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1985 AMC8真题及详解

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日 上午10:06

1985 AMC 8 真题

答案详细解析请参考文末

Problem 1

$frac{3times 5}{9times 11}times frac{7times 9times 11}{3times 5times 7}=$

$text{(A)} 1 qquad text{(B)} 0 qquad text{(C)} 49 qquad text{(D)} frac{1}{49} qquad text{(E)} 50$

Problem 2

$90+91+92+93+94+95+96+97+98+99=$

$text{(A)} 845 qquad text{(B)} 945 qquad text{(C)} 1005 qquad text{(D)} 1025 qquad text{(E)} 1045$

Problem 3

$frac{10^7}{5times 10^4}=$

$text{(A)} .002 qquad text{(B)} .2 qquad text{(C)} 20 qquad text{(D)} 200 qquad text{(E)} 2000$

Problem 4

The area of polygon?$ABCDEF$, in square units, is

$text{(A)} 24 qquad text{(B)} 30 qquad text{(C)} 46 qquad text{(D)} 66 qquad text{(E)} 74$

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Problem 5

[asy] unitsize(13); draw((0,0)--(20,0)); draw((0,0)--(0,15)); draw((0,3)--(-1,3)); draw((0,6)--(-1,6)); draw((0,9)--(-1,9)); draw((0,12)--(-1,12)); draw((0,15)--(-1,15)); fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black); fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black); fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black); fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black); fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black); label("A",(2.5,-.5),S); label("B",(4.5,-.5),S); label("C",(6.5,-.5),S); label("D",(8.5,-.5),S); label("F",(10.5,-.5),S); label("Grade",(15,-.5),S); label("$1$",(-1,3),W); label("$2$",(-1,6),W); label("$3$",(-1,9),W); label("$4$",(-1,12),W); label("$5$",(-1,15),W); [/asy]

The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory?

$text{(A)} frac{1}{2} qquad text{(B)} frac{2}{3} qquad text{(C)} frac{3}{4} qquad text{(D)} frac{4}{5} qquad text{(E)} frac{9}{10}$

Problem 6

A ream of paper containing?$500$?sheets is?$5$?cm thick. Approximately how many sheets of this type of paper would there be in a stack?$7.5$?cm high?

$text{(A)} 250 qquad text{(B)} 550 qquad text{(C)} 667 qquad text{(D)} 750 qquad text{(E)} 1250$

Problem 7

A "stair-step" figure is made of alternating black and white squares in each row. Rows?$1$?through?$4$?are shown. All rows begin and end with a white square. The number of black squares in the?$37text{th}$?row is

[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]

$text{(A)} 34 qquad text{(B)} 35 qquad text{(C)} 36 qquad text{(D)} 37 qquad text{(E)} 38$

Problem 8

If?$a = - 2$, the largest number in the set?$- 3a, 4a, frac {24}{a}, a^2, 1$?is

$text{(A)} -3a qquad text{(B)} 4a qquad text{(C)} frac {24}{a} qquad text{(D)} a^2 qquad text{(E)} 1$

Problem 9

The product of the 9 factors?$Big(1 - frac12Big)Big(1 - frac13Big)Big(1 - frac14Big)cdotsBig(1 - frac {1}{10}Big) =$

$text{(A)} frac {1}{10} qquad text{(B)} frac {1}{9} qquad text{(C)} frac {1}{2} qquad text{(D)} frac {10}{11} qquad text{(E)} frac {11}{2}$

Problem 10

The fraction halfway between?$frac{1}{5}$?and?$frac{1}{3}$?(on the number line) is

[asy] unitsize(12); draw((-1,0)--(20,0),EndArrow); draw((0,-.75)--(0,.75)); draw((10,-.75)--(10,.75)); draw((17,-.75)--(17,.75)); label("$0$",(0,-.5),S); label("$frac{1}{5}$",(10,-.5),S); label("$frac{1}{3}$",(17,-.5),S); [/asy]

