1995 AMC 12 Solutions

Compare $90$ with the average of the first three scores

Find the old and new totals before dividing by the number of exams

The first three scores total $87+83+88=258,$ so their average is $\frac{258}{3}=86.$ With the fourth score, the total is $348,$ and the new average is $\frac{348}{4}=87.$ It increases by $1,$ so the correct answer is B.

Square both sides once to isolate $\sqrt{x}$

After isolating the inner radical, square a second time

Squaring gives $2+\sqrt{x}=9,$ so $\sqrt{x}=7.$ Squaring again gives $x=49.$ Thus the correct answer is D.

Add all three payments and the one-time charge

Subtract the advertised total from the in-store price and convert dollars to cents

The advertised total is $3(29.98)+9.98=99.92$ dollars. The savings is $99.99-99.92=0.07$ dollars, or $7$ cents. Thus the correct answer is B.

Express both $M$ and $N$ as multiples of $P$

Convert the percentages to fractions before forming $\frac{M}{N}$

We have $M=(0.30)(0.20)P=0.06P$ and $N=0.50P.$ Hence $\frac{M}{N}=\frac{0.06}{0.50}=\frac{3}{25}.$ Thus the correct answer is B.

First find the field’s area in square feet

One square foot contains $12^2$ square inches

The area is $300\cdot400=120{,}000$ square feet, or $120{,}000\cdot144=17{,}280{,}000$ square inches. The estimate is $3(17{,}280{,}000)=51{,}840{,}000,$ closest to $50$ million. Thus the correct answer is C.

Hold face $A$ fixed and fold each neighboring square by $90^\circ$

Track the outward normal direction of each face as you move through the net

Hold $A$ as the front face. Folding makes $x$ the left face and $B$ the top face. The face $C,$ attached to the right of $B,$ then folds to the right face, opposite $x.$ Thus the correct answer is C.

Approximate the equator’s length with $2\pi r$

Divide the trip distance by $500$ and use a simple estimate for $\pi$

The equator is approximately $2\pi(4000)=8000\pi$ miles long. The flight time is $\frac{8000\pi}{500}=16\pi\approx50$ hours. Thus the correct answer is C.

The two right angles show that $\overline{DE}\parallel\overline{AC}$

Use similarity between $\triangle BDE$ and $\triangle BAC$

The large right triangle has $AB=\sqrt{6^2+8^2}=10.$ Since $\overline{DE}\parallel\overline{AC},$ triangles $BDE$ and $BAC$ are similar, with scale factor $\frac{DE}{AC}=\frac{4}{6}=\frac{2}{3}.$ Therefore $BD=(\frac{2}{3})(10)=\frac{20}{3}.$ Thus the correct answer is C.

Separate the triangles by size before counting

Count one orientation and use the square’s rotational symmetry

There are $8$ smallest triangles, each bounded by a half-side, a half-diagonal, and a half-midline. There are $4$ triangles whose base is a full side and vertex is the center, and $4$ half-square triangles cut off by a diagonal. The total is $8+4+4=16.$ Thus the correct answer is D.

Find where each slanted line meets $y=6$

Use the horizontal segment on $y=6$ as the triangle’s base

The vertices are $(0,0),$ $(6,6),$ and $(-6,6).$ The horizontal base has length $12,$ and the height is $6,$ so the area is $\frac12(12)(6)=36.$ Thus the correct answer is E.

Determine the possible digits $a$ and $d$ from conditions (i) and (ii)

Choose two distinct digits from $3,$ $4,$ $5,$ $6$ for $b\lt c$

The thousands digit $a$ is $4$ or $5,$ and divisibility by $5$ makes $d$ either $0$ or $5.$ The pair $(b,c)$ is obtained by choosing two of $3,$ $4,$ $5,$ $6$ in increasing order, giving $\binom42=6$ pairs. Thus the count is $2\cdot2\cdot6=24,$ and the correct answer is C.

Translate each inequality into a condition on the slope

The two slope conditions together determine the entire linear function

From $f(1)\le f(2),$ the slope is nonnegative. From $f(3)\ge f(4),$ it is nonpositive. Thus the slope is $0,$ so $f$ is constant. Since $f(5)=5,$ we have $f(0)=5.$ Thus the correct answer is D.

Add from right to left and track each carry

Remember that the same replacement must be made at every occurrence of $d$

The units, tens, and hundreds columns work with carries $0,$ $1,$ $1.$ In the thousands column, replacing every $2$ by $6$ gives $6+9+1=16,$ so the result digit is also $6.$ The next columns then give $4+6+1=11$ and $7+8+1=16,$ producing the corrected equation $746586+869430=1616016.$ Thus $d+e=2+6=8,$ and the correct answer is C.

