1970 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The fourth power of $\sqrt{1+\sqrt{1+\sqrt1}}$ is:

$\sqrt2+\sqrt3$

$\tfrac12(7+3\sqrt5)$

$1+2\sqrt3$

$3$

$3+2\sqrt2$

Concepts:radical exponent algebraic manipulation

Difficulty rating: 1740

Small Hint:

Simplify from the innermost radical outward

Big Hint:

After the innermost radical becomes $1,$ square $\sqrt{1+\sqrt2}$ twice

Solution:

The innermost radical is $\sqrt1=1,$ so the given number is $\sqrt{1+\sqrt2}.$ Its fourth power is $\begin{aligned} (1+\sqrt2)^2&=1+2\sqrt2+2\\ &=3+2\sqrt2. \end{aligned}$

Therefore, the correct answer is E.

2.

A square and a circle have equal perimeters. The ratio of the area of the circle to the area of the square is:

$\frac{4}{\pi}$

$\frac{\pi}{\sqrt2}$

$\frac{4}{1}$

$\frac{\sqrt2}{\pi}$

$\frac{\pi}{4}$

Concepts:circle area circumference area ratio square (geometry)

Difficulty rating: 1560

Small Hint:

Let the common perimeter be $p$

Big Hint:

Express the circle radius and square side in terms of $p$ , then divide their areas

Solution:

If the common perimeter is $p,$ then the circle has radius $\frac{p}{2\pi}$ and the square has side $\frac{p}{4}.$ Therefore $\frac{\pi\left(\frac{p}{2\pi}\right)^2}{\left(\frac p4\right)^2} =\frac{\frac{p^2}{4\pi}}{\frac{p^2}{16}}=\frac4\pi.$

Therefore, the correct answer is A.

3.

If $x=1+2^p$ and $y=1+2^{-p},$ then $y$ in terms of $x$ is:

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

$\dfrac{x+2}{x-1}$

$\dfrac{x}{x-1}$

$2-x$

$\dfrac{x-1}{x}$

Concepts:exponent substitution rational equation

Difficulty rating: 1360

Small Hint:

Solve the first equation for $2^p$

Big Hint:

Use $2^{-p}=\frac{1}{2^p}$ and simplify $1+\frac{1}{x-1}$

Solution:

From $x=1+2^p$ we get $2^p=x-1,$ so $\begin{aligned} y&=1+2^{-p}\\ &=1+\frac1{x-1}=\frac{x}{x-1}. \end{aligned}$

Therefore, the correct answer is C.

4.

Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that:

No member of $S$ is divisible by $2$

No member of $S$ is divisible by $3$ but some member is divisible by $11$

No member of $S$ is divisible by $3$ or by $5$

No member of $S$ is divisible by $3$ or by $7$

None of these

Concepts:modular arithmetic divisibility algebraic manipulation

Difficulty rating: 1760

Small Hint:

Write the integers as $n-1,n,n+1$

Big Hint:

Their squared sum is $3n^2+2;$ examine it modulo $3$ and test a residue modulo $11$

Solution:

The sum is $\begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2. \end{gathered}$ It is always congruent to $2\pmod3,$ so no member of $S$ is divisible by $3.$ On the other hand, taking $n=5$ gives $3(25)+2=77,$ which is divisible by $11.$

Therefore, the correct answer is B.

5.

If $f(x)=\dfrac{x^4+x^2}{x+1},$ then $f(i),$ where $i=\sqrt{-1},$ is equal to:

$1+i$

$1$

$-1$

$0$

$-1-i$

Concepts:complex number exponent function

Difficulty rating: 1560

Small Hint:

Use $i^2=-1$ and $i^4=1$

Big Hint:

The numerator vanishes while the denominator does not

Solution:

Because $i^2=-1$ and $i^4=1,$ $f(i)=\frac{i^4+i^2}{i+1}=\frac{1-1}{i+1}=0.$

Therefore, the correct answer is D.

6.