$text{(A)} frac{1}{4} qquad text{(B)} frac{2}{15} qquad text{(C)} frac{4}{15} qquad text{(D)} frac{53}{200} qquad text{(E)} frac{8}{15}$

Problem 11

A piece of paper containing six joined squares labeled?as?shown in the diagram is folded along the edges of the squares to form a cube. The label of the face opposite the face labeled?$text{X}$?is

[asy] draw((0,0)--(0,1)--(2,1)--(2,2)--(3,2)--(3,0)--(2,0)--(2,-2)--(1,-2)--(1,0)--cycle); draw((1,0)--(1,1)); draw((2,0)--(2,1)); draw((1,0)--(2,0)); draw((1,-1)--(2,-1)); draw((2,1)--(3,1)); label("U",(.5,.3),N); label("V",(1.5,.3),N); label("W",(2.5,.3),N); label("X",(1.5,-.7),N); label("Y",(2.5,1.3),N); label("Z",(1.5,-1.7),N); [/asy]

$text{(A)} text{Z} qquad text{(B)} text{U} qquad text{(C)} text{V} qquad text{(D)} text{W} qquad text{(E)} text{Y}$

Problem 12

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are?$6.2 text{ cm}$,?$8.3 text{ cm}$?and?$9.5 text{ cm}$. The area of the square is

$text{(A)} 24text{ cm}^2 qquad text{(B)} 36text{ cm}^2 qquad text{(C)} 48text{ cm}^2 qquad text{(D)} 64text{ cm}^2 qquad text{(E)} 144text{ cm}^2$

Problem 13

If you walk for?$45$?minutes at a rate of?$4 text{ mph}$?and then run for?$30$?minutes at a rate of?$10text{ mph,}$?how many miles will you have gone at the end of one hour and?$15$?minutes?

$text{(A)} 3.5text{ miles} qquad text{(B)} 8text{ miles} qquad text{(C)} 9text{ miles} qquad text{(D)} 25frac{1}{3}text{ miles} qquad text{(E)} 480text{ miles}$

Problem 14

The difference between a?$6.5%$?sales tax and a?$6%$?sales tax on an item priced at?$$20$?before tax is

$text{(A)}$?$$.01$

$text{(B)}$?$$.10$

$text{(C)}$?$$ .50$

$text{(D)}$?$$ 1$

$text{(E)}$?$$10$

Problem 15

How many whole numbers between?$100$?and?$400$?contain the digit?$2$?

$text{(A)} 100 qquad text{(B)} 120 qquad text{(C)} 138 qquad text{(D)} 140 qquad text{(E)} 148$

Problem 16

The ratio of boys to girls in Mr. Brown's math class is?$2:3$. If there are?$30$?students in the class, how many more girls than boys are in the class?

$text{(A)} 10 qquad text{(B)} 5 qquad text{(C)} 3 qquad text{(D)} 6 qquad text{(E)} 2$

Problem 17

If your average score on your first six mathematics tests was?$84$?and your average score on your first seven mathematics tests was?$85$, then your score on the seventh test was

$text{(A)} 86 qquad text{(B)} 88 qquad text{(C)} 90 qquad text{(D)} 91 qquad text{(E)} 92$

Problem 18

Nine copies of a certain pamphlet cost less than?$$10.00$?while ten copies of the same pamphlet (at the same price) cost more than?$$11.00$. How much does one copy of this pamphlet cost?

$text{(A)}$?$$1.07$

$text{(B)}$?$$1.08$

$text{(C)}$?$$1.09$

$text{(D)}$?$$1.10$

$text{(E)}$?$$1.11$

Problem 19

If the length and width of a rectangle are each increased by?$10%$, then the perimeter of the rectangle is increased by

$text{(A)} 1% qquad text{(B)} 10% qquad text{(C)} 20% qquad text{(D)} 21% qquad text{(E)} 40%$

Problem 20

In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January?$1$?fall that year?

$text{(A)} text{Monday} qquad text{(B)} text{Tuesday} qquad text{(C)} text{Wednesday} qquad text{(D)} text{Friday} qquad text{(E)} text{Saturday}$

Problem 21

Mr.Green receives a?$10%$?raise every year. His salary after four such raises has gone up by what percent?

