2003 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

$0$

$1$

$2$

$2003$

$4006$

Answer: D

Concepts:summation pairing and grouping

Difficulty rating: 890

Small Hint:

Pair each even number with the odd number just below it

Big Hint:

Each of the $2003$ pairs differs by $1$

Solution:

The $k$ th even counting number is $2k$ and the $k$ th odd counting number is $2k-1,$ which differ by $1.$

Summing this difference over all $2003$ pairs gives $2003 \cdot 1 = 2003.$

Thus, the correct answer is D.

2.

Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $\$4$ per pair and each T-shirt costs $\$5$ more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $\$2366,$ how many members are in the League?

$77$

$91$

$143$

$182$

$286$

Answer: B

Concepts:money linear equation

Difficulty rating: 1020

Small Hint:

A T-shirt costs $\$4+\$5=\$9$

Big Hint:

Each member buys two pairs of socks and two shirts

Solution:

A T-shirt costs $4+5=9$ dollars.

Each member needs two pairs of socks and two shirts, costing $2(4)+2(9)=26$ dollars.

The number of members is $2366 \div 26 = 91.$

Thus, the correct answer is B.

3.

A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?

$4.5$

$9$

$12$

$18$

$24$

Answer: D

Concepts:volume percentage

Difficulty rating: 1130

Small Hint:

The box has $8$ corners

Big Hint:

Each removed cube has volume $3^3$

Solution:

The original volume is $15 \cdot 10 \cdot 8 = 1200.$

Eight corner cubes are removed, each of volume $3^3 = 27,$ totaling $8 \cdot 27 = 216.$

The fraction removed is $\dfrac{216}{1200} = 0.18,$ which is $18\%.$

Thus, the correct answer is D.

4.

It takes Mary $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?

$3$

$3.125$

$3.5$

$4$

$4.5$

Answer: A

Concepts:distance rate and time unit conversion harmonic mean

Difficulty rating: 1130

Small Hint:

Average speed is total distance divided by total time

Big Hint:

The total time is $30+10=40$ minutes $=\dfrac23$ hour

Solution:

Mary walks a total of $2$ km in $30+10=40$ minutes, which is $\dfrac23$ hour.

Her average speed is $2 \div \dfrac23 = 3$ km/hr.

Thus, the correct answer is A.

5.

The sum of the two $5$ -digit numbers $\overline{AMC10}$ and $\overline{AMC12}$ is $123422.$ What is $A + M + C?$

$10$

$11$

$12$

$13$

$14$

Answer: E

Concepts:place value cryptarithm

Difficulty rating: 1200

Small Hint:

$\overline{AMC10}=100\cdot\overline{AMC}+10$

Big Hint:

The two numbers share the same leading three digits $\overline{AMC}$

Solution:

Write $\overline{AMC10}=100\cdot\overline{AMC}+10$ and $\overline{AMC12}=100\cdot\overline{AMC}+12.$

Their sum is $200\cdot\overline{AMC}+22=123422,$ so $\overline{AMC}=617.$

Then $A+M+C=6+1+7=14.$

Thus, the correct answer is E.

6.

Define $x\heartsuit y$ to be $|x - y|$ for all real numbers $x$ and $y.$ Which of the following statements is not true?

$x\heartsuit y = y\heartsuit x$ for all $x$ and $y$

$2(x\heartsuit y) = (2x)\heartsuit(2y)$ for all $x$ and $y$

$x\heartsuit 0 = x$ for all $x$

$x\heartsuit x = 0$ for all $x$

$x\heartsuit y \gt 0$ if $x \neq y$

Answer: C

Concepts:absolute value counterexample

Difficulty rating: 1200

Small Hint:

$x\heartsuit 0=|x|$

Big Hint:

Test a negative value of $x$

Solution:

Since $x\heartsuit 0=|x-0|=|x|,$ statement (C) claims $|x|=x$ for all $x,$ which fails when $x\lt0.$ For example, $(-1)\heartsuit 0 = 1 \neq -1.$

Every other statement follows directly from properties of the absolute value.

