InterviewSolution
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II. There are 9 multiples of 13 from 19 to 119.1). Only I2). Only II3). Both I and II4). Neither I nor II |
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Answer» Considering each statement one by one: Statement I: Multiples of 9 between 7 and 109 are 9, 18, 27, …., 108 ⇒ The above series is an AP with FIRST term = 9 and common difference = 9 and last term = 108 In an AP, An = a + (n - 1) × d $(\Rightarrow 108 = 9 + \left( {n - 1} \right)9 = 9 + 9n - 9)$ ⇒ 108 = 9n ⇒ n = 108/9 = 12 ∴ Statement I is correct. Statement II: Multiples of 13 from 19 to 119 are 26, 39, …., 117 The above series is an AP with first term = 26 and common difference = 13 and last term = 117 In an AP, An = a + (n - 1) × d ⇒ $(117 = 26 + \left( {n - 1} \right)13 = 26 + 13n - 13)$ ⇒ 117 - 13 = 13n ⇒ 13n = 104 ⇒ n = 104/13 = 8 ∴ Statement II is incorrect. |
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