1994 AMC 8 Solutions | LIVE by Po-Shen Loh

1.

Which of the following is the largest?

$\dfrac{1}{3}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

$\dfrac{7}{24}$

Concepts:fraction

Difficulty rating: 560

Small Hint:

Rewrite all five fractions over one common denominator

Big Hint:

The least common denominator of these fractions is $24$

Solution:

Over the common denominator $24$ , the fractions are $\dfrac{8}{24},$ $\dfrac{6}{24},$ $\dfrac{9}{24},$ $\dfrac{10}{24},$ and $\dfrac{7}{24}.$

The largest numerator is $10$ , so $\dfrac{5}{12} = \dfrac{10}{24}$ is the largest.

Thus, the correct answer is D .

2.

What is the value of the following expression?

$\begin{aligned} &\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} \\ &\quad {}+ \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} \\ &\quad {}+ \frac{9}{10} + \frac{55}{10} \end{aligned}$

$4\dfrac{1}{2}$

$6.4$

$9$

$10$

$11$

Concepts:fraction

Difficulty rating: 560

Small Hint:

All the terms share the denominator $10$ , so add the numerators first

Big Hint:

$1 + 2 + \cdots + 9 = 45,$ and then add the $55$

Solution:

The numerators sum to $1 + 2 + \cdots + 9 + 55$ $= 45 + 55$ $= 100.$

So the total is $\dfrac{100}{10} = 10.$

Thus, the correct answer is D .

3.

Each day Maria must work $8$ hours. This does not include the $45$ minutes she takes for lunch. If she begins working at $7{:}25$ A.M. and takes her lunch break at noon, then her working day will end at

$3{:}40$ P.M.

$3{:}55$ P.M.

$4{:}10$ P.M.

$4{:}25$ P.M.

$4{:}40$ P.M.

Concepts:clock

Difficulty rating: 660

Small Hint:

The workday spans $8$ hours of work plus the $45$ -minute lunch break

Big Hint:

Add $8$ hours to $7{:}25$ A.M., then add $45$ more minutes

Solution:

Eight hours after $7{:}25$ A.M. is $3{:}25$ P.M.

Adding the $45$ -minute lunch break gives an ending time of $4{:}10$ P.M.

Thus, the correct answer is C .

4.

Which of the following represents the result when the figure shown at the right is rotated clockwise $120°$ about its center?

Concepts:transformation

Difficulty rating: 660

Small Hint:

A clockwise turn of $120°$ moves each of the three shapes to the next position in the clockwise direction.

Big Hint:

The square must stay a square and the circle a circle; only their positions change.

Solution:

A clockwise turn of $120°$ sends each shape to the next position clockwise: the triangle at the top moves to the lower right, the square at the lower right moves to the lower left, and the circle at the lower left moves to the top.

The result therefore has a circle on top, a square at the lower left, and a triangle at the lower right, with each shape keeping its original form.

Thus, the correct answer is B .

5.

Given that $1$ mile $= 8$ furlongs and $1$ furlong $= 40$ rods, the number of rods in one mile is

$5$

$320$

$660$

$1760$

$5280$

Concepts:unit conversion

Difficulty rating: 560

Small Hint:

Convert miles to furlongs, then furlongs to rods

Big Hint:

One mile is $8$ furlongs, and each furlong is $40$ rods

Solution:

One mile is $8$ furlongs, and each furlong is $40$ rods, so one mile is $8 \times 40 = 320$ rods.

Thus, the correct answer is B .

6.

The unit’s digit (one’s digit) of the product of any six consecutive positive whole numbers is

$0$

$2$

$4$

$6$

$8$

Concepts:units digit divisibility

Difficulty rating: 730

Small Hint:

Among any six consecutive numbers there is at least one multiple of $5$ and at least one even number

Big Hint:

A multiple of $5$ times an even number is a multiple of $10$

Solution:

Any six consecutive whole numbers include at least one multiple of $5$ and at least one even number, so their product is a multiple of $5 \times 2 = 10.$

A multiple of $10$ always ends in $0$ .

