More Questions from Problems on Numbers

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
  • A
    15
  • B
    18
  • C
    20
  • D
    30

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.**
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