AMC8 Test 1

Question 1

How many subgroups of two elements may be eliminated from the set 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 so that the remaining numbers have a mean (average) of 6?

Accepted Answer

5

Answer

The correct answer:
1st option

We know their sum is 54 because there would be 9 items after elimination and their mean is 6. We also know that the pre-removal total of the collection is 66. As a result, the sum of the two eliminated items is 66 – 54 = 12. There are only five subsets of two elements that add up to 12: {1, 11}, {2, 10}, {3, 9}, {4, 8}, {5, 7}.

2nd Option

We can easily remove five subsets of two numbers, leaving only six. The answer is 5, because the average of this one-number collection is still 6.

Question 2

How many square yards of carpet are needed to cover a 12 foot long by 9 foot wide rectangular floor? (A yard is three feet long.)

Accepted Answer

12

Answer

The correct answer:
Solution

We begin by multiplying 12 by 9. To do so, you'll need to cover 108 square feet of carpet. Because a square yard is 9 square feet, you divide 108 by 9 to get 12 square yards, which is our answer.

2nd Option

Because a yard has three feet, we divide 9 by 3 to get 3, and 12 by 3 to get 4. We multiply these two figures together to get 12 to find the carpet's area.

Question 3

It's a two-digit number called N.

The remainder after dividing N by 9 is one.
The remainder after dividing N by ten is three.
When you divide N by 11, what is the result?

Accepted Answer

7

Answer

The correct answer:
We know the second digit to be 3 because of the second bullet point. Because dividing by 9 leaves a leftover of 1, the multiple of 9 must finish in a 2. This is the one we're looking for now:

9(1) = 9

9(2) = 18

9(3) = 27

9(4) = 36

9(5) = 45

9(6) = 54

9(7) = 63

9(8) = 72

Both conditions are satisfied by the number 72 + 1 = 73. To calculate the remainder, we subtract the largest multiple of 11 less than 73. As a result, 73 + 11(6) = 73 + 66 = 7.

Question 4

Curt's father takes him to school in 20 minutes during rush hour traffic. There is no traffic one day, so his father can drive him to school in 12 minutes, 18 miles per hour faster. How far is it to school in miles?

Accepted Answer

9

Answer

The correct answer:
Solution 1

We get d/v = 1/3 and d/v + 18 = 1/5

This equals d = 1/5v + 3.6 = 1/3v, which equals v = 27, which equals d = 9.

Solution 2

d = rt, d = 1/3 x r = 1/5 x (r + 18)

r/3 = r/5 + 18/5

2r/15 = 18/5

10r = 270 so r = 27, plug into the first one and it’s 9 miles to school

Solution 3

We devised a formula involving d, the distance, and x, the speed in miles per hour. We get d = (1/3)(x) = (1/5)(x + 18) as a result.

(5)(x) = (3)(x + 18)

5x = 3x + 54

2x = 54

x = 27

Hence, d = 27/3 = 9.

Solution 4

Because it takes him 3/5 of the time to get to school in no traffic, the speed must be 5/3 of the speed in traffic, or 2/3 more. Using x as the distance he can drive in an hour with traffic, we get 2x/3 = 18 miles per hour. When we solve for x, we get 27 miles per hour. The distance would be 9 miles because 20 minutes is a third of an hour.

Question 5

What is the total of 2016's distinct prime integer divisors?

Accepted Answer

12

Answer

The correct answer:
Solution 1

2016 = 25 x 32 x 7 is the prime factorization. We have 2, 3, and 7 because the problem simply asks for different prime factors. As a result, their intended total is 12.

2nd Option

Because 2 + 0 + 1 + 6 = 9, and 9|9, we observe that 9|2016. We can reach 224 by multiplying 2016 by 9. As 4|24, this is divisible by four. We get 56 when we divide 224 by four. This is plainly divisible by seven, resulting in the number eight. We have 2016 = 9 4 7 8 as our number. 4 and 8 are both multiples of 2, 9 is 32, and 7 is prime, as we know. This suggests that the prime factors 2, 3, and 7 are distinct. Their total is twelve.

AMC 8 - American Mathematics Competition 8 Practice Test

AMC 8 - American Mathematics Competition 8 Practice Test

AMC8 Test 1