Thus, the correct answer is A .

7.

If $\angle A = 60°,$ $\angle E = 40°,$ and $\angle C = 30°,$ then $\angle BDC =$

$40°$

$50°$

$60°$

$70°$

$80°$

Concepts:angle chasing angle sum

Difficulty rating: 920

Small Hint:

In triangle $ABE$ , the angles at $A$ and $E$ determine the angle $\angle ABE$

Big Hint:

Since $D$ lies on $EB$ and $A, B, C$ are collinear, $\angle DBC$ is the supplement of $\angle ABE$ ; then use triangle $BDC$

Solution:

In triangle $ABE,$ $\angle ABE = 180°$ $- (60° + 40°)$ $= 80°.$

Since $A, B, C$ are collinear and $D$ lies on segment $EB,$ the angle $\angle DBC = 180° - 80° = 100°.$

In triangle $BDC,$ $\angle BDC = 180°$ $- (100° + 30°)$ $= 50°.$

Thus, the correct answer is B .

8.

For how many three-digit whole numbers does the sum of the digits equal $25?$

$2$

$4$

$6$

$8$

$10$

Concepts:digits multiset permutations

Difficulty rating: 920

Small Hint:

The largest possible digit sum is $9 + 9 + 9 = 27$ , so the digits must be close to all $9$ s

Big Hint:

Find the digit combinations that sum to $25$ , then count how many arrangements each has

Solution:

Since the digits sum to $25$ and the maximum is $27$ , at least one digit is $9$ . The possible digit sets are $\{9, 9, 7\}$ and $\{9, 8, 8\}.$

Each set has $3$ distinct arrangements ( $997,$ $979,$ $799$ and $988,$ $898,$ $889$ ), giving $6$ numbers in all.

Thus, the correct answer is C .

9.

A shopper buys a $\$100$ coat on sale for $20\%$ off. An additional $\$5$ is taken off the sale price by using a discount coupon. A sales tax of $8\%$ is paid on the final selling price. The total amount the shopper pays for the coat is

$\$81.00$

$\$81.40$

$\$82.00$

$\$82.08$

$\$82.40$

Concepts:percentage

Difficulty rating: 860

Small Hint:

Apply the $20\%$ discount first, then subtract the $\$5$ coupon

Big Hint:

Multiply the resulting price by $1.08$ to add the $8\%$ tax

Solution:

The $20\%$ discount lowers the price to $\$80,$ and the $\$5$ coupon reduces it to $\$75.$

Adding $8\%$ tax gives $1.08 \times \$75 = \$81.00.$

Thus, the correct answer is A .

10.

For how many positive integer values of $N$ $(N \gt 0)$ is the expression

$\frac{36}{N+2}$

an integer?

$7$

$8$

$9$

$10$

$12$

Concepts:divisibility factor

Difficulty rating: 860

Small Hint:

$\dfrac{36}{N+2}$ is an integer exactly when $N+2$ is a divisor of $36$

Big Hint:

Since $N \gt 0,$ only divisors of $36$ that are greater than $2$ count

Solution:

The expression is an integer when $N+2$ divides $36.$ The divisors of $36$ are $1,$ $2,$ $3,$ $4,$ $6,$ $9,$ $12,$ $18,$ $36.$

Because $N \gt 0,$ we need $N + 2 \gt 2,$ leaving the divisors $3,$ $4,$ $6,$ $9,$ $12,$ $18,$ $36$ : that is $7$ values.

Thus, the correct answer is A .

11.

Last summer $100$ students attended basketball camp. Of those attending, $52$ were boys and $48$ were girls. Also, $40$ students were from Jones Middle School and $60$ were from Clay Middle School. Twenty of the girls were from Jones Middle School. How many of the boys were from Clay Middle School?

$20$

$32$

$40$

$48$

$52$

Concepts:logical deduction

Difficulty rating: 820

Small Hint:

First find how many girls came from Clay Middle School

Big Hint:

Subtract the Clay girls from the $60$ Clay students to get the Clay boys

Solution:

Since $48$ girls attended and $20$ were from Jones, $48 - 20 = 28$ girls were from Clay.

