Understanding the Problem: Ratios and Sum of Three Numbers

Let's explore a problem where we are given the sum of three numbers and the ratios between them. The objective is to find the value of the second number. This concept is crucial in many areas of mathematics and has practical applications in various fields, such as finance, physics, and engineering.

Tackling the Problem Step by Step

Given Information

The sum of three numbers is 98. The ratio of the first to the second number is 2:3. The ratio of the second to the third number is 5:8.

Approach 1: Direct Proportionality

First, let's express the given ratios in a common form:

A:B 2:3 B:C 5:8

To make these ratios compatible, we find the least common multiple (LCM) of 3 and 5, which is 15. We adjust the ratios:

A:B 10:15 (multiplying the first ratio by 5)

B:C 15:24 (multiplying the second ratio by 3)

Thus, the combined ratio A:B:C 10:15:24.

Let's denote the three numbers as (1), (15x), and (24x).

Since the sum of the numbers is 98, we write the equation:

1 15x 24x 98

Simplifying the equation:

49x 98

Solving for (x):

x 98 / 49 2

Now, we substitute (x 2) into the expressions for A, B, and C:

A 1 10 * 2 20 B 15x 15 * 2 30 C 24x 24 * 2 48

The second number, B, is 30.

Approach 2: Using Simultaneous Equations

Let's denote the three numbers as (x), (y), and (z).

Given:

(x : y 2 : 3) implies (y frac{3}{2}x) (y : z 5 : 8) implies (z frac{8}{5}y) (x y z 98)

Substituting (y frac{3}{2}x) and (z frac{8}{5}y frac{8}{5} times frac{3}{2}x frac{12}{5}x) into the sum equation:

(x frac{3}{2}x frac{12}{5}x 98)

Combining the fractions with a common denominator (10):

(frac{10}{10}x frac{15}{10}x frac{24}{10}x 98)

Simplifying:

(frac{49}{10}x 98)

Solving for (x):

(x 98 times frac{10}{49} 20)

Now, substituting (x 20) back into the expressions for (y) and (z):

(y frac{3}{2}x frac{3}{2} times 20 30) (z frac{12}{5}x frac{12}{5} times 20 48)

Again, the second number, (y), is 30.

Conclusion

By exploring multiple methods, we have determined that the second number, considering the given ratios and sum, is 30. This problem illustrates the power of ratios and algebraic manipulation in solving complex mathematical problems. Understanding these methods can significantly enhance one's problem-solving skills in various fields.