$text{(A)} text{less than }40% qquad text{(B)} 40% qquad text{(C)} 44% qquad text{(D)} 45% qquad text{(E)} text{more than }45%$

Problem 22

Assume every 7-digit whole number is a possible telephone number except those that begin with?$0$?or?$1$. What fraction of telephone numbers begin with?$9$?and end with?$0$?

$text{(A)} frac{1}{63} qquad text{(B)} frac{1}{80} qquad text{(C)} frac{1}{81} qquad text{(D)} frac{1}{90} qquad text{(E)} frac{1}{100}$

Note: All telephone numbers are 7-digit whole numbers.

Problem 23

King Middle School has?$1200$?students. Each pupil takes?$5$?classes a day. Each teacher teaches?$4$?classes. Each class has?$30$?students and?$1$teacher. How many teachers are there at King Middle School?

$text{(A)} 30 qquad text{(B)} 32 qquad text{(C)} 40 qquad text{(D)} 45 qquad text{(E)} 50$

Problem 24

In a magic triangle, each of the six whole numbers?$10-15$?is placed in one of the circles so that the sum,?$S$, of the three numbers on each side of the triangle is the same. The largest possible value for?$S$?is

[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]

$text{(A)} 36 qquad text{(B)} 37 qquad text{(C)} 38 qquad text{(D)} 39 qquad text{(E)} 40$

Problem 25

Five cards are lying on a table?as?shown.

[begin{matrix} & qquad & boxed{tt{P}} & qquad & boxed{tt{Q}} boxed{tt{3}} & qquad & boxed{tt{4}} & qquad & boxed{tt{6}} end{matrix}]

Each card has a letter on one side and a whole number on the other side. Jane said, "If a vowel is on one side of any card, then an even number is on the other side." Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?

$text{(A)} 3 qquad text{(B)} 4 qquad text{(C)} 6 qquad text{(D)} text{P} qquad text{(E)} text{Q}$

1985 AMC8 真题答案详细解析

1.We?could?go at it by just?multiplying?it out, dividing, etc, but there is a much more simple method.

Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by?$1$, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like[frac{3}{3} times frac{5}{5} times frac{7}{7} times frac{9}{9} times frac{11}{11}]

Notice that each number is still there, and nothing has been changed - other than the order.

Finally, since all of the fractions are equal to one, we have?$1times1times1times1times1$, which is equal to?$1$.

Thus,?$boxed{text{A}}$?is the answer.

However, if you want to multiply it out, then it would be[frac{15}{99} times frac{693}{105}].

That would be[frac{10395}{10395}], which is 1. Therefore, the answer is?$boxed{text{A}}$

2.

Solution 1

One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. We find a simpler problem in this problem, and simplify ->?$90 + 91 + ... + 98 + 99 = 90 times 10 + 1 + 2 + 3 + ... + 8 + 9$

We know?$90 times 10$, that's easy -?$900$. So how do we find?$1 + 2 + ... + 8 + 9$?

We rearrange the numbers to make?$(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5$. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding.?$4 times 10 + 5 = 45$. Adding that on to 900 makes 945.

945 is?$boxed{text{B}}$

Solution 2

Instead of breaking the sum and then rearranging, we can start by rearranging:begin{align*} 90+91+92+cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) &= 189+189+189+189+189 &= 945rightarrow boxed{text{B}} end{align*}

Solution 3

We can use the formula for finite arithmetic sequences.

It is?$frac{n}{2}times$?($a_1+a_n$) where?$n$?is the number of terms in the sequence,?$a_1$?is the first term and?$a_n$?is the last term.

Applying it here:

$frac{10}{2} times (90+99) = 945 rightarrow boxed{B}$

3.We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this:[frac{10^7}{5 times 10^4} = frac{10^3}{5}]

We know that?$10^3 = 10 times 10 times 10$, so

begin{align*} frac{10^3}{5} &= frac{10times 10times 10}{5} &= 2times 10times 10 &= 200 end{align*}

So the answer is?$boxed{text{D}}$

4.