Clay had $60$ students total, so the number of Clay boys is $60 - 28 = 32.$

Thus, the correct answer is B .

12.

Each of the three large squares shown is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?

The shaded areas in all three are equal.

Only the shaded areas of $I$ and $II$ are equal.

Only the shaded areas of $I$ and $III$ are equal.

Only the shaded areas of $II$ and $III$ are equal.

The shaded areas of $I,$ $II,$ and $III$ are all different.

Concepts:area decomposition area ratio

Difficulty rating: 820

Small Hint:

Find what fraction of each large square is shaded by cutting each figure into equal smaller pieces.

Big Hint:

Divide square $I$ into $8$ equal triangles, square $II$ into $4$ equal cells, and square $III$ into $16$ equal triangles, then count the shaded pieces in each.

Solution:

In square $II,$ $1$ of the $4$ equal cells is shaded, so $\tfrac14$ of it is shaded.

Square $I$ breaks into $8$ equal triangles with $2$ shaded, and square $III$ breaks into $16$ equal triangles with $4$ shaded; each of these equals $\tfrac{2}{8} = \tfrac{4}{16} = \tfrac14.$

Since every figure has exactly $\tfrac14$ shaded, the shaded areas are all equal.

Thus, the correct answer is A .

13.

The number halfway between $\dfrac{1}{6}$ and $\dfrac{1}{4}$ is

$\dfrac{1}{10}$

$\dfrac{1}{5}$

$\dfrac{5}{24}$

$\dfrac{7}{24}$

$\dfrac{5}{12}$

Concepts:mean fraction

Difficulty rating: 660

Small Hint:

The number halfway between two numbers is their average

Big Hint:

Add $\dfrac16$ and $\dfrac14,$ then divide the sum by $2$

Solution:

The number halfway between two values is their average: $\dfrac{\frac16 + \frac14}{2}.$

Since $\dfrac16 + \dfrac14 = \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12},$ the average is $\dfrac{5}{12} \div 2 = \dfrac{5}{24}.$

Thus, the correct answer is C .

14.

Two children at a time can play pairball. For $90$ minutes, with only two children playing at one time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

$9$

$10$

$18$

$20$

$36$

Concepts:rate

Difficulty rating: 820

Small Hint:

With two children playing for the whole $90$ minutes, the total playing time is $2 \times 90$ child-minutes

Big Hint:

Divide the total playing time equally among the $5$ children

Solution:

Two children play at every moment for $90$ minutes, so the total playing time is $2 \times 90 = 180$ child-minutes.

Split equally among $5$ children, each plays $\dfrac{180}{5} = 36$ minutes.

Thus, the correct answer is E .

15.

If this path is to continue in the same pattern, then which sequence of arrows goes from point $425$ to point $427?$

Concepts:modular arithmetic pattern recognition

Difficulty rating: 910

Small Hint:

The arrows repeat in a cycle of length $4$ , so a point’s outgoing arrows depend only on its remainder when divided by $4$ .

Big Hint:

Since $425$ and $427$ leave remainders $1$ and $3$ when divided by $4$ , the arrows match those going from point $1$ to point $3$ .

Solution:

The pattern repeats every $4$ points, so the arrows leaving a point depend only on its remainder upon division by $4.$

Because $425 = 4(106) + 1$ and $427 = 4(106) + 3,$ the path from $425$ to $427$ looks just like the path from point $1$ to point $3$ : from $1$ it goes up to $2,$ then right to $3.$

Thus, the correct answer is A .

16.

The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?

$2$

$3$

$4$

$6$

$9$

Concepts:similarity area ratio

Difficulty rating: 820

Small Hint:

If the perimeter is $3$ times as large, the side length is also $3$ times as large

Big Hint:

Area scales as the square of the ratio of side lengths

Solution:

A square’s perimeter is proportional to its side, so the larger square has side length $3$ times the smaller one.

