2006 AMC 10B — Problems & Online Timed Exam

Timed

1:15:00

1.

What is $(-1)^1 + (-1)^2 + \cdots + (-1)^{2006}?$

$-2006$

$-1$

$0$

$1$

$2006$

Answer: C

Concepts:exponent pairing and grouping

Difficulty rating: 720

Small Hint:

Group the terms in consecutive pairs

Big Hint:

In each pair, compare the signs of the odd-power and even-power terms

Solution:

There are $2006$ terms. Pairing consecutive terms gives $(-1+1)+(-1+1)+\cdots.$ Since $2006$ is even, every term pairs off and the sum is $0.$

Thus, the correct answer is C.

2.

For real numbers $x$ and $y,$ define $x \spadesuit y = (x+y)(x-y).$ What is $3 \spadesuit (4 \spadesuit 5)?$

$-72$

$-27$

$-24$

$24$

$72$

Answer: A

Concepts:custom operation difference of squares

Difficulty rating: 870

Small Hint:

Evaluate the inner operation $4 \spadesuit 5$ first

Big Hint:

$x \spadesuit y = x^2 - y^2$

Solution:

Since $x \spadesuit y = (x+y)(x-y) = x^2 - y^2,$ we have $4 \spadesuit 5 = 16 - 25 = -9.$

Then $3 \spadesuit (-9) = 9 - 81 = -72.$

Thus, the correct answer is A.

3.

A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?

$10$

$14$

$17$

$20$

$24$

Answer: A

Concepts:system of equations linear equation

Difficulty rating: 830

Small Hint:

Let the two scores be $c$ and $p$ with $c+p=34$

Big Hint:

The winning margin gives $c-p=14$

Solution:

Let $c$ and $p$ be the Cougars’ and Panthers’ scores. Then $c+p=34$ and $c-p=14.$ Subtracting gives $2p=20,$ so $p=10.$

Thus, the correct answer is A.

4.

Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?

$2$

$3$

$6$

$8$

$9$

Answer: D

Concepts:circle area area ratio

Difficulty rating: 940

Small Hint:

The radii are $\tfrac12$ and $\tfrac32$

Big Hint:

Blue is the ring $\pi(\tfrac32)^2-\pi(\tfrac12)^2$ ; red is $\pi(\tfrac12)^2$

Solution:

The red circle has area $\pi(\tfrac12)^2 = \tfrac{\pi}{4},$ and the large circle has area $\pi(\tfrac32)^2 = \tfrac{9\pi}{4}.$ The blue ring is $\tfrac{9\pi}{4}-\tfrac{\pi}{4}=2\pi.$

The ratio is $2\pi \div \tfrac{\pi}{4} = 8.$

Thus, the correct answer is D.

5.

A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any interior point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

$16$

$25$

$36$

$49$

$64$

Answer: B

Concepts:square (geometry)rectangle optimization

Difficulty rating: 1060

Small Hint:

Try stacking the rectangles so their sides of length $3$ line up

Big Hint:

The square’s side must be at least the sum of the smaller dimensions, $2+3$

Solution:

Place the rectangles side by side with their $3$ -length sides vertical. Their widths add to $2+3=5,$ and the heights $3$ and $4$ both fit within $5.$

Because the rectangles are axis-aligned and their interiors do not overlap, their horizontal projections or their vertical projections must be disjoint. In either direction, the first rectangle spans at least $2$ and the second spans at least $3,$ so the square’s side is at least $2+3=5.$ The smallest area is therefore $5^2=25.$

Thus, the correct answer is B.

6.

A region is bounded by semicircular arcs constructed on the sides of a square whose sides measure $\tfrac{2}{\pi},$ as shown. What is the perimeter of this region?

