1958 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The value of $\left[2-3(2-3)^{-1}\right]^{-1}$ is:

$5$

$-5$

$\dfrac15$

$-\dfrac15$

$\dfrac53$

Concepts:exponent order of operations fraction

Difficulty rating: 1120

Small Hint:

Evaluate the innermost difference before using the exponent $-1$

Big Hint:

An exponent of $-1$ means to take a reciprocal

Solution:

Since $2-3=-1,$ its reciprocal is also $-1.$ Therefore $\left[2-3(-1)\right]^{-1} =5^{-1}=\frac15.$

Thus, the correct answer is C.

2.

If $\dfrac1x-\dfrac1y=\dfrac1z,$ then $z$ equals:

$y-x$

$x-y$

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

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

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

Concepts:rational equation algebraic manipulation fraction

Difficulty rating: 1110

Small Hint:

Combine $\frac{1}{x}-\frac{1}{y}$ over a common denominator

Big Hint:

After finding $\frac{1}{z},$ take the reciprocal to obtain $z$

Solution:

Combining the fractions, $\frac1z=\frac1x-\frac1y =\frac{y-x}{xy}.$ Taking reciprocals gives $z=\frac{xy}{y-x}.$

Therefore, the correct answer is D.

3.

Of the following expressions, the one equal to $\dfrac{a^{-1}b^{-1}}{a^{-3}-b^{-3}}$ is:

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

$\dfrac{a^2b^2}{b^3-a^3}$

$\dfrac{ab}{b^3-a^3}$

$\dfrac{a^3-b^3}{ab}$

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

Concepts:exponent algebraic manipulation rational equation

Difficulty rating: 1360

Small Hint:

Rewrite every negative power as a reciprocal

Big Hint:

Express $a^{-3}-b^{-3}$ over the common denominator $a^3b^3$

Solution:

We have $a^{-3}-b^{-3} =\frac{b^3-a^3}{a^3b^3}.$ Hence $\frac{\frac{1}{ab}}{\frac{b^3-a^3}{a^3b^3}} =\frac{a^2b^2}{b^3-a^3}.$

Thus, the correct answer is B.

4.

In the expression $\dfrac{x+1}{x-1},$ each $x$ is replaced by $\dfrac{x+1}{x-1}.$ The resulting expression, evaluated for $x=\dfrac12,$ equals:

$3$

$-3$

$1$

$-1$

none of these

Concepts:function algebraic manipulation rational equation

Difficulty rating: 1300

Small Hint:

Let $f(x)=\frac{x+1}{x-1}$ and simplify $f(f(x))$

Big Hint:

Combine the numerator and denominator of the composed fraction separately

Solution:

Let $f(x)=\frac{x+1}{x-1}.$ Then $\begin{aligned} f(f(x)) &=\frac{\frac{x+1}{x-1}+1} {\frac{x+1}{x-1}-1}\\ &=\frac{\frac{2x}{x-1}}{\frac{2}{x-1}} =x. \end{aligned}$ At $x=\frac{1}{2},$ the value is $\frac{1}{2},$ which is not listed.

Therefore, the correct answer is E.

5.

The expression $2+\sqrt2+\dfrac1{2+\sqrt2}+\dfrac1{\sqrt2-2}$ equals:

$2$

$2-\sqrt2$

$2+\sqrt2$

$2\sqrt2$

$\dfrac{\sqrt2}{2}$

Concepts:radical rationalizing denominator algebraic manipulation

Difficulty rating: 1450

Small Hint:

Rationalize the two reciprocal terms before adding them

Big Hint:

The two reciprocals simplify to quantities whose sum is $-\sqrt2$

Solution:

Rationalizing gives $\begin{aligned} \frac1{2+\sqrt2}&=\frac{2-\sqrt2}{2},\\ \frac1{\sqrt2-2}&=-\frac{2+\sqrt2}{2}. \end{aligned}$ Their sum is $-\sqrt2.$ Therefore the whole expression is $2+\sqrt2-\sqrt2=2.$

Thus, the correct answer is A.

6.

The arithmetic mean between $\dfrac{x+a}{x}$ and $\dfrac{x-a}{x},$ when $x\ne0,$ is:

$2,$ if $a\ne0$

$1$

$1,$ only if $a=0$

$\dfrac ax$

$x$

Concepts:mean algebraic manipulation fraction

Difficulty rating: 960

Small Hint:

Add the two given expressions and divide by $2$

Big Hint:

The $a$ -terms cancel in the numerator

Solution:

The arithmetic mean is $\begin{aligned} &\frac12\left(\frac{x+a}{x}+\frac{x-a}{x}\right)\\ &\qquad=\frac12\left(\frac{2x}{x}\right)=1. \end{aligned}$

Therefore, the correct answer is B.

7.

A straight line joins the points $(-1,1)$ and $(3,9).$ Its $x$ -intercept is:

$-\dfrac32$

$-\dfrac23$

$\dfrac25$

$2$

$3$

Concepts:coordinate geometry slope linear equation

Difficulty rating: 1140

Small Hint:

First find the slope through the two given points

Big Hint:

Use one point to write the line equation, then set $y=0$

Solution:

The slope is $\frac{9-1}{3-(-1)}=2.$ Using $(-1,1),$ the line is $y=2x+3.$ Setting $y=0$ gives $x=-\frac32.$

Thus, the correct answer is A.

8.

