Problem 1

Answer: D

Solution:

Clearly the arrow marks a value between 10.25 and 10.5, so only C and D are possible.
Looking, we see that the arrow is closer to 10.3.

Problem 2

Answer: C

Solution:

Converting the decimals to fractions, this is 8* 1/4 * 2 * 1/8=1/2.

Problem 3

Answer: D

Solution:

Each of the fractions simplify to 1/10, so this sum is: 1/10+1/10+1/10=3/10=0.3.

Problem 4

Answer: B

Solution:

If, for a moment, we disregard the white squares, we notice that the number of black squares in each row increases by 1 continuously as we go down the pyramid. Thus, the number of black squares is 1+2+…+8.
Same goes for the white squares, except it starts a row later, making it 1+2+…+7.
Subtracting the number of white squares from the number of black squares, (1+2+3+…+8)-(1+2+3+…+7)=8.

Problem 5

Answer: C

Solution:

We have that 20 +<ABC+<CBD=160, since <CBD =90, we have<ABC=140-90=50.,

Problem 6

Answer: E

Solution:

=0.008/0.0004=80/4=20.

Problem 7

Answer: B

Solution:

=2.5*8.1*10=200

Problem 8

Answer: B

Solution:

The decimal point of 0.075 is three away from what Betty punched in, and that of 2.56 is two away. The decimal point is therefore 3+2=5 units to the left of where it should be, so we would want 0.19200

Problem 9

Answer: D

Solution:

The first triangle has two legs of length sqrt(5), the second has two legs of length 2, the leg lengths of the third triangle are 2, sqrt(5), sqrt(13), two legs of the fourth triangle have length sqrt(10), and two legs of the fifth triangle have length sqrt(5). Therefore all of the triangles in the diagram except the third are isosceles, and there are 4 are isosceles.

Problem 10

Answer: A

Solution:

Note that the remainder of 60/7 is 4. We count 4 days past Thursday, and arrive at Monday.

Problem 11

Answer: E

Solution:

Note that if 1<=a<b<c, then sqrt(a)<sqrt(b)<sqrt(c). since 1<144<164<169, we can say that 12<sqrt(164)<13.

Problem 12

Answer: C

Solution:

We want the cost per person, which is 20,000,000,000/250,000,000=20,000/250=80.

Problem 13

Answer: D

Solution:

The circumference of the patio is 2*Pi*12=24*Pi=75.4. Since the bushes are spaced 1 foot apart, about 75 are needed.

Problem 14

Answer: E

Solution:

36=6*6; 6+6=12; 36=4*9; 4+9=13; 26=2*18; 2+18=20; 36=1*36; 1+36=37.

Problem 15

Answer: C

Solution:

1/2+1/3=3/6+2/6=5/6; the reciprocal of 5/6 is 6/5.

Problem 16

Answer: E

Solution:

We can put in 6 by leaving out the three boxes in one of the main diagonals.

Problem 17

Answer: B

Solution:

Looking at the diagram, the shaded region is the union of two rectangles, with a small rectangle as overlap. If we just add the areas of the two rectangles, then we'll overcount the small rectangle, so we must subtract that area to get the desired area.
The areas of the two larger rectangles are 2*10=20 , 3*8=24, the area of the small rectangle is 2*3=6. The desired area is thus 20+24-6=38.

Problem 18

Answer: C

Solution:

(6*150+4*120)/10=138

Problem 19

Answer: A

Solution:

To get from the 1st term of an arithmetic sequence to the 100th term, we must add the common difference 99 times. The first term is 1 and the common difference is 5-1=9-5=13-9=…=4, so the 100th term is 1+4* 99=397.

Problem 20

Answer: C

Solution:

Let the amount of coffee the maker will hold when full be x. then 36%*x=45 => x=125.

Problem 21

Answer: C

Solution:

The possible medians after n is added are 6, n, or 9.
Case 1: the median is 6 (n<6): (3+n+6+9+10)/5=6 =>n=2;
Case 2: the median is n(6<n<9): (3+6+n+9+10)/5=n => n=7;

Problem 22

Answer: E

Solution:

Let the original price of an item be x. the sale price now is 1.25*x*0.8=x, which is the same as the original price.

Problem 23

Answer: D

Solution:

For every disks she sells, she profits 5/4-5/3=5/12. She needs to profit $100, so she needs to sell 100/(5/12)=100*12/5=240 disks.

Problem 24

Answer: A

Solution:

Alternately, we can simply keep track of the "bottom" side of the square. In the diagrams below, this bottom side is shown in red. (left)

Problem 25

Answer: A

Solution:

From 1 to 9, the times will be of the form a:ba. There are 9 choices for a and 6 choices for b, so there are 9*6=54 times in this period.
From 10 to 12, the minutes are already determined, so there are only 3 times in this case.
In total, there are 54+3=57 palindromic times.

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