The smallest value of $x^2+8x$ for real values of $x$ is:

$-16.25$

$-16$

$-15$

$-8$

None of these

Concepts:completing the square quadratic inequality

Difficulty rating: 1360

Small Hint:

Complete the square

Big Hint:

Rewrite the expression as $(x+4)^2-16$

Solution:

Completing the square gives $x^2+8x=(x+4)^2-16.$ Since $(x+4)^2\ge0,$ the minimum occurs at $x=-4$ and equals $-16.$

Therefore, the correct answer is B.

7.

Inside square $ABCD$ with side $s,$ quarter-circle arcs with radii $s$ and centers at $A$ and $B$ are drawn. These arcs intersect at a point $X$ inside the square. How far is $X$ from side $CD?$

$\tfrac12s(\sqrt3+4)$

$\tfrac12s\sqrt3$

$\tfrac12s(1+\sqrt3)$

$\tfrac12s(\sqrt3-1)$

$\tfrac12s(2-\sqrt3)$

Concepts:equilateral triangle altitude square (geometry)circle

Difficulty rating: 1980

Small Hint:

The two radii to $X$ and side $AB$ form an equilateral triangle

Big Hint:

The altitude from $X$ to $AB$ is $\frac{s\sqrt3}{2}$ ; subtract it from the square’s height

Solution:

Since $AX=BX=AB=s,$ triangle $ABX$ is equilateral. Thus the perpendicular distance from $X$ to $AB$ is $\frac{s\sqrt3}{2}.$ The distance between the parallel sides $AB$ and $CD$ is $s,$ so the requested distance is $s-\frac{s\sqrt3}{2}=\frac s2(2-\sqrt3).$

Therefore, the correct answer is E.

8.

If $a=\log_8 225$ and $b=\log_2 15,$ then:

$a=\frac{b}{2}$

$a=\frac{2b}{3}$

$a=b$

$b=\frac{a}{2}$

$a=\frac{3b}{2}$

Concepts:logarithm exponent algebraic manipulation

Difficulty rating: 1710

Small Hint:

Rewrite $225$ as $15^2$ and $8$ as $2^3$

Big Hint:

Convert both logarithms to base $2$

Solution:

Using $225=15^2$ and $8=2^3,$ $\begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3}=\frac{2b}{3}. \end{aligned}$

Therefore, the correct answer is B.

9.

Points $P$ and $Q$ are on line segment $AB,$ and both points are on the same side of the midpoint of $AB.$ Point $P$ divides $AB$ in the ratio $2:3,$ and $Q$ divides $AB$ in the ratio $3:4.$ If $PQ=2,$ then the length of segment $AB$ is:

$12$

$28$

$70$

$75$

$105$

Concepts:ratio and proportion linear equation

Difficulty rating: 1610

Small Hint:

Express $AP$ and $AQ$ as fractions of $AB$

Big Hint:

Use $AP=\frac{2AB}{5}$ and $AQ=\frac{3AB}{7}$ , then subtract

Solution:

The division ratios give $AP=\frac25AB,\qquad AQ=\frac37AB.$ Because the two points are on the same side of the midpoint, $\begin{aligned} PQ&=AQ-AP\\ &=\left(\frac37-\frac25\right)AB\\ &=\frac1{35}AB. \end{aligned}$ Since $PQ=2,$ we obtain $AB=70.$

Therefore, the correct answer is C.

10.

Let $F=0.4818181\ldots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by:

$13$

$14$

$29$

$57$

$126$

Concepts:repeating decimal geometric sequence fraction

Difficulty rating: 1590

Small Hint:

Separate the initial $0.4$ from the repeating tail

Big Hint:

The tail is $0.081(1+0.01+0.01^2+\cdots)$

Solution:

We have $\begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110}. \end{aligned}$ The denominator exceeds the numerator by $110-53=57.$

Therefore, the correct answer is D.

11.

