2007 AMC 10A Solutions | LIVE by Po-Shen Loh

1.

One ticket to a show costs $\$20$ at full price. Susan buys $4$ tickets using a coupon that gives her a $25\%$ discount. Pam buys $5$ tickets using a coupon that gives her a $30\%$ discount. How many more dollars does Pam pay than Susan?

$2$

$5$

$10$

$15$

$20$

Concepts:percentage money

Difficulty rating: 720

Small Hint:

A $25\%$ discount means paying $75\%$ of the full price

Big Hint:

Compute each person’s total, then subtract

Solution:

Susan pays $4 \cdot 0.75 \cdot 20 = 60$ dollars.

Pam pays $5 \cdot 0.70 \cdot 20 = 70$ dollars.

The difference is $70 - 60 = 10$ dollars.

Thus, the correct answer is C.

2.

Define $a@b = ab - b^2$ and $a\#b = a + b - ab^2.$ What is $\dfrac{6@2}{6\#2}?$

$-\dfrac{1}{2}$

$-\dfrac{1}{4}$

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

Concepts:custom operation substitution

Difficulty rating: 870

Small Hint:

Substitute $a = 6$ and $b = 2$ into each definition

Big Hint:

The numerator is $6 \cdot 2 - 2^2$ and the denominator is $6 + 2 - 6 \cdot 2^2$

Solution:

The numerator is $6@2 = 6 \cdot 2 - 2^2 = 12 - 4 = 8.$

The denominator is $6\#2 = 6 + 2 - 6 \cdot 2^2$ $= 8 - 24$ $= -16.$

The quotient is $\dfrac{8}{-16} = -\dfrac{1}{2}.$

Thus, the correct answer is A.

3.

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. It is filled with water to a height of $40$ cm. A brick with a rectangular base that measures $40$ cm by $20$ cm and a height of $10$ cm is placed in the aquarium. By how many centimeters does the water rise?

$0.5$

$1$

$1.5$

$2$

$2.5$

Concepts:volume rectangular prism

Difficulty rating: 960

Small Hint:

The submerged brick pushes the water up by its own volume

Big Hint:

The rise $h$ satisfies $100 \cdot 40 \cdot h = 40 \cdot 20 \cdot 10$

Solution:

The brick has volume $40 \cdot 20 \cdot 10 = 8000$ cubic centimeters.

If the water rises by $h$ centimeters, the added volume is $100 \cdot 40 \cdot h = 4000h$ cubic centimeters.

Setting these equal gives $4000h = 8000,$ so $h = 2.$

Thus, the correct answer is D.

4.

The larger of two consecutive odd integers is three times the smaller. What is their sum?

$4$

$8$

$12$

$16$

$20$

Concepts:linear equation

Difficulty rating: 870

Small Hint:

Let the smaller integer be $x,$ so the larger is $x + 2$

Big Hint:

Solve $x + 2 = 3x$

Solution:

Let the smaller integer be $x.$ Then the larger is $x + 2,$ and $x + 2 = 3x,$ so $x = 1.$

The two integers are $1$ and $3,$ and their sum is $4.$

Thus, the correct answer is A.

5.

A school store sells $7$ pencils and $8$ notebooks for $\$4.15.$ It also sells $5$ pencils and $3$ notebooks for $\$1.77.$ How much do $16$ pencils and $10$ notebooks cost?

$\$4.76$

$\$5.84$

$\$6.00$

$\$6.16$

$\$6.32$

Concepts:system of equations money

Difficulty rating: 1020

Small Hint:

Work in cents: let $p$ be a pencil and $n$ a notebook, giving $7p + 8n = 415$ and $5p + 3n = 177$

Big Hint:

Solve for $p$ and $n,$ then evaluate $16p + 10n$

Solution:

Let $p$ and $n$ be the prices in cents of a pencil and a notebook. Then $7p + 8n = 415$ $5p + 3n = 177.$

Solving this system gives $p = 9$ and $n = 44.$

So $16$ pencils and $10$ notebooks cost $16(9) + 10(44) = 584$ cents, or $\$5.84.$

Thus, the correct answer is B.

6.

At Euclid High School, the number of students taking the AMC $10$ was $60$ in $2002,$ $66$ in $2003,$ $70$ in $2004,$ $76$ in $2005,$ $78$ in $2006,$ and is $85$ in $2007.$ Between what two consecutive years was there the largest percentage increase?

