Question : The sum of the two numbers is 1215 and their HCF is 81. If the numbers lie between 500 and 700, then the sum of the reciprocals of the numbers is _____.
Option 1: $\frac{5}{702}$
Option 2: $\frac{5}{378}$
Option 3: $\frac{5}{1512}$
Option 4: $\frac{5}{1188}$
Correct Answer: $\frac{5}{1512}$
Solution :
Let the number be 81$x$ and 81$y$
81$x$ + 81$y$ = 1215
⇒ $x + y = \frac{1215}{81}$ = 15
Pair of $x+ y$ = (1,14), (2,13), (4,11), and (7,8)
⇒ $x = 7$ and $y = 8$
First number = 81 × 7 = 567
Second number = 81 × 8 = 648
The sum of reciprocals = $\frac{1}{81x}$ + $\frac{1}{81y}$ = $\frac{1}{567}$ + $\frac{1}{648}$
= $\frac{1}{81}$[$\frac{1}{7}$ + $\frac{1}{8}$]
= $\frac{1}{81}$ × $\frac{15}{56}$
= $\frac{5}{1512}$
Hence, the correct answer is $\frac{5}{1512}$.
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