Thus, the correct answer is C.

7.

How many non-congruent triangles with perimeter $7$ have integer side lengths?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:triangle inequality partitions and compositions

Difficulty rating: 1270

Small Hint:

The longest side must be less than half the perimeter

Big Hint:

The longest side is at most $3$

Solution:

Let the sides be $a\le b\le c$ with $a+b+c=7.$ The triangle inequality requires $c\lt a+b,$ so $c\lt3.5,$ forcing $c=3.$

Then $a+b=4$ with $a\le b\le3,$ giving the triangles $1\text{-}3\text{-}3$ and $2\text{-}2\text{-}3.$

Thus, the correct answer is B.

8.

What is the probability that a randomly drawn positive factor of $60$ is less than $7?$

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: E

Concepts:factor counting basic probability

Difficulty rating: 1270

Small Hint:

$60=2^2\cdot3\cdot5$ has $(2{+}1)(1{+}1)(1{+}1)=12$ factors

Big Hint:

Count the factors from $1$ through $6$

Solution:

The number $60$ has $12$ positive factors: $1,2,3,4,5,6,$ $10,12,15,20,30,60.$

Six of them are less than $7,$ namely $1,2,3,4,5,6,$ so the probability is $\dfrac{6}{12}=\dfrac12.$

Thus, the correct answer is E.

9.

A set $S$ of points in the $xy$ -plane is symmetric about the origin, both coordinate axes, and the line $y = x.$ If $(2, 3)$ is in $S,$ what is the smallest number of points in $S?$

$1$

$2$

$4$

$8$

$16$

Answer: D

Concepts:transformation symmetry

Difficulty rating: 1350

Small Hint:

Reflecting $(2,3)$ over $y=x$ gives $(3,2)$

Big Hint:

The symmetries produce every sign choice and coordinate swap

Solution:

Reflecting across $y=x$ gives $(3,2),$ and reflecting across the axes gives all points $(\pm2,\pm3)$ and $(\pm3,\pm2).$

There are $8$ such points, and this set is already symmetric about the origin, both axes, and $y=x.$

Thus, the correct answer is D.

10.

Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of $3 : 2 : 1,$ respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the first to arrive. If each takes what he believes to be his correct share of candy, what fraction of the candy goes unclaimed?

$\dfrac{1}{18}$

$\dfrac{1}{6}$

$\dfrac{2}{9}$

$\dfrac{5}{18}$

$\dfrac{5}{12}$

Answer: D

Concepts:fraction ratio and proportion

Difficulty rating: 1440

Small Hint:

The three shares are $\dfrac12,\dfrac13,\dfrac16$ of the whole

Big Hint:

Each person leaves a fixed fraction of whatever candy is present, so the order does not matter

Solution:

The shares are $\dfrac12,\dfrac13,\dfrac16$ of the pile.

Each person assumes he is first, so Al leaves $\dfrac12,$ Bert leaves $\dfrac23,$ and Carl leaves $\dfrac56$ of the candy present when he arrives.

The unclaimed fraction is $\dfrac12\cdot\dfrac23\cdot\dfrac56=\dfrac{5}{18},$ regardless of the order.

Thus, the correct answer is D.

11.