Area is the side squared, so the larger area is $3^2 = 9$ times the smaller.

Thus, the correct answer is E .

17.

Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number of hours it will take her to shovel it clean is closest to

$4$

$5$

$6$

$7$

$12$

Concepts:volume arithmetic sequence

Difficulty rating: 1000

Small Hint:

First find the total volume of snow: $4 \times 10 \times 3$

Big Hint:

Add $20 + 19 + 18 + \cdots$ hour by hour until the running total reaches the volume

Solution:

The volume of snow is $4 \times 10 \times 3 = 120$ cubic yards.

The amounts shoveled add up to $20 + 19 + 18 + 17$ $+ 16 + 15 + 14 = 119$ after $7$ hours. In the eighth hour her rate is $13$ cubic yards per hour, so the last cubic yard takes $\dfrac{1}{13}$ hour. The exact time is therefore $7\dfrac{1}{13}$ hours, which is closest to $7$ .

Thus, the correct answer is D .

18.

Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike’s trip?

Concepts:data and graph interpretation distance rate and time

Difficulty rating: 920

Small Hint:

Distance from home increases on the way out, stays constant during the hour of shopping, and decreases on the way back.

Big Hint:

Slow city driving is a gentle (shallow) slope and fast highway driving is a steep slope; the stop at the mall is a flat, horizontal segment.

Solution:

On the way out Mike drives slowly (a shallow slope) then rapidly (a steep slope), so the distance rises slowly and then more steeply. During the hour at the mall the distance stays the same, a flat segment.

On the return he drives rapidly (steep) then slowly (shallow), so the distance falls steeply and then more gently. Only the graph with a slow-then-fast rise, a flat top, and a fast-then-slow fall fits.

Thus, the correct answer is B .

19.

Around the outside of a $4$ by $4$ square, construct four semicircles with the four sides of the square as their diameters. Another square, $ABCD,$ has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is

$16$

$32$

$36$

$48$

$64$

Concepts:square (geometry)circle

Difficulty rating: 980

Small Hint:

Each semicircle has a side of the $4$ by $4$ square as its diameter, so its radius is $2$

Big Hint:

A side of $ABCD$ equals the original side length plus two radii

Solution:

Each semicircle is built on a side of length $4,$ so its radius is $2.$ A semicircle bulges out from the middle of each side by that radius.

Each side of $ABCD$ is the original side plus two radii: $4 + 2(2) = 8.$ So the area of $ABCD$ is $8^2 = 64.$

Thus, the correct answer is E .

20.

Let $W,$ $X,$ $Y,$ and $Z$ be four different digits selected from the set

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal

$\dfrac{2}{17}$

$\dfrac{3}{17}$

$\dfrac{17}{72}$

$\dfrac{25}{72}$

$\dfrac{13}{36}$

Concepts:fraction optimization

Difficulty rating: 1000

Small Hint:

To make each fraction small, use the smallest digits as numerators and the largest as denominators

Big Hint:

Compare $\dfrac18 + \dfrac29$ with $\dfrac19 + \dfrac28$ to see which is smaller

Solution:

Small numerators and large denominators make small fractions, so use $1$ and $2$ as numerators and $8$ and $9$ as denominators.

Pairing the larger numerator with the larger denominator gives $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72},$ which is smaller than $\dfrac19 + \dfrac28 = \dfrac{26}{72}.$ So the minimum sum is $\dfrac{25}{72}.$

Thus, the correct answer is D .

21.

A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

$8$

$9$

$10$

$12$

$18$

Concepts:pigeonhole principle

Difficulty rating: 980

Small Hint:

Consider the worst case: how many gumballs can you draw with at most three of each color?

Big Hint:

Each color has at least three available, so you could draw three of each before the next one forces a fourth

Solution:

In the worst case, a person could draw $3$ red, $3$ white, and $3$ blue, which is $9$ gumballs, without yet having four of any color.

The next (tenth) gumball must match one of these colors, giving four of that color. So $10$ gumballs are needed.

Thus, the correct answer is C .

22.

