1950 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $64$ is divided into three parts proportional to $2,$ $4,$ and $6,$ the smallest part is:

$5\dfrac{1}{3}$

$11$

$10\dfrac{2}{3}$

$5$

None of these answers

Concepts:ratio and proportion fraction

Difficulty rating: 840

Small Hint:

Add the three numbers in the ratio to find the total number of equal shares

Big Hint:

The smallest part is $\dfrac{2}{2+4+6}$ of $64$

Solution:

The ratio contains $2+4+6=12$ equal shares. The smallest part is therefore $64\left(\frac{2}{12}\right)=\frac{32}{3}=10\frac{2}{3}.$

Thus, the correct answer is C.

2.

Let $R=gS-4.$ When $S=8,$ $R=16.$ When $S=10,$ $R$ is equal to:

$11$

$14$

$20$

$21$

None of these answers

Concepts:linear equation substitution

Difficulty rating: 870

Small Hint:

First substitute the given values of $R$ and $S$ to determine $g$

Big Hint:

The first pair gives $16=8g-4$

Solution:

Substituting $S=8$ and $R=16$ gives $16=8g-4,$ so $g=\tfrac52.$ Hence, when $S=10,$ $R=\frac52(10)-4=21.$

Thus, the correct answer is D.

3.

The sum of the roots of the equation $4x^2+5-8x=0$ is equal to:

$8$

$-5$

$-\dfrac54$

$-2$

None of these answers

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1500

Small Hint:

Rewrite the equation in descending powers of $x$

Big Hint:

For $ax^2+bx+c=0,$ the sum of the roots is $-\frac{b}{a}$

Solution:

In standard form the equation is $4x^2-8x+5=0.$ By Vieta’s formulas, the sum of its roots is $-\frac{-8}{4}=2.$ This value is not among choices A through D.

Thus, the correct answer is E.

4.

Reduced to lowest terms, $\frac{a^2-b^2}{ab}-\frac{ab-b^2}{ab-a^2}$ is equal to:

$\dfrac{a}{b}$

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

$a^2$

$a-2b$

None of these answers

Concepts:algebraic manipulation difference of squares

Difficulty rating: 1600

Small Hint:

Factor $a^2-b^2,$ $ab-b^2,$ and $ab-a^2$

Big Hint:

The second fraction simplifies to $-\frac{b}{a}$

Solution:

Factoring the numerators and denominator gives $\begin{aligned} &\frac{a^2-b^2}{ab} -\frac{ab-b^2}{ab-a^2}\\ &=\left(\frac{a}{b}-\frac{b}{a}\right)\\ &\quad-\frac{b(a-b)}{a(b-a)}\\ &=\frac{a}{b}-\frac{b}{a} +\frac{b}{a}\\ &=\frac{a}{b}. \end{aligned}$

Thus, the correct answer is A.

5.

If five geometric means are inserted between $8$ and $5832,$ the fifth term in the geometric series is:

$648$

$832$

$1168$

$1944$

None of these answers

Concepts:geometric sequence exponent

Difficulty rating: 1670

Small Hint:

Inserting five means makes the given numbers the first and seventh terms

Big Hint:

If the common ratio is $q,$ then $8q^6=5832$

Solution:

There are six common-ratio steps from the first term $8$ to the seventh term $5832.$ Thus $q^6=\frac{5832}{8}=729=3^6,$ so the positive common ratio is $q=3.$ The fifth term is $8q^4=8\cdot3^4=648.$

Thus, the correct answer is A.

6.

The values of $y$ which will satisfy the equations

$\begin{aligned} 2x^2+6x+5y+1&=0,\\ 2x+y+3&=0 \end{aligned}$

may be found by solving:

$y^2+14y-7=0$

$y^2+8y+1=0$

$y^2+10y-7=0$

$y^2+y-12=0$

None of the above equations

Concepts:system of equations substitution quadratic

Difficulty rating: 1640

Small Hint:

Solve the linear equation for $x$ in terms of $y$

Big Hint:

Substitute $x=-\frac{(y+3)}{2}$ into the first equation and clear the denominator

Solution:

From the second equation, $x=-\tfrac{y+3}{2}.$ Substituting into the first equation and multiplying by $2$ gives $\begin{aligned} &(y+3)^2-6(y+3)\\ &\quad+10y+2=0,\\ &y^2+10y-7=0. \end{aligned}$

Thus, the correct answer is C.

7.

If the digit $1$ is placed after a two digit number whose tens’ digit is $t,$ and units’ digit is $u,$ the new number is:

$10t+u+1$

$100t+10u+1$

$1000t+10u+1$

$t+u+1$

None of these answers

Concepts:place value digits

Difficulty rating: 770

Small Hint:

The original two digit number is $10t+u$

Big Hint:

Appending a digit multiplies the original number by $10$ before adding that digit

Solution:

The two digit number is $10t+u.$ Placing $1$ after it produces $\begin{aligned} &10(10t+u)+1\\ &=100t+10u+1. \end{aligned}$

Thus, the correct answer is B.

