Solving Ratio Problems: Finding A:B:C from 3A = 4B = 5C

This problem asks us to find the ratio A : B : C given the relationship 3A = 4B = 5C. To solve this type of problem, we can use a common technique involving setting the given equality to a constant.

Understanding the Relationship 3A = 4B = 5C

The equation 3A = 4B = 5C means that three times the value of A is equal to four times the value of B, which is also equal to five times the value of C. We need to find the relative values of A, B, and C that satisfy this condition and express them as a ratio.

Solving for A, B, and C Using a Constant

Let's assume that the common value of 3A, 4B, and 5C is equal to a constant, say \(k\). So, we have:

• \(3A = k\)
• \(4B = k\)
• \(5C = k\)

Now, we can express A, B, and C in terms of this constant \(k\):

• From \(3A = k\), we get \(A = \frac{k}{3}\)
• From \(4B = k\), we get \(B = \frac{k}{4}\)
• From \(5C = k\), we get \(C = \frac{k}{5}\)

Finding the Ratio A : B : C

Now that we have expressions for A, B, and C in terms of \(k\), we can write the ratio A : B : C:

A : B : C = \(\frac{k}{3} : \frac{k}{4} : \frac{k}{5}\)

To express this ratio in its simplest form with whole numbers, we need to find a common denominator for the fractions \(\frac{1}{3}\), \(\frac{1}{4}\), and \(\frac{1}{5}\). The least common multiple (LCM) of 3, 4, and 5 is the smallest number that is a multiple of 3, 4, and 5.

LCM of 3, 4, and 5:

• Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, ...
• Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, ...
• Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, ...

The LCM of 3, 4, and 5 is 60.

Now, multiply each part of the ratio by the LCM (60) to eliminate the fractions:

\(\left(\frac{k}{3} \times 60\right) : \left(\frac{k}{4} \times 60\right) : \left(\frac{k}{5} \times 60\right)\)

Simplify each term:

• \(\frac{k}{3} \times 60 = 20k\)
• \(\frac{k}{4} \times 60 = 15k\)
• \(\frac{k}{5} \times 60 = 12k\)

So, the ratio becomes \(20k : 15k : 12k\). Since \(k\) is a common factor in all parts of the ratio (and assuming \(k \neq 0\), which must be true if A, B, and C are non-zero), we can divide by \(k\):

\(20 : 15 : 12\)

Verifying the Ratio A : B : C

Let's check if the ratio 20 : 15 : 12 satisfies the original condition 3A = 4B = 5C. Let A = 20x, B = 15x, and C = 12x for some non-zero value x. Substitute these into the equation:

• \(3A = 3 \times (20x) = 60x\)
• \(4B = 4 \times (15x) = 60x\)
• \(5C = 5 \times (12x) = 60x\)

Since \(3A = 4B = 5C = 60x\), the ratio 20 : 15 : 12 is correct.

The ratio A : B : C is 20 : 15 : 12.

Revision Table: Key Concepts for Ratio Problems

Additional Information: Generalizing the Ratio Problem

For a general problem where \(pA = qB = rC\), we can find the ratio A : B : C using the same method. Let \(pA = qB = rC = k\).

Then \(A = \frac{k}{p}\), \(B = \frac{k}{q}\), \(C = \frac{k}{r}\).

The ratio A : B : C is \(\frac{k}{p} : \frac{k}{q} : \frac{k}{r}\). Multiplying by the LCM of p, q, and r will give the ratio in whole numbers.

Alternatively, the ratio A : B : C is proportional to \(\frac{1}{p} : \frac{1}{q} : \frac{1}{r}\). You can then find the LCM of p, q, and r and multiply each term by the LCM to get integer ratios.

In our specific case, \(p=3\), \(q=4\), \(r=5\), so the ratio is proportional to \(\frac{1}{3} : \frac{1}{4} : \frac{1}{5}\).

LCM of 3, 4, 5 is 60.

Ratio \(\propto \left(\frac{1}{3} \times 60\right) : \left(\frac{1}{4} \times 60\right) : \left(\frac{1}{5} \times 60\right)\)

Ratio \(\propto 20 : 15 : 12\)

This confirms our result using the reciprocal method.