1953 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $\$1.00,$ he must sell:

$67$ oranges

$150$ oranges

$200$ oranges

an infinite number of oranges

none of these

Concepts:unit rate profit arithmetic

Difficulty rating: 890

Small Hint:

Find the buying cost and selling price per orange

Big Hint:

The profit per orange is the difference of the two unit rates

Solution:

Each orange costs $\frac{10}{3}$ cents and sells for $4$ cents, so the profit is $4-\frac{10}{3}=\frac{2}{3}$ cent per orange. To earn $100$ cents, he must sell $\frac{100}{\frac{2}{3}}=150$ oranges.

Thus, the correct answer is B.

2.

A refrigerator is offered for sale at $\$250.00$ less successive discounts of $20\%$ and $15\%.$ The sale price of the refrigerator is:

$35\%$ less than $\$250.00$

$65\%$ of $\$250.00$

$77\%$ of $\$250.00$

$68\%$ of $\$250.00$

none of these

Concepts:successive discounts percentage multiplication

Difficulty rating: 890

Small Hint:

Apply each discount to the price remaining after the preceding discount

Big Hint:

Multiply the original price by $(1-0.20)(1-0.15)$

Solution:

The two discounts leave $(0.80)(0.85)=0.68$ of the original price. The sale price is therefore $68\%$ of $\$250.00,$ or $\$170.00.$

Thus, the correct answer is D.

3.

The factors of the expression $x^2+y^2$ are:

$(x+y)(x-y)$

$(x+y)^2$

$(x^{\frac{2}{3}}+y^{\frac{2}{3}})(x^{\frac{4}{3}}+y^{\frac{4}{3}})$

$(x+iy)(x-iy)$

none of these

Concepts:complex numbers sum of squares factoring

Difficulty rating: 1340

Small Hint:

Over the complex numbers, $i^2=-1$

Big Hint:

Rewrite $y^2$ as $-(iy)^2$ and use a difference of squares

Solution:

Because $i^2=-1,$ $\begin{aligned} (x+iy)(x-iy)&=x^2-(iy)^2\\ &=x^2+y^2. \end{aligned}$

Thus, the correct answer is D.

4.

The roots of $x(x^2+8x+16)(4-x)=0$ are:

$0$

$0,$ $4$

$0,$ $4,$ $-4$

$0,$ $4,$ $-4,$ $-4$

none of these

Concepts:polynomial roots factoring multiplicity

Difficulty rating: 1290

Small Hint:

Factor the quadratic completely

Big Hint:

The factor $x^2+8x+16$ is a perfect square, so one root is repeated

Solution:

The equation factors as $x(x+4)^2(4-x)=0.$ Its roots, counted with multiplicity, are $0,$ $4,$ $-4,$ and $-4.$

Thus, the correct answer is D.

5.

If $\log_6 x=2.5,$ the value of $x$ is:

$90$

$36$

$36\sqrt6$

$0.5$

none of these

Concepts:logarithms fractional exponents radicals

Difficulty rating: 1260

Small Hint:

Rewrite the logarithmic equation in exponential form

Big Hint:

Use $2.5=2+\tfrac12$

Solution:

The equation gives $x=6^{2.5}=6^2\sqrt6=36\sqrt6.$

Thus, the correct answer is C.

6.

Charles has $5q+1$ quarters and Richard has $q+5$ quarters. The difference in their money in dimes is:

$10(q-1)$

$\dfrac25(4q-4)$

$\dfrac25(q-1)$

$\dfrac52(q-1)$

none of these

Concepts:currency conversion algebraic expression

Difficulty rating: 1070

Small Hint:

First subtract the two numbers of quarters

Big Hint:

One quarter is $\frac{5}{2}$ dimes

Solution:

The difference is $(5q+1)-(q+5)=4q-4$ quarters. Multiplying by $\frac{5}{2}$ dimes per quarter gives $\frac52(4q-4)=10(q-1).$

Thus, the correct answer is A.

7.

The fraction $\frac{\sqrt{a^2+x^2}-\dfrac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}$ reduces to:

$0$

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

$\dfrac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}$

$\dfrac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}$

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

Concepts:radicals complex fractions algebraic simplification

Difficulty rating: 1450

Small Hint:

Combine the two terms in the numerator over a common denominator

Big Hint:

The new numerator simplifies to $(a^2+x^2)-(x^2-a^2)$

Solution:

The numerator is $\begin{aligned} &\frac{a^2+x^2-(x^2-a^2)} {\sqrt{a^2+x^2}}\\ &\qquad{}=\frac{2a^2}{\sqrt{a^2+x^2}}. \end{aligned}$ Dividing by $a^2+x^2$ gives $\frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}.$

Thus, the correct answer is D.

8.

