Question

If \(r=3 s, s=5 t, t=2 u,\) and \(u \neq 0,\) what is the value of \(\frac{r s t}{u^{3}} ?\) A 30 B 60 C 150 D 300 E 600

Short answer

600

Step-by-step solution

Understand the Relationships Between Variables

Given the relationships: \(r = 3s, \; s = 5t, \; t = 2u \; \text{and} \; u eq 0\)

Express Each Variable in Terms of u

Substitute each relationship to express all variables in terms of \(u\):\(t = 2u\)\(s = 5t = 5(2u) = 10u\)\(r = 3s = 3(10u) = 30u\)

Substitute Into the Expression

We need to find the value of \(\frac{rst}{u^3}\). Substitute \(r, s\), and \(t\) in terms of \(u\) into the expression:\[\frac{r s t}{u^{3}} = \frac{(30u)(10u)(2u)}{u^3}\]

Simplify the Expression

Multiply the numerators: \((30u) (10u) (2u) = 600u^3\).Now simplify the overall expression:\[\frac{600u^3}{u^3} = 600\]

Key concepts

Algebraic Expressions

In math, algebraic expressions are a core concept crucial for problem-solving. They are combinations of numbers, variables (like x, y, u), and arithmetic operations (like addition, subtraction, multiplication, and division). For example, in the exercise, we have expressions such as \(r = 3s\).
Algebraic expressions often require simplification to solve problems. Simplification means reducing the expression to its simplest form while maintaining its value. This involves combining like terms and applying arithmetic rules. For example, in the given problem, simplifying \(r\), \(s\), and \(t\) is the key.
Understanding how to work with such expressions is essential. They form the basis of many GRE math problems, helping to establish relationships between different variables. These relationships are then used to solve complex problems more easily.

Variable Substitution

Variable substitution is another fundamental concept in algebra. It involves replacing one variable with an equivalent expression in terms of another variable. For instance, in the given problem, we start with relationships: \(r = 3s\), \(s = 5t\), and \(t = 2u\).
By substituting \(t\) as \(2u\), this allows us to express \(s\) as \(5(2u) = 10u\), and then \(r\) as \(3(10u) = 30u\). Through substitution, we transform the variables into one term, \(u\), which simplifies solving the problem.
Variable substitution helps to simplify complex expressions and equations by reducing the number of variables. This technique is pivotal in reducing potential confusion and avoiding errors when solving algebraic problems.

Simplifying Fractions

Simplifying fractions is crucial in solving algebraic equations and expressions efficiently. It involves breaking down the numerator and the denominator to their simplest forms. In the given exercise, we start by expressing \(r, s,\) and \(t\) in terms of \(u\): \(r = 30u\), \(s = 10u\), and \(t = 2u\).
We then substitute these into the given fraction: \[\frac{r s t}{u^{3}} = \frac{(30u)(10u)(2u)}{u^3} \]. Next, multiply the terms in the numerator to get \(600u^3\). Now, the fraction becomes \[\frac{600u^3}{u^3} \].
Finally, divide both the numerator and the denominator by \(u^3\), simplifying the expression to just 600. This simplification makes the expression more manageable and the solution more straightforward.
Mastering the skill of simplifying fractions is invaluable for solving complex GRE math problems effectively.