Which of these four numbers, $\sqrt{\pi^2},$ $\sqrt[3]{0.8},$ $\sqrt[4]{0.00016},$ $\sqrt[3]{-1}\cdot\sqrt{(0.09)^{-1}},$ is (are) rational?

none

all

the first and fourth

only the fourth

only the first

Concepts:radical decimal exponent

Difficulty rating: 1590

Small Hint:

Evaluate each radical separately and remember that $\pi$ is irrational

Big Hint:

The last expression uses $\sqrt[3]{-1}=-1$ and $(0.09)^{-1}=\frac{100}{9}$

Solution:

The first number is $\pi,$ which is irrational. The cube root of $\frac{4}{5}$ is irrational. Also $\sqrt[4]{0.00016} =\sqrt[4]{16\cdot10^{-5}}$ is irrational. The fourth number is $(-1)\sqrt{\frac{100}{9}}=-\frac{10}{3},$ which is rational.

Therefore, only the fourth is rational, and the correct answer is D.

9.

A value of $x$ satisfying the equation $x^2+b^2=(a-x)^2$ is:

$\dfrac{b^2+a^2}{2a}$

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

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

$\dfrac{a-b}{2}$

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

Concepts:linear equation algebraic manipulation difference of squares

Difficulty rating: 1360

Small Hint:

Expand $(a-x)^2$

Big Hint:

The $x^2$ -terms cancel, leaving a linear equation in $x$

Solution:

Expanding the right side, $x^2+b^2=a^2-2ax+x^2.$ Cancelling $x^2$ and solving gives $\begin{aligned} 2ax&=a^2-b^2,\\ x&=\frac{a^2-b^2}{2a}. \end{aligned}$

Therefore, the correct answer is C.

10.

For what real values of $k,$ other than $k=0,$ does the equation $x^2+kx+k^2=0$ have real roots?

k<0

k>0

$k\ge1$

all values of $k$

no values of $k$

Concepts:quadratic algebraic manipulation

Difficulty rating: 1280

Small Hint:

Use the discriminant condition for a quadratic to have real roots

Big Hint:

Compute $k^2-4k^2$ and use the restriction $k\ne0$

Solution:

The discriminant is $k^2-4k^2=-3k^2.$ For every nonzero real $k,$ this is negative, so the quadratic has no real roots.

Thus, the correct answer is E.

11.

The number of roots satisfying the equation $\sqrt{5-x}=x\sqrt{5-x}$ is:

unlimited

$3$

$2$

$1$

$0$

Concepts:radical zero product property substitution

Difficulty rating: 1280

Small Hint:

Move both sides to one side and factor the common radical

Big Hint:

Check separately when the radical is zero and when its remaining factor is zero

Solution:

Factoring gives $(1-x)\sqrt{5-x}=0.$ Thus either $x=1$ or $5-x=0,$ giving $x=5.$ Both values lie in the domain $x\le5$ and satisfy the original equation.

Therefore, there are two roots, and the correct answer is C.

12.

If $P=\dfrac{s}{(1+k)^n},$ then $n$ equals:

$\dfrac{\log(\frac{s}{P})}{\log(1+k)}$

$\log\dfrac{s}{P(1+k)}$

$\log\dfrac{s-P}{1+k}$

$\log\dfrac{s}{P}+\log(1+k)$

$\dfrac{\log s}{\log(P(1+k))}$

Concepts:logarithm exponent algebraic manipulation

Difficulty rating: 1300

Small Hint:

Take logarithms of both sides of the equation

Big Hint:

Use $\log P=\log s-n\log(1+k)$ and isolate $n$

Solution:

Taking logarithms, $\log P=\log s-n\log(1+k).$ Therefore $n=\frac{\log s-\log P}{\log(1+k)} =\frac{\log(\frac{s}{P})}{\log(1+k)}.$

Thus, the correct answer is A.

13.

The sum of two numbers is $10;$ their product is $20.$ The sum of their reciprocals is:

$\dfrac1{10}$

$\dfrac12$

$1$

$2$

$4$

Concepts:symmetry (algebra)fraction algebraic manipulation

Difficulty rating: 1000

Small Hint:

Call the numbers $x$ and $y$ , but do not solve for them individually

Big Hint:

Use $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$

Solution:

If the numbers are $x$ and $y,$ then $\frac1x+\frac1y =\frac{x+y}{xy} =\frac{10}{20} =\frac12.$

Therefore, the correct answer is B.

14.

At a dance party a group of boys and girls exchange dances as follows: one boy dances with $5$ girls, a second boy dances with $6$ girls, and so on, the last boy dancing with all the girls. If $b$ represents the number of boys and $g$ the number of girls, then:

$b=g$

$b=\dfrac g5$

$b=g-4$

$b=g-5$

It is impossible to determine a relation between $b$ and $g$ without knowing $b+g$

Concepts:arithmetic sequence linear equation

Difficulty rating: 1320

Small Hint:

The numbers of partners form the sequence $5,6,7,\ldots,g$

Big Hint:

There are $b$ terms, so express the last term using the arithmetic-sequence formula

Solution:

The last of the $b$ terms in the sequence $5,6,7,\ldots$ is $5+(b-1)=b+4.$ This last number equals the total number $g$ of girls. Hence $g=b+4,$ or $b=g-4.$

Thus, the correct answer is C.

15.