The value of $x$ at the intersection of $y=\dfrac8{x^2+4}$ and $x+y=2$ is:

$-2+\sqrt5$

$-2-\sqrt5$

$0$

$2$

none of these

Concepts:system of equations rational equation factoring

Difficulty rating: 1400

Small Hint:

Substitute $y=2-x$ into the rational equation

Big Hint:

After clearing the denominator, factor out $x$

Solution:

Substitution gives $(2-x)(x^2+4)=8.$ Expanding and simplifying, $x(x^2-2x+4)=0.$ The quadratic factor has negative discriminant, so the real intersection has $x=0.$

Thus, the correct answer is C.

9.

The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50\%$ alcohol to a lotion containing $30\%$ alcohol is:

$3$

$4$

$5$

$6$

$7$

Concepts:mixture percentage equation

Difficulty rating: 1070

Small Hint:

The amount of alcohol stays fixed when water is added

Big Hint:

Set $\frac{4.5}{9+w}=0.30$

Solution:

The lotion initially contains $9(0.50)=4.5$ ounces of alcohol. If $w$ ounces of water are added, then $\frac{4.5}{9+w}=0.30.$ Hence $9+w=15$ and $w=6.$

Thus, the correct answer is D.

10.

The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:

$880$

$\dfrac{440}{\pi}$

$\dfrac{880}{\pi}$

$440\pi$

none of these

Concepts:circle circumference unit conversion revolutions

Difficulty rating: 1180

Small Hint:

In one revolution the rim point travels one circumference

Big Hint:

Divide $5280$ feet by the circumference $6\pi$ feet

Solution:

The wheel’s circumference is $6\pi$ feet, so the required number of revolutions is $\frac{5280}{6\pi}=\frac{880}{\pi}.$

Thus, the correct answer is C.

11.

A running track is the ring formed by two concentric circles. It is $10$ feet wide. The circumferences of the two circles differ by about:

$10$ feet

$30$ feet

$60$ feet

$100$ feet

none of these

Concepts:circle circumference approximation

Difficulty rating: 960

Small Hint:

The outer radius is $10$ feet greater than the inner radius

Big Hint:

Subtract the circumferences; the unknown inner radius cancels

Solution:

If the inner radius is $r,$ the difference is $2\pi(r+10)-2\pi r=20\pi\approx62.8$ feet, which is about $60$ feet.

Thus, the correct answer is C.

12.

The diameters of two circles are $8$ inches and $12$ inches respectively. The ratio of the area of the smaller to the area of the larger circle is:

$\dfrac23$

$\dfrac49$

$\dfrac94$

$\dfrac12$

none of these

Concepts:circle area similarity ratio

Difficulty rating: 960

Small Hint:

Circle areas scale as the squares of their diameters

Big Hint:

Square the diameter ratio $\frac{8}{12}$

Solution:

The area ratio is the square of the diameter ratio: $\left(\frac8{12}\right)^2=\left(\frac23\right)^2=\frac49.$

Thus, the correct answer is B.

13.

A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is $18$ inches, the median of the trapezoid is:

$36$ inches

$9$ inches

$18$ inches

not obtainable from these data

none of these

Concepts:triangle area trapezoid area median

Difficulty rating: 1070

Small Hint:

The area of a trapezoid is its median times its altitude

Big Hint:

Compare $mh$ with $\tfrac12(18)h$

Solution:

If the common altitude is $h,$ the triangle has area $\frac12(18)h=9h.$ A trapezoid’s area is its median times its altitude, so its median must be $9$ inches.

Thus, the correct answer is B.

14.

Given the larger of two circles with center $P$ and radius $p$ and the smaller with center $Q$ and radius $q.$ Draw $\overline{PQ}.$ Which of the following statements is false?

$p-q$ can be equal to $\overline{PQ}$

$p+q$ can be equal to $\overline{PQ}$

$p+q$ can be less than $\overline{PQ}$

$p-q$ can be less than $\overline{PQ}$

none of these

Concepts:circles center distance tangency

Difficulty rating: 1400

Small Hint:

Interpret the center distance for internal tangency, external tangency, and disjoint circles

Big Hint:

For each of the first four statements, try to choose a valid relative position of the circles

Solution:

Internal tangency realizes $PQ=p-q,$ and external tangency realizes $PQ=p+q.$ Disjoint circles can have $PQ\gt p+q,$ and many intersecting or disjoint configurations have $PQ\gt p-q.$ Thus each of A through D can occur, so none of them is false.

Therefore, the correct answer is E.

15.

