Answer :
Certainly! Let's solve this step-by-step.
Given three numbers in the ratio 4:5:6. We will denote these numbers as \(4x\), \(5x\), and \(6x\) respectively, where \(x\) is a common multiplier.
1. Sum of the Numbers:
According to the problem, the sum of these numbers is 200.
[tex]\[ 4x + 5x + 6x = 200 \][/tex]
Simplifying this,
[tex]\[ 15x = 200 \][/tex]
2. Express x:
Solving for \(x\),
[tex]\[ x = \frac{200}{15} = \frac{40}{3} \][/tex]
3. Find Each Number:
Now, substitute the value of \(x\) back into each expression to find the actual numbers.
- The first number (smallest) is:
[tex]\[ 4x = 4 \times \frac{40}{3} = \frac{160}{3} \][/tex]
- The second number (middle) is:
[tex]\[ 5x = 5 \times \frac{40}{3} = \frac{200}{3} \][/tex]
- The third number (largest) is:
[tex]\[ 6x = 6 \times \frac{40}{3} = 80 \][/tex]
4. Verification:
- Sum of the numbers:
[tex]\[ \frac{160}{3} + \frac{200}{3} + 80 \][/tex]
First, convert 80 to a fraction over 3 to add easily:
[tex]\[ 80 = \frac{240}{3} \][/tex]
Now adding all together:
[tex]\[ \frac{160}{3} + \frac{200}{3} + \frac{240}{3} = \frac{600}{3} = 200 \][/tex]
- Check the condition that the sum of the largest and smallest equals twice the middle number:
[tex]\[ \frac{160}{3} + 80 = 2 \left( \frac{200}{3} \right) \][/tex]
Convert 80 to fraction form:
[tex]\[ 80 = \frac{240}{3} \][/tex]
So,
[tex]\[ \frac{160}{3} + \frac{240}{3} = \frac{400}{3} \][/tex]
And,
[tex]\[ 2 \left( \frac{200}{3} \right) = \frac{400}{3} \][/tex]
Both sides are equal, confirming the condition.
Thus, the three numbers are [tex]\(\frac{160}{3}\)[/tex], [tex]\(\frac{200}{3}\)[/tex], and [tex]\(80\)[/tex].
Given three numbers in the ratio 4:5:6. We will denote these numbers as \(4x\), \(5x\), and \(6x\) respectively, where \(x\) is a common multiplier.
1. Sum of the Numbers:
According to the problem, the sum of these numbers is 200.
[tex]\[ 4x + 5x + 6x = 200 \][/tex]
Simplifying this,
[tex]\[ 15x = 200 \][/tex]
2. Express x:
Solving for \(x\),
[tex]\[ x = \frac{200}{15} = \frac{40}{3} \][/tex]
3. Find Each Number:
Now, substitute the value of \(x\) back into each expression to find the actual numbers.
- The first number (smallest) is:
[tex]\[ 4x = 4 \times \frac{40}{3} = \frac{160}{3} \][/tex]
- The second number (middle) is:
[tex]\[ 5x = 5 \times \frac{40}{3} = \frac{200}{3} \][/tex]
- The third number (largest) is:
[tex]\[ 6x = 6 \times \frac{40}{3} = 80 \][/tex]
4. Verification:
- Sum of the numbers:
[tex]\[ \frac{160}{3} + \frac{200}{3} + 80 \][/tex]
First, convert 80 to a fraction over 3 to add easily:
[tex]\[ 80 = \frac{240}{3} \][/tex]
Now adding all together:
[tex]\[ \frac{160}{3} + \frac{200}{3} + \frac{240}{3} = \frac{600}{3} = 200 \][/tex]
- Check the condition that the sum of the largest and smallest equals twice the middle number:
[tex]\[ \frac{160}{3} + 80 = 2 \left( \frac{200}{3} \right) \][/tex]
Convert 80 to fraction form:
[tex]\[ 80 = \frac{240}{3} \][/tex]
So,
[tex]\[ \frac{160}{3} + \frac{240}{3} = \frac{400}{3} \][/tex]
And,
[tex]\[ 2 \left( \frac{200}{3} \right) = \frac{400}{3} \][/tex]
Both sides are equal, confirming the condition.
Thus, the three numbers are [tex]\(\frac{160}{3}\)[/tex], [tex]\(\frac{200}{3}\)[/tex], and [tex]\(80\)[/tex].
