1995 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

Walter has exactly one penny, one nickel, one dime and one quarter in his pocket. What percent of one dollar is in his pocket?

$4\%$

$25\%$

$40\%$

$41\%$

$59\%$

Answer: D

Concepts:money percentage

Difficulty rating: 370

Small Hint:

Add up the values of the coins in cents

Big Hint:

$1 + 5 + 10 + 25$ cents, and $100$ cents make one dollar

Solution:

The coins total $1 + 5 + 10 + 25 = 41$ cents.

Since one dollar is $100$ cents, this is $41\%$ of a dollar.

Thus, the correct answer is D .

2.

Jose is $4$ years younger than Zack. Zack is $3$ years older than Inez. Inez is $15$ years old. How old is Jose?

$8$

$11$

$14$

$16$

$22$

Answer: C

Concepts:ages

Difficulty rating: 370

Small Hint:

Work from Inez up to Zack, then down to Jose

Big Hint:

Zack $= 15 + 3;$ then Jose $=$ Zack $- 4$

Solution:

Zack is $15 + 3 = 18$ years old.

Jose is $4$ years younger, so Jose is $18 - 4 = 14$ .

Thus, the correct answer is C .

3.

Which of the following operations has the same effect on a number as multiplying by $\dfrac34$ and then dividing by $\dfrac35?$

dividing by $\dfrac43$

dividing by $\dfrac{9}{20}$

multiplying by $\dfrac{9}{20}$

dividing by $\dfrac54$

multiplying by $\dfrac54$

Answer: E

Concepts:fraction

Difficulty rating: 560

Small Hint:

Dividing by $\dfrac35$ is the same as multiplying by $\dfrac53$

Big Hint:

Combine $\dfrac34 \times \dfrac53$ into a single fraction

Solution:

Dividing by $\dfrac35$ is the same as multiplying by $\dfrac53,$ so the two operations together multiply by

$\dfrac34 \times \dfrac53 = \dfrac54.$

So the combined effect is multiplying by $\dfrac54$ .

Thus, the correct answer is E .

4.

A teacher tells the class, “Think of a number, add $1$ to it, and double the result. Give the answer to your partner. Partner, subtract $1$ from the number you are given and double the result to get your answer.” Ben thinks of $6,$ and gives his answer to Sue. What should Sue’s answer be?

$18$

$24$

$26$

$27$

$30$

Answer: C

Concepts:order of operations

Difficulty rating: 450

Small Hint:

First compute Ben’s answer: add $1$ to $6,$ then double

Big Hint:

Sue takes Ben’s number, subtracts $1,$ and doubles that

Solution:

Ben computes $(6 + 1) \times 2 = 14$ and gives $14$ to Sue.

Sue computes $(14 - 1) \times 2 = 26$ .

Thus, the correct answer is C .

5.

Find the smallest whole number that is larger than the sum

$2\tfrac12 + 3\tfrac13 + 4\tfrac14 + 5\tfrac15.$

$14$

$15$

$16$

$17$

$18$

Answer: C

Concepts:fraction estimation

Difficulty rating: 660

Small Hint:

Add the whole-number parts and the fraction parts separately

Big Hint:

The fractions $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ add to a little more than $1$

Solution:

The whole-number parts sum to $2 + 3 + 4 + 5 = 14$ .

The fractions $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ add to about $1.28,$ which is between $1$ and $2.$ So the total is between $15$ and $16,$ and the smallest whole number larger than it is $16$ .

Thus, the correct answer is C .

6.

Figures $I,$ $II$ and $III$ are squares. The perimeter of $I$ is $12$ and the perimeter of $II$ is $24.$ The perimeter of $III$ is

$9$

$18$

$36$

$72$

$81$

Answer: C

Concepts:perimeter square (geometry)

Difficulty rating: 770

Small Hint:

The perimeter of a square is $4$ times its side, so find the side of $I$ and of $II$

Big Hint:

From the figure, the side of $III$ equals the side of $I$ plus the side of $II$

Solution:

Square $I$ has side $12 \div 4 = 3,$ and square $II$ has side $24 \div 4 = 6.$

From the figure, the side of $III$ is $3 + 6 = 9,$ so its perimeter is $4 \times 9 = 36.$

Thus, the correct answer is C .