A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:

$\dfrac14$ the area of the original square

$\dfrac12$ the area of the original square

$\dfrac12$ the area of the circular piece

$\dfrac14$ the area of the circular piece

none of these

Concepts:inscribed figures square circle area

Difficulty rating: 1310

Small Hint:

Let the original square have side $s$

Big Hint:

The final square’s diagonal equals the circle’s diameter, which is $s$

Solution:

Let the original square have side $s.$ The inscribed circle has diameter $s.$ That diameter is the diagonal of the largest square cut from the circle, so the final square has side $\frac{s}{\sqrt2}$ and area $\frac{s^2}{2}.$ The material left after both cuts is this final square, so the total waste is $s^2-\frac{s^2}{2}=\frac{s^2}{2}.$

Thus, the correct answer is B.

16.

Adams plans a profit of $10\%$ on the selling price of an article and his expenses are $15\%$ of sales. The rate of mark-up on an article that sells for $\$5.00$ is:

$20\%$

$25\%$

$30\%$

$33\dfrac13\%$

$35\%$

Concepts:markup profit expenses percentage

Difficulty rating: 1490

Small Hint:

Subtract both profit and expenses from the selling price to recover the cost

Big Hint:

The cost is $75\%$ of the selling price

Solution:

Profit and expenses are $10\%+15\%=25\%$ of sales, so the cost is $75\%$ of the selling price. The markup as a fraction of cost is $\frac{25\%}{75\%}=\frac13=33\frac13\%.$

Thus, the correct answer is D.

17.

A man has part of $\$4500$ invested at $4\%$ and the rest at $6\%.$ If his annual return on each investment is the same, the average rate of interest which he realizes on the $\$4500$ is:

$5\%$

$4.8\%$

$5.2\%$

$4.6\%$

none of these

Concepts:simple interest weighted average equation

Difficulty rating: 1360

Small Hint:

Let $x$ be the amount invested at $4\%$ and equate the two returns

Big Hint:

Solve $0.04x=0.06(4500-x)$ , then find the total return

Solution:

Let $x$ dollars be invested at $4\%.$ Equal returns give $0.04x=0.06(4500-x),$ so $x=2700.$ Each investment returns $\$108,$ for a total of $\$216.$ The average rate is $\frac{216}{4500}=0.048=4.8\%.$

Thus, the correct answer is B.

18.

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

$x^2+2$

$x+1$

$x^2-2x+2$

$x^2-4$

none of these

Concepts:Sophie Germain identity factoring difference of squares

Difficulty rating: 1400

Small Hint:

Add and subtract $4x^2$ to create a difference of squares

Big Hint:

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

Solution:

Using a difference of squares, $\begin{gathered} x^4+4=(x^2+2)^2-(2x)^2\\ =(x^2-2x+2)(x^2+2x+2). \end{gathered}$

Thus, the correct answer is C.

19.

In the expression $xy^2,$ the values of $x$ and $y$ are each decreased $25\%;$ the value of the expression is:

decreased $50\%$

decreased $75\%$

decreased $\frac{37}{64}$ of its value

decreased $\frac{27}{64}$ of its value

none of these

Concepts:percentage change exponents multiplicative factor

Difficulty rating: 1310

Small Hint:

Each decreased variable is $\frac{3}{4}$ of its original value

Big Hint:

Account for the two powers of $y$ when finding the new multiplicative factor

Solution:

The new value is $\left(\frac34x\right)\left(\frac34y\right)^2 =\frac{27}{64}xy^2.$ Therefore the decrease is $1-\frac{27}{64}=\frac{37}{64}$ of the original value.

Thus, the correct answer is C.

20.

If $y=x+\dfrac1x,$ then $x^4+x^3-4x^2+x+1=0$ becomes:

$x^2(y^2+y-2)=0$

$x^2(y^2+y-3)=0$

$x^2(y^2+y-4)=0$

$x^2(y^2+y-6)=0$

none of these

Concepts:substitution reciprocal polynomial algebraic identity

Difficulty rating: 1740

Small Hint:

Factor out $x^2$ and group reciprocal terms

Big Hint:

Use $x^2+x^{-2}=(x+x^{-1})^2-2$

Solution:

Because $x\ne0,$ factor the expression as $\begin{aligned} x^2\Bigg[& \left(x^2+\frac1{x^2}\right)\\ &+\left(x+\frac1x\right)-4 \Bigg]. \end{aligned}$ Since $x^2+x^{-2}=y^2-2,$ this becomes $x^2(y^2+y-6)=0.$

Thus, the correct answer is D.

21.

If $\log_{10}(x^2-3x+6)=1,$ the value of $x$ is:

$10$ or $2$

$4$ or $-2$

$3$ or $-1$

$4$ or $-1$

none of these

Concepts:logarithms quadratic equation factoring

Difficulty rating: 1340

Small Hint:

Convert the logarithmic equation to $x^2-3x+6=10$

Big Hint:

Factor the resulting quadratic

Solution:

The logarithmic equation is equivalent to $x^2-3x+6=10,$ so $\begin{aligned} x^2-3x-4&=(x-4)(x+1)\\ &=0. \end{aligned}$ Thus $x=4$ or $x=-1.$

The correct answer is D.

