 # AMC 10 - Ivy League Education Center · PDF file AMC 10 Mock Test Detailed Solutions Problem 1 Answer: (E) Solution 1 Note that there is more than 1 four-legged table. So there are

Jun 15, 2020

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• MOCK EXAMINATION

AMC 10 American Mathematics Contest 10

Test Sample

Detailed Solutions

Make time to take the practice test.

It’s one of the best ways to get ready for the AMC.

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AMC 10 Mock Test

Detailed Solutions

Problem 1

Solution 1

Note that there is more than 1 four-legged table. So there are at least 2 four-legged tables. Since

there are 23 legs in total, there must be fewer than 6 four-legged tables, which have 6 × 4 = 24 legs. Thus, there are between 2 and 5 four-legged tables.

If there are 2 four-legged tables, then these tables account for 2 × 4 = 8 legs, leaving 23 − 8 = 15 legs for the three-legged tables, which implies that there are 5 three-legged tables.

(We can check that if there are 3 or 4 four-legged tables, then the number of remaining legs is

not divisible by 3, and if there are 5 four-legged tables, then there is only 1 three-legged table,

which is not allowed.)

Solution 2

Since there is more than 1 table of each type, then there are at least 2 three-legged tables and

2 four-legged tables. These tables account for 2(3) + 2(4) = 14 legs.

There are 23 − 14 = 9

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more legs that need to be accounted for. These must come from a combination of three-legged

and four-legged tables. The only way to make 9 from 3s and 4s is to use three 3s.

Therefore, there are 2 + 3 = 5 three-legged tables and 2 four-legged tables.

Problem 2

Traveling at a constant speed of 10 miles per hour, in 4 hours the bicycle will travel 9 × 4 = 36 miles.

At the start, the bicycle was 200 miles ahead of the car.

Therefore, in order to catch up to the bicycle, the car must travel 200 miles plus the additional

36 miles that the bicycle travels, or 200 + 36 = 236 miles.

To do this in 4 hours, the car must travel at an average speed of 2364 = 59 miles per hour.

Problem 3

Solution 1

The total area of the rectangle is 3 × 4 = 12. The unshaded region is a triangle with base 1 and height 4, and its area is

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1 × 42 = 2. The total area of the shaded regions equals the total area of the rectangle minus the area of the

unshaded region: 12 − 2 = 10. Solution 2

The shaded triangle on the left has base of length 2 and height of length 4, so has an area of 2 × 42 = 4. The shaded triangle on the right has base of length 3 (at the top) and height of length 4, so has an

area of 3 × 42 = 6. Therefore, the total area of the shaded regions is 4 + 6 = 10.

Problem 4

Note that 𝑎 ∙ 𝑏 + 𝑏 ∙ 𝑐 + 𝑑 ∙ 𝑏 = 𝑏(𝑎 + 𝑐 + 𝑑). Thus, to make 𝑏(𝑎 + 𝑐 + 𝑑) as large as possible, we choose 𝑏 = 5, the largest value possible. Hence, the maximum possible value of 𝑏(𝑎 + 𝑐 + 𝑑) is 5(2 + 3 + 4) = 45.

Problem 5

The ratio of boys to girls at Einstein High School is 3 : 2, so 33 + 2 = 35

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of the students are boys. Thus, there are 35 × 400 = 240 boys at Einstein H.S.

The ratio of boys to girls at Edison High School is 2 : 3, so 23 + 2 = 25 of the students are boys. Thus, there are 25 × 600 = 240 boys at Edison High School.

There are 400 + 600 = 1000 students in total at the two schools.

Of these, 240 + 240 = 480 are boys, and so the remaining 1000 − 480 = 520 students are girls.

Therefore, the overall ratio of boys to girls is 480 ∶ 520 = 48 ∶ 52 = 12 ∶ 13.

Problem 6

Solution 1

Since the sale price has been reduced by 20%, then the sale price of \$112 is 80% or 45 of the

regular price. Thus, the regular price is \$112 ÷ 45 = \$140.

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If the regular price is reduced by 30%, the new sale price would be 70% of the regular price, or 70% × \$140 = \$98.