Solution 1

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(6,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we?do?know how to find the area of.

If we continue segment?$overline{FE}$?until it reaches the right side at?$G$, we create two rectangles - one on the top and one on the bottom.

We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of?$ABGF$?is?$6times5 = 30$. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.

Note that?$GC+GB=9$, and?$GB=AF=5$, so we must have[GC+5=9Rightarrow GC=4]

The area of the bottom rectangle is then[(DC)(GC)=4times 4=16]

Finally, we just add the areas of the rectangles together to get?$16 + 30 = 46$.

$boxed{text{C}}$

Solution 2

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(0,0),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Let?$langle ABCDEF rangle$?be the area of polygon?$ABCDEF$. Also, let?$G$?be the intersection of?$DC$?and?$AF$?when both are extended.

Clearly,[langle ABCDEF rangle = langle ABCG rangle - langle GFED rangle]

Since?$AB=6$?and?$BC=9$,?$langle ABCG rangle =6times 9=54$.

To compute the area of?$GFED$, note that[AB=GD+DC][BC=GF+FA]

We know that?$AB=6$,?$DC=4$,?$BC=9$, and?$FA=5$, so[6=GD+4Rightarrow GD=2][9=GF+5Rightarrow GF=4]

Thus?$langle GFED rangle = 4times 2=8$

Finally, we havebegin{align*} langle ABCDEF rangle &= langle ABCG rangle - langle GFED rangle &= 54-8 &= 46 end{align*}

This is answer choice?$boxed{text{C}}$

5.To get the fraction, we need to find the number of people who got grades that are "satisfactory" over the total number of people.

Finding the number of people who got acceptable grades is pretty easy. 5 people got A's, 4 people got B's, 3 people got C's and 3 people got D's. Adding this up, we just have?$5+4+3+3 = 15$.

So we know the top of the fraction is 15. Only 5 people got "unacceptable" scores, so there are?$15 + 5 = 20$?scores.

$frac{15}{20}=frac{3}{4}$?is our fraction, so?$boxed{text{C}}$?is the answer.

6.

Solution 1

We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.

Let's say that?$500text{ sheets}=5text{ cm}Rightarrow frac{500 text{ sheets}}{5 text{ cm}} = 1$. So by multiplying?$7.5 text{ cm}$?by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets

begin{align*} frac{7.5 times 500}{5} &= 7.5 times 100 &= 750 text{ sheets} end{align*}

$750$?is?$boxed{text{D}}$

Solution 2

We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness. Our proportion is[frac{5}{500}=frac{7.5}{x}]where?$x$?is the number we are looking for. Next, we cross-multiply to get?$5x=500 times 7.5$?so?$x=750$?which is?$boxed{text{D}}$

7.

Solution 1

The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - but not if your method is to continue the diagram...

So hopefully there's a pattern. We find a pattern by noticing what is changing from row 1 to row 2. Basically, for the next row, we are just adding 2 squares (1 on each side) to the number of squares we had in the previous row. So each time we're adding 2. So how can we find?$N$, if?$N$?is the number of squares in the?$a^text{th}$?row of this diagram? We can't just say that?$N = 1 + 2a$, because it doesn't work for the first row. But since 1 is the first term, we have to EXCLUDE the first term, meaning that we must subtract 1 from a. Thus,?$N = 1 + 2times(a - 1) = 2a - 1$. So in the 37th row we will have?$2 times 37 - 1 = 74 - 1 = 73$.

You may now be thinking - aha, we're finished. But we're only half finished. We still need to find how many black squares there are in these 73 squares. Well let's see - they alternate white-black-white-black... but we can't divide by two - there aren't exactly?as?many white squares?as?black squares... there's always 1 more white square... aha! If we subtract 1 from the number of squares (1 white square), we will have exactly 2 times the number of black squares.