A quadrilateral is inscribed in a circle. If an angle is inscribed into each of the four segments outside the quadrilateral, the sum of these four angles, expressed in degrees, is:

$1080$

$900$

$720$

$540$

$360$

Concepts:cyclic quadrilateral inscribed angle arc angle sum

Difficulty rating: 1630

Small Hint:

Let the four arcs cut off by the quadrilateral’s sides have measures summing to $360^\circ$

Big Hint:

An angle in the outside segment of a side subtends the complementary major arc

Solution:

Let the four minor arcs cut off by the sides have measures $\alpha_1,\ldots,\alpha_4,$ whose sum is $360^\circ.$ The angle in the outside segment corresponding to arc $\alpha_i$ subtends the other arc, so its measure is $\frac{360^\circ-\alpha_i}{2}.$ The sum of the four angles is therefore $\frac{4(360^\circ)-360^\circ}{2}=540^\circ.$

Thus, the correct answer is D.

16.

The area of a circle inscribed in a regular hexagon is $100\pi.$ The area of the hexagon is:

$600$

$300$

$200\sqrt2$

$200\sqrt3$

$120\sqrt5$

Concepts:incircle, incenter, and inradius regular polygon area

Difficulty rating: 1630

Small Hint:

The circle’s radius is the apothem of the regular hexagon

Big Hint:

For apothem $r,$ the hexagon side is $2r\tan30^\circ$

Solution:

The circle has radius $r=10,$ which is the hexagon’s apothem. Thus the side length is $2r\tan30^\circ=\frac{20}{\sqrt3}.$ Using one-half the product of perimeter and apothem, the hexagon’s area is $\frac12\left(6\cdot\frac{20}{\sqrt3}\right)(10) =200\sqrt3.$

Therefore, the correct answer is D.

17.

If $x$ is positive and $\log x\ge\log2+\dfrac12\log x,$ then:

$x$ has no minimum or maximum value

the maximum value of $x$ is $1$

the minimum value of $x$ is $1$

the maximum value of $x$ is $4$

the minimum value of $x$ is $4$

Concepts:logarithm inequality algebraic manipulation

Difficulty rating: 1280

Small Hint:

Subtract $\tfrac12\log x$ from both sides

Big Hint:

Double the resulting inequality and combine $2\log2$

Solution:

The inequality gives $\frac12\log x\ge\log2,$ so $\log x\ge2\log2=\log4.$ Hence $x\ge4,$ and the minimum possible value of $x$ is $4.$

Therefore, the correct answer is E.

18.

The area of a circle is doubled when its radius $r$ is increased by $n.$ Then $r$ equals:

$n(\sqrt2+1)$

$n(\sqrt2-1)$

$n$

$n(2-\sqrt2)$

$\dfrac{n\pi}{\sqrt2+1}$

Concepts:circle area radical algebraic manipulation

Difficulty rating: 1280

Small Hint:

Translate the doubled-area condition into $\pi(r+n)^2=2\pi r^2$

Big Hint:

After taking positive square roots, solve $r+n=\sqrt2\,r$

Solution:

The doubled-area condition is $\pi(r+n)^2=2\pi r^2.$ Since the radii are positive, $r+n=\sqrt2\,r.$ Thus $r=\frac{n}{\sqrt2-1} =n(\sqrt2+1).$

Thus, the correct answer is A.

19.

The sides of a right triangle are $a$ and $b$ and the hypotenuse is $c.$ A perpendicular from the vertex divides $c$ into segments $r$ and $s,$ adjacent respectively to $a$ and $b.$ If $a:b=1:3,$ then the ratio of $r$ to $s$ is:

$1:3$

$1:9$

$1:10$

$3:10$

$1:\sqrt{10}$

Concepts:right triangle similarity altitude

Difficulty rating: 1360

Small Hint:

Use the projection relations $a^2=cr$ and $b^2=cs$

Big Hint:

Divide the two projection equations to express $\frac{r}{s}$ in terms of $\frac{a}{b}$

Solution:

Similarity from the altitude-to-hypotenuse construction gives $a^2=cr,\qquad b^2=cs.$ Therefore $\frac rs=\frac{a^2}{b^2} =\left(\frac13\right)^2=\frac19.$

Thus, the correct answer is B.

20.

If $4^x-4^{x-1}=24,$ then $(2x)^x$ equals:

$5\sqrt5$

$\sqrt5$

$25\sqrt5$

$125$

$25$

Concepts:exponent factoring substitution

Difficulty rating: 1590

Small Hint:

Factor $4^{x-1}$ from the left side

Big Hint:

Once $4^x=32,$ rewrite both sides as powers of $2$

Solution:

We have $4^x-4^{x-1} =\frac34\,4^x=24,$ so $4^x=32.$ Thus $2^{2x}=2^5,$ giving $x=\frac{5}{2}.$ Therefore $(2x)^x=5^{\frac{5}{2}}=25\sqrt5.$

Therefore, the correct answer is C.

21.

In the accompanying figure $\overline{CE}$ and $\overline{DE}$ are equal chords of a circle with center $O.$ Arc $AB$ is a quarter-circle. Then the ratio of the area of triangle $CED$ to the area of triangle $AOB$ is:

$\sqrt2:1$

$\sqrt3:1$

$4:1$

$3:1$

$2:1$

Concepts:chord circle triangle area area ratio

Difficulty rating: 1590

Small Hint:

The equal chords from $E$ make the pictured triangle symmetric about the perpendicular diameter

Big Hint:

Compare each triangle’s base-height area in terms of the circle radius

Solution:

Let the circle have radius $r.$ In the pictured configuration, $\overline{CD}$ is a diameter and $E$ is the endpoint of the perpendicular radius, so $[CED]=\frac12(2r)(r)=r^2.$ Since arc $AB$ is a quarter-circle, $\angle AOB=90^\circ,$ and $[AOB]=\frac12r^2.$ The requested ratio is therefore $2:1.$

Thus, the correct answer is E.