$\dfrac{4}{\pi}$

$2$

$\dfrac{8}{\pi}$

$4$

$\dfrac{16}{\pi}$

Answer: D

Concepts:arc perimeter circumference

Difficulty rating: 1060

Small Hint:

Each side of the square is the diameter of one semicircular arc

Big Hint:

A semicircle on diameter $d$ has arc length $\tfrac12\pi d$

Solution:

Each side has length $\tfrac{2}{\pi},$ the diameter of a semicircular arc, so each arc has length $\tfrac12\pi\cdot\tfrac{2}{\pi}=1.$

The boundary consists of four such arcs, so the perimeter is $4\cdot1=4.$

Thus, the correct answer is D.

7.

Which of the following is equivalent to

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

when $x \lt 0?$

$-x$

$x$

$1$

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

$x\sqrt{-1}$

Answer: A

Concepts:radical absolute value algebraic manipulation

Difficulty rating: 1240

Small Hint:

Combine the denominator $1-\dfrac{x-1}{x}$ into a single fraction

Big Hint:

$\sqrt{x^2}=|x|,$ and $|x|=-x$ when $x \lt 0$

Solution:

The denominator simplifies: $1-\dfrac{x-1}{x}=\dfrac{x-(x-1)}{x}=\dfrac{1}{x}.$

So the expression is $\sqrt{\dfrac{x}{\frac{1}{x}}}=\sqrt{x^2}=|x|.$ Since $x \lt 0,$ this equals $-x.$

Thus, the correct answer is A.

8.

A square of area $40$ is inscribed in a semicircle as shown. What is the area of the semicircle?

$20\pi$

$25\pi$

$30\pi$

$40\pi$

$50\pi$

Answer: B

Concepts:circle area Pythagorean Theorem square (geometry)

Difficulty rating: 1260

Small Hint:

Let the square have side $s$ with $s^2=40,$ base centered on the diameter

Big Hint:

The radius satisfies $r^2=\left(\tfrac{s}{2}\right)^2+s^2$

Solution:

Let the square have side $s,$ so $s^2=40.$ Its base lies centered on the diameter, and a top corner at $\left(\tfrac{s}{2},s\right)$ lies on the circle.

Then $r^2=\left(\tfrac{s}{2}\right)^2+s^2=\tfrac{40}{4}+40=50.$ The semicircle area is $\tfrac12\pi r^2=\tfrac12\pi(50)=25\pi.$

Thus, the correct answer is B.

9.

Francesca uses $100$ grams of lemon juice, $100$ grams of sugar, and $400$ grams of water to make lemonade. There are $25$ calories in $100$ grams of lemon juice and $386$ calories in $100$ grams of sugar. Water contains no calories. How many calories are in $200$ grams of her lemonade?

$129$

$137$

$174$

$223$

$411$

Answer: B

Concepts:ratio and proportion rate

Difficulty rating: 1000

Small Hint:

The whole batch weighs $600$ grams and holds $25+386$ calories

Big Hint:

Scale the total calories by $\tfrac{200}{600}$

Solution:

The lemonade totals $100+100+400=600$ grams containing $25+386=411$ calories.

In $200$ grams there are $411\cdot\tfrac{200}{600}=\tfrac{411}{3}=137$ calories.

Thus, the correct answer is B.

10.

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15.$ What is the greatest possible perimeter of the triangle?

$43$

$44$

$45$

$46$

$47$

Answer: A

Concepts:triangle inequality perimeter optimization

Difficulty rating: 1190

Small Hint:

Let the sides be $x,$ $3x,$ and $15$

Big Hint:

The binding triangle inequality is $x+15 \gt 3x$

Solution:

Let the sides be $x,$ $3x,$ and $15.$ The triangle inequality $x+15 \gt 3x$ gives $x \lt 7.5.$

The largest integer is $x=7,$ giving sides $7,$ $21,$ $15$ and perimeter $7+21+15=43.$

Thus, the correct answer is A.

11.

What is the tens digit in the sum $7! + 8! + 9! + \cdots + 2006!\,?$

$1$

$3$

$4$

$6$

$9$

Answer: C

Concepts:factorial modular arithmetic digits

Difficulty rating: 1280

Small Hint:

For $n\ge 10,$ $n!$ ends in at least two zeros

Big Hint:

Only $7!+8!+9!$ affects the last two digits

Solution:

For $n\ge 10,$ $n!$ is divisible by $100,$ so it does not affect the last two digits.