8.

If the radius of a circle is increased $100\%,$ the area is increased:

$100\%$

$200\%$

$300\%$

$400\%$

By none of these

Concepts:circle area percentage power scaling of length, area, and volume

Difficulty rating: 1220

Small Hint:

An increase of $100\%$ doubles the radius

Big Hint:

Circle area is proportional to the square of the radius

Solution:

Increasing $r$ by $100\%$ changes it to $2r.$ The area changes from $\pi r^2$ to $\pi(2r)^2=4\pi r^2,$ an increase of $3\pi r^2.$ This is $300\%$ of the original area.

Thus, the correct answer is C.

9.

The area of the largest triangle that can be inscribed in a semicircle whose radius is $r$ is:

$r^2$

$r^3$

$2r^2$

$2r^3$

$\dfrac12r^2$

Concepts:triangle area circle optimization

Difficulty rating: 1600

Small Hint:

The longest possible base is the diameter of the semicircle

Big Hint:

With the diameter as base, the greatest possible altitude is the radius

Solution:

A triangle in the semicircle has base at most the diameter $2r$ and altitude at most $r.$ Therefore its area is at most $\frac12(2r)(r)=r^2.$ This bound is attained by using the diameter as the base and the topmost point of the semicircle as the third vertex.

Thus, the correct answer is A.

10.

After rationalizing the numerator of $\dfrac{\sqrt3-\sqrt2}{\sqrt3},$ the denominator in simplest form is:

$\sqrt3(\sqrt3+\sqrt2)$

$\sqrt3(\sqrt3-\sqrt2)$

$3-\sqrt3\sqrt2$

$3+\sqrt6$

None of these answers

Concepts:rationalizing denominator radical

Difficulty rating: 1530

Small Hint:

Multiply the numerator and denominator by the conjugate of the numerator

Big Hint:

The numerator becomes $(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1$

Solution:

Multiplying by the conjugate of the numerator gives $\begin{aligned} &\frac{\sqrt3-\sqrt2}{\sqrt3} \cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ &=\frac{1}{\sqrt3(\sqrt3+\sqrt2)}\\ &=\frac{1}{3+\sqrt6}. \end{aligned}$

Thus, the correct answer is D.

11.

If in the formula $C=\dfrac{en}{R+nr},$ $n$ is increased while $e,$ $R,$ and $r$ are kept constant, then $C:$

Decreases

Increases

Remains constant

Increases and then decreases

Decreases and then increases

Concepts:function algebraic manipulation

Difficulty rating: 1580

Small Hint:

Divide the numerator and denominator by $n$

Big Hint:

As positive $n$ grows, $\frac{R}{n}$ decreases

Solution:

Rewrite the formula as $C=\frac{e}{\frac{R}{n}+r}.$ As positive $n$ increases, $\frac{R}{n}$ decreases, so the positive denominator decreases while the numerator remains fixed. Therefore $C$ increases.

Thus, the correct answer is B.

12.

As the number of sides of a polygon increases from $3$ to $n,$ the sum of the exterior angles formed by extending each side in succession:

Increases

Decreases

Remains constant

Cannot be predicted

Becomes $(n-3)$ straight angles

Concepts:angle sum

Difficulty rating: 1310

Small Hint:

Imagine walking once around the boundary of the polygon

Big Hint:

The exterior angles record one complete turn

Solution:

Traversing the polygon and turning through each exterior angle makes one complete turn. Hence the sum is always $360^\circ,$ independent of the number of sides.

Thus, the correct answer is C.

13.

The roots of $(x^2-3x+2)(x)(x-4)=0$ are:

$4$

$0$ and $4$

$1$ and $2$

$0,$ $1,$ $2,$ and $4$

$1,$ $2,$ and $4$

Concepts:factoring zero product property quadratic

Difficulty rating: 1410

Small Hint:

Factor the quadratic factor

Big Hint:

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

Solution:

Since $x^2-3x+2=(x-1)(x-2),$ the equation becomes $(x-1)(x-2)x(x-4)=0.$ Its roots are $0,$ $1,$ $2,$ and $4.$

Thus, the correct answer is D.

14.

For the simultaneous equations

$\begin{aligned} 2x-3y&=8,\\ 6y-4x&=9, \end{aligned}$

$x=4,$ $y=0$

$x=0,$ $y=\dfrac32$

$x=0,$ $y=0$

There is no solution

There are an infinite number of solutions

Concepts:system of equations linear equation

Difficulty rating: 1510

Small Hint:

Compare the left side of the second equation with the left side of the first

Big Hint:

Multiplying the first equation by $-2$ would require the second right side to be $-16$

Solution:

The left side of the second equation is $-2(2x-3y).$ If the first equation holds, that expression must equal $-2\cdot8=-16,$ but the second equation says it equals $9.$ This contradiction means that no ordered pair satisfies both equations.

