The product of two numbers is 120 and their sum is 22. The larger number is:
Answer: B
\[ (a-b)^{2} = (a+b)^{2} - 4ab = 484 - 480 = 4 \]
\[ a - b = 2 \quad\Longrightarrow\quad a = 12, \ b = 10 \]
So the larger is 12.
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Two positive integers have a product of 156. If one of them is one more than the other, the larger integer is:
Answer: C
Consecutive integers multiplying to 156 satisfy \(n(n+1) = 156\), so \(n^2 + n - 156 = 0\) and \((n + 13)(n - 12) = 0\), giving \(n = 12\). The pair is 12 and 13, so the larger is 13. Since \(\sqrt{156} \approx 12.5\), the two integers must straddle that value, which locates them at once.
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Seven added to three times a number gives 43. The number is:
Answer: D
\[ 3x + 7 = 43 \quad\Longrightarrow\quad 3x = 36 \]
\[ x = 12 \]
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The difference between the squares of two consecutive natural numbers is 35. The larger number is:
Answer: A
For consecutive numbers, \((n+1)^2 - n^2 = 2n + 1\). Setting \(2n + 1 = 35\) gives \(n = 17\), so the larger number is 18. Check: \(18^2 - 17^2 = 324 - 289 = 35\). The difference of consecutive squares is always odd, and equals the sum of the two numbers.
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The sum of the squares of two consecutive even integers is 340. The smaller integer is:
Answer: B
Let the integers be \(n\) and \(n + 2\). Then \(n^2 + (n+2)^2 = 340\), so \(2n^2 + 4n - 336 = 0\) and \(n^2 + 2n - 168 = 0\), which factors as \((n + 14)(n - 12) = 0\). Taking the positive root, \(n = 12\). Check: \(144 + 196 = 340\).
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The square of a number exceeds the number itself by 132. The positive value of the number is:
Answer: C
The condition is \(x^2 - x = 132\), so \(x^2 - x - 132 = 0\) and \((x - 12)(x + 11) = 0\). The positive root is 12. Check: \(144 - 12 = 132\). Looking for two factors of 132 that differ by 1 finds 11 and 12 immediately.
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The square of a number is 25 more than 24 times the number. The positive value of the number is:
Answer: D
\[ x^{2} = 24x + 25 \quad\Longrightarrow\quad x^{2} - 24x - 25 = 0 \]
\[ (x - 25)(x + 1) = 0 \]
\[ x = 25 \text{ (taking the positive root)} \]
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The difference between the squares of two numbers is 256 and their sum is 32. The difference between the numbers is:
Answer: A
\[ a^{2} - b^{2} = (a+b)(a-b) \]
\[ 256 = 32(a - b) \quad\Longrightarrow\quad a - b = 8 \]
The numbers are 20 and 12.
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If \(x^2 + \dfrac{1}{x^2} = 34\) and \(x\) is positive, then \(x + \dfrac{1}{x}\) equals:
Answer: B
Squaring gives \(\left(x + \dfrac{1}{x}\right)^2 = x^2 + 2 + \dfrac{1}{x^2} = 34 + 2 = 36\), so \(x + \dfrac{1}{x} = 6\) for positive \(x\). The cross term is exactly 2 because \(x\) and \(\dfrac{1}{x}\) multiply to 1, which is what makes this identity so useful.
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A number increased by 20% of itself becomes 96. The original number is:
Answer: C
Increasing by 20% multiplies the number by 1.2, so \(1.2x = 96\) and \(x = 80\). Check: 20% of 80 is 16, and \(80 + 16 = 96\). Subtracting 20% from 96 gives 76.8, which is not the same thing - percentages are taken of the original, not the result.
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