Thus, the number of black squares is?$frac{73 - 1}{2} = frac{72}{2} = 36$

36 is?$boxed{text{C}}$

Solution 2

Note that each row adds one black and one white square to the end of the previous one, and then shifts the new row over a little. Since the first row had no black squares, the number of black squares in any row is one less than the row number.

Now that this has been established, we just have?$37-1=36rightarrow boxed{text{C}}$

8.Since all the numbers are small, we can just evaluate the set to be[{ (-3)(-2), 4(-2), frac{24}{-2}, (-2)^2, 1 }= { 6, -8, -12, 4, 1 }]

The largest number is?$6$, which corresponds to?$-3a$.

$boxed{text{A}}$

9.First doing the subtraction, we get[frac{1}{2}timesfrac{2}{3}timesfrac{3}{4}timescdotstimesfrac{9}{10}]

We notice a lot of terms cancel. In fact, every term in the numerator except for the?$1$?and every term in the denominator except for the?$10$?will cancel, so the answer is?$frac{1}{10}$, or?$boxed{text{A}}$

If you don't believe this, then rearrange the factors in the denominator to get[frac{1}{10}timesfrac{2}{2}timesfrac{3}{3}timescdotstimesfrac{9}{9}]

Everything except for the first term is?$1$, so the product is?$textbf{(A)}frac{1}{10}$

10.The fraction halfway between?$frac{1}{5}$?and?$frac{1}{3}$?is simply their?average, which isbegin{align*} frac{frac{1}{5}+frac{1}{3}}{2} &= frac{frac{3}{15}+frac{5}{15}}{2} &= frac{frac{8}{15}}{2} &= frac{4}{15} end{align*}

$boxed{text{C}}$

  • For any two fractions that have the property?$frac{a}$?and?$frac{a}{c}$?where a is equal for both fractions;

Set aside the a. Take the average of b and c. For this example, the average of 3 and 5 is 4. Then multiply those numbers. 3*5=15. The number that is exactly in between?$frac{a}$?and?$frac{a}{c}$?will be?$frac{a*((b+c)/2)}{b*c}$.

In this example, it would be?$frac{1((3+5)/2)}{3*5}$, which simplifies to?$frac{4}{15}$.

Therefore, the answer is?$boxed{text{C}}$

11.To find the face opposite?$text{X}$, we can find the faces sharing an edge with?$text{X}$, so the only face remaining will be the opposite face.

Clearly,?$text{V}$?and?$text{Z}$?share an edge with?$text{X}$. Also, the faces?$text{V}$,?$text{X}$, and?$text{W}$?share a common vertex, therefore?$text{X}$?shares an edge with?$text{W}$. Similarly, the faces?$text{U}$,?$text{V}$, and?$text{X}$?share a common vertex, so?$text{X}$?shares an edge with?$text{U}$.

The only face?$text{X}$?doesn't share an edge with is?$text{Y}$, which is choice?$boxed{text{E}}$

12.We are given the three side lengths of the triangle, so we can compute the perimeter of the triangle to be?$6.2+8.3+9.5=24$. The square has the same perimeter as the triangle, so its side length is?$frac{24}{4}=6$. Finally, the area of the square is?$6^2=36$, which is choice?$boxed{text{B}}$

13.$45$?minutes is?$frac{3}{4}$?of an hour, so the walking contributes?$frac{3text{ hr}}{4}times frac{4 text{ miles}}{1text{ hr}}=3text{ miles}$.

Similarly,?$30$?minutes is?$frac{1}{2}$?of an hour, so the running adds?$frac{1text{ hr}}{2}timesfrac{10text{ miles}}{1text{ hr}}=5text{ miles}$.

Their total is?$3text{ miles}+5text{ miles}=8text{ miles}$, which is?$boxed{text{B}}$

 

14.The most straightforward method would be to calculate both prices, and subtract. But there's a better method...

Before we start, it's always good to convert the word problems into?expressions, we can solve.