Thus, the correct answer is D.

15.

The real factors of $x^2+4$ are:

$(x^2+2)(x^2+2)$

$(x^2+2)(x^2-2)$

$x^2(x^2+4)$

$(x^2-2x+2)(x^2+2x+2)$

Non-existent

Concepts:factoring complex number quadratic

Difficulty rating: 1400

Small Hint:

A real linear factor would correspond to a real root

Big Hint:

Solving $x^2+4=0$ requires $x^2=-4$

Solution:

A real linear factor would give a real root. But $x^2+4=0$ implies $x^2=-4,$ which has no real solution. Therefore the polynomial has no real linear factors.

Thus, the correct answer is E.

16.

The number of terms in the expansion of $\big[(a+3b)^2(a-3b)^2\big]^2$ when simplified is:

$4$

$5$

$6$

$7$

$8$

Concepts:difference of squares binomial theorem

Difficulty rating: 1580

Small Hint:

Combine $a+3b$ and $a-3b$ before expanding

Big Hint:

The expression simplifies to $(a^2-9b^2)^4$

Solution:

Using the difference of squares, $\begin{aligned} &\big[(a+3b)^2(a-3b)^2\big]^2\\ &=\big[(a^2-9b^2)^2\big]^2\\ &=(a^2-9b^2)^4. \end{aligned}$ Its binomial expansion has one nonzero term for each exponent choice $0,1,2,3,4,$ so it has five terms.

Thus, the correct answer is B.

17.

The formula which expresses the relationship between $x$ and $y$ as shown in the accompanying table is:

$\begin{array}{|c|c|c|c|c|c|} \hline x&0&1&2&3&4\\ \hline y&100&90&70&40&0\\ \hline \end{array}$

$y=100-10x$

$y=100-5x^2$

$y=100-5x-5x^2$

$y=20-x-x^2$

None of these

Concepts:function substitution quadratic

Difficulty rating: 1360

Small Hint:

Substitute a small nonzero table value such as $x=1$ into each proposed formula

Big Hint:

Then verify the surviving formula with $x=2$ or $x=3$

Solution:

At $x=1,$ choices A and C both give $90,$ while B gives $95$ and D gives $18.$ Testing the two survivors at $x=2,$ choice A gives $80,$ whereas choice C gives $100-5(2)-5(2^2)=70,$ matching the table.

Thus, the correct answer is C.

18.

Of the following

$(1)\quad a(x-y)=ax-ay$

$(2)\quad a^{x-y}=a^x-a^y$

$(3)\quad \log(x-y)=\log x-\log y$

$(4)\quad \dfrac{\log x}{\log y}=\log x-\log y$

$(5)\quad a(xy)=ax\times ay$

Only $1$ and $4$ are true

Only $1$ and $5$ are true

Only $1$ and $3$ are true

Only $1$ and $2$ are true

Only $1$ is true

Concepts:distributive property exponent logarithm

Difficulty rating: 1470

Small Hint:

Compare each statement with the standard rules for exponents and logarithms

Big Hint:

In particular, subtraction inside an exponent or logarithm does not distribute as written

Solution:

Statement $1$ is the distributive property. Statement $2$ is false because $a^{x-y}=\frac{a^x}{a^y}.$ Statement $3$ is false because $\log x-\log y=\log(\frac{x}{y}).$ Statement $4$ confuses the change-of-base quotient with a difference, and statement $5$ has right side $a^2xy,$ not $axy.$

Thus, only statement $1$ is true, so the correct answer is E.

19.

If $m$ men can do a job in $d$ days, then $m+r$ men can do the job in:

$d+r$ days

$d-r$ days

$\dfrac{md}{m+r}$ days

$\dfrac{d}{m+r}$ days

None of these

Concepts:rate ratio and proportion

Difficulty rating: 1220

Small Hint:

Measure the total job in man-days

Big Hint:

The original crew performs $md$ man-days of work

Solution:

The job requires $m\cdot d=md$ man-days. With $m+r$ men working at the same individual rate, the required number of days is $\frac{md}{m+r}.$

Thus, the correct answer is C.

20.

When $x^{13}+1$ is divided by $x-1,$ the remainder is:

$1$

$-1$

$0$

$2$

None of these answers

Concepts:polynomial substitution

Difficulty rating: 1520

Small Hint:

Use the Remainder Theorem

Big Hint:

For division by $x-1,$ evaluate the polynomial at $x=1$

Solution:

By the Remainder Theorem, the remainder upon division by $x-1$ is the value of the polynomial at $x=1.$ This value is $1^{13}+1=2.$

Thus, the correct answer is D.

21.