A square and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the square and $B$ be the area of the circle circumscribed about the triangle. Find $\frac{A}{B}.$

$\dfrac{9}{16}$

$\dfrac{3}{4}$

$\dfrac{27}{32}$

$\dfrac{3\sqrt{6}}{8}$

$1$

Answer: C

Concepts:circumcircle, circumcenter, and circumradius area ratio

Difficulty rating: 1500

Small Hint:

Choose a convenient common perimeter, such as $12$

Big Hint:

A square’s circumradius is half its diagonal; an equilateral triangle’s is $\dfrac{\text{side}}{\sqrt3}$

Solution:

Let the common perimeter be $12,$ so the square has side $3$ and the triangle has side $4.$

The square’s circumradius is $\dfrac{3\sqrt2}{2},$ so $A=\pi\left(\dfrac{3\sqrt2}{2}\right)^2=\dfrac{9\pi}{2}.$

The triangle’s circumradius is $\dfrac{4}{\sqrt3},$ so $B=\pi\left(\dfrac{4}{\sqrt3}\right)^2=\dfrac{16\pi}{3}.$

Then $\dfrac{A}{B}=\dfrac{\frac{9}{2}}{\frac{16}{3}}=\dfrac{27}{32}.$

Thus, the correct answer is C.

12.

Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6.$ She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

$8$

$9$

$10$

$11$

$12$

Answer: E

Concepts:divisibility logical deduction

Difficulty rating: 1500

Small Hint:

Among the blue numbers $3\text{-}6,$ the red $4$ divides only $4$ and the red $5$ divides only $5$

Big Hint:

Work inward from the forced ends $R4,B4$ and $B5,R5$

Solution:

Since $4$ divides only $4$ and $5$ divides only $5$ among $3,4,5,6,$ the two ends must be $R4,B4,\dots,B5,R5.$

Because $2$ divides only $4$ and $6,$ the next card is $R2,B6,$ and since $3$ divides only $3$ and $6,$ the full stack is $R4,$ $B4,$ $R2,$ $B6,$ $R3,$ $B3,$ $R1,$ $B5,$ $R5.$

The middle three cards are $6,3,3,$ which sum to $12.$

Thus, the correct answer is E.

13.

The polygon enclosed by the solid lines in the figure consists of $4$ congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?

$2$

$3$

$4$

$5$

$6$

Answer: E

Concepts:net (3D geometry)cube geometry

Difficulty rating: 1530

Small Hint:

A cube with one face missing needs $5$ squares

Big Hint:

Fold the four-square piece partway around a cube, then check which attached square lands on a still-open face

Solution:

A cube missing one face has $5$ faces, so the fifth square must fold onto a face not already occupied by the four-square piece. Tracing the face reached by each attachment, positions $1,$ $2,$ and $3$ fold onto a face already occupied and overlap it. Each of positions $4,$ $5,$ $6,$ $7,$ $8,$ and $9$ folds onto an unoccupied face. Therefore $6$ of the nine positions work.

Thus, the correct answer is E.

14.

Points $K,$ $L,$ $M,$ and $N$ lie in the plane of the square $ABCD$ so that $AKB,$ $BLC,$ $CMD,$ and $DNA$ are equilateral triangles. If $ABCD$ has an area of $16,$ find the area of $KLMN.$

$32$

$16 + 16\sqrt{3}$

$48$

$32 + 16\sqrt{3}$

$64$

Answer: D

Concepts:equilateral triangle square (geometry)symmetry

Difficulty rating: 1570

Small Hint:

$KLMN$ is a square by the rotational symmetry of the figure

Big Hint:

Its diagonal $KM$ equals the square’s side plus twice the triangle height $2\sqrt3$

Solution:

The square $ABCD$ has side $4.$ By the $90^\circ$ rotational symmetry, $KLMN$ is also a square.

Each equilateral triangle on a side of length $4$ has height $2\sqrt3,$ so the diagonal $KM=4+2(2\sqrt3)=4+4\sqrt3.$

A square with diagonal $d$ has area $\dfrac12 d^2,$ so $[KLMN]=\dfrac12(4+4\sqrt3)^2$ $=32+16\sqrt3.$

Thus, the correct answer is D.

15.