So we know that the price of the object after a?$6.5%$?increase will be?$20 times 6.5%$, and the price of it after a?$6%$?increase will be?$20 times 6%$. And what we're trying to find is?$6.5% times 20 - 6% times 20$, and if you have at least a little experience in the field of?algebra, you'll notice that both of the items have a common?factor,?$20$, and we can?factor?the expression intobegin{align*} (6.5% - 6% ) times 20 &= (.5% )times 20 &= frac{.5}{100}times 20 &= frac{1}{200}times 20 &= .10 end{align*}

$.10$?is choice?$boxed{text{B}}$

15.

Solution 1

This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.

If you ever learned about?complementary counting, this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?

So let's find the number of numbers. Obviously, we'd start by?subtracting?100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.

So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit.?$2 times 9 times 9 = 162$. However, one of the numbers we counted is?$100$, which isn't allowed, so there are?$162-1=161$?numbers without a 2.

Since there are 299 numbers in total and 161 that DON'T have any 2's,?$299 - 161 = 138$?numbers WILL have at least one two.

$boxed{text{C}}$

Solution using PIE

As?in the previous solution, get rid of the?$400$?to make things easier. This gives us the numbers from?$100$?to?$399$.

Let?$A$?be the event that the first digit is?$2$,?$B$?be the event that the second digit is?$2$, and?$C$?be the event that the third digit is?$2$. Then, PIE says that our answer will be?$|A|+|B|+|C|-|Acup B|-|Acup C|-|Bcup C|+|Acup Bcup C|$.

We have that?$|A|$?is just?$1times10times10=100$,?$|B|=3times1times10=30$, and?$|C|=3times10times1=30$.

Next,?$|Acup B|$?is just having something in the form?$22_$, so there are?$10$?ways. Similarly,?$|Acup C|$?means that we have?$2_2$, so there are again?$10$ways. Finally,?$|Bcup C|$?means that our number is like?$_22$, so there are?$3$?ways.

Finally,?$|Acup Bcup C|$?counts the number of three digit numbers with all three digits?$2$, which there is only?$1$?of:?$222$.

Putting this together, our answer will be?$100+30+30-10-10-3+1=boxed{text{(C)} 138}$.

 

16.Let the number of boys be?$2x$. It follows that the number of girls is?$3x$. These two values add up to?$30$?students, so[2x+3x=5x=30Rightarrow x=6]

The?difference?between the number of girls and the number of boys is?$3x-2x=x$, which is?$6$, so the answer is?$boxed{text{D}}$.

17.

Solution 1

If the average score of the first six is?$84$, then the?sum?of those six scores is?$6times 84=504$.

The average score of the first seven is?$85$, so the sum of the seven is?$7times 85=595$

Taking the difference leaves us with just the seventh score, which is?$595-504=91$, so the answer is?$boxed{text{D}}$

Solution 2

Let's remove the condition that the average of the first seven tests is?$85$, and say the 7th test score was a?$0$. Then, the average of the first seven tests would be[frac{6times 84}{7}=72]

If we increase the seventh test score by?$7$, the average will increase by?$frac{7}{7}=1$. We need the average to increase by?$85-72=13$, so the seventh test score is?$7times 13=91$?more than?$0$, which is clearly?$91$. This is choice?$boxed{text{D}}$

 

18.Let?$p$?be the cost of the first pamphlet, in dollars. The first part tells us[9p<10 Rightarrow p<1.overline{1}]

The second part tells us that[10p>11Rightarrow p>1.1]

Combining these two parts, the only possible value for?$p$?is?$1.11$, since?$p$?represents a monetary value.

$boxed{text{E}}$

Also, just by quickly going over the answers, it is seen that all of the answers satisfy the first statement: The cost of 9 pamphlet is less that 10 dollars.

The second statement states that 10 pamphlets cost more than 11 dollars. Therefore, since there are only two statements and the first one can be satisfied with all of the answer choices, the answer choice that is the most expensive of them all will be the right answer choice, thus the answer is?$boxed{text{E}}$

 

19.

Solution #1

Let the width be?$w$?and the length be?$l$. Then, the original perimeter is?$2(w+l)$.

After the increase, the new width and new length are?$1.1w$?and?$1.1l$, so the new perimeter is?$2(1.1w+1.1l)=2.2(w+l)$.