The tens digit comes from $7!+8!+9!$ $=5040+40320+362880$ $=408240,$ whose tens digit is $4.$

Thus, the correct answer is C.

12.

The lines $x=\tfrac14 y+a$ and $y=\tfrac14 x+b$ intersect at the point $(1,2).$ What is $a+b?$

$0$

$\dfrac{3}{4}$

$1$

$2$

$\dfrac{9}{4}$

Answer: E

Concepts:substitution system of equations

Difficulty rating: 1140

Small Hint:

Substitute $(1,2)$ into both equations

Big Hint:

Add the two resulting equations to get $a+b$ directly

Solution:

Substituting $(1,2)$ : from $1=\tfrac14(2)+a$ we get $a=\tfrac12,$ and from $2=\tfrac14(1)+b$ we get $b=\tfrac74.$

Then $a+b=\tfrac12+\tfrac74=\tfrac94.$

Thus, the correct answer is E.

13.

Joe and JoAnn each bought $12$ ounces of coffee in a $16$ -ounce cup. Joe drank $2$ ounces of his coffee and then added $2$ ounces of cream. JoAnn added $2$ ounces of cream, stirred the coffee well, and then drank $2$ ounces. What is the resulting ratio of the amount of cream in Joe’s coffee to that in JoAnn’s coffee?

$\dfrac{6}{7}$

$\dfrac{13}{14}$

$1$

$\dfrac{14}{13}$

$\dfrac{7}{6}$

Answer: E

Concepts:mixture ratio and proportion

Difficulty rating: 1340

Small Hint:

Joe simply has $2$ ounces of cream

Big Hint:

JoAnn drinks $2$ of $14$ ounces of a well-mixed drink holding $2$ ounces of cream

Solution:

Joe adds $2$ ounces of cream and drinks nothing afterward, so he has $2$ ounces of cream.

JoAnn has $12$ ounces of coffee plus $2$ ounces of cream, making $14$ ounces of uniform mixture. After drinking $2$ ounces she keeps $\tfrac{12}{14}=\tfrac67$ of her cream, which is $\tfrac67\cdot2=\tfrac{12}{7}$ ounces.

The ratio is $2\div\tfrac{12}{7}=\tfrac{7}{6}.$

Thus, the correct answer is E.

14.

Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0.$ Suppose that $a+\tfrac1b$ and $b+\tfrac1a$ are the roots of the equation $x^2-px+q=0.$ What is $q?$

$\dfrac{5}{2}$

$\dfrac{7}{2}$

$4$

$\dfrac{9}{2}$

$8$

Answer: D

Concepts:Vieta’s Formulas symmetry (algebra)

Difficulty rating: 1480

Small Hint:

By Vieta’s formulas, $ab=2$

Big Hint:

$q$ is the product $\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)$

Solution:

Since $a$ and $b$ are roots of $x^2-mx+2,$ we have $ab=2.$

The value $q$ is the product of the new roots: $\begin{aligned} q&=\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)\\ &=ab+1+1+\tfrac{1}{ab}\\ &=2+2+\tfrac12=\tfrac92. \end{aligned}$

Thus, the correct answer is D.

15.

Rhombus $ABCD$ is similar to rhombus $BFDE.$ The area of rhombus $ABCD$ is $24,$ and $\angle BAD=60^\circ.$ What is the area of rhombus $BFDE?$

$6$

$4\sqrt{3}$

$8$

$9$

$6\sqrt{3}$

Answer: C

Concepts:rhombus equilateral triangle congruence (geometry)

Difficulty rating: 1460

Small Hint:

Since $AB=AD$ and $\angle BAD=60^\circ,$ triangle $ABD$ is equilateral

Big Hint:

$E$ and $F$ split $ABCD$ into six congruent triangles

Solution:

Because $AB=AD$ and $\angle BAD=60^\circ,$ triangle $ABD$ is equilateral, and so is triangle $CBD.$

Points $E$ and $F$ split the rhombus into six congruent triangles, each of area $\tfrac{24}{6}=4.$

Rhombus $BFDE$ is the union of triangles $BED$ and $BFD,$ so its area is $2\cdot4=8.$

Thus, the correct answer is C.