The volume of a rectangular solid each of whose side, front, and bottom faces are $12$ square inches, $8$ square inches, and $6$ square inches respectively is:

$576$ cubic inches

$24$ cubic inches

$9$ cubic inches

$104$ cubic inches

None of these

Concepts:rectangular prism volume area

Difficulty rating: 1600

Small Hint:

Let the three edge lengths be $x,$ $y,$ and $z$

Big Hint:

Multiplying the three face areas gives $(xyz)^2$

Solution:

If the edge lengths are $x,$ $y,$ and $z,$ then the three face areas are $xy,$ $yz,$ and $xz.$ Therefore $\begin{aligned} (xyz)^2&=(xy)(yz)(xz)\\ &=12\cdot8\cdot6=576. \end{aligned}$ Since the volume is positive, $xyz=\sqrt{576}=24$ cubic inches.

Thus, the correct answer is B.

22.

Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:

$30\%$

$15\%$

$72\%$

$28\%$

None of these

Concepts:percentage

Difficulty rating: 1100

Small Hint:

Apply both discounts to a price of $100$

Big Hint:

After the first discount, the second discount is taken from $90,$ not from $100$

Solution:

Starting from a price of $100,$ the first discount leaves $90.$ The second discount leaves $0.8(90)=72.$ The total reduction is therefore $100-72=28,$ or $28\%.$

Thus, the correct answer is D.

23.

A man buys a house for $\$10{,}000$ and rents it. He puts $12\dfrac12\%$ of each month’s rent aside for repairs and upkeep; pays $\$325$ a year taxes and realizes $5\dfrac12\%$ on his investment. The monthly rent is:

$\$64.82$

$\$83.33$

$\$72.08$

$\$45.83$

$\$177.08$

Concepts:money percentage linear equation

Difficulty rating: 1650

Small Hint:

Let $r$ be the monthly rent and express the annual rent as $12r$

Big Hint:

After upkeep and taxes, the annual return must be $5.5\%$ of $\$10,000$

Solution:

Let the monthly rent be $r$ dollars. The yearly rent is $12r,$ of which $1-0.125=\tfrac78$ remains after setting aside the upkeep money. The desired yearly return is $0.055(10,000)=550$ dollars, so $\frac78(12r)-325=550.$ Hence $10.5r=875,$ giving $r=83.\overline3,$ which is $\$83.33$ to the nearest cent.

Thus, the correct answer is B.

24.

The equation $x+\sqrt{x-2}=4$ has:

$2$ real roots

$1$ real and $1$ imaginary root

$2$ imaginary roots

No roots

$1$ real root

Concepts:radical quadratic substitution

Difficulty rating: 1470

Small Hint:

Isolate the square root before squaring

Big Hint:

Check every root of the resulting quadratic in the original equation

Solution:

Isolating and squaring gives $\begin{aligned} \sqrt{x-2}&=4-x,\\ x-2&=(4-x)^2,\\ x^2-9x+18&=0. \end{aligned}$ The quadratic roots are $3$ and $6.$ The value $3$ satisfies the original equation, but $6$ gives $6+\sqrt4=8,$ so it is extraneous.

Thus, the correct answer is E.

25.

The value of $\log_5\dfrac{(125)(625)}{25}$ is equal to:

$725$

$6$

$3125$

$5$

None of these answers

Concepts:logarithm exponent

Difficulty rating: 1320

Small Hint:

Write $125,$ $625,$ and $25$ as powers of $5$

Big Hint:

The logarithm’s argument simplifies to $5^{3+4-2}$

Solution:

Since $125=5^3,$ $625=5^4,$ and $25=5^2,$ $\begin{aligned} &\log_5\frac{(125)(625)}{25}\\ &=\log_5\left(5^{3+4-2}\right)\\ &=\log_5(5^5)=5. \end{aligned}$

Thus, the correct answer is D.

26.

If $\log_{10}m=b-\log_{10}n,$ then $m=$

$\dfrac{b}{n}$

$bn$

$10^bn$

$b-10^n$

$\dfrac{10^b}{n}$

Concepts:logarithm algebraic manipulation

Difficulty rating: 1400

Small Hint:

Move $\log_{10}n$ to the left side

Big Hint:

Combine the two logarithms and then exponentiate with base $10$

Solution:

Rearranging and using the product rule for logarithms gives $\begin{aligned} &\log_{10}m+\log_{10}n\\ &=\log_{10}(mn)=b. \end{aligned}$ Therefore $mn=10^b,$ so $m=\dfrac{10^b}{n}.$

Thus, the correct answer is E.

27.

A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:

$33$ mph

$34$ mph

$35$ mph

$36$ mph

$37$ mph

Concepts:distance rate and time estimation

Difficulty rating: 1410

Small Hint:

Average speed is total distance divided by total time

Big Hint:

The two legs take $\frac{120}{30}$ and $\frac{120}{40}$ hours

Solution:

The total distance is $240$ miles. The travel times are $4$ hours and $3$ hours, so the average speed, in miles per hour, is $\frac{240}{4+3}=\frac{240}{7}\approx34.29$ which is closest to $34$ mph.