The terms with even powers are unchanged when $x$ is replaced by $-x$

Compare $f(3)$ and $f(-3)$ without solving for $a$ or $b$

The even-power terms and constant are the same at $3$ and $-3,$ while the linear term changes from $-3$ to $3.$ Hence $f(3)-f(-3)=6,$ so $f(3)=2+6=8.$ Thus the correct answer is E.

Write down the landing points for the first few jumps

Once a point repeats, reduce the remaining number of jumps modulo the cycle length

Starting at $5,$ the landing points are $1,2,4,1,2,4,\ldots$ Thus the cycle $1,2,4$ has length $3.$ Since $1995$ is divisible by $3,$ the $1995$th landing point is $4.$ Thus the correct answer is D.

In the two statements, the $10\%$ is taken of different quantities

Find both attendance intervals, then maximize the larger endpoint minus the smaller endpoint

Atlanta’s actual attendance $A$ satisfies $45{,}000\le A\le55{,}000.$ For Boston’s actual attendance $B,$ the condition is $|60{,}000-B|\le0.1B,$ so $\frac{60{,}000}{1.1}\le B\le\frac{60{,}000}{0.9}.$ The largest difference is therefore $$\frac{60{,}000}{0.9}-45{,}000 =21{,}666.\overline6,$$ which rounds to $22{,}000.$ Thus the correct answer is E.

The radii to $A$ and $D$ are perpendicular to the tangent sides

Use the $72^\circ$ exterior angle of a regular pentagon to compare $\overline{AB}$ and $\overline{DC}$

Let $O$ be the circle’s center. The direction changes by $72^\circ$ at each vertex of the pentagon, so the acute angle between lines $AB$ and $DC$ is $36^\circ.$ The radii to their tangency points are perpendicular to these lines; for the circle shown, the minor central angle $\angle AOD$ is the supplementary angle $180^\circ-36^\circ=144^\circ.$ Therefore minor arc $\overset{\frown}{AD}$ measures $144^\circ,$ and the correct answer is E.

Apply the Law of Cosines to $\triangle AOB$

Treat the resulting relation as a quadratic in $OA$ and require a real solution

Let $OA=x$ and $OB=y.$ The Law of Cosines gives $$1=x^2+y^2-\sqrt3xy.$$ As a quadratic in $x,$ this has discriminant $3y^2-4(y^2-1)=4-y^2.$ A real $x$ requires $y\le2,$ and equality is attained when $x=\sqrt3.$ Hence the maximum is $2,$ and the correct answer is D.

Let the outer triangle have side $1$ and the inner triangle have side $s$

Use the $60^\circ$ side slopes together with the fact that $\overline{DE}$ is vertical

Let $ABC$ have side $1,$ with $B=(0,0)$ and $C=(1,0).$ If the inner side is $s,$ write $D=(d,0)$ and $E=(d,s).$ Since $E$ lies on $AC,$ $s=\sqrt3(1-d).$ The third inner vertex is $F=(d-\frac{\sqrt3}{2}s,\frac12s),$ and $F$ lies on $AB,$ so $$\frac{s}{2}=\sqrt3\left(d-\frac{\sqrt3}{2}s\right),$$ giving $d=\frac{2s}{\sqrt3}.$ Substitution yields $s=\frac{1}{\sqrt3}.$ Areas of equilateral triangles scale as the square of their sides, so the ratio is $s^2=\frac{1}{3}.$ Thus the correct answer is C.

The sum is even when $ab$ and $c$ have the same parity

The product $ab$ is odd only when both selected factors are odd

There are $3$ odd and $2$ even choices. The product $ab$ is odd with probability $\frac{9}{25}$ and even with probability $\frac{16}{25}.$ Therefore $$\begin{aligned} P(ab+c\text{ even}) &=\frac9{25}\cdot\frac35\\ &\quad{}+\frac{16}{25}\cdot\frac25\\ &=\frac{59}{125}. \end{aligned}$$ Thus the correct answer is B.

A rectangle’s diagonals have equal length and the same midpoint

Enumerate integer vectors of length $5$ from the origin, identifying antipodal pairs

The given diagonal has midpoint $(0,0)$ and half-length $5.$ The other diagonal must therefore have endpoints $(u,v)$ and $(-u,-v)$ with $u^2+v^2=25.$ Conversely, two equal diagonals with the same midpoint form a rectangle. There are $12$ integer points on this circle: $$\begin{gathered} (\pm5,0),(0,\pm5),\\ (\pm3,\pm4),(\pm4,\pm3). \end{gathered}$$ Antipodal points determine the same diagonal, giving $6$ possibilities. One is the given diagonal itself, which is degenerate, so $6-1=5$ rectangles remain. Thus the correct answer is E.