Solution 2

Let 𝑥 be the original price, in dollars. Since the price has been reduced by 20%, the sale price is 80% of the original price. So 80%𝑥 = 112. Solving for 𝑥 gives: 𝑥 = 140. If it were on sale for 30% off, then the price would be 70% of the original price: 70%𝑥 = 70% × 140 = 98.

Problem 7

Note that 1 ≤ 𝑥 ≤ 7. If 𝑥 = 7, then 𝑦2 = 1, so 𝑦 = 1. If 𝑥 = 6, then 𝑦2 = 14, which is not possible since 𝑦 is an integer. If 𝑥 = 5, then 𝑦2 = 25, so 𝑦 = 5. If 𝑥 = 4, then 𝑦2 = 34, which is not possible. If 𝑥 = 3, then 𝑦2 = 41, which is not possible. If 𝑥 = 2, then 𝑦2 = 46, which is not possible. If 𝑥 = 1, then 𝑦2 = 49, so 𝑦 = 7. Therefore, there are 3 pairs (𝑥, 𝑦) that satisfy the equation, namely (7, 1), (5, 5), (1, 7).

Problem 8

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Note that the sum of two even numbers is even, the sum of two odd numbers is even, and the

sum of an odd number and an even number is odd. Thus, for the sum 𝑎 + 𝑏 to be even, both 𝑎 and 𝑏 must be even, or they must both be odd. If both 𝑎 and 𝑏 must be even, then the number of different ordered pairs is equal to the number of combinations of two elements out of the set {2, 4, 6, ⋯ , 30}, which is (152 ) = 105. If both 𝑎 and 𝑏 must be odd, then the number of different ordered pairs is equal to the number of combinations of two elements out of the set {1, 3, 5, ⋯ , 29}, which is (152 ) = 105. So the total number of different ordered pairs (𝑎, 𝑏) is equal to 105 + 105 = 210.

Problem 9

Solution 1

Let 𝑥1, 𝑥2, 𝑥3, ⋯ , 𝑥2017, 𝑥2018 be the sequence. Then 𝑥1 + 𝑥2 + 𝑥3 + ⋯ + 𝑥2018 = 5311. Let 𝑆 and 𝑇 denote, respectively, the sums of the odd-numbered terms and even-numbered terms in the sequence. That is, 𝑆 = 𝑥1 + 𝑥3 + 𝑥5 + ⋯ , 𝑥2017, 𝑇 = 𝑥2 + 𝑥4 + 𝑥6 + ⋯ , 𝑥2018. Then 𝑆 + 𝑇 = 5311. Since 𝑥2 − 𝑥1 = 𝑥4 − 𝑥3 = ⋯ = 𝑥2018 − 𝑥2017 = 1, we have:

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𝑇 − 𝑆 = 1009. Solving the above two equation simultaneously gives: 𝑆 = 2151. Hence, the required sum is 2151.

Solution 2

Let 𝑎 be the first term of the sequence, and 𝑋 be the sum of all odd-numbered terms. Note that the sequence is arithmetic with common difference 𝑑 = 1. The sum of the terms of the sequence is 𝑎 + (𝑎 + 2017)2 ∙ 2018 = 5311. Thus, 𝑎 + (𝑎 + 2017)2 ∙ 1009 = 53112 . Since all odd-numbered terms consists of a 1009-term arithmetic sequence with the first term 𝑎 and common difference 𝑑 = 2, the sum of all odd-numbered terms is: 𝑋 = 𝑎 + (𝑎 + 1008 ∙ 2)2 ∙ 1009 = 𝑎 + (𝑎 + 2016)2 ∙ 1009 = 𝑎 + (𝑎 + 2017)2 ∙ 1009 − 10092 = 53112 − 10092 = 2051.

Problem 10

Since Alex needs 12 hours to shovel all of his snow, he shovels 112 of his snow per hour.

Since Bob needs 8 hours to shovel all of Alex's snow, he shovels 18 of Alex's snow per hour.

Similarly, Carl shovels 16 of Alex's snow every hour, and Dick shovels 14 of Alex's snow per hour.

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Together, Alex, Bob, Carl, and Dick can shovel 112 + 18 + 16 + 14 = 1524 of Alex's snow per hour.

Therefore, together they can shovel 152460 =

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