Thus, the correct answer is B.

28.

Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B.$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:

$4$ mph

$8$ mph

$12$ mph

$16$ mph

$20$ mph

Concepts:distance rate and time relative speed ratio and proportion

Difficulty rating: 1640

Small Hint:

By the meeting time, $A$ has traveled $48$ miles

Big Hint:

In the same time, $B$ has traveled $60+12=72$ miles

Solution:

At the meeting point, $A$ has traveled $60-12=48$ miles and $B$ has traveled $60+12=72$ miles. Since their travel times are equal, their speeds are in the ratio $48:72=2:3.$ If $A$ ’s speed is $v,$ then $B$ ’s is $v+4,$ so $\frac{v}{v+4}=\frac23.$ Thus $3v=2v+8,$ and $v=8$ mph.

Thus, the correct answer is B.

29.

A manufacturer built a machine which will address $500$ envelopes in $8$ minutes. He wishes to build another machine so that when both are operating together they will address $500$ envelopes in $2$ minutes. The equation used to find how many minutes $x$ it would require the second machine to address $500$ envelopes alone is:

$8-x=2$

$\dfrac18+\dfrac1x=\dfrac12$

$\dfrac{500}{8}+\dfrac{500}{x}=500$

$\dfrac{x}{2}+\dfrac{x}{8}=1$

None of these answers

Concepts:rate rational equation

Difficulty rating: 1440

Small Hint:

Measure each machine’s rate in batches of $500$ envelopes per minute

Big Hint:

The two individual rates must add to the combined rate

Solution:

The first machine completes $\tfrac18$ of a $500$ -envelope batch per minute, and the second completes $\tfrac1x$ of a batch per minute. Together they must complete $\tfrac12$ of a batch per minute. Therefore the required equation is $\frac18+\frac1x=\frac12.$

Thus, the correct answer is B.

30.

From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:

$40$

$43$

$29$

$50$

None of these

Concepts:system of equations ratio and proportion

Difficulty rating: 1600

Small Hint:

Let $G$ and $B$ be the original numbers of girls and boys

Big Hint:

Translate the two ratios as $B=2(G-15)$ and $G-15=5(B-45)$

Solution:

Let the original counts be $G$ girls and $B$ boys. The two conditions give $\begin{aligned} B&=2(G-15),\\ G-15&=5(B-45). \end{aligned}$ Substituting the first into the second yields $\begin{aligned} G-15 &=5\bigl(2(G-15)-45\bigr)\\ &=10G-375. \end{aligned}$ Hence $9G=360,$ so $G=40.$

Thus, the correct answer is A.

31.

John ordered $4$ pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by $50\%.$ The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:

$4:1$

$2:1$

$1:4$

$1:2$

$1:8$

Concepts:money ratio and proportion linear equation

Difficulty rating: 1560

Small Hint:

Let a blue pair cost $p$ and let $n$ be the original number of blue pairs

Big Hint:

Compare the original bill $(8+n)p$ with the interchanged bill $(2n+4)p$

Solution:

Let a blue pair cost $p,$ so a black pair costs $2p,$ and let $n$ be the original number of blue pairs. The original bill is $(8+n)p.$ After the quantities are interchanged, the bill is $(2n+4)p.$ The latter is $50\%$ greater, so $2n+4=\frac32(n+8).$ Thus $n=16,$ and the original black-to-blue ratio is $4:16=1:4.$

Thus, the correct answer is C.

32.

A $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:

$9$ ft

$15$ ft

$5$ ft

$8$ ft

$4$ ft

Concepts:Pythagorean Theorem right triangle

Difficulty rating: 1310

Small Hint:

Find the ladder’s original height on the wall using a right triangle

Big Hint:

After the top slips, the new height is $4$ feet less while the hypotenuse stays $25$

Solution:

The initial height, in feet, is $\sqrt{25^2-7^2}=\sqrt{576}=24.$ After the top slips, the height is $20$ feet, so the new horizontal distance is $\sqrt{25^2-20^2}=15$ feet. The foot therefore slides $15-7=8$ feet.

Thus, the correct answer is D.

33.

The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:

$6\pi$

$6$

$12$

$36$

$36\pi$

Concepts:circle area power scaling of length, area, and volume

Difficulty rating: 1240

Small Hint:

Water-carrying capacity is proportional to cross-sectional area

Big Hint:

Circular area scales as the square of the diameter

Solution:

The ratio of the diameters is $6:1,$ so the ratio of cross-sectional areas is $6^2:1^2=36:1.$ Thus $36$ of the smaller pipes have the same total cross-sectional area as the larger pipe.

Thus, the correct answer is D.

34.