The two wheels shown at the right are spun and the two resulting numbers are added. The probability that the sum of the two numbers is even is

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

$\dfrac{4}{9}$

Concepts:basic probability parity

Difficulty rating: 1060

Small Hint:

On the first wheel $3$ covers half and $1$ and $2$ each cover a quarter; the second wheel is divided into equal thirds

Big Hint:

The sum is even when both numbers are even or both are odd

Solution:

On the first wheel, $P(1) = \dfrac14,$ $P(2) = \dfrac14,$ and $P(3) = \dfrac12.$ On the second wheel, each of $4,$ $5,$ $6$ has probability $\dfrac13.$

The sum is even when both numbers are odd or both are even. Both odd: the first is $1$ or $3$ $\left(\dfrac34\right)$ and the second is $5$ $\left(\dfrac13\right),$ giving $\dfrac34 \cdot \dfrac13 = \dfrac14.$ Both even: the first is $2$ $\left(\dfrac14\right)$ and the second is $4$ or $6$ $\left(\dfrac23\right),$ giving $\dfrac14 \cdot \dfrac23 = \dfrac16.$

The total probability is $\dfrac14 + \dfrac16 = \dfrac{5}{12}.$

Thus, the correct answer is D .

23.

If $X,$ $Y,$ and $Z$ are different digits, then the largest possible $3$ -digit sum for

$\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}$

has the form

$XXY$

$XYZ$

$YYX$

$YYZ$

$ZZY$

Concepts:cryptarithm place value

Difficulty rating: 1090

Small Hint:

Write the addition in columns; if $X$ is too large the hundreds place carries and the answer gains a fourth digit.

Big Hint:

Take the largest $X$ that keeps the sum to three digits, then choose $Y$ to make the total as large as possible and see which letters the resulting digits match.

Solution:

The hundreds digit of the sum comes from $X$ plus any carry, so if $X = 9$ the sum would spill over into four digits. The largest allowed value is $X = 8.$

To make the sum as large as possible, take $Y = 9:$ then $888 + 98 + 8 = 994.$ Its digits are $9,$ $9,$ $4;$ since $9 = Y$ and $4$ is a new digit $Z,$ the sum has the form $YYZ.$

Thus, the correct answer is D .

24.

A $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.

$4$

$6$

$7$

$8$

$16$

Concepts:arrangements with restrictions casework

Difficulty rating: 1150

Small Hint:

If a square is green, then the square directly above it and the square directly to its right must also be green

Big Hint:

This forces the green squares to cluster toward the top-right; count by how many squares are green

Solution:

The rule says a green square cannot have a red square on its top or right side, so any green square forces the squares above and to its right to be green as well. The green squares must therefore cluster toward the top-right corner.

The valid colorings are: all four red; only the top-right green; the whole top row green; the whole right column green; all green except the bottom-left; and all four green. That is $6$ colorings.

Thus, the correct answer is B .

25.

Find the sum of the digits in the answer to

$\underbrace{9999\cdots99}_{94} \times \underbrace{4444\cdots44}_{94}$

where a string of $94$ nines is multiplied by a string of $94$ fours.

$846$

$855$

$945$

$954$

$1072$

Concepts:digits pattern recognition small cases

Difficulty rating: 1200

Small Hint:

Try small cases first: $9 \times 4,$ $99 \times 44,$ $999 \times 444$

Big Hint:

The product is a block of $4$ s, then a $3,$ then a block of $5$ s, then a $6$ ; count how many of each

Solution:

Small cases show the pattern: $99 \times 44 = 4356$ and $999 \times 444 = 443556.$ In general, a string of $n$ nines times a string of $n$ fours gives $(n-1)$ fours, then a $3,$ then $(n-1)$ fives, then a $6.$

For $n = 94,$ the product has $93$ fours, one $3,$ $93$ fives, and one $6.$ The digit sum is $93(4) + 3$ $+ 93(5) + 6$ $= 93(9) + 9$ $= 94(9)$ $= 846.$

Thus, the correct answer is A .