7.

At Clover View Junior High, one half of the students go home on the school bus. One fourth go home by automobile. One tenth go home on their bicycles. The rest walk home. What fractional part of the students walk home?

$\dfrac{1}{16}$

$\dfrac{3}{20}$

$\dfrac13$

$\dfrac{17}{20}$

$\dfrac{9}{10}$

Answer: B

Concepts:fraction

Difficulty rating: 730

Small Hint:

Give the fractions a common denominator to add them

Big Hint:

Subtract the total who ride from $1$ (the whole group)

Solution:

The students who ride make up

$\begin{aligned} \dfrac12 + \dfrac14 + \dfrac{1}{10} &= \dfrac{10}{20} + \dfrac{5}{20} + \dfrac{2}{20} \\ &= \dfrac{17}{20}. \end{aligned}$

So the fraction who walk is $1 - \dfrac{17}{20} = \dfrac{3}{20}.$

Thus, the correct answer is B .

8.

An American traveling in Italy wishes to exchange American money (dollars) for Italian money (lire). If $3000$ lire $= \$1.60,$ how many lire will the traveler receive in exchange for $\$1.00?$

$180$

$480$

$1800$

$1875$

$4875$

Answer: D

Concepts:ratio and proportion

Difficulty rating: 820

Small Hint:

Find how many lire equal one dollar using the given rate

Big Hint:

$\$1.00$ is $\dfrac{1.00}{1.60}$ of $\$1.60,$ so take that fraction of $3000$

Solution:

Since $\$1.00$ is $\dfrac{1.00}{1.60} = \dfrac58$ of $\$1.60,$ the traveler gets $\dfrac58$ of $3000$ lire.

That is $\dfrac58 \times 3000 = 1875$ lire.

Thus, the correct answer is D .

9.

Three congruent circles with centers $P,$ $Q$ and $R$ are tangent to the sides of rectangle $ABCD$ as shown. The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R.$ The area of the rectangle is

$16$

$24$

$32$

$64$

$128$

Answer: C

Concepts:tangent circles rectangle

Difficulty rating: 960

Small Hint:

The diameter of each circle is $4,$ so the short side of the rectangle equals $4$

Big Hint:

The long side spans two full diameters

Solution:

Each circle has diameter $4.$ The short side of the rectangle equals one diameter, so it is $4.$

Since the circle at $Q$ passes through $P$ and $R,$ all three circles have radius $2,$ and the long side spans two full diameters: $4 + 4 = 8.$ The area is $8 \times 4 = 32.$

Thus, the correct answer is C .

10.

A jacket and a shirt originally sold for $\$80$ and $\$40,$ respectively. During a sale Chris bought the $\$80$ jacket at a $40\%$ discount and the $\$40$ shirt at a $55\%$ discount. The total amount saved was what percent of the total of the original prices?

$45\%$

$47\tfrac12\%$

$50\%$

$79\tfrac16\%$

$95\%$

Answer: A

Concepts:percentage

Difficulty rating: 930

Small Hint:

Find the dollars saved on each item separately

Big Hint:

Divide the total saved by the total original price of $\$120$

Solution:

The jacket discount saves $40\%$ of $\$80 = \$32,$ and the shirt discount saves $55\%$ of $\$40 = \$22.$ The total saved is $\$32 + \$22 = \$54.$

The original total is $\$80 + \$40 = \$120,$ so the percent saved is $\dfrac{54}{120} = 0.45 = 45\%.$

Thus, the correct answer is A .

11.

Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the $18$ -block area as shown. When they meet for the first time, they will be closest to

$A$

$B$

$C$

$D$

$E$

Answer: D

Concepts:distance rate and time perimeter relative speed

Difficulty rating: 1030

Small Hint:

When they first meet, together they have covered the full $18$ -block loop

Big Hint:

Jane walks twice as fast, so she covers $12$ blocks while Hector covers $6$

Solution:

The perimeter of the region is $18$ blocks, so when Jane and Hector meet they have together walked $18$ blocks. Since Jane walks twice as fast, she covers $12$ blocks and Hector covers $6.$

Starting from the middle of the bottom edge, Hector walks $6$ blocks (to $E,$ then up to $D$ ), and Jane walks $12$ blocks (to $A,$ up to $B,$ then across the top to $D$ ). They meet at $D.$

Thus, the correct answer is D .

