Question
The sum of five consecutive odd numbers of set p is 435. What is the sum of five consecutive numbers of another set q. Whose largest number is 45 more than the largest number of set p?
Answer: Option B
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Let the five consecutive odd numbers of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5.Sum of these five numbers= 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5= 10n + 5 = 435 => n = 43Largest number of set p = 2(43) + 5 = 91The largest number of set q = 91 + 45 = 136=> The five numbers of set q are 132, 133, 134, 135, 136.Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.
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