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2,$ as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

$\dfrac{1}{6}\pi - \dfrac{\sqrt{3}}{4}$

$\dfrac{\sqrt{3}}{4} - \dfrac{1}{12}\pi$

$\dfrac{\sqrt{3}}{4} - \dfrac{1}{24}\pi$

$\dfrac{\sqrt{3}}{4} + \dfrac{1}{24}\pi$

$\dfrac{\sqrt{3}}{4} + \dfrac{1}{12}\pi$

Answer: C

Concepts:sector area decomposition equilateral triangle

Difficulty rating: 1630

Small Hint:

The small diameter is a chord of length $1$ in the large radius- $1$ circle, spanning a $60^\circ$ arc

Big Hint:

Lune $=$ (equilateral triangle $+$ small semicircle) $-$ the $60^\circ$ sector of the large circle

Solution:

The small semicircle’s diameter is a chord of length $1$ in the large circle of radius $1,$ so it subtends a $60^\circ$ angle at the center.

The region bounded by that chord and the small arc is an equilateral triangle of area $\dfrac{\sqrt3}{4}$ topped by the small semicircle of area $\dfrac12\pi\left(\dfrac12\right)^2=\dfrac{\pi}{8}.$

Subtracting the $60^\circ$ sector of the large circle, $\dfrac16\pi(1)^2=\dfrac{\pi}{6},$ gives the lune area $\dfrac{\sqrt3}{4}+\dfrac{\pi}{8}-\dfrac{\pi}{6}=\dfrac{\sqrt3}{4}-\dfrac{\pi}{24}.$

Thus, the correct answer is C.

16.

A point $P$ is chosen at random in the interior of equilateral triangle $ABC.$ What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP?$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Answer: C

Concepts:geometric probability triangle area symmetry

Difficulty rating: 1660

Small Hint:

The three triangles share equal bases, so their areas depend only on $P$ ’s distances to the sides

Big Hint:

By the threefold symmetry each side yields the largest triangle equally often

Solution:

The triangles $\triangle ABP,$ $\triangle ACP,$ and $\triangle BCP$ have equal bases (the sides of the equilateral triangle), so their areas are proportional to the distances from $P$ to those sides.

By the threefold symmetry of the equilateral triangle, $\triangle ABP$ is the largest with the same probability as each of the other two, so that probability is $\dfrac13.$

Thus, the correct answer is C.

17.

Square $ABCD$ has sides of length $4,$ and $M$ is the midpoint of $\overline{CD}.$ A circle with radius $2$ and center $M$ intersects a circle with radius $4$ and center $A$ at points $P$ and $D.$ What is the distance from $P$ to $\overline{AD}?$

$3$

$\dfrac{16}{5}$

$\dfrac{13}{4}$

$2\sqrt{3}$

$\dfrac{7}{2}$

Answer: B

Concepts:coordinate geometry circle

Difficulty rating: 1730

Small Hint:

Put $D$ at the origin with $A$ on the $y$ -axis and $C$ on the $x$ -axis

Big Hint:

The distance from $P$ to $\overline{AD}$ is the $x$ -coordinate of $P$

Solution:

Place $D=(0,0),$ $C=(4,0),$ and $A=(0,4).$ The circle centered at $M=(2,0)$ is $(x-2)^2+y^2=4,$ and the circle centered at $A$ is $x^2+(y-4)^2=16.$

Solving these equations gives the intersection $P=\left(\dfrac{16}{5},\dfrac85\right).$

Since $\overline{AD}$ lies on the $y$ -axis, the distance from $P$ to $\overline{AD}$ is its $x$ -coordinate, $\dfrac{16}{5}.$

Thus, the correct answer is B.

18.

Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by $100.$ For how many values of $n$ is $q + r$ divisible by $11?$

$8180$

$8181$

$8182$

$9000$

$9090$

Answer: B

Concepts:modular arithmetic multiple

Difficulty rating: 1800

Small Hint:

$n=100q+r=(q+r)+99q$

Big Hint:

Since $99$ is a multiple of $11,$ $q+r\equiv n\pmod{11}$

Solution:

Since $n=100q+r=(q+r)+99q$ and $99$ is divisible by $11,$ we have $q+r\equiv n\pmod{11}.$

So $(q+r)$ is a multiple of $11$ exactly when $n$ is a multiple of $11.$

Among the $5$ -digit numbers, the count of multiples of $11$ is $\left\lfloor\dfrac{99999}{11}\right\rfloor$ $-\left\lfloor\dfrac{9999}{11}\right\rfloor$ $=9090-909=8181.$

Thus, the correct answer is B.