22.

The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:

$8\dfrac12$

$4\dfrac16$

$5$

$3$

none of these

Concepts:logarithms nested radicals fractional exponents

Difficulty rating: 1630

Small Hint:

Rewrite $27$ and both occurrences of $9$ as powers of $3$

Big Hint:

Add the exponents $3,$ $\frac{1}{2},$ and $\frac{2}{3}$

Solution:

The printed expression is a product of two separate radicals: $27\sqrt[4]{9}\sqrt[3]{9} =3^{3+\frac{1}{2}+\frac{2}{3}}=3^{\frac{25}{6}}.$ Its base- $3$ logarithm is $\frac{25}{6}=4\dfrac16.$

Thus, the correct answer is B.

23.

The equation $\sqrt{x+10}-\dfrac6{\sqrt{x+10}}=5$ has:

an extraneous root between $-5$ and $-1$

an extraneous root between $-10$ and $-6$

a true root between $20$ and $25$

two true roots

two extraneous roots

Concepts:radical equation extraneous root domain

Difficulty rating: 1590

Small Hint:

Multiply by $\sqrt{x+10}$ before eliminating the radical

Big Hint:

After solving the resulting quadratic, substitute both candidates into the original equation

Solution:

Multiplying by $\sqrt{x+10}$ gives $x+4=5\sqrt{x+10}.$ Squaring produces $\begin{gathered} x^2-17x-234\\ =(x-26)(x+9)\\ =0. \end{gathered}$ The value $x=26$ satisfies the original equation, while $x=-9$ makes its left side $1-6=-5,$ not $5.$ Thus $-9,$ which lies between $-10$ and $-6,$ is extraneous.

The correct answer is B.

24.

If $a,$ $b$ and $c$ are positive integers less than $10,$ then $(10a+b)(10a+c)$ equals $100a(a+1)+bc$ if:

$b+c=10$

$b=c$

$a+b=10$

$a=b$

$a+b+c=10$

Concepts:algebraic identity place value expansion

Difficulty rating: 1470

Small Hint:

Expand both sides and cancel their common terms

Big Hint:

The only unmatched terms are $10a(b+c)$ and $100a$

Solution:

Expanding and canceling $100a^2+bc$ from both sides leaves $10a(b+c)=100a.$ Since $a$ is positive, this is equivalent to $b+c=10.$

Thus, the correct answer is A.

25.

In a geometric progression whose terms are positive, any term is equal to the sum of the next two following terms. Then the common ratio is:

$1$

about $\dfrac{\sqrt5}{2}$

$\dfrac{\sqrt5-1}{2}$

$\dfrac{1-\sqrt5}{2}$

$\dfrac2{\sqrt5}$

Concepts:geometric progression quadratic equation positive root

Difficulty rating: 1400

Small Hint:

Divide the relation among three consecutive terms by the first of them

Big Hint:

The common ratio satisfies $1=r+r^2$ ; choose its positive root

Solution:

If a term is $t,$ the next two are $tr$ and $tr^2.$ Thus $t=tr+tr^2$ and $r^2+r-1=0.$ Because the terms are positive, $r=\frac{\sqrt5-1}{2}.$

Thus, the correct answer is C.

26.

The base of a triangle is $15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:

$5\sqrt6$ inches

$10$ inches

$4\sqrt3$ inches

$7.5$ inches

none of these

Difficulty rating: 1420

Small Hint:

The triangle above the lower parallel contains two-thirds of the total area

Big Hint:

For similar triangles, the area ratio is the square of the corresponding-length ratio

Solution:

The parallel closer to the base bounds a smaller triangle above it whose area is $\frac{2}{3}$ of the whole triangle. If its length is $x,$ similarity gives $\frac{x^2}{15^2}=\frac23.$ Hence $x^2=150$ and $x=5\sqrt6.$

Thus, the correct answer is A.

27.

The radius of the first circle is $1$ inch, that of the second $\dfrac12$ inch, that of the third $\dfrac14$ inch and so on indefinitely. The sum of the areas of the circles is:

$\dfrac{3\pi}{4}$

$1.3\pi$

$2\pi$

$\dfrac{4\pi}{3}$

none of these

Concepts:circle area infinite geometric series

Difficulty rating: 1420

Small Hint:

Squaring each radius changes the common ratio

Big Hint:

The areas form a geometric series with first term $\pi$ and ratio $\frac{1}{4}$

Solution:

The areas are $\pi,\ \frac{\pi}{4},\ \frac{\pi}{16},\ldots$ Therefore their sum is $\frac{\pi}{1-\frac{1}{4}}=\frac{4\pi}{3}.$

Thus, the correct answer is D.