22.

A particle is placed on the parabola $y=x^2-x-6$ at a point $P$ whose ordinate is $6.$ It is allowed to roll along the parabola until it reaches the nearest point $Q$ whose ordinate is $-6.$ The horizontal distance traveled by the particle (the numerical value of the difference in the abscissas of $P$ and $Q$ ) is:

$5$

$4$

$3$

$2$

$1$

Concepts:parabola quadratic coordinate geometry

Difficulty rating: 1360

Small Hint:

Find the two $x$ -coordinates where the parabola has ordinate $6$ and the two where it has ordinate $-6$

Big Hint:

Pair each upper point with the nearer lower point on the same branch

Solution:

For $y=6,$ $x^2-x-12=0,$ so $x=4$ or $-3.$ For $y=-6,$ $x^2-x=0,$ so $x=0$ or $1.$ The nearest lower point to $x=4$ is $x=1,$ and the nearest to $x=-3$ is $x=0.$ Either horizontal distance is $3.$

Therefore, the correct answer is C.

23.

If, in the expression $x^2-3,$ $x$ increases or decreases by a positive amount $a,$ the expression changes by an amount:

$\pm2ax+a^2$

$2ax\pm a^2$

$\pm a^2-3$

$(x+a)^2-3$

$(x-a)^2-3$

Concepts:polynomial algebraic manipulation substitution

Difficulty rating: 1150

Small Hint:

Compute $f(x+a)-f(x)$ and $f(x-a)-f(x)$ for $f(x)=x^2-3$

Big Hint:

The constant term cancels in both differences

Solution:

For an increase, $\begin{aligned} &(x+a)^2-3-(x^2-3)\\ &\qquad=2ax+a^2. \end{aligned}$ For a decrease, $\begin{aligned} &(x-a)^2-3-(x^2-3)\\ &\qquad=-2ax+a^2. \end{aligned}$ Together these changes are $\pm2ax+a^2.$

Thus, the correct answer is A.

24.

A man travels $m$ feet due north at $2$ minutes per mile. He returns due south to his starting point at $2$ miles per minute. The average rate in miles per hour for the entire trip is:

$75$

$48$

$45$

$24$

impossible to determine without knowing the value of $m$

Concepts:distance rate and time unit conversion harmonic mean

Difficulty rating: 1360

Small Hint:

Convert both stated rates to miles per hour

Big Hint:

For equal distances, divide twice the one-way distance by the sum of the two travel times

Solution:

The northbound rate is $30$ mph, and the southbound rate is $120$ mph. For equal distances, the round-trip average is $\frac{2(30)(120)}{30+120}=48\text{ mph}.$ The distance $m$ cancels.

Therefore, the correct answer is B.

25.

If $\log_kx\cdot\log_5k=3,$ then $x$ equals:

$k^6$

$5k^3$

$k^3$

$243$

$125$

Concepts:logarithm algebraic manipulation exponent

Difficulty rating: 1280

Small Hint:

Use the chain identity $\log_kx\cdot\log_5k=\log_5x$

Big Hint:

Convert the resulting logarithmic equation to exponential form

Solution:

By the change-of-base identity, $\log_kx\cdot\log_5k=\log_5x.$ Hence $\log_5x=3,$ so $x=5^3=125.$

Thus, the correct answer is E.

26.

A set of $n$ numbers has the sum $s.$ Each number of the set is increased by $20,$ then multiplied by $5,$ and then decreased by $20.$ The sum of the numbers in the new set thus obtained is:

$s+20n$

$5s+80n$

$s$

$5s$

$5s+4n$

Concepts:distributive property summation

Difficulty rating: 1390

Small Hint:

Apply all three operations to a typical original number $u$

Big Hint:

After simplifying the transformed number, sum over all $n$ entries

Solution:

An original number $u$ becomes $5(u+20)-20=5u+80.$ Summing over the $n$ original numbers therefore gives $5s+80n.$

Thus, the correct answer is B.

27.

The points $(2,-3),$ $(4,3),$ and $(5,\frac{k}{2})$ are on the same straight line. The value(s) of $k$ is (are):

$12$

$-12$

$\pm12$

$12$ or $6$

$6$ or $6\frac23$

Concepts:coordinate geometry slope linear equation

Difficulty rating: 1320

Small Hint:

Compute the slope through the first two points

Big Hint:

Equate that slope to the slope from $(4,3)$ to $(5,\frac{k}{2})$

Solution:

The slope through the first two points is $\frac{3-(-3)}{4-2}=3.$ Thus $\frac{\frac{k}{2}-3}{5-4}=3,$ which gives $\frac{k}{2}=6$ and $k=12.$

Therefore, the correct answer is A.

28.

A $16$ -quart radiator is filled with water. Four quarts are removed and replaced with pure antifreeze liquid. Then four quarts of the mixture are removed and replaced with pure antifreeze. This is done a third and a fourth time. The fractional part of the final mixture that is water is:

$\dfrac14$

$\dfrac{81}{256}$

$\dfrac{27}{64}$

$\dfrac{37}{64}$

$\dfrac{175}{256}$

Concepts:mixture geometric sequence process simulation

Difficulty rating: 1280

Small Hint:

Each removal takes away the same fraction of whatever water remains

Big Hint:

After one replacement, $\frac{3}{4}$ of the previous water remains

Solution:

Each removal takes one-fourth of the well-mixed radiator contents, so it leaves three-fourths of the water then present. After four replacements, the water fraction is $\left(\frac34\right)^4=\frac{81}{256}.$

Therefore, the correct answer is B.

