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AMC 10 Problem Solutions

1. Problem: Simple Algebra - Solutions

What is the value of \( x \) if \( 3x + 7 = 19 \)?

Solution:

\[ 3x + 7 = 19 \quad \Rightarrow \quad 3x = 19 - 7 = 12 \quad \Rightarrow \quad x = \frac{12}{3} = 4 \]

Thus, \( x = \boxed{4} \).


2. Problem: Geometry of a Triangle - Solutions

The perimeter of an equilateral triangle is \( 24 \). What is the length of one side?

Solution: The perimeter of an equilateral triangle is \( 3s \), where \( s \) is the side length. Given that the perimeter is \( 24 \):

\[ 3s = 24 \quad \Rightarrow \quad s = \frac{24}{3} = 8 \]

Thus, the length of one side is \( \boxed{8} \).


3. Problem: Percentage Increase - Solutions

A store originally sells a product for \( 40 \) dollars. The price is increased by \( 20\% \). What is the new price?

Solution: The increase is \( 20\% \) of \( 40 \), which is:

\[ 0.20 \times 40 = 8 \]

Thus, the new price is: [ 40 + 8 = 48 ]

The new price is \( \boxed{48} \) dollars.


4. Problem: Factors - Solutions

How many positive divisors does \( 60 \) have?

Solution: The prime factorization of \( 60 \) is:

\[ 60 = 2^2 \times 3 \times 5 \]

The number of divisors is given by \( (2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12 \). Thus, \( 60 \) has \( \boxed{12} \) divisors.


5. Problem: Time Problem - Solutions

A clock shows \( 4:30 \). What is the measure of the smaller angle between the hour and minute hands?

Solution: The minute hand at 30 minutes is \( 180^\circ \) from the 12 o'clock position. The hour hand at 4:30 is halfway between the 4 and 5 positions. Since each hour represents \( 30^\circ \), the hour hand is \( 4 \times 30 + 15 = 135^\circ \) from the 12 o'clock position.

The difference between the hands is:

\[ 180^\circ - 135^\circ = 45^\circ \]

Thus, the smaller angle is \( \boxed{45^\circ} \).


6. Problem: Consecutive Integers - Solutions

The sum of three consecutive integers is \( 33 \). What are the integers?

Solution: Let the integers be \( x-1 \), \( x \), and \( x+1 \). Their sum is:

\[ (x-1) + x + (x+1) = 3x = 33 \quad \Rightarrow \quad x = \frac{33}{3} = 11 \]

Thus, the integers are \( 10, 11, 12 \), and the answer is \( \boxed{10, 11, 12} \).


7. Problem: Arithmetic Sequence - Solutions

In an arithmetic sequence, the first term is \( 5 \), and the common difference is \( 3 \). What is the sum of the first 20 terms?

Solution: The sum of the first \( n \) terms of an arithmetic sequence is:

\[ S_n = \frac{n}{2} \times (a_1 + a_n) \]

The 20th term is \( a_{20} = a_1 + 19d = 5 + 19 \times 3 = 62 \). Thus, the sum is:

\[ S_{20} = \frac{20}{2} \times (5 + 62) = 10 \times 67 = 670 \]

Thus, the sum of the first 20 terms is \( \boxed{670} \).


8. Problem: Probability - Solutions

Two dice are rolled. What is the probability that the sum of the numbers rolled is \( 8 \)?

Solution: There are \( 6 \times 6 = 36 \) possible outcomes. The combinations that sum to \( 8 \) are:

\[ (2,6), (3,5), (4,4), (5,3), (6,2) \]

There are 5 favorable outcomes, so the probability is:

\[ \frac{5}{36} \]

Thus, the probability is \( \boxed{\frac{5}{36}} \).


9. Problem: Quadratic Equation - Solutions

Solve the equation \( x^2 - 5x + 6 = 0 \).

Solution: Factor the quadratic:

\[ x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \]

Thus, \( x = 2 \) or \( x = 3 \). The solutions are \( \boxed{2} \) and \( \boxed{3} \).


10. Problem: Right Triangle - Solutions

In a right triangle, the legs are \( 6 \) and \( 8 \). What is the length of the hypotenuse?

Solution: Using the Pythagorean theorem:

\[ c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

Thus, the hypotenuse is \( \boxed{10} \).


11. Problem: Circles and Geometry - Solutions

Two concentric circles have radii of \( 3 \) and \( 5 \). What is the area of the ring-shaped region between the two circles?

Solution: The area of the ring is the difference between the areas of the larger and smaller circles:

[ \text{Area} = \pi \times 5^2 - \p i \times 3^2 = 25\pi - 9\pi = 16\pi ] Thus, the area of the ring is \( \boxed{16\pi} \).


12. Problem: Sum of Angles in a Polygon - Solutions

What is the sum of the interior angles of a pentagon?

Solution: The sum of the interior angles of a polygon with \( n \) sides is given by:

\[ \text{Sum of angles} = 180(n-2) \]

For a pentagon (\( n = 5 \)):

\[ \text{Sum of angles} = 180(5-2) = 180 \times 3 = 540 \]

Thus, the sum of the interior angles is \( \boxed{540^\circ} \).