If two factors of $2x^3-hx+k$ are $x+2$ and $x-1,$ the value of $|2h-3k|$ is:

$4$

$3$

$2$

$1$

$0$

Concepts:polynomial substitution system of equations

Difficulty rating: 1590

Small Hint:

Apply the factor theorem at $x=-2$ and $x=1$

Big Hint:

Solve $-16+2h+k=0$ and $2-h+k=0$

Solution:

The factor theorem gives $\begin{aligned} -16+2h+k&=0,\\ 2-h+k&=0. \end{aligned}$ Subtracting the second equation from the first gives $3h=18,$ so $h=6$ and then $k=4.$ Hence $|2h-3k|=|12-12|=0.$

Therefore, the correct answer is E.

12.

A circle with radius $r$ is tangent to sides $AB,$ $AD,$ and $CD$ of rectangle $ABCD$ and passes through the midpoint of diagonal $AC.$ The area of the rectangle, in terms of $r,$ is:

$4r^2$

$6r^2$

$8r^2$

$12r^2$

$20r^2$

Concepts:rectangle tangent circles midpoint area

Difficulty rating: 2090

Small Hint:

Tangency to the two opposite sides makes one side of the rectangle equal to $2r$

Big Hint:

The line from the third tangency point through the center bisects the diagonal, making the corresponding chord a diameter

Solution:

Let the circle be tangent to the parallel sides $AB$ and $CD.$ Their separation is its diameter, so $AD=BC=2r.$ Let $R$ be the tangency point on $AD,$ let $Q$ be the center, and let $M$ be the midpoint of $AC.$

The line $RQ$ is parallel to $AB$ and halfway between $AB$ and $CD,$ so $R$ is the midpoint of $AD$ and the line contains $M.$ Since $M$ lies on the circle and $R,Q,M$ are collinear, $RM=2r.$ Thus $RM$ is a midsegment of triangle $ADC,$ so $DC=2RM=4r.$ Therefore the area is $(2r)(4r)=8r^2.$

Therefore, the correct answer is C.

13.

Given the binary operation $*$ defined by $a*b=a^b$ for all positive numbers $a$ and $b.$ Then for all positive $a,$ $b,$ $c,$ $n,$ we have:

$a*b=b*a$

$a*(b*c)=(a*b)*c$

$a*(b^n)=(a*n)*b$

$(a*b)^n=a*(bn)$

None of these

Concepts:custom operation exponent algebraic manipulation

Difficulty rating: 1710

Small Hint:

Replace every $*$ using the definition before comparing the two sides

Big Hint:

For choice D, both sides simplify to a power with exponent $bn$

Solution:

For choice D, $\begin{aligned} (a*b)^n&=(a^b)^n=a^{bn},\\ a*(bn)&=a^{bn}. \end{aligned}$ Thus that identity always holds. The other choices would assert false general identities such as $a^b=b^a$ or $a^{b^c}=a^{bc}.$

Therefore, the correct answer is D.

14.

Consider $x^2+px+q=0,$ where $p$ and $q$ are positive numbers. If the roots of this equation differ by $1,$ then $p$ equals:

$\sqrt{4q+1}$

$q-1$

$-\sqrt{4q+1}$

$q+1$

$\sqrt{4q-1}$

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1670

Small Hint:

Use the quadratic formula to express the difference of the roots

Big Hint:

The difference has magnitude $\sqrt{p^2-4q}$

Solution:

The two roots differ in absolute value by $\sqrt{p^2-4q}.$ Hence $p^2-4q=1,$ so $p^2=4q+1.$ Since $p$ is positive, $p=\sqrt{4q+1}.$

Therefore, the correct answer is A.

15.

Lines in the $xy$ -plane are drawn through the point $(3,4)$ and the trisection points of the line segment joining the points $(-4,5)$ and $(5,-1).$ One of these lines has the equation:

$3x-2y-1=0$

$4x-5y+8=0$

$5x+2y-23=0$

$x+7y-31=0$

$x-4y+13=0$

Concepts:coordinate geometry slope linear equation

Difficulty rating: 1620

Small Hint:

Find the two trisection points by moving one-third and two-thirds of the displacement vector

Big Hint:

The trisection points are $(-1,3)$ and $(2,1)$ ; find the line from either one to $(3,4)$

Solution:

The displacement from $(-4,5)$ to $(5,-1)$ is $(9,-6).$ The trisection points are therefore $(-4,5)+\tfrac13(9,-6)=(-1,3)$ and $(2,1).$ The line through $(-1,3)$ and $(3,4)$ has slope $\frac{1}{4},$ so $y-4=\frac14(x-3),$ or $x-4y+13=0.$

Therefore, the correct answer is E.