16.

Leap Day, February $29,$ $2004,$ occurred on a Sunday. On what day of the week will Leap Day, February $29,$ $2020,$ occur?

Tuesday

Wednesday

Thursday

Friday

Saturday

Answer: E

Concepts:date and time modular arithmetic

Difficulty rating: 1340

Small Hint:

Count the days from one Leap Day to the next: $3\cdot365+366$

Big Hint:

That total is $1461\equiv 5\pmod 7,$ so each $4$ -year cycle advances the weekday by $5$

Solution:

From one Leap Day to the next is $3\cdot365+366=1461$ days, and $1461\equiv 5\pmod 7.$

Over the four cycles from $2004$ to $2020,$ the weekday advances $4\cdot5=20\equiv 6\pmod 7,$ that is, $6$ days forward, which is one day back from Sunday.

So Leap Day $2020$ falls on a Saturday.

Thus, the correct answer is E.

17.

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob’s bag. Bob then randomly selects one ball from his bag and puts it into Alice’s bag. What is the probability that after this process the contents of the two bags are the same?

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: D

Concepts:basic probability symmetry

Difficulty rating: 1460

Small Hint:

After Alice’s move, Bob’s bag has $6$ balls with one color doubled

Big Hint:

The bags match only if Bob returns a ball of that doubled color

Solution:

Alice moves one ball to Bob, so Bob’s bag holds $6$ balls with exactly one color appearing twice.

The two bags end up identical exactly when Bob returns one of that duplicated pair. Two of the six balls qualify, so the probability is $\tfrac26=\tfrac13.$

Thus, the correct answer is D.

18.

Let $a_1,$ $a_2,$ $\ldots$ be a sequence for which $a_1=2,$ $a_2=3,$ and $a_n=\dfrac{a_{n-1}}{a_{n-2}}$ for each positive integer $n\ge 3.$ What is $a_{2006}?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{2}$

$2$

$3$

Answer: E

Concepts:recursion pattern recognition

Difficulty rating: 1280

Small Hint:

Compute the first several terms to detect a repeating cycle

Big Hint:

The sequence repeats every $6$ terms; reduce $2006$ modulo $6$

Solution:

The terms are $2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23,$ then $2,\,3,\ldots,$ a cycle of length $6.$

Since $2006=6\cdot334+2,$ we have $a_{2006}=a_2=3.$

Thus, the correct answer is E.

19.

A circle of radius $2$ is centered at $O.$ Square $OABC$ has side length $1.$ Sides $\overline{AB}$ and $\overline{CB}$ are extended past $B$ to meet the circle at $D$ and $E,$ respectively. What is the area of the shaded region in the figure, which is bounded by $\overline{BD},$ $\overline{BE},$ and the minor arc connecting $D$ and $E?$

$\dfrac{\pi}{3}+1-\sqrt{3}$

$\dfrac{\pi}{2}(2-\sqrt{3})$

$\pi(2-\sqrt{3})$

$\dfrac{\pi}{6}+\dfrac{\sqrt{3}-1}{2}$

$\dfrac{\pi}{3}-1+\sqrt{3}$

Answer: A

Concepts:sector triangle area angle chasing

Difficulty rating: 1820

Small Hint:

Use $OC=1$ and $OE=2$ to find $\angle EOA=30^\circ$

Big Hint:

Shaded area equals sector $DOE$ minus the two right triangles $OBD$ and $OBE$

Solution:

Since $OA=1$ and $OD=2$ with $D$ on the line $x=1,$ we get $\angle AOD=60^\circ,$ and likewise $\angle COE=60^\circ,$ so $\angle DOE=30^\circ.$

The sector $DOE$ has area $\tfrac{30}{360}\pi(2^2)=\tfrac{\pi}{3}.$

The region is this sector minus triangles $OBD$ and $OBE.$ With $BD=BE=\sqrt3-1,$ each triangle has area $\tfrac12(\sqrt3-1)(1),$ totaling $\sqrt3-1.$

So the shaded area is $\tfrac{\pi}{3}-(\sqrt3-1)=\tfrac{\pi}{3}+1-\sqrt3.$

Thus, the correct answer is A.

