If the sum of two numbers is 27 and their hcf and lcm are 3 and 60 then sum of their reciprocal is
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Let the no. be a and b
• a + b = 27
a = 27 - b .....( i )
Since ,
HCF × LCM = product of two nos.
3 × 60 = ab
180 = ab .....( ii )
Putting in ( ii ) value of a from ( i )
ab = 180
( 27 - b ) b = 180
27b - b² = 180
b² - 27b + 180 = 0
b² - 12b - 15b + 180 = 0
b ( b - 12 ) - 15 ( b - 15 ) = 0
( b - 12 ) ( b - 15 ) = 0
• ( b - 12 ) = 0b = 12
• ( b - 15 ) = 0b = 15
If we take b = 12
a = 27 - b
a = 27 - 12
a = 15
If we take b = 15
a = 27 - b
a = 12
• a + b = 27
a = 27 - b .....( i )
Since ,
HCF × LCM = product of two nos.
3 × 60 = ab
180 = ab .....( ii )
Putting in ( ii ) value of a from ( i )
ab = 180
( 27 - b ) b = 180
27b - b² = 180
b² - 27b + 180 = 0
b² - 12b - 15b + 180 = 0
b ( b - 12 ) - 15 ( b - 15 ) = 0
( b - 12 ) ( b - 15 ) = 0
• ( b - 12 ) = 0b = 12
• ( b - 15 ) = 0b = 15
If we take b = 12
a = 27 - b
a = 27 - 12
a = 15
If we take b = 15
a = 27 - b
a = 12
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