A number exceeds its two-fifths part by 24. The number is:
Answer: A
Let the number be \(x\). Then \(x - \dfrac{2}{5}x = 24\), so \(\dfrac{3}{5}x = 24\) and \(x = 40\). Check: two-fifths of 40 is 16, and \(40 - 16 = 24\). The common slip is to solve \(\dfrac{2}{5}x = 24\), which gives 60 and does not satisfy the condition.
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Two-thirds of a number is 14 less than the number itself. The number is:
Answer: B
Let the number be \(x\). Then \(x - \dfrac{2}{3}x = 14\), which gives \(\dfrac{1}{3}x = 14\) and \(x = 42\). Check: two-thirds of 42 is 28, and \(42 - 28 = 14\). Notice the leftover fraction is always one minus the given fraction.
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If three-sevenths of a number is 42, then five-sevenths of the same number is:
Answer: C
\[ \frac{3}{7}x = 42 \quad\Longrightarrow\quad x = 98 \]
\[ \frac{5}{7} \times 98 = 70 \]
Or directly: five sevenths is \( \tfrac{5}{3} \) of three sevenths, and \( 42 \times \tfrac{5}{3} = 70 \).
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A number decreased by 30% of itself equals 84. The number is:
Answer: D
Decreasing by 30% leaves 70% of the number, so \(0.7x = 84\) and \(x = 120\). Check: 30% of 120 is 36, and \(120 - 36 = 84\). Adding 30% back to 84 gives 109.2, not 120, which is why the reverse percentage cannot simply be added on.
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Five-eighths of a number exceeds one-third of the same number by 21. The number is:
Answer: A
Set up \(\dfrac{5}{8}x - \dfrac{1}{3}x = 21\). The common denominator is 24, giving \(\dfrac{15x - 8x}{24} = 21\), so \(\dfrac{7x}{24} = 21\) and \(x = 72\). Check: \(\dfrac{5}{8}\) of 72 is 45 and \(\dfrac{1}{3}\) of 72 is 24, differing by 21.
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If one-fifth of one-fourth of a number is 12, then one-half of that number is:
Answer: B
\[ \frac{1}{5} \times \frac{1}{4} x = \frac{x}{20} = 12 \quad\Longrightarrow\quad x = 240 \]
\[ \frac{240}{2} = 120 \]
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Three-fourths of a number is 60 more than one-fourth of it. The number is:
Answer: C
\[ \frac{3x}{4} - \frac{x}{4} = \frac{x}{2} = 60 \]
\[ x = 120 \]
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A number multiplied by seven-ninths gives 63. The number is:
Answer: D
From \(\dfrac{7}{9}x = 63\), multiply both sides by \(\dfrac{9}{7}\) to get \(x = 63 \times \dfrac{9}{7} = 81\). Check: seven-ninths of 81 is \(7 \times 9 = 63\). Multiplying by the fraction instead of its reciprocal gives 49, the first distractor.
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The sum of two numbers is 45 and their difference is 9. The greater number is:
Answer: A
For any two numbers, the greater is \(\dfrac{\text{sum} + \text{difference}}{2}\), here \(\dfrac{45 + 9}{2} = 27\), and the smaller is \(\dfrac{45 - 9}{2} = 18\). Check: \(27 + 18 = 45\) and \(27 - 18 = 9\). This shortcut is worth memorising for any sum-and-difference pair.
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The sum of two numbers is 72 and one of them is twice the other. The smaller number is:
Answer: B
Let the smaller be \(x\), so the larger is \(2x\) and \(x + 2x = 72\). Then \(3x = 72\) and \(x = 24\), with the larger being 48. Check: \(24 + 48 = 72\). Answering 48 means giving the larger number rather than the smaller one asked for.
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