AMC 8 Problem Solutions¶

1. Problem: Age Puzzle - Solution¶

Corinne’s age is \(3\) times as old as her brother, Zander. In \(4\) years, Corinne’s age will be twice Zander’s age. How old is Corinne now?

Solution: Let \( m \) be Corinne’s current age and \( b \) be her Zander's current age. We are given two pieces of information:

\( m = 3b \) (Corinne is 3 times as old as her brother).
\( m + 4 = 2(b + 4) \) (In 4 years, Corinne’s age will be twice her brother’s).

Substitute \( m = 3b \) into the second equation:

\[ 3b + 4 = 2(b + 4) \]

Expand and simplify:

\[ 3b + 4 = 2b + 8 \]

\[ 3b - 2b = 8 - 4 \]

\[ b = 4 \]

Since \( m = 3b \), Corinne’s current age is:

\[ m = 3 \times 4 = 12 \]

Thus, Corinne is \( \boxed{12} \) years old.

2. Problem: Sum of Angles - Solution¶

A triangle has angles that measure \( 40^\circ \), \( 60^\circ \), and \( x^\circ \). What is the value of \( x \)?

Solution: The sum of the interior angles of any triangle is always \( 180^\circ \). So we can set up the equation:

\[ 40^\circ + 60^\circ + x^\circ = 180^\circ \]

Solving for \( x \):

\[ 100^\circ + x^\circ = 180^\circ \]

\[ x^\circ = 180^\circ - 100^\circ = 80^\circ \]

Thus, \( x = \boxed{80^\circ} \).

3. Problem: Number of Rectangles - Solution¶

How many rectangles are in a grid that is \( 3 \times 2 \) (3 rows and 2 columns)?

Solution: To count the number of rectangles, note that a rectangle is determined by selecting two horizontal lines and two vertical lines. In a \( 3 \times 2 \) grid, there are \( 4 \) horizontal lines (top and bottom of each row) and \( 3 \) vertical lines (left and right of each column).

The number of ways to choose two horizontal lines from the \( 4 \) available is:

\[ \binom{4}{2} = \frac{4 \times 3}{2} = 6 \]

The number of ways to choose two vertical lines from the \( 3 \) available is:

\[ \binom{3}{2} = \frac{3 \times 2}{2} = 3 \]

Thus, the total number of rectangles is:

\[ 6 \times 3 = 18 \]

So, there are \( \boxed{18} \) rectangles in the grid.

4. Problem: Divisibility - Solution¶

What is the smallest integer divisible by \( 5 \), \( 6 \), and \( 7 \)?

Solution: To find the smallest integer divisible by \( 5 \), \( 6 \), and \( 7 \), we need to find the least common multiple (LCM) of these numbers.

First, find the prime factorizations of the numbers:

\( 5 \) is prime.
\( 6 = 2 \times 3 \)
\( 7 \) is prime.

The LCM is the product of the highest powers of all prime factors:

\[ \text{LCM}(5, 6, 7) = 2^1 \times 3^1 \times 5^1 \times 7^1 = 210 \]

Thus, the smallest integer divisible by \( 5 \), \( 6 \), and \( 7 \) is \( \boxed{210} \).

5. Problem: Probability with Dice - Solution¶

Two six-sided dice are rolled. What is the probability that the sum of the numbers rolled is \( 7 \)?

Solution: There are a total of \( 6 \times 6 = 36 \) possible outcomes when rolling two dice (since each die has 6 faces). Now, list the outcomes where the sum of the dice is \( 7 \):

\[ (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \]

There are \( 6 \) outcomes where the sum is \( 7 \). Therefore, the probability is:

\[ \frac{6}{36} = \frac{1}{6} \]

So, the probability that the sum is \( 7 \) is \( \boxed{\frac{1}{6}} \).

6. Problem: Arithmetic Sequence - Solution¶

The first term of an arithmetic sequence is \( 3 \), and the fifth term is \( 19 \). What is the common difference of the sequence?

Solution: In an arithmetic sequence, each term is given by:

\[ a_n = a_1 + (n-1)d \]

where \( a_n \) is the \( n \)-th term, \( a_1 \) is the first term, and \( d \) is the common difference.

We know:

\( a_1 = 3 \)
\( a_5 = 19 \)

Substitute \( n = 5 \) into the formula for the fifth term:

\[ a_5 = a_1 + (5-1)d = 19 \]

\[ 3 + 4d = 19 \]

\[ 4d = 16 \]

\[ d = 4 \]

Thus, the common difference is \( \boxed{4} \).