12.

A lucky year is one in which at least one date, when written in the form month/day/year, has the following property: the product of the month times the day equals the last two digits of the year. For example, $1956$ is a lucky year because it has the date $7/8/56$ and $7 \times 8 = 56.$ Which of the following is NOT a lucky year?

$1990$

$1991$

$1992$

$1993$

$1994$

Answer: E

Concepts:factor casework

Difficulty rating: 1120

Small Hint:

For each year, try to write its last two digits as (month) $\times$ (day) with a valid month and day

Big Hint:

A year fails only when the last two digits cannot be a product of a number at most $12$ and a number at most $31$

Solution:

Each of the other years works: $90 = 9 \times 10,$ $91 = 7 \times 13,$ $92 = 4 \times 23,$ and $93 = 3 \times 31,$ each a valid month/day.

For $1994,$ the last two digits factor only as $94 = 2 \times 47,$ and $47$ is too large for a day (and $94$ or $47$ is too large for a month). So $1994$ has no lucky date.

Thus, the correct answer is E .

13.

In the figure, $\angle A,$ $\angle B$ and $\angle C$ are right angles. If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE,$ then $\angle CDE =$

$75^\circ$

$80^\circ$

$85^\circ$

$90^\circ$

$95^\circ$

Answer: E

Concepts:angle chasing angle sum

Difficulty rating: 1150

Small Hint:

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ$

Big Hint:

$\angle AED = \angle AEB + \angle BED;$ then use quadrilateral $AEDC,$ whose angles sum to $360^\circ$

Solution:

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ,$ so $\angle BED = \angle BDE = 45^\circ.$

Then $\angle AED = \angle AEB + \angle BED$ $= 40^\circ + 45^\circ$ $= 85^\circ.$ In quadrilateral $AEDC,$ the angles at $A$ and $C$ are $90^\circ,$ so

$\begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ. \end{aligned}$

Thus, the correct answer is E .

14.

A team won $40$ of its first $50$ games. How many of the remaining $40$ games must this team win so it will have won exactly $70\%$ of its games for the season?

$20$

$23$

$28$

$30$

$35$

Answer: B

Concepts:percentage

Difficulty rating: 930

Small Hint:

The season has $50 + 40 = 90$ games; find $70\%$ of that

Big Hint:

Subtract the $40$ games already won from the needed total

Solution:

The season has $50 + 40 = 90$ games, and $70\%$ of $90$ is $63$ wins.

The team already has $40$ wins, so it needs $63 - 40 = 23$ more.

Thus, the correct answer is B .

15.

What is the $100$ th digit to the right of the decimal point in the decimal form of $\dfrac{4}{37}?$

$0$

$1$

$2$

$7$

$8$

Answer: B

Concepts:repeating decimal modular arithmetic

Difficulty rating: 1060

Small Hint:

Write $\dfrac{4}{37}$ as a repeating decimal and find the length of the repeating block

Big Hint:

The block has length $3;$ use the remainder of $100$ divided by $3$

Solution:

$\dfrac{4}{37} = 0.\overline{108},$ repeating with block length $3.$ The digits in positions $3,$ $6,$ $9,$ $\ldots$ (multiples of $3$ ) are $8.$

Since $99$ is a multiple of $3,$ the $99$ th digit is $8,$ so the $100$ th digit starts the next block: it is $1.$

Thus, the correct answer is B .

16.

Students from three middle schools worked on a summer project. Seven students from Allen School worked for $3$ days. Four students from Balboa School worked for $5$ days. Five students from Carver School worked for $9$ days. The total amount paid for the students’ work was $\$774.$ Assuming each student received the same amount for a day’s work, how much did the students from Balboa School earn altogether?

