If the sum of one-half and one-fifth of a number exceeds one-third of that number by $7 \frac{1}{3}$, the number is
Aptitude
Problems on Numbers
Difficulty: Hard
Choose an option
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A15
-
B18
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C20
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D30
Answer
Correct Answer: 20
Explanation
### Concept & Formula
This problem tests your ability to add and subtract fractional parts of a single variable and handle mixed fractions appropriately.
$$ \left(\frac{1}{2}x + \frac{1}{5}x\right) - \frac{1}{3}x = 7 \frac{1}{3} $$
### Step-by-Step Solution
* Let the unknown number be $x$.
* Convert the mixed fraction $7 \frac{1}{3}$ into an improper fraction:
$$ 7 \frac{1}{3} = \frac{22}{3} $$
* Set up the equation based on the given conditions:
$$ \left(\frac{x}{2} + \frac{x}{5}\right) - \frac{x}{3} = \frac{22}{3} $$
* Simplify the sum in the parenthesis first (common denominator is 10):
$$ \frac{5x + 2x}{10} - \frac{x}{3} = \frac{22}{3} $$
$$ \frac{7x}{10} - \frac{x}{3} = \frac{22}{3} $$
* Find a common denominator (30) for the remaining terms on the left:
$$ \frac{21x - 10x}{30} = \frac{22}{3} $$
$$ \frac{11x}{30} = \frac{22}{3} $$
* Solve for $x$:
$$ 11x = \frac{22}{3} \times 30 $$
$$ 11x = 22 \times 10 $$
$$ 11x = 220 $$
$$ x = 20 $$
### Exam Strategy & Shortcut
To avoid dealing with large fractions mid-way, find the Least Common Multiple (LCM) of all denominators involved (2, 5, and 3), which is 30.
Assume the number is $30k$.
One-half = $15k$. One-fifth = $6k$. One-third = $10k$.
The equation becomes: $(15k + 6k) - 10k = 7 \frac{1}{3}$
$$ 21k - 10k = \frac{22}{3} $$
$$ 11k = \frac{22}{3} $$
$$ k = \frac{2}{3} $$
Since the number is $30k$, the result is $30 \times \frac{2}{3} = 20$. This method is highly resistant to arithmetic errors.
### Common Pitfall
Forgetting to convert the mixed fraction $7 \frac{1}{3}$ into an improper fraction before attempting algebraic manipulation will almost certainly lead to an incorrect result.
### Final Answer
**Therefore, the correct answer is 20.**