Problem 1

Answer: D

Solution:

Walter has 1+5+10+25=41 cents in his pocket. There are 100 cents in a dollar. Therefore, Walter has 41/100=41% of a dollar in his pocket.

Problem 2

Answer: C

Solution:

Zack is 3 years older than Inez, who is 15. Therefore, Zack is 15+3=18 years old.
Jose is 4 years younger than Zack, who is 18. Therefore, Jose is 18-4=14 years old, and the answer is C.

Problem 3

Answer: E

Solution:

Dividing by 3/5 is the same as multiplying by its reciprocal, 5/3. Multiplying a number by two numbers is the same as multiplying the number by the product of those two numbers. Thus, multiplying by 3/4 and then 5/3 is the same as multiplying by 3/4*5/3=5/4. , and the answer is E.

Problem 4

Answer: C

Solution:

Ben adds 1 to 6, getting 7. Then Ben doubles it, to get 14. He gives this number to Sue. Then Sue subtracts 1 from the number, to get 13. Finally, Sue doubles the result to get 26, and the answer is C.

Problem 5

Answer: C

Solution:

Adding the whole numbers gives 2+3+4+5=14. Adding the fractions gives 1/2+1/3+1/4+1/5=30/60+20/60+15/60+12/60=77/60. This will create one more whole, and a fraction that is less than 1. Thus, the smallest whole number that is less than 15 plus some fractional part is 16, and the answer is C.

Problem 6

Answer: C

Solution:

Since the perimeter of I is 12, each side is 3.
Since the perimeter of II is 24, each side is 6.
The side of III is equal to the sum of the sides of I and II. Therefore, the side of III is 3+6=9.
Since III is also a square, it has an perimeter of 9*4=36, and the answer is C.

Problem 7

Answer: B

Solution:

The students who walk home are: 1-1/2-1/4-1/10=3/20.

Problem 8

Answer: D

Solution:

3000 lire/ 1.6dollars= x lire/ 1 dollar =>x=1875.

Problem 9

Answer: C

Solution:

If circle Q has diameter 4, then so do congruent circles P and R. Draw a diameter through P parallel to AD. The diameter will be congruent to AD, and thus AD=4, which is the height of the rectangle.
Draw a horizontal line PQR that extends to the sides of the rectangle. This line is 2 diameters long, so it has length 8. It is parallel and congruent to AB, so the width of the rectangle is 8.
Thus, the area of the rectangle is 4*8=32.

Problem 10

Answer: A

Solution:

Chris would have spent 40+80=120 dollars if there were no sale. The sale means that he saved 80*0.4+40*0.55=54 dollars. He saved54/120=45% off the total of the original prices.

Problem 11

Answer: D

Solution:

Counting around, when Jane walks 12 steps, she will be at D. When Hector walks 6 steps, he will also be at D. Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is D.

Problem 12

Answer: E

Solution:

We examine only the factors of 90,91,92, 93, and 94 that are less than 13, because for a year to be lucky, it must have at least one factor between 1 and 12 to represent the month. 90 =9*10; 91=7*13; 92=4*23; 93=3*31; 94 has factors of 1 and 2. Dividing 94 by 2 gives 47, and no month has 47 days. Thus, 1994 is not a lucky year.

Problem 13

Answer: E

Solution:

Because <BED=<BDE, <B=90, we get<BED=<BDE=45. <AED=<AEB+<BED=40+45=85. Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: <A+<C+<AED+<CDE+360, =>90+90+85+<CDE=360, => <CDE=95.

Problem 14

Answer: B

Solution:

we can note that they will play a total of 50+40=90 games and must win 0.7*90=63 games. Since they won 40 games already they need 63-40=23 more games.

Problem 15

Answer: B

Solution:

4/37=0.108108108… Since this repeats every three digits, the 100th digit is 1.

Problem 16

Answer: C

Solution:

Altogether, the summer project totaled 7*3+4*5+5*9=86 days of work for a single student. This equals744/86=9 dollars per day per student. The students from Balboa school earned 9*4*5=180 dollars.