28.

In triangle $ABC,$ sides $a,$ $b$ and $c$ are opposite angles $A,$ $B$ and $C$ respectively. $\overline{AD}$ bisects angle $A$ and meets $\overline{BC}$ at $D.$ Then if $x=CD$ and $y=BD$ the correct proportion is:

$\dfrac xa=\dfrac a{b+c}$

$\dfrac xb=\dfrac a{a+c}$

$\dfrac yc=\dfrac c{b+c}$

$\dfrac yc=\dfrac a{b+c}$

$\dfrac xy=\dfrac cb$

Concepts:angle bisector theorem triangle proportion

Difficulty rating: 1470

Small Hint:

Apply the angle bisector theorem to $\frac{BD}{DC}$

Big Hint:

Use $x+y=a$ after writing $\frac{y}{c}=\frac{x}{b}$

Solution:

The angle bisector theorem gives $\frac{y}{x}=\frac{c}{b},$ or $\frac{y}{c}=\frac{x}{b}.$ Since $x+y=a,$ $\frac yc=\frac xb=\frac{x+y}{b+c}=\frac a{b+c}.$

Thus, the correct answer is D.

29.

The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to the nearest ten-thousandth of a square inch is:

$2$

$3$

$4$

$5$

$1$

Concepts:measurement error significant digits square root

Difficulty rating: 1810

Small Hint:

The true area lies from $1.10245$ up to $1.10255$ square inches

Big Hint:

Take square roots of the error interval and see how many digits of the side are fixed

Solution:

The reported area means the true area lies in $1.10245\le A\lt1.10255.$ Taking square roots gives approximately $1.049976\le\sqrt A\lt1.050024.$ Every possible side length therefore rounds to $1.0500$ to the nearest ten-thousandth. The trailing zeros after the decimal are significant, so $1.0500$ has five significant digits.

Thus, the correct answer is D.

30.

A house worth $\$9000$ is sold by Mr. $A$ to Mr. $B$ at a $10\%$ loss. Mr. $B$ sells the house back to Mr. $A$ at a $10\%$ gain. The result of the two transactions is:

Mr. $A$ breaks even

Mr. $B$ gains $\$900$

Mr. $A$ loses $\$900$

Mr. $A$ loses $\$810$

Mr. $B$ gains $\$1710$

Concepts:percentage profit and loss transactions

Difficulty rating: 1070

Small Hint:

Find the price of each sale separately

Big Hint:

The second $10\%$ is taken from the first sale price, not from $\$9000$

Solution:

Mr. $B$ first pays $9000(0.90)=8100$ dollars. He then sells the house back for $8100(1.10)=8910$ dollars. Mr. $A$ receives $\$8100$ and pays $\$8910,$ so he loses $\$810.$

Thus, the correct answer is D.

31.

The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click. The speed of the train in miles per hour is approximately the number of clicks heard in:

$20$ seconds

$2$ minutes

$1\dfrac12$ minutes

$5$ minutes

none of these

Concepts:unit conversion rate approximation

Difficulty rating: 1540

Small Hint:

Convert one mile per hour to feet per second, then divide by $30$ feet per click

Big Hint:

Find the time interval for which the click count is approximately the numerical speed in miles per hour

Solution:

A speed of $v$ miles per hour is $\frac{22v}{15}$ feet per second. Since each click represents $30$ feet, the click rate is $\frac{\frac{22v}{15}}{30}=\frac{11v}{225}$ clicks per second. In $\frac{225}{11}\approx20.45$ seconds, the number of clicks is $v.$ The closest listed interval is $20$ seconds.

Thus, the correct answer is A.

32.

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:

a square

a rectangle

a parallelogram with unequal sides

a rhombus

a quadrilateral with no special properties

Concepts:rectangle angle trisectors rhombus symmetry

Difficulty rating: 1830

Small Hint:

Use the horizontal and vertical symmetry axes of the rectangle

Big Hint:

The four intersection points have perpendicular diagonals that bisect each other

Solution:

For each side, the two trisectors adjacent to it meet on that side’s perpendicular bisector. Opposite such intersection points are reflections across the center of the rectangle, so the diagonals of the resulting quadrilateral bisect each other. One diagonal lies on the horizontal symmetry axis and the other on the vertical symmetry axis, so they are perpendicular.

A quadrilateral whose diagonals bisect each other is a parallelogram; if those diagonals are perpendicular, its four sides are equal. Hence the quadrilateral is a rhombus. It need not be a square because the two diagonals need not have equal length.

Thus, the correct answer is D.

33.