16.

If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1,$ and such that $F(n+1)=\frac{F(n)F(n-1)+1}{F(n-2)}$ for $n\ge3,$ then $F(6)$ is equal to:

$2$

$3$

$7$

$11$

$26$

Concepts:recursion function substitution

Difficulty rating: 1450

Small Hint:

Compute the terms one at a time beginning with $F(4)$

Big Hint:

Use the recurrence successively for $n=3,4,5$

Solution:

Applying the recurrence successively, $\begin{aligned} F(4)&=\frac{1\cdot1+1}{1}=2,\\ F(5)&=\frac{2\cdot1+1}{1}=3, \end{aligned}$ and $F(6)=\frac{3\cdot2+1}{1}=7.$

Therefore, the correct answer is C.

17.

If $r\gt0,$ then for all $p$ and $q$ such that $pq\ne0$ and $pr\gt qr,$ we have:

$-p\gt-q$

$-p\gt q$

$1\gt-\frac{q}{p}$

$1\lt \frac{q}{p}$

None of these

Concepts:inequality counterexample ratio and proportion

Difficulty rating: 1850

Small Hint:

Because $r\gt0,$ the hypothesis gives only $p\gt q$

Big Hint:

Test the remaining claims with both positive and negative choices of $p,q$

Solution:

Since $r\gt0,$ the hypothesis is equivalent to $p\gt q.$ Then $-p\lt-q,$ so A is false. Taking $(p,q)=(2,1)$ makes B and D false. Taking $(p,q)=(1,-2)$ makes C false because $-\frac{q}{p}=2.$ Thus none of A-D follows in every case.

Therefore, the correct answer is E.

18.

$\sqrt{3+2\sqrt2}-\sqrt{3-2\sqrt2}$ is equal to:

$2$

$2\sqrt3$

$4\sqrt2$

$\sqrt6$

$2\sqrt2$

Concepts:radical perfect square algebraic manipulation

Difficulty rating: 1740

Small Hint:

Recognize each radicand as a squared binomial

Big Hint:

Use $3\pm2\sqrt2=(\sqrt2\pm1)^2$

Solution:

Since $\begin{gathered} 3+2\sqrt2=(\sqrt2+1)^2,\\ 3-2\sqrt2=(\sqrt2-1)^2, \end{gathered}$ and both $\sqrt2+1$ and $\sqrt2-1$ are positive, the difference is $(\sqrt2+1)-(\sqrt2-1)=2.$

Therefore, the correct answer is A.

19.

The sum of an infinite geometric series with common ratio $r$ such that $|r|\lt1$ is $15,$ and the sum of the squares of the terms of this series is $45.$ The first term of the series is:

$12$

$10$

$5$

$3$

$2$

Concepts:geometric sequence system of equations algebraic manipulation

Difficulty rating: 1990

Small Hint:

If the first term is $a,$ write one equation for each infinite sum

Big Hint:

Use $\frac{a}{1-r}=15$ and $\frac{a^2}{1-r^2}=45$ , then cancel a factor of $1-r$

Solution:

Let the first term be $a.$ Then $\frac{a}{1-r}=15,\qquad \frac{a^2}{1-r^2}=45.$ Since $1-r^2=(1-r)(1+r),$ dividing the second equation by the first gives $\frac{a}{1+r}=3.$ Thus $a=15(1-r)=3(1+r).$ Solving gives $r=\frac{2}{3}$ and $a=5.$

Therefore, the correct answer is C.

20.

