SOLVING WORD PROBLEMS INVOLVING TWO NUMBERS

Problem 1 :

Divide 29 into two parts so that sum of their squares is 425.

Solution :

Let x and y be the two parts of 29.

Then, 

x + y  =  29

y  =  29 - x -----(1)

Sum of their squares is 425.

 x² + y²  =  425

 x² + (29 - x)²  =  425

Expanding (29 - x)² using the algebraic identity (a - b)², 

(a - b)²  =  a² + 2ab + b²

(29 - x)²  =  29² + 2(x) (29) + x²

(29 - x)²   =  841 + 58x + x²

 x² + 841 + 58x + x²  =  425

2x² + 841 - 58x - 425  =  0

2x² - 58x + 416  =  0

 x² - 29 x + 208 = 0

 x² - 13 x - 16 x + 208 = 0

 x (x - 13) - 16 (x - 13) = 0

(x - 13) (x - 16) = 0 

Therefore the required numbers are 13 and 16.

Verification :

Sum of two numbers  =  29

13 + 16  =  29

29  =  29

Sum of their squares  =  425

13² + 16²  =  425

169 + 256  =  425

425  =  425

Problem 2 :

Divide 25 into two parts so that sum of their reciprocals is 1/6.

Solution :

Let x and y be the two parts of 25.

Then, 

x + y  =  25

y  =  25 - x -----(1)

Sum of their reciprocals is 1/6.

1/x + 1/y  =  1/6

 1/x + 1/(25 - x)  =  1/6

Simplify. 

(25 - x + x) / x(25 - x)  =  1/6

25 / x(25 - x)  =  1/6

6(25)  =  x(25 - x)

150  =  25x - x²

x² - 25x + 150  =  0

Factor. 

x² - 25x + 150  =  0

x² - 15x - 10x + 150  =  0

x(x - 15) - 10(x - 15)  =  0

(x - 15)(x - 10)  =  0

If x  =  15, 

(1)-----> y  =  25 - 15  =  10

If x  =  10, 

(1)-----> y  =  25 - 10  =  15

So, the two parts of 25 are 10 and 15.

Verification :

Sum of their reciprocals  =  1/6

1/10 + 1/15  =  1/6

(15 + 10) / 150  =  1/6

25/150  =  1/6

1/6  =  16

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