19.

A parabola with equation $y = ax^2 + bx + c$ is reflected about the $x$ -axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of $y = f(x)$ and $y = g(x),$ respectively. Which of the following describes the graph of $y = (f + g)(x)?$

a parabola tangent to the $x$ -axis

a parabola not tangent to the $x$ -axis

a horizontal line

a non-horizontal line

the graph of a cubic function

Answer: D

Concepts:parabola transformation function

Difficulty rating: 1840

Small Hint:

Write the parabola as $a(x-h)^2+k;$ its reflection is $-a(x-h)^2-k$

Big Hint:

After the opposite $\pm5$ shifts, add $f$ and $g$ and watch the squared terms cancel

Solution:

Write the parabola in vertex form $y=a(x-h)^2+k.$ Its reflection about the $x$ -axis is $y=-a(x-h)^2-k.$

Shifting in opposite directions gives $f(x)=a(x-h+5)^2+k$ and $g(x)=-a(x-h-5)^2-k.$

Adding, the squared terms cancel and $(f+g)(x)=20a(x-h),$ which is a non-horizontal line since $a\neq0.$

Thus, the correct answer is D.

20.

How many $15$ -letter arrangements of $5$ A’s, $5$ B’s, and $5$ C’s have no A’s in the first $5$ letters, no B’s in the next $5$ letters, and no C’s in the last $5$ letters?

$\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3$

$3^5 \cdot 2^5$

$2^{15}$

$\dfrac{15!}{(5!)^3}$

$3^{15}$

Answer: A

Concepts:multiset permutations combinations casework

Difficulty rating: 1910

Small Hint:

In the first block (no A’s), suppose there are $k$ B’s and $5-k$ C’s

Big Hint:

Fixing $k$ determines the letter counts in all three blocks, each chosen with $\binom{5}{k}$ placements

Solution:

Suppose the first block holds $k$ B’s and $5-k$ C’s. The remaining $k$ C’s must go in the second block (since the third has no C’s), forcing $5-k$ A’s there.

Then the third block contains the remaining $k$ A’s and $5-k$ B’s.

For each $k,$ the $k$ B’s in the first block, $k$ C’s in the second, and $k$ A’s in the third can be placed in $\binom{5}{k}^3$ ways, so the total is $\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3.$

Thus, the correct answer is A.

21.

The graph of the polynomial $\begin{aligned} &P(x) = x^5 + ax^4 + bx^3 \\ &\quad {}+ cx^2 + dx + e \end{aligned}$

has five distinct $x$ -intercepts, one of which is at $(0, 0).$ Which of the following coefficients cannot be zero?

$a$

$b$

$c$

$d$

$e$

Answer: D

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1990

Small Hint:

$P(0)=0$ forces $e=0$

Big Hint:

Factor out $x;$ the constant term of the remaining quartic is the product of the four nonzero roots

Solution:

Since $(0,0)$ is an intercept, $P(0)=e=0,$ so $P(x)$ $=x\left(x^4+ax^3+bx^2+cx+d\right).$

The four remaining intercepts are nonzero and distinct, and $d$ equals their product, which is therefore nonzero.

Any of $a,b,c$ can be zero for suitable choices of those roots, but $d\neq0.$

Thus, the correct answer is D.

22.

Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0, 0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5, 7)$ and each of its steps is either left or down, both equally likely. Which of the following is closest to the probability that the objects meet?