Lines $HK$ and $BC$ lie in a plane. $M$ is the midpoint of line segment $BC,$ and $BH$ and $CK$ are perpendicular to $HK.$ Then we:

always have $MH=MK$

always have $MH\gt BK$

sometimes have $MH=MK$ but not always

always have $MH\gt MB$

always have $BH\lt BC$

Concepts:coordinate geometry midpoint perpendicular bisector

Difficulty rating: 1790

Small Hint:

Put $HK$ on the $x$ -axis

Big Hint:

Use $H=(h,0),$ $B=(h,u),$ $K=(k,0),$ and $C=(k,v)$ ; compare squared distances from their midpoint

Solution:

Choose coordinates $\begin{gathered} H=(h,0),\quad B=(h,u),\\ K=(k,0),\quad C=(k,v). \end{gathered}$ Then $M=\left(\frac{h+k}{2},\frac{u+v}{2}\right).$ The horizontal displacements from $M$ to $H$ and $K$ are opposites, while the vertical displacements are equal. Therefore $MH^2=MK^2,$ so $MH=MK$ always.

Therefore, the correct answer is A.

21.

On an auto trip, the distance read from the instrument panel was $450$ miles. With snow tires on for the return trip over the same route, the reading was $440$ miles. Find, to the nearest hundredth of an inch, the increase in radius of the wheels if the original radius was $15$ inches.

$0.33$

$0.34$

$0.35$

$0.38$

$0.66$

Concepts:ratio and proportion circumference distance rate and time

Difficulty rating: 1650

Small Hint:

For a fixed true distance, the odometer reading is inversely proportional to wheel radius

Big Hint:

If the radii are $r_1,r_2,$ use $450r_1=440r_2$

Solution:

For a fixed true distance, the odometer reading is proportional to the number of wheel revolutions. Thus $450r_1=440r_2, \qquad \frac{r_2}{r_1}=\frac{45}{44}.$ With $r_1=15,$ the increase is $\begin{aligned} r_2-r_1 &=15\left(\frac{45}{44}-1\right)\\ &=\frac{15}{44}\\ &\approx0.3409. \end{aligned}$ To the nearest hundredth this is $0.34$ inch.

Therefore, the correct answer is B.

22.

If the sum of the first $3n$ positive integers is $150$ more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is:

$300$

$350$

$400$

$450$

$600$

Concepts:triangular number quadratic summation

Difficulty rating: 1590

Small Hint:

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

Big Hint:

The given difference simplifies to $4n^2+n$

Solution:

Let $S_m=\frac{m(m+1)}{2}.$ Then $\begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n. \end{aligned}$ Since $S_{3n}-S_n=150,$ this gives $4n^2+n=150.$ Hence $(n-6)(4n+25)=0,$ and the positive solution is $n=6.$ Therefore $S_{4n}=S_{24}=\frac{24\cdot25}{2}=300.$

Therefore, the correct answer is A.

23.

The number $10!$ ( $10$ is written in base $10$ ), when written in the base $12$ system, ends in exactly $k$ zeros. The value of $k$ is:

$1$

$2$

$3$

$4$

$5$

Concepts:number base trailing zeros prime factorization factorial

Difficulty rating: 1830

Small Hint:

A trailing base- $12$ zero contributes one factor of $12=2^2\cdot3$

Big Hint:

Compare the exponents of $2$ and $3$ in $10!$

Solution:

The prime exponents in $10!$ are $\begin{gathered} v_2(10!)=5+2+1=8,\\ v_3(10!)=3+1=4. \end{gathered}$ Each factor of $12=2^2\cdot3$ uses two factors of $2$ and one of $3.$ Hence the number of trailing base- $12$ zeros is $\min\left(\left\lfloor\frac82\right\rfloor,4\right)=4.$

Therefore, the correct answer is D.

24.