Therefore, the percent change isbegin{align*} frac{2.2(w+l)-2(w+l)}{2(w+l)} &= frac{.2(w+l)}{2(w+l)} &= frac{.2}{2} &= 10% end{align*}

$boxed{text{B}}$

Solution #2 (Quick Fakesolve)

Assume WLOG that the rectangle is a square with length?$10$?and width?$10$. Thus, the square has a perimeter of?$40$. Increasing the length and width by?$10%$?will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is?$44$, and because?$44$?is?$110%$?of?$40$, our answer is?$boxed{text{(B) } 10%}$.

20.January has four full weeks and then three extra consecutive days. Each full week contributes one Tuesday and one Saturday, so the three extra days do not contain a Tuesday and Saturday. Therefore, those three days are Wednesday, Thursday, and Friday.

Wednesday is the 29th day of January, therefore the 22nd, 15th, 8th, and 1st of January are all Wednesdays, so the answer is?$boxed{text{C}}$

21.Assume his salary is?$100$?dollars. Then in the next year, he would have?$110$?dollars, and in the next he would have?$121$?dollars. The next year he would have?$133.1$?dollars and in the final year, he would have?$146.41$.?As?the total increase is greater than?$45%$, the answer is?$boxed{text{E}}$

Note that we could have generalized this for any integer?$n$.

22.An equivalent problem is finding the?probability?that a randomly selected telephone number begins with?$9$?and ends with?$0$.

There are?$10-2=8$?possibilities for the first digit in total, and only?$1$?that works, so the probability the number begins with?$9$?is?$frac{1}{8}$

There are?$10$?possibilities for the last digit, and only?$1$?that works?$(0)$, so the probability the number ends with?$0$?is?$frac{1}{10}$

Since these are?independent events, the probability both happens is[frac{1}{8}cdot frac{1}{10}=frac{1}{80}]

$boxed{text{B}}$

23.If each student has?$5$?classes, and there are?$1200$?students, then they have a total of?$5times 1200=6000$?classes among them.

Each class has?$30$?students, so there must be?$frac{6000}{30}=200$?classes. Each class has?$1$?teachers, so the teachers have a total of?$200$?classes among them.

Each teacher teaches?$4$?classes, so if there are?$t$?teachers, they have?$4t$?classes among them. This was found to be?$200$, so[4t=200Rightarrow t=50]

This is answer choice?$boxed{text{E}}$

24.Let the number in the top circle be?$a$?and then?$b$,?$c$,?$d$,?$e$, and?$f$, going in clockwise order. Then, we have[S=a+b+c][S=c+d+e][S=e+f+a]

Adding these?equations?together, we get

begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) &= 75+(a+c+e) end{align*}

where the last step comes from the fact that since?$a$,?$b$,?$c$,?$d$,?$e$, and?$f$?are the numbers?$10-15$?in some order, their?sum?is?$10+11+12+13+14+15=75$

The left hand side is?divisible?by?$3$?and?$75$?is divisible by?$3$, so?$a+c+e$?must be divisible by?$3$. The largest possible value of?$a+c+e$?is then?$15+14+13=42$, and the corresponding value of?$S$?is?$frac{75+42}{3}=39$, which is choice?$boxed{text{D}}$.

It turns out this sum is attainable if you let[a=15][b=10][c=14][d=12][e=13][f=11]

25.

Solution 1

Logically, Jane's statement is equivalent to its?contrapositive,

If an even number is not on one side of any card, then a vowel is not on the other side.

For Mary to show Jane wrong, she must find a card with an?odd number?on one side, and a vowel on the other side. The only card that could possibly have this property is the card with?$3$, which is answer choice?$boxed{text{A}}$

Solution 2

Using the answer choices, we see that P and Q are logically equivalent (both are non vowel letters) and so are?$4$?and?$6$?(both are even numbers). Thus, if one of these are correct, then the other option in its pair must also be correct. The only remaining answer choice is?$boxed{text{A}}$

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