29.

In a general triangle $ADE$ (as shown), lines $EB$ and $EC$ are drawn. Which of the following angle relations is true?

$x+z=a+b$

$y+z=a+b$

$m+x=w+n$

$x+z+n=w+c+m$

$x+y+n=a+b+m$

Concepts:angle sum angle chasing

Difficulty rating: 1830

Small Hint:

Write the angle sum in triangle $AEC$

Big Hint:

Write the angle sum in triangle $BED$ , then compare the two equations

Solution:

Triangle $AEC$ gives $x+y+w+n=180^\circ.$ Triangle $BED$ gives $m+a+b+w=180^\circ.$ Equating the left sides and cancelling $w$ yields $x+y+n=a+b+m.$

Thus, the correct answer is E.

30.

If $xy=b$ and $\dfrac1{x^2}+\dfrac1{y^2}=a,$ then $(x+y)^2$ equals:

$(a+2b)^2$

$a^2+b^2$

$b(ab+2)$

$ab(b+2)$

$\dfrac1a+2b$

Concepts:rational equation symmetry (algebra)algebraic manipulation

Difficulty rating: 1590

Small Hint:

Combine $\frac{1}{x^2}+\frac{1}{y^2}$ using the common denominator $x^2y^2$

Big Hint:

Use $xy=b$ to find $x^2+y^2$ , then add $2xy$

Solution:

Since $xy=b,$ $a=\frac{x^2+y^2}{x^2y^2} =\frac{x^2+y^2}{b^2},$ so $x^2+y^2=ab^2.$ Therefore $\begin{aligned} (x+y)^2 &=x^2+y^2+2xy\\ &=ab^2+2b\\ &=b(ab+2). \end{aligned}$

Therefore, the correct answer is C.

31.

The altitude drawn to the base of an isosceles triangle is $8,$ and the perimeter is $32.$ The area of the triangle is:

$56$

$48$

$40$

$32$

$24$

Concepts:isosceles triangle Pythagorean Theorem triangle area

Difficulty rating: 1550

Small Hint:

Let each equal side be $a$ and half the base be $b$

Big Hint:

Use $a+b=16$ and $a^2-b^2=8^2$

Solution:

Let each equal side be $a$ and half the base be $b.$ The perimeter gives $a+b=16.$ The altitude bisects the base, so $a^2-b^2=8^2.$ Thus $(a-b)(a+b)=64,$ giving $a-b=4.$ Hence $b=6,$ so the base is $12.$ The area is $\frac12(12)(8)=48.$

Therefore, the correct answer is B.

32.

With $\$1000$ a rancher is to buy steers at $\$25$ each and cows at $\$26$ each. If the number of steers $s$ and the number of cows $c$ are both positive integers, then:

this problem has no solution

there are two solutions with $s$ exceeding $c$

there are two solutions with $c$ exceeding $s$

there is one solution with $s$ exceeding $c$

there is one solution with $c$ exceeding $s$

Concepts:Diophantine Equation modular arithmetic bounding to limit cases

Difficulty rating: 1630

Small Hint:

Write $25s+26c=1000$ and reduce it modulo $25$

Big Hint:

The congruence and positivity leave only one possible positive value of $c$

Solution:

The purchase equation is $25s+26c=1000.$ Reducing modulo $25$ gives $c\equiv0\pmod{25}.$ Since $c$ is positive and $26c<1000,$ the only possibility is $c=25.$ Then $s=\frac{1000-26(25)}{25}=14.$ There is one solution, and $c>s.$

Thus, the correct answer is E.

33.

For one root of $ax^2+bx+c=0$ to be double the other, the coefficients $a,$ $b,$ $c$ must be related as follows:

$4b^2=9c$

$2b^2=9ac$

$2b^2=9a$

$b^2-8ac=0$

$9b^2=2ac$

Concepts:quadratic Vieta’s Formulas algebraic manipulation

Difficulty rating: 1630

Small Hint:

Represent the two roots as $q$ and $2q$

Big Hint:

Apply Vieta’s formulas to their sum and product, then eliminate $q$

Solution:

Let the roots be $q$ and $2q.$ Vieta’s formulas give $3q=-\frac ba,\qquad 2q^2=\frac ca.$ Squaring the first relation yields $9q^2=\frac{b^2}{a^2}.$ Dividing this by the second relation gives $\frac92=\frac{b^2}{ac},$ or $2b^2=9ac.$

Therefore, the correct answer is B.

34.

The numerator of a fraction is $6x+1,$ the denominator is $7-4x,$ and $x$ can have any value between $-2$ and $2,$ both included. The values of $x$ for which the numerator is greater than the denominator are:

$\dfrac35\lt x\le2$

$\dfrac35\le x\le2$

$0\lt x\le2$

$0\le x\le2$

$-2\le x\le2$

Concepts:inequality linear equation

Difficulty rating: 1110

Small Hint:

Compare the numerator and denominator directly by solving $6x+1\gt7-4x$

Big Hint:

Intersect the resulting inequality with the given interval $[-2,2]$

Solution:

The required comparison gives $6x+1\gt7-4x,$ so $10x\gt6$ and $x\gt\frac{3}{5}.$ Intersecting this with $-2\le x\le2$ gives $\frac35\lt x\le2.$

Thus, the correct answer is A.