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $2,$ then the area of the hexagon is:

$2$

$3$

$4$

$6$

$12$

Concepts:equilateral triangle regular polygon area ratio perimeter

Difficulty rating: 1400

Small Hint:

Let the hexagon side be $s$ ; equal perimeters make the triangle side $2s$

Big Hint:

The triangle splits into four small equilateral triangles of side $s$ , while the hexagon splits into six

Solution:

Let the hexagon side be $s.$ Its perimeter is $6s,$ so the equilateral triangle has side $2s.$ The large triangle consists of $4$ equilateral triangles of side $s,$ while the hexagon consists of $6.$ Their area ratio is therefore $\frac{6}{4}=\frac{3}{2}.$ Since the triangle’s area is $2,$ the hexagon’s area is $3.$

Therefore, the correct answer is B.

25.

For every real number $x,$ let $\lfloor x\rfloor$ be the greatest integer which is less than or equal to $x.$ If the postal rate for first class mail is six cents for every ounce or portion thereof, then the cost in cents of first-class postage on a letter weighing $W$ ounces is always:

$6W$

$6\lfloor W\rfloor$

$6(\lfloor W\rfloor-1)$

$6(\lfloor W\rfloor+1)$

$-6\lfloor-W\rfloor$

Concepts:floor and ceiling functions money function

Difficulty rating: 1850

Small Hint:

The number of charged ounces is the least integer at least $W$

Big Hint:

Use the identity $\lceil W\rceil=-\lfloor-W\rfloor$

Solution:

Charging for every ounce or portion thereof means the number of charged ounces is $\lceil W\rceil.$ The floor and ceiling functions satisfy $\lceil W\rceil=-\lfloor-W\rfloor.$ Thus the cost is $6\lceil W\rceil=-6\lfloor-W\rfloor.$

Therefore, the correct answer is E.

26.

The number of distinct points in the $xy$ -plane common to the graphs of $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is:

$0$

$1$

$2$

$3$

$4$

infinite

Concepts:system of equations zero product property counting intersections

Difficulty rating: 1640

Small Hint:

Each product equation represents a union of two lines

Big Hint:

Check the intersection of the two lines in each pair before considering all four pairings

Solution:

The first graph is the union of $x+y=5,\qquad 2x-3y=-5,$ and those two lines meet at $(2,3).$ The second graph is the union of $x-y=-1,\qquad 3x+2y=12,$ and these also meet at $(2,3).$ In fact, substituting $(2,3)$ satisfies all four line equations. Since the four lines have distinct slopes, no other point can belong to a line from each graph. There is exactly one common point.

Therefore, the correct answer is B.

27.

In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?

$2$

$3$

$4$

$5$

$6$

Concepts:incircle, incenter, and inradius triangle area perimeter

Difficulty rating: 1830

Small Hint:

Express the area in terms of the inradius and semiperimeter

Big Hint:

Use $K=rs$ and perimeter $=2s$

Solution:

If $r$ is the inradius and $s$ the semiperimeter, the triangle’s area is $K=rs.$ Its perimeter is $2s.$ The given equality yields $rs=2s.$ Since $s\gt0,$ we obtain $r=2.$

Therefore, the correct answer is A.

28.

In triangle $ABC,$ the median from vertex $A$ is perpendicular to the median from vertex $B.$ If the lengths of sides $AC$ and $BC$ are $6$ and $7,$ respectively, then the length of side $AB$ is:

$\sqrt{17}$

$4$

$4\tfrac12$

$2\sqrt5$

$4\tfrac14$

Concepts:centroid median (geometry)vector Pythagorean Theorem

Difficulty rating: 2300

Small Hint:

Place the centroid at the origin and let the position vectors of $A,B$ be perpendicular vectors $u,v$

Big Hint:

Then $C=-u-v,$ so $AC^2=|2u+v|^2$ and $BC^2=|u+2v|^2$

Solution:

Place the centroid at the origin. Let the position vectors of $A$ and $B$ be $u$ and $v.$ The two medians lie along $u$ and $v,$ so $u\cdot v=0.$ Also the centroid condition gives $C=-u-v.$

Let $U=|u|^2$ and $V=|v|^2.$ Then $AC^2=|2u+v|^2=4U+V=36$ and $\begin{aligned} BC^2&=|u+2v|^2\\ &=U+4V=49. \end{aligned}$ Adding appropriate multiples gives $U+V=17.$ Finally, $AB^2=|u-v|^2=U+V=17,$ so $AB=\sqrt{17}.$

Therefore, the correct answer is A.

