Understanding the Ratio and Proportion Problem

This problem asks us to find the ratio of four variables, \(a:b:c:d\), given three proportional relationships involving sums of these variables. We are given:

• \((a + b + c): d = 5: 1\)
• \((b + c + d): a = 7: 1\)
• \((c + d + a): b = 3: 1\)

To find the ratio \(a:b:c:d\), we need to express each variable in terms of a common value or find their relative values.

Step-by-Step Solution to Find the Ratio

Let's convert the given ratios into equations:

1. From \((a + b + c): d = 5: 1\), we have \(\frac{a + b + c}{d} = \frac{5}{1}\). This gives us the equation: \(a + b + c = 5d\)
2. From \((b + c + d): a = 7: 1\), we have \(\frac{b + c + d}{a} = \frac{7}{1}\). This gives us the equation: \(b + c + d = 7a\)
3. From \((c + d + a): b = 3: 1\), we have \(\frac{c + d + a}{b} = \frac{3}{1}\). This gives us the equation: \(c + d + a = 3b\)

Notice that the sum of all four variables \(a + b + c + d\) is present in different combinations in these equations. Let's denote the sum \(a + b + c + d\) as \(S\).

We can rewrite the equations by adding the missing variable to both sides to get the total sum \(S\):

• For the first equation, \(a + b + c = 5d\), add \(d\) to both sides: \((a + b + c) + d = 5d + d \implies S = 6d\)
• For the second equation, \(b + c + d = 7a\), add \(a\) to both sides: \((b + c + d) + a = 7a + a \implies S = 8a\)
• For the third equation, \(c + d + a = 3b\), add \(b\) to both sides: \((c + d + a) + b = 3b + b \implies S = 4b\)

Now we have expressed \(a\), \(b\), and \(d\) in terms of \(S\):

• \(S = 6d \implies d = \frac{S}{6}\)
• \(S = 8a \implies a = \frac{S}{8}\)
• \(S = 4b \implies b = \frac{S}{4}\)

We know that \(S = a + b + c + d\). We can substitute the expressions for \(a\), \(b\), and \(d\) into this equation to find \(c\) in terms of \(S\):

\(S = \frac{S}{8} + \frac{S}{4} + c + \frac{S}{6}\)

To solve for \(c\), isolate \(c\) on one side:

\(c = S - \frac{S}{8} - \frac{S}{4} - \frac{S}{6}\)

Find a common denominator for 8, 4, and 6. The least common multiple (LCM) of 8, 4, and 6 is 24.

\(c = \frac{24S}{24} - \frac{3S}{24} - \frac{6S}{24} - \frac{4S}{24}\)

Combine the terms:

\(c = \frac{24S - 3S - 6S - 4S}{24}\)

\(c = \frac{(24 - 3 - 6 - 4)S}{24}\)

\(c = \frac{11S}{24}\)

Now we have all four variables expressed in terms of \(S\):

• \(a = \frac{S}{8}\)
• \(b = \frac{S}{4}\)
• \(c = \frac{11S}{24}\)
• \(d = \frac{S}{6}\)

To find the ratio \(a : b : c : d\), we write:

\(a : b : c : d = \frac{S}{8} : \frac{S}{4} : \frac{11S}{24} : \frac{S}{6}\)

To simplify this ratio and express it using integers, we multiply each term by the LCM of the denominators (8, 4, 24, 6), which is 24:

\(a : b : c : d = \left(\frac{S}{8} \times 24\right) : \left(\frac{S}{4} \times 24\right) : \left(\frac{11S}{24} \times 24\right) : \left(\frac{S}{6} \times 24\right)\)

\(a : b : c : d = 3S : 6S : 11S : 4S\)

Assuming \(S \ne 0\), we can divide all terms by \(S\):

\(a : b : c : d = 3 : 6 : 11 : 4\)

This gives us the final ratio for \(a:b:c:d\).

Ratio a:b:c:d Final Result

Based on our calculations, the ratio \(a:b:c:d\) is \(3:6:11:4\).

Revision Table for Ratio Problem

Additional Information on Ratio and Proportion

A ratio is a comparison of two quantities by division. For example, the ratio \(a:b\) means \(\frac{a}{b}\). A proportion is an equation stating that two ratios are equal.

In this problem, we used the property that if a set of quantities (\(a, b, c, d\)) have a constant sum \(S\), and parts of the sum are in proportion to one of the quantities, we can express each quantity relative to the total sum \(S\).

If \((a+b+c)/d = k\), this means \(a+b+c = kd\). Adding \(d\) to both sides gives \(a+b+c+d = kd+d = (k+1)d\). If we let \(S = a+b+c+d\), then \(S = (k+1)d\), so \(d = S/(k+1)\). We applied this concept for each given ratio to express \(a, b, c, d\) in terms of \(S\).

Once each variable is expressed as a fraction of the common sum \(S\), finding the ratio involves comparing these fractions and scaling them up to the smallest possible integers by multiplying by the LCM of their denominators.