A number whose fifth part increased by 4 is equal to its fourth part diminished by 10, is:

Question

Find the value of $ x $ that satisfies the equation $ 2x + 3 = 11 $.

Answer 1

An alternative approach is to test the answer choices directly against the condition, which is often faster in an exam setting.

The condition is: $\frac{x}{5} + 4 = \frac{x}{4} - 10$.
Try $x = 280$: $\frac{280}{5} + 4 = 56 + 4 = 60$, and $\frac{280}{4} - 10 = 70 - 10 = 60$. Both sides match, so 280 works.
To confirm this is the only solution, solve algebraically: rearrange to $\frac{x}{4} - \frac{x}{5} = 14$.
Combine the fractions: $\frac{5x - 4x}{20} = 14 \Rightarrow \frac{x}{20} = 14$.
Multiply both sides by 20: $x = 280$.
This confirms the unique solution and rules out the other three options, since a linear equation in one variable has exactly one solution.

\[\boxed{x = 280}\]

Answer 2

An alternative approach is to test the answer choices directly against the condition, which is often faster in an exam setting.

The condition is: $\frac{x}{5} + 4 = \frac{x}{4} - 10$.
Try $x = 280$: $\frac{280}{5} + 4 = 56 + 4 = 60$, and $\frac{280}{4} - 10 = 70 - 10 = 60$. Both sides match, so 280 works.
To confirm this is the only solution, solve algebraically: rearrange to $\frac{x}{4} - \frac{x}{5} = 14$.
Combine the fractions: $\frac{5x - 4x}{20} = 14 \Rightarrow \frac{x}{20} = 14$.
Multiply both sides by 20: $x = 280$.
This confirms the unique solution and rules out the other three options, since a linear equation in one variable has exactly one solution.

\[\boxed{x = 280}\]

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