29.

It is now between $10{:}00$ and $11{:}00$ o’clock, and six minutes from now, the minute hand of a watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?

$10{:}05\tfrac5{11}$

$10{:}07\tfrac12$

$10{:}10$

$10{:}15$

$10{:}17\tfrac12$

Concepts:clock linear equation

Difficulty rating: 1890

Small Hint:

Let $x$ be the number of minutes after $10{:}00$ now

Big Hint:

In minute-space units, the future minute hand is at $x+6$ , while the point opposite the past hour hand is at $20+\frac{x-3}{12}$

Solution:

Let $x$ be the number of minutes after $10{:}00$ now. Six minutes from now, the minute hand is $x+6$ minute spaces clockwise from $12.$ Three minutes ago, the hour hand was $\frac{x-3}{12}$ minute spaces past the $10,$ so the point opposite it was $20+\frac{x-3}{12}$ minute spaces past $12.$ Hence $x+6=20+\frac{x-3}{12}.$ Solving gives $x=15,$ so the time is $10{:}15.$

Therefore, the correct answer is D.

30.

In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B,$ and the measures of segments $AD$ and $CD$ are $a$ and $b,$ respectively. Then the measure of $AB$ is equal to:

$\tfrac12a+2b$

$\tfrac32b+\tfrac34a$

$2a-b$

$4b-\tfrac12a$

$a+b$

Concepts:angle bisector parallel lines isosceles triangle parallelogram

Difficulty rating: 2190

Small Hint:

Bisect angle $D$ and let the bisector meet $AB$ at $P$

Big Hint:

Angle chasing makes $\triangle APD$ isosceles, while $PBCD$ is a parallelogram

Solution:

Let the bisector of $\angle D$ meet $AB$ at $P.$ Because $\angle D=2\angle B,$ each half has measure $\angle B.$ Since $AB\parallel CD,$ the alternate interior angle $\angle APD$ also equals $\angle PDC.$ Thus triangle $APD$ is isosceles and $AP=AD=a.$ Also $PB\parallel CD$ and $BC\parallel PD,$ so $PBCD$ is a parallelogram. Hence $PB=CD=b.$ Therefore $AB=AP+PB=a+b.$

Therefore, the correct answer is E.

31.

If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to $43,$ what is the probability that this number will be divisible by $11?$

$\frac{2}{5}$

$\frac{1}{5}$

$\frac{1}{6}$

$\frac{1}{11}$

$\frac{1}{15}$

Concepts:digits divisibility basic counting basic probability

Difficulty rating: 2300

Small Hint:

Start from $99999$ ; a digit sum of $43$ means a total deficit of $2$

Big Hint:

The possibilities are one $7$ or two $8$ s; apply the divisibility-by- $11$ alternating-sum test

Solution:

The maximum five-digit digit sum is $45,$ so a sum of $43$ has total deficit $2$ from $99999.$ Either one digit is $7$ and the others are $9$ , giving $5$ numbers, or two digits are $8$ and the others are $9$ , giving $\binom52=10$ numbers. Thus there are $15$ numbers total.

Using the divisibility-by- $11$ test, exactly $97999,\qquad 99979,\qquad 98989$ are divisible by $11.$ The probability is therefore $\frac{3}{15}=\frac{1}{5}.$

Therefore, the correct answer is B.

32.