35.

A triangle is formed by joining three points whose coordinates are integers. If the $x$ -unit and the $y$ -unit are each $1$ inch, then the area of the triangle, in square inches:

must be an integer

may be irrational

must be irrational

must be rational

will be an integer only if the triangle is equilateral

Concepts:lattice point determinant triangle area

Difficulty rating: 1360

Small Hint:

Translate one vertex to the origin and use the coordinate determinant for twice the area

Big Hint:

The determinant of integer coordinates is an integer

Solution:

After translating one vertex to the origin, write the other two as $(a,c)$ and $(b,d),$ with all coordinates integers. The area is $\frac12|ad-bc|.$ Since $ad-bc$ is an integer, the area is an integer or a half-integer, and in either case it is rational.

Therefore, the correct answer is D.

36.

The sides of a triangle are $30,$ $70,$ and $80$ units. If an altitude is dropped upon the side of length $80,$ the larger segment cut off on this side is:

$62$

$63$

$64$

$65$

$66$

Concepts:altitude Pythagorean Theorem right triangle

Difficulty rating: 1550

Small Hint:

Let the altitude divide the side of length $80$ into $x$ and $80-x$

Big Hint:

Equate the two expressions for the square of the altitude

Solution:

Let $x$ be the segment adjacent to the side of length $30.$ If the altitude has length $h,$ then $\begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2. \end{aligned}$ Cancelling $x^2$ and solving gives $x=15.$ The other segment is $80-15=65,$ which is the larger one.

Thus, the correct answer is D.

37.

The first term of an arithmetic series of consecutive integers is $k^2+1.$ The sum of $2k+1$ terms of this series may be expressed as:

$k^3+(k+1)^3$

$(k-1)^3+k^3$

$(k+1)^3$

$(k+1)^2$

$(2k+1)(k+1)^2$

Concepts:arithmetic sequence summation algebraic manipulation

Difficulty rating: 1630

Small Hint:

Find the last term after $2k$ increases from the first term

Big Hint:

Use the arithmetic-series average of the first and last terms

Solution:

The last term is $k^2+1+2k=(k+1)^2.$ The average of the first and last terms is $k^2+k+1.$ Therefore the sum is $\begin{aligned} S&=(2k+1)(k^2+k+1)\\ &=k^3+(k+1)^3. \end{aligned}$

Therefore, the correct answer is A.

38.

Let $r$ be the distance from the origin to a point $P$ with coordinates $x$ and $y.$ Designate the ratio $\frac{y}{r}$ by $s$ and the ratio $\frac{x}{r}$ by $c.$ Then the values of $s^2-c^2$ are limited to the numbers:

less than $-1$ and greater than $+1,$ both excluded

less than $-1$ and greater than $+1,$ both included

between $-1$ and $+1,$ both excluded

between $-1$ and $+1,$ both included

$-1$ and $+1$ only

Concepts:coordinate geometry Pythagorean Theorem inequality

Difficulty rating: 1280

Small Hint:

Use $r^2=x^2+y^2$ to relate $s^2+c^2$

Big Hint:

Rewrite $s^2-c^2$ as $2s^2-1$

Solution:

Since $r^2=x^2+y^2,$ $s^2+c^2=\frac{y^2+x^2}{r^2}=1.$ Thus $s^2-c^2=2s^2-1.$ Because $0\le s^2\le1,$ this expression ranges from $-1$ through $1,$ with both endpoints included.

Thus, the correct answer is D.

39.

We may say concerning the solution of $|x|^2+|x|-6=0$ that:

there is only one root

the sum of the roots is $1$

the sum of the roots is $0$

the product of the roots is $4$

the product of the roots is $-6$

Concepts:absolute value quadratic zero product property

Difficulty rating: 1280

Small Hint:

Set $u=|x|$ , so $u\ge0$

Big Hint:

Factor the resulting quadratic in $u$ , then translate the admissible value back to $x$

Solution:

Let $u=|x|\ge0.$ Then $\begin{aligned} u^2+u-6&=0,\\ (u+3)(u-2)&=0. \end{aligned}$ The nonnegative solution is $u=2,$ so $|x|=2$ and $x=\pm2.$ Their sum is $0.$

Therefore, the correct answer is C.

40.

Given $a_0=1,$ $a_1=3,$ and the general relation $a_n^2-a_{n-1}a_{n+1}=(-1)^n$ for $n\ge1.$ Then $a_3$ equals:

$\dfrac{13}{27}$

$33$

$21$

$10$

$-17$

Concepts:recursion substitution

Difficulty rating: 1210

Small Hint:

Use the recurrence first with $n=1$ to find $a_2$

Big Hint:

Then use it with $n=2$ to solve for $a_3$

Solution:

For $n=1,$ $3^2-(1)a_2=-1,$ so $a_2=10.$ For $n=2,$ $10^2-3a_3=1,$ which gives $a_3=33.$

Thus, the correct answer is B.

41.

