two real roots

Answer: A

Concepts:radical equation inequalities real domain

Difficulty rating: 1280

Small Hint:

For real square roots the domain requires $x\ge3$

Big Hint:

On that domain, compare $\sqrt{x+4}$ directly with $\sqrt{x-3}$

Solution:

For a real solution, $x\ge3.$ Then $x+4>x-3,$ so $\sqrt{x+4}>\sqrt{x-3}.$ Consequently $\sqrt{x+4}-\sqrt{x-3}+1>1,$ and it cannot equal zero. The original equation is a real radical expression, so nonreal roots are outside its domain.

The correct answer is A.

43.

The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:

$1$

$2$

$3$

$4$

$18$

Answer: C

Concepts:integer triangles triangle inequality enumeration

Difficulty rating: 2030

Small Hint:

Order the distinct integer sides as $a\lt b\lt c$ and use $a+b\gt c$

Big Hint:

List possibilities by the largest side; the perimeter bound leaves only small values of $c$

Solution:

Write the distinct integer sides in increasing order. Checking the small possibilities under the triangle inequality and perimeter bound gives $\begin{aligned} &(2,3,4),\ (2,4,5),\\ &(3,4,5). \end{aligned}$ For $c\ge6,$ the smallest new scalene candidate satisfying $a+b\gt c$ is $(3,4,6),$ whose perimeter is already $13,$ and larger choices cannot qualify. Thus there are $3$ triangles.

The correct answer is C.

44.

If $x\lt a\lt0$ means that $x$ and $a$ are numbers such that $x$ is less than $a$ and $a$ is less than zero, then:

$x^2\lt ax\lt0$

$x^2\gt ax\gt a^2$

$x^2\lt a^2\lt0$

$x^2\gt ax$ but $ax\lt0$

$x^2\gt a^2$ but $a^2\lt0$

Answer: B

Concepts:inequalities negative numbers multiplication

Difficulty rating: 1340

Small Hint:

Both numbers are negative, but $x$ has the larger absolute value

Big Hint:

Multiply $x\lt a$ once by $x\lt0$ and once by $a\lt0,$ reversing each inequality

Solution:

Since $x\lt a\lt0,$ multiplying $x\lt a$ by the negative number $x$ reverses the inequality and gives $x^2\gt ax.$ Multiplying the same inequality by the negative number $a$ gives $ax\gt a^2.$ Therefore $x^2\gt ax\gt a^2.$

Thus, the correct answer is B.

45.

A wheel with a rubber tire has an outside diameter of $25$ in. When the radius has been decreased a quarter of an inch, the number of revolutions in one mile will:

be increased about $2\%$

be increased about $1\%$

be increased about $20\%$

be increased $\dfrac12\%$

remain the same

Answer: A

Concepts:circumference inverse proportion percent change

Difficulty rating: 1550

Small Hint:

For a fixed distance, the revolution count is inversely proportional to the wheel’s radius

Big Hint:

Compare the original radius $12.5$ with the new radius $12.25$

Solution:

The original radius is $12.5$ inches and the new radius is $12.25$ inches. For a fixed distance, the number of revolutions varies inversely with radius, so the relative increase is $\begin{aligned} \frac{12.5}{12.25}-1 &=\frac{50}{49}-1\\ &=\frac1{49}\\ &\approx2.04\%. \end{aligned}$ This is about $2\%.$

Thus, the correct answer is A.

46.

For the equation $\dfrac{1+x}{1-x}=\dfrac{N+1}{N}$ to be true where $N$ is positive, $x$ can have:

any positive value less than $1$

any value less than $1$

the value zero only

any non-negative value

any value

Answer: A

Concepts:rational equation parameter range inequalities

Difficulty rating: 1810

Small Hint:

Cross-multiply and solve for $x$ in terms of $N$

Big Hint:

The equation reduces to $x=\frac{1}{2N+1};$ also solve this relation for $N$

Solution:

Cross-multiplication gives $N(1+x)=(N+1)(1-x),$ so $x(2N+1)=1$ and $x=\frac1{2N+1}.$ Every positive $N$ gives $0\lt x\lt1.$ Conversely, for any $0\lt x\lt1,$ $N=\frac{1-x}{2x}>0,$ so such an $N$ exists.

Thus, $x$ may be any positive value less than $1,$ and the correct answer is A.

47.

An engineer said he could finish a highway section in $3$ days with his present supply of a certain type of machine. However, with $3$ more of these machines the job could be done in $2$ days. If the machines all work at the same rate, how many days would it take to do the job with one machine?