$A$ and $B$ travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after $B$ has travelled $100$ yards, and meet a second time $60$ yards before $A$ completes one lap, then the circumference of the track in yards is:

$400$

$440$

$480$

$560$

$880$

Concepts:relative speed circumference ratio and proportion system of equations

Difficulty rating: 2190

Small Hint:

Let the circumference be $2C$ and compare the two runners’ traveled-distance ratios at each meeting

Big Hint:

At the first meeting their distances are $C-100$ and $100$ ; at the second they are $2C-60$ and $C+60$

Solution:

Let the circumference be $2C.$ At the first meeting, $B$ has traveled $100$ yards and $A$ has traveled $C-100,$ so their speed ratio is $\frac{v_A}{v_B}=\frac{C-100}{100}.$ At the second meeting, $A$ is $60$ yards short of one lap, so it has traveled $2C-60.$ By then their combined distance is $3C,$ so $B$ has traveled $C+60.$ Hence $\frac{C-100}{100}=\frac{2C-60}{C+60}.$ Cross-multiplying gives $C=240,$ so the circumference is $2C=480$ yards.

Therefore, the correct answer is C.

33.

Find the sum of the digits of all the numerals in the sequence $1,$ $2,$ $3,$ $4,$ $\ldots,$ $10000.$

$180{,}001$

$154{,}756$

$45{,}001$

$154{,}755$

$270{,}001$

Concepts:digits place value symmetry (algebra)summation

Difficulty rating: 1890

Small Hint:

First sum the digits from $0000$ through $9999$

Big Hint:

In each of the four positions, every digit appears exactly $1000$ times

Solution:

Write the integers from $0$ through $9999$ using four digits with leading zeros. In each of the four positions, every digit $0,$ $1,$ $\ldots,$ $9$ appears $1000$ times. Thus their total digit sum is $\begin{gathered} 4\cdot1000(0+1+\cdots+9)\\ =4000\cdot45\\ =180000. \end{gathered}$ Replacing $0$ by $10000$ adds $1,$ so the requested sum is $180001.$

Therefore, the correct answer is A.

34.

The greatest integer that will divide $13{,}511,$ $13{,}903,$ and $14{,}589$ and leave the same remainder is:

$28$

$49$

$98$

an odd multiple of $7$ greater than $49$

an even multiple of $7$ greater than $98$

Concepts:greatest common divisor divisibility algebraic manipulation

Difficulty rating: 1650

Small Hint:

A common remainder means the divisor divides every pairwise difference

Big Hint:

Find the GCD of $13903-13511$ and $14589-13903$

Solution:

A divisor leaves the same remainder on all three numbers exactly when it divides their differences. The two successive differences are $\begin{gathered} 13903-13511=392,\\ 14589-13903=686. \end{gathered}$ Therefore the greatest possible divisor is $\begin{gathered} \gcd(392,686)\\ =\gcd(392,294)\\ =\gcd(294,98)=98. \end{gathered}$

Therefore, the correct answer is C.

35.

A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served $a$ years more, his pension would have been $p$ dollars greater, whereas, had he served $b$ years more ( $b\ne a$ ), his pension would have been $q$ dollars greater than the original annual pension. Find his annual pension in terms of $a,$ $b,$ $p,$ and $q.$

$\dfrac{p^2-q^2}{2(a-b)}$

$\dfrac{(p-q)^2}{2\sqrt{ab}}$

$\dfrac{ap^2-bq^2}{2(ap-bq)}$

$\dfrac{aq^2-bp^2}{2(bp-aq)}$

$\sqrt{(a-b)(p-q)}$

Concepts:ratio and proportion radical system of equations algebraic manipulation

Difficulty rating: 2440

Small Hint:

Let the current pension be $X=k\sqrt n$ and write the two hypothetical pensions

Big Hint:

Square $X+p=k\sqrt{n+a}$ and $X+q=k\sqrt{n+b}$ , then eliminate both $n$ and $k^2$

Solution:

Let the current pension be $X=k\sqrt n.$ Since $X+p=k\sqrt{n+a}$ and $X+q=k\sqrt{n+b},$ squaring and using $X^2=k^2n$ gives the first line below. Multiplying its equations by $b$ and $a,$ respectively, and subtracting gives the second: $\begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2. \end{gathered}$ Hence $X=\dfrac{aq^2-bp^2}{2(bp-aq)}.$

Therefore, the correct answer is D.