The sum of two numbers is 55 and the greater exceeds the smaller by 11. The smaller number is:
Answer: C
The smaller is \(\dfrac{\text{sum} - \text{difference}}{2} = \dfrac{55 - 11}{2} = 22\), and the greater is 33. Check: \(22 + 33 = 55\) and \(33 - 22 = 11\). Halving the sum alone gives 27.5, which is the average rather than either number.
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Two numbers are in the ratio 5 : 3 and they differ by 18. The larger number is:
Answer: D
Let the numbers be \(5k\) and \(3k\). Their difference is \(2k = 18\), so \(k = 9\). The larger is \(5 \times 9 = 45\) and the smaller is 27. Check: \(45 - 27 = 18\) and \(45 : 27 = 5 : 3\). Using the difference of the ratio terms, not their sum, is the key step.
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The sum of three consecutive integers is 87. The middle integer is:
Answer: A
Write the integers as \(n-1\), \(n\), \(n+1\); their sum is \(3n = 87\), so \(n = 29\) and the three are 28, 29, 30. For any odd count of consecutive integers the middle one is simply the sum divided by the count, which makes this a one-step problem.
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The sum of three consecutive even numbers is 96. The largest of them is:
Answer: B
Write them as \(n-2\), \(n\), \(n+2\). Their sum is \(3n = 96\), so \(n = 32\) and the numbers are 30, 32 and 34. The largest is 34. The spacing of 2 cancels in the sum, so the middle term is still the sum divided by three.
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The sum of four consecutive odd numbers is 80. The smallest of them is:
Answer: C
Write them as \(n\), \(n+2\), \(n+4\), \(n+6\). Their sum is \(4n + 12 = 80\), so \(4n = 68\) and \(n = 17\). The four numbers are 17, 19, 21 and 23. With an even count there is no middle term, so the constant must be subtracted before dividing.
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The sum of two numbers is 30 and their product is 221. The positive difference between them is:
Answer: D
\[ (a-b)^{2} = (a+b)^{2} - 4ab = 900 - 884 = 16 \]
\[ a - b = 4 \]
The numbers are 13 and 17.
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The digits of a two-digit number add up to 9, and reversing the digits increases the number by 27. The number is:
Answer: A
Let the tens digit be \(t\) and the units digit \(u\). Reversing changes the value by \(9(u - t) = 27\), so \(u - t = 3\). With \(t + u = 9\) we get \(u = 6\) and \(t = 3\), making the number 36. Check: \(63 - 36 = 27\). Any reversal difference is always a multiple of 9.
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A two-digit number is three times the sum of its digits. The number is:
Answer: B
Write the number as \(10t + u\). Then \(10t + u = 3(t + u)\), which simplifies to \(7t = 2u\). Since digits run from 0 to 9, the only fit is \(t = 2\) and \(u = 7\), giving 27. Check: the digit sum is 9 and \(3 \times 9 = 27\).
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A two-digit number exceeds the sum of its digits by 18. Its tens digit is:
Answer: C
The excess is \((10t + u) - (t + u) = 9t\), so \(9t = 18\) and \(t = 2\). The units digit is unconstrained, so 20 through 29 all satisfy the condition - the question can only ask for the tens digit, which is fixed at 2.
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A two-digit number is seven times its units digit. The number is:
Answer: D
Write the number as \(10t + u = 7u\), which gives \(10t = 6u\) and so \(5t = 3u\). The only single digits that fit are \(t = 3\) and \(u = 5\), giving 35. Check: \(7 \times 5 = 35\). Testing each option against its own units digit is just as quick here.
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