When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:

$5$ in

$2\dfrac12$ in

$\dfrac5\pi$ in

$\dfrac{5}{2\pi}$ in

$\dfrac{\pi}{5}$ in

Concepts:circumference algebraic manipulation

Difficulty rating: 1310

Small Hint:

Use $C=2\pi r$ for each circumference

Big Hint:

The change in circumference is $2\pi$ times the change in radius

Solution:

Because $C=2\pi r,$ the changes satisfy $25-20=2\pi\mathopen{}\left(r_{\mathrm{new}}-r_{\mathrm{old}}\right)\mathclose{}.$ Hence the radius increases by $\dfrac{5}{2\pi}$ inches.

Thus, the correct answer is D.

35.

In triangle $ABC,$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:

$26$ in

$4$ in

$13$ in

$8$ in

None of these

Concepts:right triangle incircle, incenter, and inradius Pythagorean Triple

Difficulty rating: 1510

Small Hint:

The side lengths form a right triangle

Big Hint:

For a right triangle with legs $a,b$ and hypotenuse $c,$ the inradius is $\frac{a+b-c}{2}$

Solution:

Since $10^2+24^2=26^2,$ the triangle is right. Its area is $\tfrac12(10)(24)=120,$ and its semiperimeter is $\tfrac{10+24+26}{2}=30.$ Using $K=rs,$ the inradius, in inches, is $r=\frac{K}{s}=\frac{120}{30}=4.$

Thus, the correct answer is B.

36.

A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What per cent of the list price must he mark the goods?

$125\%$

$100\%$

$120\%$

$80\%$

$75\%$

Concepts:percentage money linear equation

Difficulty rating: 1560

Small Hint:

Take the list price to be $L$ and the marked price to be $M$

Big Hint:

The cost is $0.75L,$ the selling price is $0.8M,$ and the cost is $75\%$ of the selling price

Solution:

Let the list and marked prices be $L$ and $M.$ The merchant’s cost is $0.75L,$ while the selling price after the discount is $0.8M.$ A profit equal to $25\%$ of the selling price means that the cost is the remaining $75\%$ of that price. Therefore $0.75L=0.75(0.8M)=0.6M.$ Thus $M=1.25L,$ or $125\%$ of the list price.

Thus, the correct answer is A.

37.

If $y=\log_a x,$ $a>1,$ which of the following statements is incorrect?

If $x=1,$ $y=0$

If $x=a,$ $y=1$

If $x=-1,$ $y$ is imaginary (complex)

If 0<x<1, $y$ is always less than $0$ and decreases without limit as $x$ approaches zero

Only some of the above statements are correct

Concepts:logarithm function complex number

Difficulty rating: 1530

Small Hint:

Check choices A and B directly from the definition of a logarithm

Big Hint:

For $a>1,$ the real logarithm is increasing and tends to $-\infty$ as $x$ approaches $0$ from the right

Solution:

We have $\log_a1=0$ and $\log_aa=1.$ A real logarithm is not defined at $-1,$ though its complex values are nonreal. For $a>1,$ $\log_a x<0$ on $0<x<1,$ and it tends to $-\infty$ as $x$ approaches $0$ from the right. Thus statements A through D are all correct, making the claim that only some are correct the incorrect statement.

Thus, the correct answer is E.

38.

If the expression $\begin{vmatrix}a&c\\d&b\end{vmatrix}$ has the value $ab-cd$ for all values of $a,$ $b,$ $c,$ and $d,$ then the equation $\begin{vmatrix}2x&1\\x&x\end{vmatrix}=3$

Is satisfied for only $1$ value of $x$

Is satisfied for $2$ values of $x$

Is satisfied for no values of $x$

Is satisfied for an infinite number of values of $x$

None of these

Concepts:determinant quadratic

Difficulty rating: 1600

Small Hint:

Apply the given rule to turn the determinant into a quadratic equation

Big Hint:

The equation is $2x^2-x=3$

Solution:

The given rule turns the equation into $(2x)(x)-(1)(x)=3,$ or $2x^2-x-3=0.$ Factoring gives $(2x-3)(x+1)=0,$ whose two distinct solutions are $x=\tfrac32$ and $x=-1.$

Thus, the correct answer is B.

39.

Given the series $2+1+\dfrac12+\dfrac14+\cdots$ and the following five statements:

$(1)$ the sum increases without limit.

$(2)$ the sum decreases without limit.

$(3)$ the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small.

$(4)$ the difference between the sum and $4$ can be made less than any positive quantity no matter how small.

$(5)$ the sum approaches a limit.

Of these statements, the correct ones are:

Only $3$ and $4$

Only $5$

Only $2$ and $4$

Only $2,$ $3,$ and $4$

Only $4$ and $5$

Concepts:geometric sequence summation calculus

Difficulty rating: 1670

Small Hint:

Distinguish the terms of the sequence from its partial sums

Big Hint:

A geometric series with first term $2$ and ratio $\frac{1}{2}$ has partial sums approaching $4$

Solution:

The partial sums increase toward $\frac{2}{1-\frac12}=4,$ so they neither increase nor decrease without limit. This makes statements $1$ and $2$ false, while statements $4$ and $5$ are true.