The perimeter of an isosceles right triangle is $2p.$ Its area is:

$(2+\sqrt2)p$

$(2-\sqrt2)p$

$(3-2\sqrt2)p^2$

$(1-2\sqrt2)p^2$

$(3+2\sqrt2)p^2$

Concepts:isosceles right triangle perimeter area radicals

Difficulty rating: 1630

Small Hint:

Let each leg have length $s$ , so the hypotenuse is $s\sqrt2$

Big Hint:

Solve $s(2+\sqrt2)=2p$ , then use area $\frac{s^2}{2}$

Solution:

If each leg is $s,$ then $s(2+\sqrt2)=2p,$ so $s=p(2-\sqrt2).$ Therefore the area is $\begin{gathered} \frac{s^2}{2} =\frac{p^2(2-\sqrt2)^2}{2}\\ =(3-2\sqrt2)p^2. \end{gathered}$

Thus, the correct answer is C.

34.

If one side of a triangle is $12$ inches and the opposite angle is $30$ degrees, then the diameter of the circumscribed circle is:

$18$ inches

$30$ inches

$24$ inches

$20$ inches

none of these

Concepts:extended law of sines circumcircle triangle

Difficulty rating: 1340

Small Hint:

Relate a side, its opposite angle, and the circumdiameter

Big Hint:

The extended law of sines gives $d=\frac{a}{\sin A}$

Solution:

By the extended law of sines, the circumdiameter is $d=\frac{12}{\sin30^\circ}=\frac{12}{\frac{1}{2}}=24$ inches.

Thus, the correct answer is C.

35.

If $f(x)=\dfrac{x(x-1)}2,$ then $f(x+2)$ equals:

$f(x)+f(2)$

$(x+2)f(x)$

$x(x+2)f(x)$

$\dfrac{xf(x)}{x+2}$

$\dfrac{(x+2)f(x+1)}x$

Concepts:function notation algebraic identity

Difficulty rating: 1450

Small Hint:

Write explicit formulas for both $f(x+2)$ and $f(x+1)$

Big Hint:

The factor $x$ in $f(x+1)=\frac{x(x+1)}{2}$ cancels in one choice

Solution:

Directly, $f(x+2)=\frac{(x+2)(x+1)}2.$ Also $f(x+1)=\frac{x(x+1)}{2},$ so $\begin{gathered} \frac{(x+2)f(x+1)}x\\ =\frac{(x+2)(x+1)}2. \end{gathered}$

Thus, the correct answer is E.

36.

Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3.$ The obtained value, $m,$ is an exact divisor of:

$12$

$20$

$36$

$48$

$64$

Concepts:factor theorem divisibility polynomial

Difficulty rating: 1280

Small Hint:

Use the factor theorem at $x=3$

Big Hint:

After finding $m,$ test which listed number is divisible by it

Solution:

Divisibility by $x-3$ requires $4(3)^2-6(3)+m=0,$ so $18+m=0$ and $m=-18.$ Of the listed numbers, only $36$ is an exact multiple of $-18.$

Thus, the correct answer is C.

37.

The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches. The radius of the circle through the vertices of the triangle is:

$\dfrac{7\sqrt{15}}5$

$4\sqrt3$

$3\sqrt5$

$6\sqrt3$

none of these

Concepts:isosceles triangle circumradius Heron formula

Difficulty rating: 1740

Small Hint:

Drop the altitude to split the base into two segments of length $3$

Big Hint:

Find the area, then use $R=\frac{abc}{4K}$

Solution:

The altitude is $\sqrt{12^2-3^2}=3\sqrt{15},$ so the area is $K=\tfrac12(6)(3\sqrt{15})=9\sqrt{15}.$ Hence the circumradius is $R=\frac{(12)(12)(6)}{4(9\sqrt{15})} =\frac{8\sqrt{15}}5.$ This value is not listed.

Thus, the correct answer is E.

38.

If $f(a)=a-2$ and $F(a,b)=b^2+a,$ then $F[3,f(4)]$ is:

$a^2-4a+7$

$28$

$7$

$8$

$11$

Concepts:function notation composition substitution

Difficulty rating: 1390

Small Hint:

Evaluate the inner function $f(4)$ first

Big Hint:

Then substitute $a=3$ and the resulting value for $b$ into $F(a,b)$

Solution:

First $f(4)=4-2=2.$ Therefore $\begin{aligned} F[3,f(4)]&=F(3,2)\\ &=2^2+3=7. \end{aligned}$

Thus, the correct answer is C.

39.

The product, $\log_a b\cdot\log_b a$ is equal to:

$1$

$a$

$b$

$ab$

none of these

Concepts:logarithms change of base

Difficulty rating: 1180

Small Hint:

Use the change-of-base formula on both logarithms

Big Hint:

The two resulting fractions are reciprocals

Solution:

For permissible bases and arguments, $\begin{aligned} \log_a b\cdot\log_b a &=\frac{\log b}{\log a}\\ &\quad{}\cdot\frac{\log a}{\log b}\\ &=1. \end{aligned}$

Thus, the correct answer is A.