The roots of $Ax^2+Bx+C=0$ are $r$ and $s.$ For the roots of

$x^2+px+q=0$

to be $r^2$ and $s^2,$ $p$ must equal:

$\dfrac{B^2-4AC}{A^2}$

$\dfrac{B^2-2AC}{A^2}$

$\dfrac{2AC-B^2}{A^2}$

$B^2-2C$

$2C-B^2$

Concepts:Vieta’s Formulas quadratic symmetry (algebra)

Difficulty rating: 1590

Small Hint:

For the new monic quadratic, $p$ is the negative of the sum $r^2+s^2$

Big Hint:

Write $r^2+s^2=(r+s)^2-2rs$ and use Vieta’s formulas for the original quadratic

Solution:

For the original quadratic, $r+s=-\frac BA,\qquad rs=\frac CA.$ Hence $\begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=\frac{B^2-2AC}{A^2}. \end{aligned}$ The coefficient $p$ is the negative of this sum, so $p=\frac{2AC-B^2}{A^2}.$

Therefore, the correct answer is C.

42.

In a circle with center $O,$ chord $\overline{AB}$ equals chord $\overline{AC}.$ Chord $\overline{AD}$ cuts $\overline{BC}$ in $E.$ If $AC=12$ and $AE=8,$ then $AD$ equals:

$27$

$24$

$21$

$20$

$18$

Concepts:chord inscribed angle similarity

Difficulty rating: 1790

Small Hint:

Compare triangles $AEC$ and $ACD$

Big Hint:

Angles $ACE$ and $ADC$ subtend the equal chords $AB$ and $AC$

Solution:

Because $E$ lies on $BC$ and $AD,$ triangles $AEC$ and $ACD$ share angle $A.$ Also $\angle ACE=\angle ACB$ subtends chord $AB,$ while $\angle ADC$ subtends chord $AC.$ Since $AB=AC,$ these angles are equal. Thus $\triangle AEC\sim\triangle ACD.$ Corresponding sides give $\frac{AE}{AC}=\frac{AC}{AD}.$ Therefore $\frac{8}{12}=\frac{12}{AD},$ so $AD=18.$

Thus, the correct answer is E.

43.

$\overline{AB}$ is the hypotenuse of a right triangle $ABC.$ Median $AD=7$ and median $BE=4.$ The length of $AB$ is:

$10$

$5\sqrt3$

$5\sqrt2$

$2\sqrt{13}$

$2\sqrt{15}$

Concepts:median (geometry)Stewart’s Theorem Pythagorean Theorem

Difficulty rating: 1990

Small Hint:

Let the legs opposite $A$ and $B$ have squared lengths $u$ and $v$

Big Hint:

Use the two median formulas together with the Pythagorean relation for the hypotenuse

Solution:

Let $a=BC,$ $b=CA,$ and $c=AB.$ Since the right angle is at $C,$ $c^2=a^2+b^2.$ The median formulas give $\begin{aligned} 4(7^2) &=a^2+4b^2,\\ 4(4^2) &=4a^2+b^2. \end{aligned}$ Solving yields $a^2=4$ and $b^2=48.$ Therefore $\begin{aligned} AB=c &=\sqrt{a^2+b^2}\\ &=\sqrt{52}=2\sqrt{13}. \end{aligned}$

Thus, the correct answer is D.

44.

Given the true statements:

$1.$ If $a$ is greater than $b,$ then $c$ is greater than $d.$

$2.$ If $c$ is less than $d,$ then $e$ is greater than $f.$

A valid conclusion is:

If $a$ is less than $b,$ then $e$ is greater than $f$

If $e$ is greater than $f,$ then $a$ is less than $b$

If $e$ is less than $f,$ then $a$ is greater than $b$

If $a$ is greater than $b,$ then $e$ is less than $f$

none of these

Concepts:logical deduction counterexample

Difficulty rating: 1360

Small Hint:

Represent the comparisons as propositions and distinguish each implication from its converse

Big Hint:

The second premise’s hypothesis is incompatible with the first premise’s conclusion, but that alone does not chain the implications

Solution:

Let $P$ mean $a>b,$ $Q$ mean $c>d,$ $S$ mean $c<d,$ and $R$ mean $e>f.$ The premises are $P\Rightarrow Q,\qquad S\Rightarrow R.$ The statements $Q$ and $S$ cannot both hold. The contrapositives are $\neg Q\Rightarrow\neg P$ and $\neg R\Rightarrow\neg S.$ None of the four proposed implications follows. For example, knowing $P$ gives $Q$ and hence $\neg S,$ but says nothing about $R;$ knowing $\neg R$ gives $\neg S,$ not $P.$

Therefore, the correct answer is E.

45.

A check is written for $x$ dollars and $y$ cents, $x$ and $y$ both two-digit numbers. In error it is cashed for $y$ dollars and $x$ cents, the incorrect amount exceeding the correct amount by $\$17.82.$ Then:

$x$ cannot exceed $70$

$y$ can equal $2x$

the amount of the check cannot be a multiple of $5$

the incorrect amount can equal twice the correct amount

the sum of the digits of the correct amount is divisible by $9$

Concepts:money digits linear equation

Difficulty rating: 1790

Small Hint:

Write the correct and incorrect amounts in cents

Big Hint:

Their difference simplifies to $99(y-x)$

Solution:

In cents, the incorrect amount minus the correct amount is $\begin{aligned} &(100y+x)-(100x+y)\\ &\qquad=99(y-x). \end{aligned}$ Since $\$17.82$ is $1782$ cents, $99(y-x)=1782,$ so $y-x=18.$ The two-digit values $x=18,\ y=36$ satisfy this relation and have $y=2x.$ Thus that equality can occur.

Therefore, the correct answer is B.

46.

