7th Grade Common Core Math Worksheets - Free Printable

Let's solve each problem step by step:

---

Problem 1:

Six balls numbered from 1 to 6 are placed in an urn. If one ball is selected at random, find the probability that it is an odd-numbered ball.

#### Solution:
- The total number of balls is 6.
- The odd-numbered balls are: 1, 3, 5. So, there are 3 odd-numbered balls.
- The probability of selecting an odd-numbered ball is:
\[
P(\text{odd}) = \frac{\text{Number of odd-numbered balls}}{\text{Total number of balls}} = \frac{3}{6} = \frac{1}{2}
\]

#### Answer:
\[
\boxed{B}
\]

---

Problem 2:

A single six-sided fair die is tossed. Find the probability of obtaining a number greater than 4.

#### Solution:
- The possible outcomes when rolling a six-sided die are: 1, 2, 3, 4, 5, 6.
- The numbers greater than 4 are: 5 and 6. So, there are 2 favorable outcomes.
- The probability of rolling a number greater than 4 is:
\[
P(\text{number > 4}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}
\]

#### Answer:
\[
\boxed{A}
\]

---

Problem 3:

You are one of 30 people entering a contest. What is the probability that your name will be drawn first?

#### Solution:
- There are 30 people in total.
- Each person has an equal chance of being drawn first.
- The probability that your name will be drawn first is:
\[
P(\text{your name}) = \frac{1}{30}
\]

#### Answer:
\[
\boxed{C}
\]

---

Problem 4:

Given the set of numbers \(\{0, 1, 2, 3, 4, 5, 6, 7, 8\}\), if one of the numbers of the set is chosen at random, find the probability that the number is a solution of \(3x + 1 < 13\).

#### Solution:
1. Solve the inequality \(3x + 1 < 13\):
\[
3x + 1 < 13 \implies 3x < 12 \implies x < 4
\]
So, the solutions to the inequality are \(x = 0, 1, 2, 3\).

2. The total number of elements in the set is 9.
3. The favorable outcomes (numbers less than 4) are: 0, 1, 2, 3. So, there are 4 favorable outcomes.
4. The probability is:
\[
P(\text{solution}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{9}
\]

#### Answer:
\[
\boxed{B}
\]

---

Problem 5:

What is the probability of drawing a spade from a deck of 52 playing cards?

#### Solution:
- A standard deck of 52 playing cards contains 4 suits: hearts, diamonds, clubs, and spades.
- Each suit has 13 cards. Therefore, there are 13 spades in the deck.
- The probability of drawing a spade is:
\[
P(\text{spade}) = \frac{\text{Number of spades}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}
\]

#### Answer:
\[
\boxed{D}
\]

---

Problem 6:

This is a spinner used in a board game Helen invented. What is the probability that the spinner will land on a multiple of 3 and 4?

#### Solution:
1. Identify the numbers on the spinner: 10, 12, 18, 22, 24, 36, 54, 64.
2. Determine which numbers are multiples of both 3 and 4. A number that is a multiple of both 3 and 4 is a multiple of their least common multiple (LCM). The LCM of 3 and 4 is 12.
3. The multiples of 12 in the list are: 12, 24, 36.
4. Count the total number of sections on the spinner: 8.
5. The probability is:
\[
P(\text{multiple of 3 and 4}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}
\]

#### Answer:
\[
\boxed{C}
\]

---

Problem 7:

Donato’s little brother Joseph is too small to see inside his sock drawer. Joseph has 2 pairs of white socks, 4 pairs of black socks, and 1 pair of blue socks inside his drawer. If the socks are not paired together, what is the probability that Joseph will reach inside his drawer and pick a black sock?

#### Solution:
1. Calculate the total number of socks:
- 2 pairs of white socks = \(2 \times 2 = 4\) white socks.
- 4 pairs of black socks = \(4 \times 2 = 8\) black socks.
- 1 pair of blue socks = \(1 \times 2 = 2\) blue socks.
- Total number of socks = \(4 + 8 + 2 = 14\).

2. The number of black socks is 8.
3. The probability of picking a black sock is:
\[
P(\text{black sock}) = \frac{\text{Number of black socks}}{\text{Total number of socks}} = \frac{8}{14} = \frac{4}{7}
\]

#### Answer:
\[
\boxed{C}
\]

---

Final Answers:

1. \(\boxed{B}\)
2. \(\boxed{A}\)
3. \(\boxed{C}\)
4. \(\boxed{B}\)
5. \(\boxed{D}\)
6. \(\boxed{C}\)
7. \(\boxed{C}\)