20.

In rectangle $ABCD,$ we have $A=(6,-22),$ $B=(2006,178),$ and $D=(8,y)$ for some integer $y.$ What is the area of rectangle $ABCD?$

$4000$

$4040$

$4400$

$40{,}000$

$40{,}400$

Answer: E

Concepts:coordinate geometry slope distance formula

Difficulty rating: 1580

Small Hint:

$\overline{AB}\perp\overline{AD},$ so their slopes multiply to $-1$

Big Hint:

The area is $AB\cdot AD$

Solution:

The slope of $\overline{AB}$ is $\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}.$ Since $\overline{AD}\perp\overline{AB},$ its slope is $-10,$ so $\tfrac{y+22}{8-6}=-10$ gives $y=-42.$

Then $AB=\sqrt{2000^2+200^2}$ $=200\sqrt{101}$ and $AD=\sqrt{2^2+20^2}=2\sqrt{101}.$

The area is $200\sqrt{101}\cdot2\sqrt{101}$ $=400\cdot101$ $=40{,}400.$

Thus, the correct answer is E.

21.

For a particular peculiar pair of dice, the probabilities of rolling $1,$ $2,$ $3,$ $4,$ $5,$ and $6$ on each die are in the ratio $1:2:3:4:5:6.$ What is the probability of rolling a total of $7$ on the two dice?

$\dfrac{4}{63}$

$\dfrac{1}{8}$

$\dfrac{8}{63}$

$\dfrac{1}{6}$

$\dfrac{2}{7}$

Answer: C

Concepts:dice (probability)basic probability

Difficulty rating: 1630

Small Hint:

The probability of rolling $k$ is $\tfrac{k}{21}$

Big Hint:

Sum $P(i)P(7-i)$ over the six ordered pairs with $i+j=7$

Solution:

Each die shows $k$ with probability $\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}.$

For a total of $7,$ the ordered pairs $(1,6),(2,5),\ldots,(6,1)$ contribute $\begin{aligned} &\scriptsize\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}\\ &=\dfrac{56}{441}=\dfrac{8}{63}. \end{aligned}$

Thus, the correct answer is C.

22.

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4$ ¢ per glob and $J$ blobs of jam at $5$ ¢ per blob. The cost of the peanut butter and jam to make all the sandwiches is $\$2.53.$ Assume that $B,$ $J,$ and $N$ are positive integers with $N \gt 1.$ What is the cost of the jam Elmo uses to make the sandwiches?

$\$1.05$

$\$1.25$

$\$1.45$

$\$1.65$

$\$1.85$

Answer: D

Concepts:Diophantine Equation prime factorization casework

Difficulty rating: 1860

Small Hint:

The total cost in cents is $N(4B+5J)=253$

Big Hint:

$253=11\cdot23,$ and $N \gt 1,$ so test each factor

Solution:

The total cost is $N(4B+5J)=253$ cents $=11\cdot23.$ Since $N \gt 1,$ $N\in\{11,23,253\}.$

If $N=253$ or $N=23,$ then $4B+5J$ equals $1$ or $11,$ impossible for positive integers.

So $N=11$ and $4B+5J=23,$ whose only positive solution is $B=2,$ $J=3.$ The jam costs $11\cdot3\cdot5=165$ cents, or $\$1.65.$

Thus, the correct answer is D.

23.

A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are $3,$ $7,$ and $7,$ as shown. What is the area of the shaded quadrilateral?

