Question

The sum of two number a and b is 15, and the sum of their reciprocals $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{3}{10}$. Find the numbers a and b.

Answer 1

SOLUTION :  

Given : Sum of two number a and b is 15 and Sum of their reciprocals is 3/10 .

a + b = 15  

b = 15 - a …………………..(1)

1/a +  1/b  = 3/10

(b + a) / ab = 3/10

[By taking LCM]

(a + b) / ab = 3/10

10(a + b) = 3(ab)  

[By cross multiplication]

10a + 10b = 3ab …………..(2)

Put the value of b from eq 1 in eq 2  

10a + 10(15 - a) = 3a(15 - a)

10a + 150 - 10a = 45a - 3a²

150 = 45a - 3a²

3a² - 45a + 150 = 0

3(a² - 15a + 50) = 0

a² - 15a + 50 = 0

a² - 10a - 5a + 50 = 0

[By middle term splitting]

a(a - 10) - 5(a - 10) = 0

(a - 5) (a - 10) = 0

(a - 5)  = 0  or (a - 10) = 0

a = 5 or a = 10

Case 1 :  

When a = 5 then b= (15 - a) = 15 - 5 = 10

Case 1 :  

When a = 10 then b= (15 - a) = 15 - 10 = 5

Hence , the two numbers are (a = 5 ,b = 10) & (a = 10 , b = 5) .

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Answer 2

The sum of two numbers a and b is 15.

( a + b ) = 15

a = 15 - b

Sum of their reciprocals =$\frac{1} {a}$+ $\frac{1} {b}$ = $\frac{3} {10}$

$\frac{b + a } {ab}$ = $\frac{3} {10}$


10 ( b + a ) = 3 ab

10 b + 10a = 3 ab

10b + 10( 15 - b ) = 3 ( 15 - b ) b

10b + 150 - 10b = 45b - 3b ^2

150 = - 3b^2 + 45b

3b^2 - 45b +150 = 0

3 ( b^2 - 15b + 50 ) = 0

b^2 - 15b + 50 = 0

b ^2 - 10b - 5b + 50 = 0

b ( b - 10 ) - 5 ( b - 10 ) = 0

( b - 5 ) ( b - 10 ) = 0

b = 5, 10

Putting value of b in a,

a = 15 - b = 15 - 5 = 10 or

a = 15 - 10 = 5

$<h3>So, Values of a are 5 and 10 and values of b are 5 and 10.</h3>$