The average of three numbers is 49. If the difference between the largest and the smallest number is 31 and the addition of these two numbers is 99, what is the middle number?
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Correct Answer: Option B
Explanation:
To find the middle number, follow these steps:
1. Determine the Sum of the Three Numbers:
- The average of the three numbers is 49.
- The total sum \( S \) of the three numbers is:
\[
S = \text{Average} \times \text{Number of Numbers} = 49 \times 3 = 147
\]
2. Set Up the Equations:
- Let the three numbers be \( a \), \( b \), and \( c \) where \( a \leq b \leq c \).
- We are given:
\[
c - a = 31
\]
\[
a + c = 99
\]
3. Solve for \( a \) and \( c \):
- Add the two equations to eliminate \( b \):
\[
(a + c) + (c - a) = 99 + 31
\]
\[
2c = 130 \implies c = 65
\]
- Substitute \( c = 65 \) into \( a + c = 99 \):
\[
a + 65 = 99 \implies a = 34
\]
4. Find the Middle Number \( b \):
- Use the total sum \( S = 147 \):
\[
a + b + c = 147
\]
\[
34 + b + 65 = 147
\]
\[
b + 99 = 147 \implies b = 48
\]
The middle number is 48.
The correct answer is B. 48.
To find the middle number, follow these steps:
1. Determine the Sum of the Three Numbers:
- The average of the three numbers is 49.
- The total sum \( S \) of the three numbers is:
\[
S = \text{Average} \times \text{Number of Numbers} = 49 \times 3 = 147
\]
2. Set Up the Equations:
- Let the three numbers be \( a \), \( b \), and \( c \) where \( a \leq b \leq c \).
- We are given:
\[
c - a = 31
\]
\[
a + c = 99
\]
3. Solve for \( a \) and \( c \):
- Add the two equations to eliminate \( b \):
\[
(a + c) + (c - a) = 99 + 31
\]
\[
2c = 130 \implies c = 65
\]
- Substitute \( c = 65 \) into \( a + c = 99 \):
\[
a + 65 = 99 \implies a = 34
\]
4. Find the Middle Number \( b \):
- Use the total sum \( S = 147 \):
\[
a + b + c = 147
\]
\[
34 + b + 65 = 147
\]
\[
b + 99 = 147 \implies b = 48
\]
The middle number is 48.
The correct answer is B. 48.