$15$

$17$

$\dfrac{35}{2}$

$18$

$\dfrac{55}{3}$

Answer: D

Concepts:area ratio system of equations

Difficulty rating: 1950

Small Hint:

Split the quadrilateral into two triangles with areas $R$ and $S$

Big Hint:

Triangles sharing an altitude have areas proportional to their bases

Solution:

Split the quadrilateral into two triangles of areas $R$ and $S,$ so the shaded area is $T=R+S.$

Comparing triangles that share an altitude, base ratios give $\tfrac{R}{3}=\tfrac{T+7}{10}$ and $\tfrac{S}{7}=\tfrac{T+3}{14}.$

Then $T=R+S=3\cdot\tfrac{T+7}{10}+7\cdot\tfrac{T+3}{14},$ so $10T=3(T+7)$ $+5(T+3)$ $=8T+36,$ giving $T=18.$

Thus, the correct answer is D.

24.

Circles with centers at $O$ and $P$ have radii $2$ and $4,$ respectively, and are externally tangent. Points $A$ and $B$ on the circle with center $O$ and points $C$ and $D$ on the circle with center $P$ are such that $\overline{AD}$ and $\overline{BC}$ are common external tangents to the circles. What is the area of the concave hexagon $AOBCPD?$

$18\sqrt{3}$

$24\sqrt{2}$

$36$

$24\sqrt{3}$

$32\sqrt{2}$

Answer: B

Concepts:tangent circles trapezoid right triangle

Difficulty rating: 2010

Small Hint:

Draw $OF\parallel AD$ meeting $PD$ at $F$ to form a rectangle and a right triangle

Big Hint:

Then $DF=2,$ $FP=2,$ and $OF=\sqrt{OP^2-FP^2}$ with $OP=6$

Solution:

The hexagon is symmetric about $\overline{OP},$ so its area is twice that of trapezoid $AOPD.$

Draw $OF\parallel AD$ with $F$ on $\overline{PD}.$ Then $AOFD$ is a rectangle, so $DF=OA=2$ and $FP=PD-DF=4-2=2.$

Since the circles are externally tangent, $OP=2+4=6,$ so in right triangle $OFP,$ $OF=\sqrt{36-4}=4\sqrt2.$

Trapezoid $AOPD$ has parallel sides $OA=2$ and $PD=4$ with height $OF=4\sqrt2,$ giving area $\tfrac12(2+4)(4\sqrt2)=12\sqrt2.$ The hexagon area is $2\cdot12\sqrt2=24\sqrt2.$

Thus, the correct answer is B.

25.

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is $9,$ spots a license plate with a $4$ -digit number in which each of two digits appears two times. “Look, daddy!” she exclaims. “That number is evenly divisible by the age of each of us kids!” “That’s right,” replies Mr. Jones, “and the last two digits just happen to be my age.” Which of the following is not the age of one of Mr. Jones’s children?

$4$

$5$

$6$

$7$

$8$

Answer: B

Concepts:divisibility casework

Difficulty rating: 2120

Small Hint:

The oldest child is $9,$ so the number is divisible by $9$

Big Hint:

With two digits each appearing twice, the digit sum $2(a+b)$ is a multiple of $9,$ forcing $a+b=9$

Solution:

Since a child is $9,$ the number is divisible by $9,$ so its digit sum $2(a+b)$ is a multiple of $9,$ which forces $a+b=9.$

The eight distinct ages are eight of the nine integers from $1$ through $9,$ so at least one of $4$ and $8$ is an age. Hence the plate number is divisible by $4.$ Up to interchanging the two digits, its pattern is $aabb,$ $abab,$ or $baab.$ Combining these patterns with $a+b=9$ and divisibility by $4$ leaves $\begin{gathered} 1188,\ 2772,\ 3636,\ 5544,\\ 6336,\ 7272,\ 9900. \end{gathered}$

The last candidate would make Mr. Jones’s age $00,$ so it is impossible. None of the other six candidates is divisible by $5,$ so $5$ cannot be one of the children’s ages. The conditions are attainable: $5544$ is divisible by each age in $\{1,2,3,4,6,7,8,9\}$ and its last two digits give Mr. Jones’s age as $44.$

Thus, the correct answer is B.