In statement $3,$ “any term” refers to an already selected term, whose distance from zero is fixed and cannot be made smaller; one can instead choose a sufficiently late term below any prescribed positive bound. Thus statement $3$ , as worded, is false.

Therefore only statements $4$ and $5$ are correct, so the correct answer is E.

40.

The limit of $\dfrac{x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:

$0$

Indeterminate

$x-1$

$2$

$1$

Concepts:calculus difference of squares

Difficulty rating: 1470

Small Hint:

Factor the numerator as a difference of squares

Big Hint:

For $x\ne1,$ cancel the common factor $x-1$

Solution:

For $x\ne1,$ $\begin{aligned} \frac{x^2-1}{x-1} &=\frac{(x-1)(x+1)}{x-1}\\ &=x+1. \end{aligned}$ Therefore the limit as $x$ approaches $1$ is $1+1=2.$

Thus, the correct answer is D.

41.

The least value of the function $ax^2+bx+c$ with $a>0$ is:

$-\dfrac ba$

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

$b^2-4ac$

$\dfrac{4ac-b^2}{4a}$

None of these

Concepts:completing the square quadratic optimization

Difficulty rating: 1580

Small Hint:

Complete the square in $ax^2+bx+c$

Big Hint:

Since $a>0,$ the squared term is minimized when it equals zero

Solution:

Completing the square gives $\begin{aligned} ax^2+bx+c &=a\left(x+\frac{b}{2a}\right)^2\\ &\quad+c-\frac{b^2}{4a}. \end{aligned}$ Because $a>0,$ the squared term has minimum $0.$ The least value is therefore $c-\frac{b^2}{4a}=\frac{4ac-b^2}{4a}.$

Thus, the correct answer is D.

42.

The equation $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=2$ is satisfied when $x$ is equal to:

Infinity

$2$

$\sqrt[4]{2}$

$\sqrt2$

None of these

Concepts:exponent recursion

Difficulty rating: 1830

Small Hint:

The exponent above the first $x$ is the entire infinite tower again

Big Hint:

Replace that repeated tower by its given value $2$

Solution:

Let the value of the infinite tower be $T.$ Removing its bottom $x$ leaves the same tower as the exponent, so $T=x^T.$ Since $T=2,$ we obtain $2=x^2.$ The positive base is therefore $x=\sqrt2,$ for which the tower is convergent.

Thus, the correct answer is D.

43.

The sum to infinity of $\dfrac17+\dfrac{2}{7^2}+\dfrac{1}{7^3}+\dfrac{2}{7^4}+\cdots$ is:

$\dfrac15$

$\dfrac1{24}$

$\dfrac5{48}$

$\dfrac1{16}$

None of these

Concepts:geometric sequence summation

Difficulty rating: 1670

Small Hint:

Group the series into consecutive pairs of terms

Big Hint:

Each pair is $\frac{1}{49}$ times the preceding pair

Solution:

Grouping consecutive terms gives a geometric series whose first grouped term is $\frac17+\frac2{49}=\frac9{49}$ and whose ratio is $\tfrac1{49}.$ Hence the sum is $\frac{\frac9{49}}{1-\frac1{49}} =\frac9{48} =\frac3{16}.$ This is not among choices A through D.

Thus, the correct answer is E.

44.

The graph of $y=\log x$

Cuts the $y$ -axis

Cuts all lines perpendicular to the $x$ -axis

Cuts the $x$ -axis

Cuts neither axis

Cuts all circles whose center is at the origin

Concepts:logarithm function coordinate geometry

Difficulty rating: 1340

Small Hint:

Find where $\log x=0$

Big Hint:

The logarithm is defined only for $x>0$

Solution:

Since $\log1=0,$ the graph passes through $(1,0)$ and therefore cuts the $x$ -axis. It cannot cut the $y$ -axis because $x=0$ is outside its domain.

Thus, the correct answer is C.

45.

The number of diagonals that can be drawn in a polygon of $100$ sides is:

$4850$

$4950$

$9900$

$98$

$8800$

Concepts:diagonal combinations

Difficulty rating: 1410

Small Hint:

Every pair of vertices determines a segment

Big Hint:

Subtract the $100$ sides from the $\binom{100}{2}$ vertex pairs

Solution:

There are $\binom{100}{2}$ segments joining pairs of vertices. Exactly $100$ of these are sides, so the number of diagonals is $\begin{aligned} \binom{100}{2}-100 &=4950-100\\ &=4850. \end{aligned}$

Thus, the correct answer is A.

46.

