## SOLVING WORD PROBLEMS INVOLVING TWO NUMBERS

Solving Word Problems Involving Two Numbers :

In the section, we will learn how to solve word problems involving two numbers.

Example 1 :

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

Solution :

Let x and y be the required two numbers

Divide 29 into two parts

So, let us take it as sum of two numbers is

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) =  a2 + 2ab + b2

(29 - x) =  292 + 2(x) (29) + x2

(29 - x)  =  841 + 58x + x2

x2841 + 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

 x - 13  =  0x  =  13 x - 16  =  0x  =  16

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

Example 2 :

Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

Solution :

Let x be the required number

Whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

x - 20  =  69 (1/x)

x - 20  =  69/x

x(x - 20)  =  69

x2 - 20x  =  69

x2 - 20x - 69  =  0

(x + 3) (x - 23) = 0

 x + 3  =  0x  =  -3 x - 23  =  0x  =  23

Therefore the required number is 23.

Verification :

whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

23 - 20  =  69 (1/23)

3  =  69/23

3  =  3

After having gone through the stuff given above, we hope that the students would have understood how to solve word problems involving two numbers.

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