$0.10$

$0.15$

$0.20$

$0.25$

$0.30$

Answer: C

Concepts:lattice paths random walk bijection

Difficulty rating: 2010

Small Hint:

They can meet only after each has taken $6$ steps, along the line $x+y=6$

Big Hint:

The number of meeting path-pairs equals $\binom{12}{5}$ out of $2^{12}$ total pairs

Solution:

The objects are $12$ steps apart, so they can only meet after each takes $6$ steps, on the anti-diagonal $x+y=6.$

Pairing $A$ ’s six-step path with $B$ ’s reversed six-step path matches meeting pairs one-to-one with the $\binom{12}{5}$ monotone walks from $(0,0)$ to $(5,7).$

The probability is $\dfrac{\binom{12}{5}}{2^{12}}=\dfrac{792}{4096}\approx0.19,$ which is closest to $0.20.$

Thus, the correct answer is C.

23.

How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdots 9!\,?$

$504$

$672$

$864$

$936$

$1008$

Answer: B

Concepts:prime factorization perfect square factor counting

Difficulty rating: 2110

Small Hint:

Combine the factorials into $2^{30}\cdot3^{13}\cdot5^{5}\cdot7^{3}$

Big Hint:

A square divisor uses an even exponent of each prime

Solution:

The product is $1! \cdot 2! \cdots 9! = 2^{30}\cdot3^{13}\cdot5^{5}\cdot7^{3}.$

A perfect-square divisor has the form $2^{2a}3^{2b}5^{2c}7^{2d}$ with $0\le a\le15,$ $0\le b\le6,$ $0\le c\le2,$ and $0\le d\le1.$

The number of choices is $16\cdot7\cdot3\cdot2=672.$

Thus, the correct answer is B.

24.

If $a \ge b \gt 1,$ what is the largest possible value of $\log_a(\frac{a}{b}) + \log_b(\frac{b}{a})?$

$-2$

$0$

$2$

$3$

$4$

Answer: B

Concepts:logarithm AM-GM Inequality

Difficulty rating: 2170

Small Hint:

Simplify to $2-\log_a b-\log_b a$

Big Hint:

With $c=\log_a b\gt0,$ use $c+\dfrac1c\ge2$

Solution:

Expand: $\begin{aligned} &\log_a\dfrac ab+\log_b\dfrac ba \\ &\quad {}=(1-\log_a b)+(1-\log_b a) \\ &\quad {}=2-\left(\log_a b+\log_b a\right). \end{aligned}$

Let $c=\log_a b\gt0.$ Since $c+\dfrac1c\ge2$ by AM-GM, the expression is at most $0.$

Equality holds when $c=1,$ that is, when $a=b,$ so the largest value is $0.$

Thus, the correct answer is B.

25.

Let $f(x) = \sqrt{ax^2 + bx}.$ For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set?

$0$

$1$

$2$

$3$

infinitely many

Answer: C

Concepts:function radical casework

Difficulty rating: 2380

Small Hint:

If $a=0,$ both the domain and range equal $[0,\infty)$ for every $b\gt0$

Big Hint:

For $a\lt0$ the domain is $[0,-\frac{b}{a}]$ and the maximum of $f$ is $\dfrac{b}{2\sqrt{-a}}$

Solution:

If $a=0,$ then $f(x)=\sqrt{bx}$ has domain and range both $[0,\infty),$ so $a=0$ works.

If $a\gt0,$ the domain is $({-}\infty,-b/a]\cup[0,\infty),$ while the range is $[0,\infty),$ so no such $b$ exists.

If $a\lt0,$ the domain is $[0,-\frac{b}{a}]$ and the range is $\left[0,\dfrac{b}{2\sqrt{-a}}\right].$ Equating the right endpoints gives $-\dfrac ba=\dfrac{b}{2\sqrt{-a}},$ so $2\sqrt{-a}=-a,$ giving $a=-4.$

Thus there are $2$ values of $a,$ and the correct answer is C.