For values of $x$ less than $1$ but greater than $-4,$ the expression

$\frac{x^2-2x+2}{2x-2}$

has:

no maximum or minimum value

a minimum value of $1$

a maximum value of $1$

a minimum value of $-1$

a maximum value of $-1$

Concepts:AM-GM Inequality inequality optimization

Difficulty rating: 1830

Small Hint:

Set $t=x-1,$ which is negative on the given interval

Big Hint:

Rewrite the expression as $\tfrac12(t+\frac{1}{t})$ and use the negative- $t$ form of AM-GM

Solution:

Set $t=x-1.$ Then $-5\lt t\lt0,$ and the expression becomes $\frac{t^2+1}{2t} =\frac12\left(t+\frac1t\right).$ For $t\lt0,$ $t+\frac{1}{t}\le-2,$ with equality when $t=-1.$ That value is allowed and corresponds to $x=0.$ Hence the expression is at most $-1,$ and its maximum is $-1.$

Thus, the correct answer is E.

47.

$ABCD$ is a rectangle (see the accompanying diagram) with $P$ any point on $\overline{AB}.$ $PS\perp BD$ and $PR\perp AC.$ $AF\perp BD$ and $PQ\perp AF.$ Then $PR+PS$ is equal to:

$PQ$

$AE$

$PT+AT$

$AF$

$EF$

Concepts:rectangle parallel lines similarity

Difficulty rating: 2070

Small Hint:

Because $PQ\parallel BD$ and $PS\parallel AF,$ identify the small parallelogram near $F$ and $S$

Big Hint:

Use the intersection $T=PQ\cap AC$ to compare the right triangles $PTR$ and $ATQ$

Solution:

Since $AF\perp BD$ and $PQ\perp AF,$ we have $PQ\parallel BD.$ Also $PS\perp BD,$ so $PS\parallel AF.$ Thus quadrilateral $QPSF$ is a rectangle, and $PS=QF.$ Let $T=PQ\cap AC.$ The diagonals of a rectangle make equal angles with side $AB,$ so $\angle PAT=\angle APT,$ giving $AT=PT.$ The right triangles $ATQ$ and $PTR$ are similar because they share the angle at $T.$ Since their hypotenuses $AT$ and $PT$ are equal, $AQ=PR.$ Therefore $PR+PS=AQ+QF=AF.$

Thus, the correct answer is D.

48.

Diameter $\overline{AB}$ of a circle with center $O$ is $10$ units. $C$ is a point $4$ units from $A,$ and on $\overline{AB}.$ $D$ is a point $4$ units from $B,$ and on $\overline{AB}.$ $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D:$

has the same length for all positions of $P$

exceeds $10$ units for all positions of $P$

cannot exceed $10$ units

is shortest when $CPD$ is a right triangle

is longest when $P$ is equidistant from $C$ and $D$

Concepts:circle distance formula optimization symmetry

Difficulty rating: 1990

Small Hint:

Place the center at the origin and the diameter on the $x$ -axis, so $C=(-1,0)$ and $D=(1,0)$

Big Hint:

For $P=(u,v)$ on the circle, compare $CP^2=26+2u$ and $DP^2=26-2u$

Solution:

Place $O=(0,0),$ $C=(-1,0),$ $D=(1,0),$ and $P=(u,v)$ with $u^2+v^2=25.$ Then $\begin{aligned} CP^2&=26+2u,\\ DP^2&=26-2u. \end{aligned}$ Therefore $\begin{aligned} (CP+DP)^2 &=52\\ &\quad+2\sqrt{676-4u^2}. \end{aligned}$ This is largest when $u=0,$ exactly when $CP=DP.$

Therefore, the correct answer is E.

49.

In the expansion of $(a+b)^n$ there are $n+1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a+b+c)^{10}$ is:

$11$

$33$

$55$

$66$

$132$

Concepts:stars and bars combinations polynomial

Difficulty rating: 1340

Small Hint:

A term is determined by nonnegative exponents $i,j,k$ whose sum is $10$

Big Hint:

Count the solutions of $i+j+k=10$ by stars and bars

Solution:

Each distinct term is $a^ib^jc^k$ for nonnegative integers satisfying $i+j+k=10.$ By stars and bars, the number of such triples is $\binom{10+3-1}{3-1}=\binom{12}{2}=66.$

Thus, the correct answer is D.

50.

In this diagram a scheme is indicated for associating all the points of segment $AB$ with those of segment $A'B',$ and reciprocally. To describe this association scheme analytically, let $x$ be the distance from a point $P$ on $AB$ to $D$ and let $y$ be the distance from the associated point $P'$ of $A'B'$ to $D'.$ Then for any pair of associated points, if $x=a,$ $x+y$ equals:

$13a$

$17a-51$

$17-3a$

$\dfrac{17-3a}{4}$

$12a-34$

Difficulty rating: 1830

Small Hint:

The perspective lines in the diagram associate $x=3$ with $y=5$ and $x=4$ with $y=1$

Big Hint:

Because the two numbered segments are parallel, the induced relation between $x$ and $y$ is linear

Solution:

The two numbered segments are parallel, so projection through the fixed intersection point gives a linear relation between $x$ and $y.$ The endpoint associations shown are $(x,y)=(3,5)$ and $(x,y)=(4,1).$ The slope is $\frac{1-5}{4-3}=-4,$ so $y=-4x+17.$ If $x=a,$ then $\begin{aligned} x+y &=a+(-4a+17)\\ &=17-3a. \end{aligned}$

Therefore, the correct answer is C.