$2002$ and $2003$

$2003$ and $2004$

$2004$ and $2005$

$2005$ and $2006$

$2006$ and $2007$

Concepts:percentage fraction

Difficulty rating: 960

Small Hint:

Percentage increase equals the increase divided by the earlier value

Big Hint:

Compare all five fractional increases using a common benchmark or cross-multiplication

Solution:

From $2002$ to $2003,$ the increase is $\dfrac{6}{60} = \dfrac{1}{10} = 10\%.$

The other increases are $\dfrac{4}{66},$ $\dfrac{6}{70},$ $\dfrac{2}{76},$ and $\dfrac{7}{78},$ each less than $\dfrac{1}{10}.$

So the largest percentage increase was between $2002$ and $2003.$

Thus, the correct answer is A.

7.

Last year Mr. John Q. Public received an inheritance. He paid $20\%$ in federal taxes on the inheritance, and paid $10\%$ of what he had left in state taxes. He paid a total of $\$10{,}500$ for both taxes. How many dollars was the inheritance?

$30{,}000$

$32{,}500$

$35{,}000$

$37{,}500$

$40{,}000$

Concepts:percentage linear equation

Difficulty rating: 1070

Small Hint:

After federal tax, $80\%$ of the inheritance remains

Big Hint:

The state tax is $10\%$ of that $80\%,$ so the total tax is $20\% + 8\%$ of the inheritance

Solution:

After federal taxes, Mr. Public keeps $80\%$ of his inheritance.

State taxes take $10\%$ of that, which is $8\%$ of the inheritance.

The total tax is $20\% + 8\% = 28\%$ of the inheritance, so the inheritance is $\dfrac{10{,}500}{0.28} = 37{,}500.$

Thus, the correct answer is D.

8.

Triangles $ABC$ and $ADC$ are isosceles with $AB = BC$ and $AD = DC.$ Point $D$ is inside $\triangle ABC,$ $\angle ABC = 40^\circ,$ and $\angle ADC = 140^\circ.$ What is the degree measure of $\angle BAD?$

$20$

$30$

$40$

$50$

$60$

Concepts:isosceles triangle angle chasing

Difficulty rating: 1170

Small Hint:

In an isosceles triangle the two base angles are equal

Big Hint:

Find $\angle BAC$ and $\angle DAC,$ then subtract

Solution:

Since $\triangle ABC$ is isosceles, $\angle BAC = \tfrac12(180^\circ - 40^\circ) = 70^\circ.$

Since $\triangle ADC$ is isosceles, $\angle DAC = \tfrac12(180^\circ - 140^\circ) = 20^\circ.$

Therefore $\angle BAD = \angle BAC - \angle DAC$ $= 70^\circ - 20^\circ$ $= 50^\circ.$

Thus, the correct answer is D.

9.

Real numbers $a$ and $b$ satisfy the equations $3^a = 81^{b+2}$ and $125^b = 5^{a-3}.$ What is $ab?$

$-60$

$-17$

$9$

$12$

$60$

Concepts:exponent system of equations

Difficulty rating: 1240

Small Hint:

Rewrite $81 = 3^4$ and $125 = 5^3$ to match the bases

Big Hint:

Equating exponents gives $a = 4(b + 2)$ and $3b = a - 3$

Solution:

The equations become $3^a = 3^{4(b+2)}$ and $5^{3b} = 5^{a-3}.$

So $a = 4(b + 2)$ and $3b = a - 3.$

Solving gives $a = -12$ and $b = -5,$ so $ab = 60.$

Thus, the correct answer is E.

10.

The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20,$ the father is $48$ years old, and the average age of the mother and children is $16.$ How many children are in the family?

$2$

$3$

$4$

$5$

$6$

Concepts:mean linear equation

Difficulty rating: 1240

Small Hint:

Let $N$ be the number of children; the family has $N + 2$ members

Big Hint:

The total age is $20(N + 2);$ removing the father leaves $16(N + 1)$

Solution:

Let $N$ be the number of children and $T$ the total age of the family.

Then $20 = \dfrac{T}{N + 2}$ and $16 = \dfrac{T - 48}{N + 1}.$

These give $20N + 40 = T$ and $16N + 64 = T,$ so $20N + 40 = 16N + 64.$

Hence $4N = 24$ and $N = 6.$

Thus, the correct answer is E.

11.

