Question
The HCF and LCM of 2 numbers are 2 and 60 respectively.
If one of the numbers is 14 more than the other, find the smaller number.Solution
Let the first number be 'a' and second number be 'b'. LCM × HCF = a x b  60 x 2 = (b + 14) x b 120= b² + 14b  b² +14b-120 = 0 (b-6) (b+20) = 0 b = 6. The second number is 6.
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