In triangle $ABC,$ $AB=12,$ $AC=7,$ and $BC=10.$ If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:

The area is doubled

The altitude is doubled

The area is four times the original area

The median is unchanged

The area of the triangle is $0$

Concepts:triangle inequality area

Difficulty rating: 1450

Small Hint:

Write down the three new side lengths

Big Hint:

Compare the largest new side with the sum of the other two

Solution:

The new side lengths are $24,$ $14,$ and $10.$ Because $14+10=24,$ they form a degenerate triangle: all three vertices are collinear. Its area is therefore $0.$

Thus, the correct answer is E.

47.

A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h,$ and the altitude $x$ of the rectangle is half the base of the rectangle, then:

$x=\dfrac12h$

$x=\dfrac{bh}{b+h}$

$x=\dfrac{bh}{2h+b}$

$x=\sqrt{\dfrac{hb}{2}}$

$x=\dfrac12b$

Concepts:similarity rectangle altitude

Difficulty rating: 1800

Small Hint:

The segment across the triangle at height $x$ has length $b(1-\frac{x}{h})$

Big Hint:

The rectangle’s base is $2x$

Solution:

By similarity, the width of the triangle at height $x$ above its base is $b(1-\tfrac{x}{h}).$ This is the base of the inscribed rectangle. Since the rectangle’s altitude is half its base, its base is $2x.$ Hence $2x=b\left(1-\frac{x}{h}\right).$ Multiplying by $h$ and solving gives $x(2h+b)=bh,$ so $x=\dfrac{bh}{2h+b}.$

Thus, the correct answer is C.

48.

A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:

Least when the point is the center of gravity of the triangle

Greater than the altitude of the triangle

Equal to the altitude of the triangle

One-half the sum of the sides of the triangle

Greatest when the point is the center of gravity

Concepts:Viviani’s Theorem equilateral triangle area decomposition

Difficulty rating: 1920

Small Hint:

Join the interior point to all three vertices

Big Hint:

Add the areas of the three smaller triangles using the common side length as their bases

Solution:

Let the equilateral triangle have side length $s$ and altitude $h,$ and let the three perpendicular distances be $d_1,d_2,d_3.$ Splitting the triangle at the interior point gives $\frac12s(d_1+d_2+d_3)=\frac12sh.$ Therefore $d_1+d_2+d_3=h,$ independent of the selected point.

Thus, the correct answer is C.

49.

A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $1\dfrac12$ inches long and can have any position emanating from $A.$ The locus of the vertex $C$ of the triangle is:

A straight line $AB,$ $1\dfrac12$ inches from $A$

A circle with $A$ as center and radius $2$ inches

A circle with $A$ as center and radius $3$ inches

A circle with radius $3$ inches and center $4$ inches from $B$ along $BA$

An ellipse with $A$ as focus

Concepts:median (geometry)circle transformation

Difficulty rating: 2170

Small Hint:

Let $M$ be the midpoint of $BC$ ; then $M$ moves on a circle centered at $A$

Big Hint:

In vector form, the midpoint relation gives $\vec C=2\vec M-\vec B$ when $A$ is the origin

Solution:

Put $A$ at the origin and regard $B,$ $C,$ and the midpoint $M$ of $BC$ as vectors. Since $AM=\tfrac32,$ the point $M$ moves on a circle of radius $\tfrac32$ centered at $A.$ The midpoint relation gives $C=2M-B.$ Thus the locus of $C$ is the image of that circle under a dilation by $2$ followed by translation by $-B.$ It is a circle of radius $3$ centered at $-B.$

Because $AB=2,$ the point $-B$ is $4$ inches from $B$ along the ray $BA.$ Hence the correct answer is D.

50.

A privateer discovers a merchantman $10$ miles to leeward at $11{:}45$ a.m. and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away: she can now make only $17$ miles while the merchantman makes $15.$ The privateer will overtake the merchantman at:

$3{:}45$ p.m.

$3{:}30$ p.m.

$5{:}00$ p.m.

$2{:}45$ p.m.

$5{:}30$ p.m.

Concepts:distance rate and time relative speed ratio and proportion

Difficulty rating: 1900

Small Hint:

Find the remaining gap after the first two hours

Big Hint:

After the sail is lost, the speed ratio is $17:15$ while the merchantman still travels at $8$ mph

Solution:

During the first two hours, the number of miles the privateer gains is $2(11-8)=6,$ reducing the gap from $10$ miles to $4$ miles at $1{:}45$ p.m.

After the damage, the ships’ speeds are in the ratio $17:15.$ Since the merchantman still travels at $8$ mph, the privateer’s new speed is $8\cdot\tfrac{17}{15}=\tfrac{136}{15}$ mph. The closing speed is therefore $\tfrac{136}{15}-8=\tfrac{16}{15}$ mph, so the number of hours needed to close the remaining gap is $\frac{4}{\frac{16}{15}}=\frac{15}{4}=3\frac34.$ Adding $3$ hours $45$ minutes to $1{:}45$ p.m. gives $5{:}30$ p.m.

Thus, the correct answer is E.