The numbers from $1$ to $8$ are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

$14$

$16$

$18$

$20$

$24$

Concepts:double counting cube geometry

Difficulty rating: 1280

Small Hint:

Each vertex of a cube lies on exactly three faces

Big Hint:

Adding all six face sums counts each vertex $3$ times: $3(1 + 2 + \cdots + 8)$

Solution:

Each vertex belongs to exactly three faces, so summing the numbers over all six faces gives $\begin{aligned} 3(1 + 2 + \cdots + 8) &= 3 \cdot 36 \\ &= 108. \end{aligned}$

There are six faces, so the common sum is $108 \div 6 = 18.$

Thus, the correct answer is C.

12.

Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

$56$

$58$

$60$

$62$

$64$

Concepts:multiplication principle complementary counting

Difficulty rating: 1330

Small Hint:

Each of the six tourists independently picks one of the two guides

Big Hint:

Subtract the two arrangements in which a guide ends up with no tourists

Solution:

Each tourist independently chooses one of the two guides, giving $2^6 = 64$ arrangements.

Exactly two of these leave a guide with no tourists, so the answer is $64 - 2 = 62.$

Thus, the correct answer is D.

13.

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides $7$ times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan’s distance from his home to his distance from the stadium?

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$\dfrac{5}{6}$

$\dfrac{6}{7}$

Concepts:distance rate and time ratio and proportion

Difficulty rating: 1420

Small Hint:

Let $x$ be his distance from home and $y$ his distance from the stadium, with walking speed $w$

Big Hint:

Set the times equal: $\dfrac{y}{w} = \dfrac{x}{w} + \dfrac{x + y}{7w}$

Solution:

Let $x$ and $y$ be Yan’s distances from home and from the stadium, and let $w$ be his walking speed.

Walking to the stadium takes $\dfrac{y}{w}.$ Walking home and biking takes $\dfrac{x}{w} + \dfrac{x + y}{7w}.$

Setting these equal gives $7y = 8x + y,$ so $8x = 6y$ and $\dfrac{x}{y} = \dfrac{3}{4}.$

Thus, the correct answer is B.

14.

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle of radius $3.$ What is the area of the triangle?

$8.64$

$12$

$5\pi$

$17.28$

$18$

Concepts:inscribed angle right triangle triangle area

Difficulty rating: 1420

Small Hint:

A $3 : 4 : 5$ triangle is right-angled, so its longest side is a diameter

Big Hint:

With sides $3x, 4x, 5x,$ the hypotenuse $5x$ equals the diameter $6$

Solution:

Let the sides be $3x, 4x, 5x.$ The triangle is right-angled, so its hypotenuse is a diameter.

Thus $5x = 2 \cdot 3 = 6,$ giving $x = \tfrac65.$

The area is $\begin{aligned} \tfrac12 (3x)(4x) &= 6x^2 = 6 \cdot \tfrac{36}{25} \\ &= 8.64. \end{aligned}$

Thus, the correct answer is A.

15.

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2,$ as shown. What is the area of the square?

$32$

$22 + 12\sqrt{2}$

$16 + 16\sqrt{3}$

$48$

$36 + 16\sqrt{2}$

Concepts:tangent circles special right triangle square (geometry)

Difficulty rating: 1540

Small Hint:

Connect the center of the large circle to the centers of two adjacent small circles

Big Hint:

That triangle is isosceles right with legs $2 + 1 = 3,$ so its hypotenuse is $3\sqrt2$

Solution:

Consider the isosceles right triangle joining the center of the radius- $2$ circle to the centers of two adjacent small circles. Its legs have length $2 + 1 = 3,$ so its hypotenuse is $3\sqrt2.$

The side of the square exceeds this hypotenuse by $2$ (one radius on each end), so $s = 2 + 3\sqrt2.$

The area is $\begin{aligned} (2 + 3\sqrt2)^2 &= 4 + 12\sqrt2 + 18 \\ &= 22 + 12\sqrt2. \end{aligned}$

Thus, the correct answer is B.

16.

Integers $a,$ $b,$ $c,$ and $d,$ not necessarily distinct, are chosen independently and at random from $0$ to $2007,$ inclusive. What is the probability that $ad - bc$ is even?

$\dfrac{3}{8}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{5}{8}$

Concepts:parity basic probability independent events

Difficulty rating: 1540

Small Hint:

$ad - bc$ is even exactly when $ad$ and $bc$ have the same parity

Big Hint:

A product is odd only if both factors are odd, which happens with probability $\tfrac14$

Solution:

Half the integers from $0$ to $2007$ are odd, so each of $ad$ and $bc$ is odd with probability $\tfrac12 \cdot \tfrac12 = \tfrac14$ and even with probability $\tfrac34.$

The difference $ad - bc$ is even when both products have the same parity: $\tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58.$

Thus, the correct answer is E.

17.

