# SOLVING WORD PROBLEMS WITH QUADRATIC EQUATIONS

Problem 1 :

Two water taps together can fill a tank is 9 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.

Then, time taken by the larger pipe to fill the part of the tank is (x - 10).

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/( 3/8)

[1/x] + [1/(x - 10)]  =  8/75

[(x - 10 + x)] / [x(x -10)]  =  8/75

(2x - 10)/(x2 - 10 x)  =  8/75

75(2x - 10)  =  8(x2 - 10x)

150x - 750  =  8x2 - 80x

8x2 - 80x - 150x + 750  =  0

8x2 - 230x + 750  =  0

8x2 - 200x + 30x + 750  =  0

8x(x - 25) + 30(x - 25)  =  0

(8x + 30) (x - 25)  =  0

8x + 30  =  0    or    x - 25  =  0

x  =  -15/4    or    x  =  25

Because x represents amount of time, it can not take negative value.

Then,

x  =  25

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

Time taken by larger pipe alone to fill the tank is

=  25 - 10

=  15 hours

Problem 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.

Then, the average speed of express train is (x + 11).

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 + 11x)]  =  1

1452/(x2 + 11x)  =  1

1452  =  x2 + 11x

x2 + 11x - 1452  =  0

x2 + 44x - 33x -1452  =  0

x(x + 44) - 33(x + 44)  =  0

(x + 44) (x - 33)  =  0

x  =  -44    or    x  =  33

Because x represents the average speed, it can not take negative value.

Then,

x  =  33

So, the average speed of passenger train is 33 km/hr.

Average speed of express train is

=  x + 11

=  33 + 11

=  44 km/hr

Problem 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)

4x - 4y  =  24

4x  =  24 + 4y

x  =  (24 + 4y)/4

x  =  6 + y -----(2)

Substitute x = 6 + y in (1).

(1)-----> (6 + y)2 + y2  =  468

62 + 2(6)y + y2 + y2  =  468

36 + 12y + 2y2  =  468

2y2 + 12y - 468 + 36  =  0

2y2 + 12y - 432  =  0

Divide the whole equation by 2.

y2 + 6 y - 216  =  0

y2 + 18y - 12y - 216  =  0

y(y + 18) - 12(y + 18)  =  0

(y - 12) (y + 18)  =  0

y - 12  =  0    or    y + 18  =  0

y  =  12    or    y  =  - 18

Because y represents the length, it can not take negative value. .

Then,

y  =  12

Substitute 12 for y in (2).

(2)-----> x  =  6 + 12

x  =  18

So, the side length of two squares are 18 cm and 12 cm. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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