SOLVING WORD PROBLEMS ON QUADRATIC EQUATIONS

Problem 1 :


Two numbers are such that thrice the smaller number exceeds twice the greater one by 18 and 1/3 of the smaller and 1/5 of the greater number are together 21. The numbers are -

(a) (45,36)     (b) (50, 38)      (c) (54,45)       (d) (55,41)

Solution :

Let x and y be smaller and larger numbers respectively.

3x = 2y + 18

3x - 2y = 18 ---(1)

1/3 of x + 1/5 of y = 21

x/3 + y/5 = 21

5x + 3y = 21(15)

5x + 3y = 315 -----(2)

3(1) + 2(2) ==>

9x - 6y + 10x + 6y = 54 + 630

19x = 684

x = 36

By applying the value x = 36 in (1)

3(36) - 2y = 18

108 - 2y = 18

108 - 18 = 2y

2y = 90

y = 45

So, the value of and y is 36 and 45.

Problem 2 :

Two sides of an equilateral triangle are shortened by 12 units 13 units and 14 units respectively and a right angle is formed. The sides of the equilateral triangle is 

(a) 17 units      (b) 16 units      (c) 15 units      (d) 18 units

Solution :

Let x be the side length of equilateral triangle.

Side lengths of right triangle are (x - 12), (x - 13) and (x - 14).

Using Pythagorean theorem,

(x - 12)2 = (x - 13)2 + (x - 14)2

x2 - 24x + 144 = x2 - 26x + 169 + x2 - 28x + 196

Combing the like terms, we get

x2 - 30x + 221 = 0

(x - 13) (x - 17) = 0

x = 13 and x = 17

So, the required side is 17 units.

Problem 3 :

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4cm. The sides are

(a) (11 cm, 15 cm)       (b) (12 cm, 16 cm)

(c) (20 cm, 24 cm)       (d) None of these

Solution :

If one side is x, then other side be (x + 4)

202 = x2 + (x + 4)2

400 = x2 + x2 + 8x + 16

2x2 + 8x + 16 - 400 = 0

2x2 + 8x - 384 = 0

x2 + 4x - 192 = 0

(x + 16)(x - 12) = 0

x = -16 and x = 12

So, 12 and 16 are the required sides.

Problem 4 :

The sum of two numbers is 45 and the mean proportional between them is 18. The numbers are

(a) (15, 30)     (b) (32, 13)     (c) (36, 9)        (d) (25, 20)

Solution :

Let x and y be the numbers.

x + y = 45  -------(1)

y = 45 - x

√xy = 18 -------(2)

xy = 324

x(45 - x) = 324

45x - x2 = 324

x2 - 45x + 324 = 0

(x - 9) (x - 36) = 0

x = 9 and x = 36

Problem 5 :

Solving the equation

7x1-x + 81-xx = 15

(a) 64/113, 1/2       (b) 1/50, 1/65

(c) 49/50, 1/65      (d) 1/50, 64/65

Solution :

7[√(x/(1-x)] + 8[√(1-x)/x] = 15

7(x) + 8(1-x)/x√(1-x) = 15

(7x + 8 - 8x)/x(1-x) = 15

(8-x)2 = 15x(1-x)

Taking squares on both sides.

64 - 16x + x2 = 225(x - x2)

64 - 16x + x2 = 225x - 225x2

226x2 - 16x - 225x + 64 = 0

226x2 - 241x + 64 = 0

Using factoring 

(113x - 64) (2x - 1) = 0

x = 64/113 and x = 1/2

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