Question

1/a + 1/b = 1/c and a + b = 20 ,where a,b and c are positive numbers. What might be the value of c?

Answer 1

Answer:

Take a = 2, b = 2. In this case c = 1

Take a = 4, b = 4. In this case,

1/4 + 1/4 = 1/2

c = 2

Take a = 3, b = 6. In this case,

1/3 + 1/6 = 1/2

c = 2

Take a = 15, b = 30. In this case

1/15 + 1/30 = 1/10

Answer 2

$\green{\large\underline{\sf{Solution-}}}$

Given that

$\rm :\longmapsto\:a,b,c \: are \: positive \: integers \: such \: that \: a + b = 20$

and

$\rm :\longmapsto\:\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{c}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{a} + \dfrac{1}{20 - a} = \dfrac{1}{c}$

$\rm :\longmapsto\:\dfrac{20 - a + a}{a(20 - a)} = \dfrac{1}{c}$

$\rm :\longmapsto\:\dfrac{20}{a(20 - a)} = \dfrac{1}{c}$

$\rm \implies\:c = \dfrac{a(20 - a)}{20}$

Let we assign the values of a, corresponding values of c are given below

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf c = \dfrac{a(20 - a)}{20} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf \dfrac{19}{20} \\ \\ \sf 2 & \sf \dfrac{9}{5} \\ \\ \sf 3 & \sf \dfrac{51}{20}\\ \\ \sf 4 & \sf \dfrac{16}{5}\\ \\ \sf 5 & \sf \dfrac{15}{4}\\ \\ \sf 6 & \sf \dfrac{21}{5}\\ \\ \sf 7 & \sf \dfrac{91}{20}\\ \\ \sf 8 & \sf \dfrac{24}{5}\\ \\ \sf 9 & \sf \dfrac{99}{20}\\ \\ \sf 10 & \sf 5\\ \\ \sf 11 & \sf \dfrac{99}{20}\\ \\ \sf 12 & \sf \dfrac{24}{5}\\ \\ \sf 13 & \sf \dfrac{91}{20}\\ \\ \sf 14 & \sf \dfrac{21}{5}\\ \\ \sf 15 & \sf \dfrac{15}{4}\\ \\ \sf 16 & \sf \dfrac{16}{5}\\ \\ \sf 17 & \sf \dfrac{51}{20}\\ \\ \sf 18 & \sf \dfrac{9}{5}\\ \\ \sf 19 & \sf \dfrac{19}{20} \end{array}} \\ \end{gathered}$

So,

$\bf\implies \:c = 5$