$\$9.00$

$\$48.38$

$\$180.00$

$\$193.50$

$\$258.00$

Answer: C

Concepts:rate

Difficulty rating: 1090

Small Hint:

Compute the total number of student-days worked across all three schools

Big Hint:

Divide $\$774$ by the total student-days to get the pay per student-day

Solution:

The total student-days are $7 \times 3 + 4 \times 5 + 5 \times 9$ $= 21 + 20 + 45$ $= 86.$

So each student-day pays $\$774 \div 86 = \$9.$ Balboa worked $20$ student-days, earning $20 \times \$9 = \$180.$

Thus, the correct answer is C .

17.

The table below gives the percent of students in each grade at Annville and Cleona elementary schools:

	$K$	$1$	$2$	$3$	$4$	$5$	$6$
Annville	$16\%$	$15\%$	$15\%$	$14\%$	$13\%$	$16\%$	$11\%$
Cleona	$12\%$	$15\%$	$14\%$	$13\%$	$15\%$	$14\%$	$17\%$

Annville has $100$ students and Cleona has $200$ students. In the two schools combined, what percent of the students are in grade $6?$

$12\%$

$13\%$

$14\%$

$15\%$

$28\%$

Answer: D

Concepts:percentage mean

Difficulty rating: 980

Small Hint:

Find the number of grade- $6$ students at each school separately

Big Hint:

Divide the combined grade- $6$ count by the $300$ total students

Solution:

Annville has $11\%$ of $100 = 11$ sixth graders, and Cleona has $17\%$ of $200 = 34$ sixth graders.

Combined, that is $11 + 34 = 45$ out of $300$ students, which is $\dfrac{45}{300} = 15\%.$

Thus, the correct answer is D .

18.

The area of each of the four congruent L-shaped regions of this $100$ -inch by $100$ -inch square is $\dfrac{3}{16}$ of the total area. How many inches long is the side of the center square?

$25$

$44$

$50$

$62$

$75$

Answer: C

Concepts:square (geometry)area

Difficulty rating: 980

Small Hint:

The four L-shaped regions together make up $4 \times \dfrac{3}{16}$ of the square

Big Hint:

The center square is the remaining fraction of the area; its side is the square root of its area

Solution:

The four L-shaped regions cover $4 \times \dfrac{3}{16} = \dfrac34$ of the square, so the center square is the remaining $\dfrac14$ of the total area.

The total area is $100 \times 100 = 10000$ square inches, so the center square has area $\dfrac14 \times 10000 = 2500,$ and its side is $\sqrt{2500} = 50$ inches.

Thus, the correct answer is C .

19.

The graph shows the distribution of the number of children in the families of the students in Ms. Jordan’s English class. The median number of children in the family for this distribution is

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:median (data)data and graph interpretation

Difficulty rating: 960

Small Hint:

The median is the middle value when all the family sizes are listed in order

Big Hint:

Count the total number of families, then find which position is the middle one

Solution:

The graph gives $2$ families with $1$ child, $1$ with $2,$ $2$ with $3,$ $2$ with $4,$ and $6$ with $5,$ for $2 + 1 + 2 + 2 + 6 = 13$ families.

The median is the $7$ th value in order. Listing the family sizes, the $7$ th value is $4.$

Thus, the correct answer is D .

20.

Diana and Apollo each roll a standard die obtaining a number at random from $1$ to $6.$ What is the probability that Diana’s number is larger than Apollo’s number?

$\dfrac13$

$\dfrac{5}{12}$

$\dfrac49$

$\dfrac{17}{36}$

$\dfrac12$

Answer: B

Concepts:dice (probability)symmetry

Difficulty rating: 1120

Small Hint:

Of the $36$ equally likely outcomes, some are ties; remove those

Big Hint:

By symmetry, Diana is larger in exactly half of the non-tie outcomes

Solution:

There are $6 \times 6 = 36$ equally likely outcomes, of which $6$ are ties, leaving $30$ outcomes with different numbers.

By symmetry, Diana is larger in exactly half of those, or $15,$ so the probability is $\dfrac{15}{36} = \dfrac{5}{12}.$

Thus, the correct answer is B .

21.

A plastic snap-together cube has a protruding snap on one side and receptacle holes on the other five sides. What is the smallest number of these cubes that can be snapped together so that only receptacle holes are showing?

