Question
Sum of the two numbers is 40 and their HCF and LCM are 8
and 48, respectively. Find the sum of reciprocal of the given two numbers.Solution
Let the two numbers be '8a' and '8b', where 'a' and 'b' are co-prime numbers. 8a + 8b = 40 Or, 8 X (a + b) = 40 Or, a + b = (40/8) So, a + b = 5 We know that product of two numbers = HCF of the two numbers X LCM of the two numbers Or, 8a X 8b = 8 X 48 Or, ab = (8 X 48)/(8 X 8) = 6 therefore, sum of reciprocals = (a+b)/(8ab) = 5/48 Hence, option a.
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