SOLVING WORD PROBLEMS WITH QUADRATIC EQUATIONS

Solving Word Problems with Quadratic Equations :

In this section, let us learn how to solve word problems with quadratic equations.

Solving Word Problems with Quadratic Equations

Example 1 :

Two water taps together can fill a tank is 9 ⅜ hours. The tap of the larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution :

Let x be the time taken by the smaller pipe to fill the part of tank

let x -10 be the time taken by the larger pipe to fill the part of the tank

Total time taken by two pipes  =  9  3/8

part of tank filled by the smaller pipe in 1 hour  =  1/x

part of tank filled by the larger pipe in 1 hour  =  1/(x-10)

[1/x] + [1/(x - 10)]  =  1/(9  3/8)

[1/x] + [1/(x - 10)]  =  8/75

[(x - 10 + x)]/[x (x -10)]  =  8/75

(2 x - 10)/(x² - 10 x)  =  8/75

75(2 x - 10)  =  8(x² - 10 x)

150 x - 750  =  8 x² - 80 x

8 x² - 80 x - 150 x + 750  =  0

8 x² - 230 x + 750  =  0

8 x² - 200 x + 30 x + 750  =  0

8 x (x - 25) + 30 (x - 25)  =  0

(8x + 30) (x -25)  =  0

 8x + 30  =  0      x - 25  =  0

8x  =  -30           x  =  25

x  =  -30/8  ==> x  =  -15/4

So, time taken by smaller pipe alone to fill the tank  =  25 hours.

Time taken by larger pipe alone to fill the tank  =  25 - 10

= 15 hours

Example 2 :

A express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.

Solution :

Let x be the average speed of passenger train

let x + 11 be the average speed of express train

Distance to be covered  =  132 km

Time  =  Distance/speed

 T₁  =  132/x

 T₂  =  132/(x + 11)

 [132/x] - [132/(x + 11)]  =  1

 132[(1/x) - (1/(x + 11))]  =  1

 132 [ (x + 11 - x)/(x(x + 11))]  =  1

 132[11/(x² + 11 x)]  =  1

 1452/(x² + 11 x)  =  1

 1452  =  x² + 11 x

x² + 11 x - 1452  =  0

x² + 44 x - 33 x -1452  =  0

x(x + 44) - 33 (x + 44)  =  0

(x + 44) (x - 33)  =  0

 x  =  -44       x  =  33

Speed should not be negative

So speed of passenger train  =  33 km/hr

speed of express train  =  x + 11  =  33 + 11  =  44 km/hr

Example 3 :

Sum of area of two squares is 468 m². If the difference of their perimeter is 24 m,find the sides of the two squares.

Solution :

Let 'x' be the side length of one square

let 'y' be the side length of another square

Sum of area of two squares = 468

 x² + y²  =  468 ------- (1)

 4 x - 4 y  =  24

 4x  =  24 + 4y 

  x  =  (24 + 4y)/4

  x  =  (6 + y)

Substitute x = 6 + y in the first equation

(6 + y)² + y²  =  468

 6² + 2 (6)y + y² + y²  =  468

 36 + 12 y + 2 y²  =  468

 2 y² + 12 y - 468 + 36  =  0

 2 y² + 12 y - 432  =  0

Dividing the whole equation by 2 =>

y² + 6 y - 216  =  0

y² + 18 y - 12 y - 216  =  0

y(y + 18) - 12(y + 18)  =  0

(y - 12) (y + 18)  =  0

y - 12  =  0         y + 18  =  0

 y  =  12 cm            y  =  - 18

 x  =  6 + 12

x  =  18 cm

So, the side length of two squares are 18 cm and 12 cm.

After having gone through the stuff given above, we hope that the students would have understood, solving word problems with quadratic equations.

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