Suppose that $m$ and $n$ are positive integers such that $75m = n^3.$ What is the minimum possible value of $m + n?$

$15$

$30$

$50$

$60$

$5700$

Concepts:perfect power prime factorization

Difficulty rating: 1480

Small Hint:

In a perfect cube, every prime’s exponent is a multiple of $3$

Big Hint:

Since $75 = 3 \cdot 5^2,$ find the smallest $m$ making all exponents multiples of $3$

Solution:

Since $n^3 = 75m = 3 \cdot 5^2 \cdot m,$ every prime factor must occur a multiple of three times.

The smallest such $m$ is $3^2 \cdot 5 = 45,$ giving $n^3 = 3^3 \cdot 5^3$ and $n = 15.$

Then $m + n = 45 + 15 = 60.$

Thus, the correct answer is D.

18.

Consider the $12$ -sided polygon $ABCDEFGHIJKL,$ as shown. Each of its sides has length $4,$ and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M.$ What is the area of quadrilateral $ABCM?$

$\dfrac{44}{3}$

$16$

$\dfrac{88}{5}$

$20$

$\dfrac{62}{3}$

Concepts:coordinate geometry shoelace formula

Difficulty rating: 1790

Small Hint:

Place the cross on coordinates using the side length $4$

Big Hint:

Find $M$ as the intersection of lines $AG$ and $CH,$ then apply the shoelace formula to $ABCM$

Solution:

Put the figure on coordinates with $A = (-2, 6),$ $B = (2, 6),$ $C = (2, 2),$ $G = (2, -6),$ and $H = (-2, -6).$

Line $AG$ is $y = -3x,$ and line $CH$ is $y = 2x - 2.$

Their intersection is $M = \left(\tfrac25, -\tfrac65\right).$

Applying the shoelace formula to $A, B, C, M$ gives $\begin{aligned} [ABCM] &=\dfrac12\left|{-8}-\dfrac{136}{5}\right|\\ &=\dfrac{88}{5}. \end{aligned}$

Thus, the correct answer is C.

19.

A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?

$2\sqrt{2} + 1$

$3\sqrt{2}$

$2\sqrt{2} + 2$

$3\sqrt{2} + 1$

$3\sqrt{2} + 2$

Concepts:area decomposition special right triangle rationalizing denominator

Difficulty rating: 1820

Small Hint:

The four unpainted corner regions are congruent isosceles right triangles totaling half the square

Big Hint:

Each triangle has leg $x = \tfrac{s}{2},$ and $x + w$ equals half the diagonal $\tfrac{\sqrt2}{2}s$

Solution:

Let $s$ be the side, $w$ the brush width, and $x$ the leg of one unpainted isosceles right triangle. Each triangle has area $\tfrac18 s^2,$ so $\tfrac12 x^2 = \tfrac18 s^2$ and $x = \tfrac{s}{2}.$

The leg plus the brush width is half the diagonal: $x + w = \tfrac{\sqrt2}{2} s.$ Thus $w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}.$

Therefore $\dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2.$

Thus, the correct answer is C.

20.

Suppose that the number $a$ satisfies the equation $4 = a + a^{-1}.$ What is the value of $a^4 + a^{-4}?$

$164$

$172$

$192$

$194$

$212$

Concepts:algebraic manipulation symmetry (algebra)

Difficulty rating: 1460

Small Hint:

Squaring $a + a^{-1}$ gives $a^2 + 2 + a^{-2}$

Big Hint:

Square $a^2 + a^{-2}$ and subtract the middle term $2$

Solution:

Squaring $a + a^{-1} = 4$ gives $a^2 + 2 + a^{-2} = 16,$ so $a^2 + a^{-2} = 14.$

Squaring again gives $a^4 + 2 + a^{-4} = 196,$ so $a^4 + a^{-4} = 194.$

Thus, the correct answer is D.

21.

A sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

$3$

$6$

$8$

$9$

$12$

Concepts:cube geometry sphere surface area

Difficulty rating: 1580

Small Hint:

The sphere’s diameter equals the outer cube’s side and the inner cube’s space diagonal

Big Hint:

For inner cube side $l,$ the space diagonal is $l\sqrt3 = 2$

Solution:

Each face of the outer cube has area $24 \div 6 = 4,$ so its side is $2,$ and the sphere has diameter $2.$

This diameter is the space diagonal of the inner cube, so $l\sqrt3 = 2,$ giving $l^2 = \tfrac43.$

The inner cube’s surface area is $6 l^2 = 6 \cdot \tfrac43 = 8.$

Thus, the correct answer is C.