$6$

$12$

$15$

$18$

$36$

Answer: D

Concepts:work rates inverse proportion linear equation

Difficulty rating: 1490

Small Hint:

Let $m$ be the present number of machines and measure work in machine-days

Big Hint:

Equate $3m$ with $2(m+3)$

Solution:

If there are presently $m$ machines, the job requires $3m$ machine-days. With three more machines it requires $2(m+3)$ machine-days, so $3m=2(m+3),$ giving $m=6.$ The job therefore requires $3m=18$ machine-days, so one machine would take $18$ days.

Thus, the correct answer is D.

48.

If $p$ is a positive integer, then $\dfrac{3p+25}{2p-5}$ can be a positive integer, if and only if $p$ is:

at least $3$

equal to $3,$ $5,$ $9,$ or $35$

no more than $35$

equal to $35$

equal to $3$ or $35$

Answer: B

Concepts:integer divisibility rational expression factor divisors

Difficulty rating: 2210

Small Hint:

Let $q=2p-5;$ positivity forces $q$ to be a positive odd integer

Big Hint:

Rewrite the quotient as $\frac{3+\frac{65}{q}}{2},$ so $q$ must be a positive divisor of $65$

Solution:

Set $q=2p-5.$ A positive quotient requires $q>0,$ and $q$ is odd. Since $p=\frac{q+5}{2},$ $\begin{aligned} \frac{3p+25}{2p-5} &=\frac{3q+65}{2q}\\ &=\frac{3+\frac{65}{q}}{2}. \end{aligned}$ Thus $q$ must be a positive divisor of $65.$ The possibilities $q=1,$ $q=5,$ $q=13,$ and $q=65$ give $p=3,$ $p=5,$ $p=9,$ and $p=35,$ respectively, and each works. No other positive integer $p$ works.

Therefore the mathematically correct set is $\{3,5,9,35\},$ shown in the adjusted choice B.

49.

Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB=40^\circ;$ then angle $AOB$ equals:

$45^\circ$

$50^\circ$

$55^\circ$

$60^\circ$

$70^\circ$

Answer: E

Concepts:excircle angle bisectors triangle angles

Difficulty rating: 2030

Small Hint:

The circle is tangent to one side of triangle $PAB$ and the extensions of the other two, so $O$ is the excenter opposite $P$

Big Hint:

The external angle bisectors at $A$ and $B$ form an angle of $90^\circ-\frac12\angle P$

Solution:

The circle lies opposite $P$ across side $AB,$ so $O$ is the excenter opposite $P.$ Thus $AO$ and $BO$ bisect the exterior angles at $A$ and $B.$ If the interior angles at $A$ and $B$ are $\alpha$ and $\beta,$ then triangle $AOB$ has angles $90^\circ-\frac{\alpha}{2}$ and $90^\circ-\frac{\beta}{2}$ at $A$ and $B.$ Hence $\begin{aligned} \angle AOB &=180^\circ-\left(90^\circ-\frac\alpha2\right)\\ &\mathrel{\phantom{=}}-\left(90^\circ-\frac\beta2\right)\\ &=\frac{\alpha+\beta}{2} =\frac{180^\circ-40^\circ}{2}\\ &=70^\circ. \end{aligned}$

Thus, the correct answer is E.

50.

In triangle $ABC,$ $CA=CB.$ On $CB$ square $BCDE$ is constructed away from the triangle. If $x$ is the number of degrees in angle $DAB,$ then

$x$ depends upon triangle $ABC$

$x$ is independent of the triangle

$x$ may equal angle $CAD$

$x$ can never equal angle $CAB$

$x$ is greater than $45^\circ$ but less than $90^\circ$

Answer: B

Concepts:isosceles triangle square construction coordinate geometry

Difficulty rating: 2070

Small Hint:

Scale $CA=CB$ to $1,$ and place $C=(0,0),$ $B=(1,0),$ $A=(\cos\theta,\sin\theta)$

Big Hint:

Because the square is outside the triangle, take $D=(0,-1)$ and compare vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$

Solution:

Scale the equal sides to $1$ and place $\begin{aligned} C&=(0,0),\qquad B=(1,0),\\ A&=(\cos\theta,\sin\theta). \end{aligned}$ Since square $BCDE$ is constructed away from the triangle, $D=(0,-1).$ Put $c=\cos\theta$ and $s=\sin\theta.$ Then $\begin{aligned} \overrightarrow{AB}&=(1-c,-s),\\ \overrightarrow{AD}&=(-c,-1-s). \end{aligned}$ Their dot product is $1+s-c,$ while the absolute value of their two-dimensional cross product is also $1+s-c.$ Therefore $\tan\angle DAB=1,$ so $\angle DAB=45^\circ,$ independent of $\theta$ and hence independent of the triangle’s shape.

Thus, the correct answer is B.