Problem 17

Answer: D

Solution:

By the tables, Annville has 11 6th graders and Cleona has 34. Together they have 45 6th graders and 300 total students, so the percent is 45/300=15%.

Problem 18

Answer: C

Solution:

The area taken up by the L's is 4*(3/16)=3/4 of the area of the whole square. What remains has 1/4 of the area of the larger square. 1/4* 100*100=50*50 is the area of the smaller square, so its side length is 50.

Problem 19

Answer: D

Solution:

Counting, there are thirteen total families. The middle number is 7th in either direction, and it is easy to see from the right side that this number is 4 .

Problem 20

Answer: B

Solution:

Note that the probability of Diana rolling a number larger than Apollo's is the same as the probability of Apollo's being more than Diana's. If we denote this common probability D, then 2D+P(Apollo=Diana)=1. Now all we need to do is find P(Apollo=Diana). There are 6*6=36 possibilities total, and 6 of those have Apollo=Diana, so P(Apollo=Diana)=6/36=1/6. Going back to our first equation and solving for D, we get 2D+1/6=1 =>D=5/12.
Method 2:
if Apollo’s number is 1, it is not possible large than Diana’s number;
if Apollo’s is 2, and Dianna’s is 1 then Apollo’s number is larger than Diana’s. the probability is 1/6*1/6=1/36;
if Apollo’s is 3, and Dianna’s is 1 or 2, then Apollo’s number is larger than Diana’s. the probability is 1/6*2/6=2/36;
if Apollo’s is 4, and Dianna’s is 1, 2 or 3, then Apollo’s number is larger than Diana’s. the probability is 1/6*3/6=3/36;
if Apollo’s is 5, and Dianna’s is 1, 2 3,or 4, then Apollo’s number is larger than Diana’s. the probability is 1/6*4/6=4/36;
if Apollo’s is 6, and Dianna’s is 1, 2 3,4,or 5, Apollo’s number is larger than Diana’s. the probability is 1/6*5/6=5/36;
Totally, the probability is 1/36+2/36+3/36+4/36+5/36=15/36=5/12.

Problem 21

Answer: B

Solution:

It is rather easy to see that the only way that you can possibly have three cubes without protruding snaps is to have them in a triangle formation. However, in this case, the exterior angles are 120, whereas the outside angle of the cubes are 90 each. A square like formation will work in the case of 4 cubes, so the answer is B.

Problem 22

Answer: A

Solution:

The prime factorization of 6545 is 5*7*11*17, so every two-digit number pair has to be two number of the form pq. Now we do trial and error: 5*7=35, 11*17=185; (3 digit) ; 5*11=55, 7*17=119(3 digit) ; 5*17=85, 7*11=77(correct) , so 85+77=162.

Problem 23

Answer: B

Solution:

Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are 5*5*8*7=1400 numbers.

Problem 24

Answer: C

Solution:

Note that DE*AB=DF*BC=area(ABCD).
DE=6; AB=DC=12; so Area(ABCD)=6*12=72;
Since <DEA is a right angle, we can use the pythagorean theorem:AD=sqrt(AE*AE+DE*DE)=sqrt((AB-BE)^2+6*6)=sqrt((12-4)^2+6*6))=10;
BC=AD=10;
Now we can finally substitute: area(ABCD)=72=BC*DF=10*DF;=>DF=7.2.

Problem 25

Answer: D

Solution:

Say you are on the Houston-bound bus that left at 12:30, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away from Dallas, Your bus is also 15 minutes from Houston. At 1:15 pm, the 9:00 am Dallas bus meets you again, being 45 minutes from Houston. you might notice that the buses meet you in 30 minute intervals after 12:45 pm; They are 12:45, 1:15, 1:45, 2:15, 2:45, 3:15, 3:45, 4:15, 4:45, 5:15, and 5:30, your arrival time.

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