40.

The negation of the statement “all men are honest,” is:

no men are honest

all men are dishonest

some men are dishonest

no men are dishonest

some men are honest

Concepts:logic quantifiers negation

Difficulty rating: 1360

Small Hint:

To disprove a universal statement, only one counterexample is needed

Big Hint:

Negating “every man is honest” asserts that at least one man is not honest

Solution:

The negation of “every man is honest” is “there exists a man who is not honest.” In the language of the choices, some men are dishonest.

Thus, the correct answer is C.

41.

A girls’ camp is located $300$ rods from a straight road. On this road, a boys’ camp is located $500$ rods from the girls’ camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is:

$400$ rods

$250$ rods

$87.5$ rods

$200$ rods

none of these

Concepts:coordinate geometry Pythagorean theorem equidistance

Difficulty rating: 1470

Small Hint:

The perpendicular and camp-to-camp distances form a $300$ - $400$ - $500$ right triangle

Big Hint:

Place the camps at $(0,300)$ and $(400,0)$ , and put the canteen at $(t,0)$

Solution:

Let the foot of the perpendicular from the girls’ camp be $(0,0).$ The camps can be placed at $G=(0,300)$ and $B=(400,0).$ If the canteen is $C=(t,0),$ equidistance gives $t^2+300^2=(400-t)^2,$ so $t=87.5.$ The common distance is $BC=400-87.5=312.5$ rods, which is not listed.

Thus, the correct answer is E.

42.

The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. The length of the common internal tangent is:

$41$ inches

$39$ inches

$39.8$ inches

$40.1$ inches

$40$ inches

Concepts:circles common internal tangent Pythagorean theorem

Difficulty rating: 1400

Small Hint:

For an internal common tangent, the perpendicular separation of the centers from the tangent is the sum of the radii

Big Hint:

Use a right triangle with hypotenuse $41$ and one leg $4+5$

Solution:

The center segment, the tangent segment, and a perpendicular leg of length $4+5=9$ form a right triangle. Thus the tangent length is $\sqrt{41^2-9^2}=\sqrt{1600}=40$ inches.

Thus, the correct answer is E.

43.

If the price of an article is increased by per cent $p,$ then the decrease in per cent of sales must not exceed $d$ in order to yield the same income. The value of $d$ is:

$\dfrac1{1+p}$

$\dfrac1{1-p}$

$\dfrac p{1+p}$

$\dfrac p{p-1}$

$\dfrac{1-p}{1+p}$

Concepts:revenue percentage change algebraic equation

Difficulty rating: 1590

Small Hint:

Represent the new price and number sold by factors $1+p$ and $1-d$

Big Hint:

Equal revenue requires $(1+p)(1-d)=1$

Solution:

Writing the percentage rates as fractions of the original quantities, equal income requires $(1+p)(1-d)=1.$ Hence $p-d-pd=0,$ so $d=\frac p{1+p}.$

Thus, the correct answer is C.

44.

In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $8$ and $2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and finds $-9$ and $-1$ for the roots. The correct equation was:

$x^2-10x+9=0$

$x^2+10x+9=0$

$x^2-10x+16=0$

$x^2-8x-9=0$

none of these

Concepts:quadratic roots Vieta formulas coefficients

Difficulty rating: 1660

Small Hint:

The first student’s correct linear coefficient is determined by the sum $8+2$

Big Hint:

The second student’s correct constant term is determined by the product $(-9)(-1)$

Solution:

The first student changed only the constant term, so the correct coefficient of $x$ is the one in the monic quadratic with roots $8$ and $2:$ it is $-(8+2)=-10.$ The second student changed only that linear coefficient, so the correct constant is $(-9)(-1)=9.$ Therefore the correct equation is $x^2-10x+9=0.$

Thus, the correct answer is A.

45.

The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:

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

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

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

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

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

Concepts:AM-GM inequality equality case

Difficulty rating: 1180

Small Hint:

Start with the nonnegative square $(a-b)^2$

Big Hint:

Equality must remain possible when the two segment lengths are equal

Solution:

Since $(a-b)^2\ge0,$ we have $(a+b)^2\ge4ab.$ Both sides are nonnegative, so $\frac{a+b}{2}\ge\sqrt{ab}.$ Equality occurs when $a=b.$

Thus, the correct answer is E.

46.

