A number is greater than its 2/5  th by 24. The number is

Let the number be \( x \).

According to the problem, the number \( x \) is greater than its \(\frac{2}{5}\)th by 24. This can be written as:

\[
x = \frac{2}{5}x + 24
\]

To solve for \( x \), subtract \(\frac{2}{5}x\) from both sides:

\[
x - \frac{2}{5}x = 24
\]

Factor out \( x \) on the left side:

\[
\left(1 - \frac{2}{5}\right)x = 24
\]

Simplify the expression in the parentheses:

\[
\frac{3}{5}x = 24
\]

Now, solve for \( x \) by multiplying both sides by \(\frac{5}{3}\):

\[
x = 24 \times \frac{5}{3}
\]

\[
x = 40
\]

So, the number is \( 40 \).

Would you like more details, or do you have any questions about this solution?

Here are 5 related questions you might find interesting:
1. What would the equation look like if the number was greater than its \(\frac{3}{4}\)th by 30?
2. If a number is greater than its \(\frac{3}{5}\)th by 36, what is the number?
3. How can you check if the solution to the problem is correct?
4. What happens to the equation if the number was less than its \(\frac{2}{5}\)th by 24?
5. Can you generalize this method to find a number that is greater than its \(\frac{m}{n}\)th by any given value?

**Tip:** When solving problems involving fractions, it's often helpful to first isolate the term involving the fraction before solving for the variable.

Learn how to solve a problem where a number is greater than its 2/5th by 24. Understand the step-by-step solution involving linear equations and fractions. Suitable for Grades 7-9 students and anyone interested in basic algebraic problem-solving.

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