22.

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with terms $247,$ $475,$ and $756$ and end with the term $824.$ Let $S$ be the sum of all the terms in the sequence. What is the largest prime number that always divides $S?$

$3$

$7$

$13$

$37$

$43$

Concepts:place value divisibility

Difficulty rating: 1920

Small Hint:

Because of the cyclic shifting, each digit spends equal time in the hundreds, tens, and units places

Big Hint:

So $S = 111k$ for some integer $k,$ and $111 = 3 \cdot 37$

Solution:

Each digit appears as a hundreds digit, a tens digit, and a units digit the same number of times across the sequence.

If $k$ is the sum of the units digits of all terms, then $S = 111k = 3 \cdot 37 \cdot k,$ so $S$ is always divisible by $37.$

The sequence $123,$ $231,$ $312$ gives $S = 666 = 2 \cdot 3^2 \cdot 37,$ which has no larger prime factor forced, so $37$ is the answer.

Thus, the correct answer is D.

23.

How many ordered pairs $(m, n)$ of positive integers, with $m \gt n,$ have the property that their squares differ by $96?$

$3$

$4$

$6$

$9$

$12$

Concepts:difference of squares Diophantine Equation parity

Difficulty rating: 1400

Small Hint:

Write $m^2 - n^2 = (m + n)(m - n) = 96$

Big Hint:

$m + n$ and $m - n$ have the same parity, so both must be even

Solution:

The factors $m + n$ and $m - n$ have the same parity. Since their product is $96,$ they cannot both be odd, so both must be even.

The even factor pairs are $(48, 2),$ $(24, 4),$ $(16, 6),$ and $(12, 8),$ giving $(m, n) = (25, 23),$ $(14, 10),$ $(11, 5),$ and $(10, 2).$

So there are $4$ ordered pairs.

Thus, the correct answer is B.

24.

Circles centered at $A$ and $B$ each have radius $2,$ as shown. Point $O$ is the midpoint of $\overline{AB},$ and $OA = 2\sqrt{2}.$ Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B,$ respectively, and $\overline{EF}$ is a common tangent. What is the area of the shaded region $ECODF?$

$\dfrac{8\sqrt{2}}{3}$

$8\sqrt{2} - 4 - \pi$

$4\sqrt{2}$

$4\sqrt{2} + \dfrac{\pi}{8}$

$8\sqrt{2} - 2 - \dfrac{\pi}{2}$

Concepts:tangent line sector area decomposition

Difficulty rating: 1960

Small Hint:

The region is rectangle $ABFE$ with two right triangles and two circular sectors removed

Big Hint:

Each triangle formed by tangency is isosceles right with legs $2;$ each sector has a $45^\circ$ angle and radius $2$

Solution:

Rectangle $ABFE$ has area $AE \cdot AB = 2 \cdot 4\sqrt2 = 8\sqrt2.$

Right triangles $ACO$ and $BDO$ each have hypotenuse $2\sqrt2$ and a leg of $2,$ so each is isosceles right with area $2.$

Angles $CAE$ and $DBF$ are each $45^\circ,$ so sectors $CAE$ and $DBF$ each have area $\tfrac18 \pi \cdot 2^2 = \tfrac{\pi}{2}.$

The shaded area is $\begin{gathered} 8\sqrt2 - 2 \cdot 2 - 2 \cdot \tfrac{\pi}{2} \\ = 8\sqrt2 - 4 - \pi. \end{gathered}$

Thus, the correct answer is B.

25.

For each positive integer $n,$ let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$

$1$

$2$

$3$

$4$

$5$

Concepts:digits modular arithmetic bounding to limit cases

Difficulty rating: 2200

Small Hint:

For $n \le 2007,$ $S(n) \le 28$ and $S(S(n)) \le 10,$ so $n \ge 1969$

Big Hint:

$n,$ $S(n),$ and $S(S(n))$ are all congruent modulo $9,$ and $2007$ is a multiple of $9$

Solution:

If $n \le 2007,$ then $S(n) \le 28$ and $S(S(n)) \le 10,$ so $n \ge 2007 - 28 - 10 = 1969.$

Since $n,$ $S(n),$ and $S(S(n))$ all leave the same remainder modulo $9$ and $2007$ is a multiple of $9,$ each must be a multiple of $3.$

Checking the multiples of $3$ between $1969$ and $2007,$ the condition holds for $1977,$ $1980,$ $1983,$ and $2001.$

So there are $4$ values of $n.$

Thus, the correct answer is D.