$3$

$4$

$5$

$6$

$8$

Answer: B

Concepts:3D geometry

Difficulty rating: 1170

Small Hint:

Each cube has one snap that must be hidden inside another cube’s hole

Big Hint:

Try arranging the cubes in a small ring so every snap plugs into a neighbor

Solution:

Every cube’s single snap must be plugged into another cube’s hole to be hidden. Two cubes can share only one face, so two cubes cannot hide both snaps. With three cubes, hiding all three snaps would require a three-cube loop in which every pair shares a face, but three unit cubes cannot be pairwise face-adjacent. Thus one, two, or three cubes cannot work.

Four cubes can be arranged in a square ring, each snap fitting into the neighbor’s hole, so only receptacle holes show. The smallest number is $4.$

Thus, the correct answer is B .

22.

The number $6545$ can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?

$162$

$172$

$173$

$174$

$222$

Answer: A

Concepts:prime factorization factor

Difficulty rating: 1170

Small Hint:

Find the prime factorization of $6545$

Big Hint:

Group the prime factors into two numbers that are each between $10$ and $99$

Solution:

The prime factorization is $6545 = 5 \times 7 \times 11 \times 17.$ To split into two two-digit factors, pair the primes: $5 \times 17 = 85$ and $7 \times 11 = 77.$

These are the only two-digit pair, and their sum is $85 + 77 = 162.$

Thus, the correct answer is A .

23.

How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?

$1120$

$1400$

$1800$

$2025$

$2500$

Answer: B

Concepts:multiplication principle

Difficulty rating: 1220

Small Hint:

Count choices for each digit position in turn: first, second, third, fourth

Big Hint:

First digit: $5$ odd options; second: $5$ even options; then $8$ and $7$ for the distinct remaining digits

Solution:

The first digit is odd: $5$ choices. The second is even: $5$ choices (none of which repeats the odd first digit).

The third digit is any of the $8$ unused digits, and the fourth is any of the $7$ remaining. In total, $5 \times 5 \times 8 \times 7 = 1400.$

Thus, the correct answer is B .

24.

In parallelogram $ABCD,$ $\overline{DE}$ is the altitude to the base $\overline{AB}$ and $\overline{DF}$ is the altitude to the base $\overline{BC}.$ (Both pictures represent the same parallelogram.) If $DC = 12,$ $EB = 4,$ and $DE = 6,$ then $DF =$

$6.4$

$7$

$7.2$

$8$

$10$

Answer: C

Concepts:parallelogram Pythagorean Theorem

Difficulty rating: 1150

Small Hint:

Opposite sides are equal, so $AB = 12;$ find $AE,$ then $AD$ with the right triangle $ADE$

Big Hint:

The area equals base $\times$ height for either base: $AB \cdot DE = BC \cdot DF$

Solution:

Since $AB = DC = 12,$ we get $AE = 12 - 4 = 8.$ In right triangle $ADE,$ $AD = \sqrt{8^2 + 6^2} = 10,$ so $BC = AD = 10.$

The area is $AB \cdot DE = 12 \cdot 6 = 72,$ and also $BC \cdot DF = 10 \cdot DF.$ So $DF = \dfrac{72}{10} = 7.2.$

Thus, the correct answer is C .

25.

Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass on the highway (not in the station)?

$5$

$6$

$9$

$10$

$11$

Answer: D

Concepts:distance rate and time

Difficulty rating: 1260

Small Hint:

Track one Houston-bound bus over its $5$ -hour trip and see which oncoming buses share the road with it

Big Hint:

Count the Dallas-bound buses already on the highway when it starts, plus those that leave before it arrives

Solution:

Consider a bus leaving Dallas at $6\!:\!00,$ arriving in Houston at $11\!:\!00.$ It meets every Dallas-bound bus that is on the highway during that window.

Dallas-bound buses leave Houston on the half hour and take $5$ hours. The ones sharing the road (meeting away from a station) are those that left Houston at $1\!:\!30,$ $2\!:\!30,$ $\ldots,$ $10\!:\!30,$ which is $10$ buses.

Thus, the correct answer is D .