Answer :
To solve the problem, we need to find the number of marbles Leah has given two conditions:
1. Leah has 28 more marbles than Dan.
2. One third of Leah's marbles is equal to four-fifths of Dan's marbles.
Let's denote:
- Leah's number of marbles as \( L \).
- Dan's number of marbles as \( D \).
Step-by-step solution:
Step 1: Establish the equations based on the conditions provided.
From the first condition:
[tex]\[ L = D + 28 \tag{1} \][/tex]
From the second condition:
[tex]\[ \frac{L}{3} = \frac{4}{5} D \tag{2} \][/tex]
Step 2: Solve the equations simultaneously.
Start by substituting \( L \) from equation (1) into equation (2):
[tex]\[ \frac{D + 28}{3} = \frac{4}{5} D \][/tex]
To eliminate the fractions, multiply every term by 15 (the least common multiple of 3 and 5):
[tex]\[ 15 \cdot \frac{D + 28}{3} = 15 \cdot \frac{4}{5} D \][/tex]
This simplifies to:
[tex]\[ 5(D + 28) = 12D \][/tex]
Expand and simplify:
[tex]\[ 5D + 140 = 12D \][/tex]
Move terms involving \( D \) to one side:
[tex]\[ 140 = 12D - 5D \][/tex]
[tex]\[ 140 = 7D \][/tex]
Solve for \( D \):
[tex]\[ D = \frac{140}{7} \][/tex]
[tex]\[ D = 20 \][/tex]
So, Dan has 20 marbles.
Step 3: Find Leah's number of marbles using equation (1).
[tex]\[ L = D + 28 \][/tex]
Substitute \( D = 20 \):
[tex]\[ L = 20 + 28 \][/tex]
[tex]\[ L = 48 \][/tex]
So, Leah has 48 marbles.
Final Answer: Leah has 48 marbles.
1. Leah has 28 more marbles than Dan.
2. One third of Leah's marbles is equal to four-fifths of Dan's marbles.
Let's denote:
- Leah's number of marbles as \( L \).
- Dan's number of marbles as \( D \).
Step-by-step solution:
Step 1: Establish the equations based on the conditions provided.
From the first condition:
[tex]\[ L = D + 28 \tag{1} \][/tex]
From the second condition:
[tex]\[ \frac{L}{3} = \frac{4}{5} D \tag{2} \][/tex]
Step 2: Solve the equations simultaneously.
Start by substituting \( L \) from equation (1) into equation (2):
[tex]\[ \frac{D + 28}{3} = \frac{4}{5} D \][/tex]
To eliminate the fractions, multiply every term by 15 (the least common multiple of 3 and 5):
[tex]\[ 15 \cdot \frac{D + 28}{3} = 15 \cdot \frac{4}{5} D \][/tex]
This simplifies to:
[tex]\[ 5(D + 28) = 12D \][/tex]
Expand and simplify:
[tex]\[ 5D + 140 = 12D \][/tex]
Move terms involving \( D \) to one side:
[tex]\[ 140 = 12D - 5D \][/tex]
[tex]\[ 140 = 7D \][/tex]
Solve for \( D \):
[tex]\[ D = \frac{140}{7} \][/tex]
[tex]\[ D = 20 \][/tex]
So, Dan has 20 marbles.
Step 3: Find Leah's number of marbles using equation (1).
[tex]\[ L = D + 28 \][/tex]
Substitute \( D = 20 \):
[tex]\[ L = 20 + 28 \][/tex]
[tex]\[ L = 48 \][/tex]
So, Leah has 48 marbles.
Final Answer: Leah has 48 marbles.