Instead of walking along two adjacent sides of a rectangular field, a boy took a short-cut along the diagonal of the field and saved a distance equal to $\frac{1}{2}$ the longer side. The ratio of the shorter side of the rectangle to the longer side was:

$\dfrac12$

$\dfrac25$

$\dfrac14$

$\dfrac34$

$\dfrac25$

Concepts:rectangle Pythagorean theorem ratio

Difficulty rating: 1590

Small Hint:

Let the longer and shorter sides be $L$ and $W$

Big Hint:

The condition is $L+W-\sqrt{L^2+W^2}=\frac{L}{2}$

Solution:

The saving condition gives $\sqrt{L^2+W^2}=\frac L2+W.$ Squaring and canceling $W^2$ yields $L^2=\frac{L^2}{4}+LW,$ so $\frac{W}{L}=\frac{3}{4}.$

Thus, the correct answer is D.

47.

If $x$ is greater than zero, then the correct relationship is:

$\log(1+x)=\dfrac{x}{1+x}$

$\log(1+x)\lt\dfrac{x}{1+x}$

$\log(1+x)\gt x$

$\log(1+x)\lt x$

none of these

Concepts:logarithmic inequality exponential function

Difficulty rating: 1450

Small Hint:

Compare $1+x$ with an exponential function for $x\gt0$

Big Hint:

The standard inequality $\ln(1+x)\lt x$ is even stronger for common logarithms

Solution:

For $x\gt0,$ the standard exponential inequality gives $1+x\lt e^x.$ Taking natural logarithms yields $\ln(1+x)\lt x.$ If $\log$ denotes the common logarithm, then $\begin{aligned} \log(1+x)&=\frac{\ln(1+x)}{\ln10}\\ &\lt\ln(1+x), \end{aligned}$ so the same listed inequality holds.

Thus, the correct answer is D.

48.

If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, then the ratio of the smaller base to the larger base is:

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{3}{4}$

$\frac{3}{5}$

$\frac{2}{5}$

Concepts:isosceles trapezoid Pythagorean theorem ratio

Difficulty rating: 1740

Small Hint:

A diagonal’s horizontal projection is half the sum of the two bases

Big Hint:

Scale the larger base to $1$ , and let the smaller base and altitude both be $r$

Solution:

Let the larger base be $1,$ and let the smaller base and altitude both be $r.$ A diagonal has horizontal projection $\frac{1+r}{2}$ and length $1.$ Thus $1=r^2+\left(\frac{1+r}{2}\right)^2.$ This simplifies to $\begin{aligned} 5r^2+2r-3&=(5r-3)(r+1)\\ &=0. \end{aligned}$ The positive ratio is $r=\frac{3}{5}.$

Thus, the correct answer is D.

49.

The coordinates of $A,$ $B$ and $C$ are $(5,5),$ $(2,1)$ and $(0,k)$ respectively. The value of $k$ that makes $\overline{AC}+\overline{BC}$ as small as possible is:

$3$

$4\dfrac12$

$3\dfrac67$

$4\dfrac56$

$2\dfrac17$

Concepts:reflection shortest path coordinate geometry

Difficulty rating: 1910

Small Hint:

Reflect $B$ across the $y$ -axis so that $BC$ becomes the distance from $C$ to the reflected point

Big Hint:

The shortest broken path occurs where the straight line from $A$ to the reflected point meets the $y$ -axis

Solution:

Reflect $B=(2,1)$ across the $y$ -axis to $B'=(-2,1).$ For $C$ on the $y$ -axis, $BC=B'C,$ so $AC+BC$ is minimized when $A,C,B'$ are collinear. The line from $A=(5,5)$ to $B'=(-2,1)$ has slope $\frac{4}{7}.$ At $x=0,$ its height is $5-\frac47(5)=\frac{15}{7}=2\frac17.$ Hence $k=\frac{15}{7}.$

Thus, the correct answer is E.

50.

One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4,$ then the length of the shortest side of the triangle is:

$12$ units

$13$ units

$14$ units

$15$ units

$16$ units

Concepts:incircle tangent lengths Heron formula triangle

Difficulty rating: 1950

Small Hint:

Equal tangent segments from a vertex make the three sides $14,$ $6+z,$ and $8+z$ for some $z$

Big Hint:

Use both $K=rs$ and Heron’s formula with semiperimeter $s=14+z$

Solution:

Let the two tangent segments from the third vertex each have length $z.$ Equal tangents from a common vertex make the side lengths $14,\qquad 6+z,\qquad 8+z,$ with semiperimeter $s=14+z.$ Since the inradius is $4,$ $K=rs=4(14+z).$ Heron’s formula gives $\begin{aligned} K^2 &=(14+z)z\cdot8\cdot6\\ &=48z(14+z). \end{aligned}$ Equating the squares yields $16(14+z)=48z,$ so $z=7.$ The sides are $14,$ $13,$ and $15,$ and the shortest is $13.$

Thus, the correct answer is B.