QUADRATIC EQUATION WORD PROBLEMS WORKSHEET WITH ANSWERS

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Quadratic equation word problems worksheet with answers is much useful to the students who would like to practice problems on quadratic equations.

Quadratic equation word problems worksheet

1) Difference between a number and its positive square root is 12. Find the number.

2) A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod?

3) Divide 25 in two parts so that sum of their reciprocals is 1/6

4) The hypotenuse of a right angled triangle is 20 cm. The difference between its other two side is 4 cm. Find the length of the sides.

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

6) The area of a rectangular field is 2000 sq.m and its perimeter is 180 m. Find the length and width of the field.

7) A distributor of apple juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D is given by D = -2000p² + 2000p + 17000. Find the price per bottle that will result zero inventory.

8) Two squares have sides p cm and (p+5) cm. The sum of their squares is 625 sq.cm. Find the sides of the squares.

9) A ball is thrown upwards from a rooftop which is above from the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball "h" from the ground at time "t" seconds is given by, h = -16t² + 64t + 80.How long will the ball take to hit the ground?

10) A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

Quadratic equation word problems worksheet - Answers

Problem 1 :

Difference between a number and its positive square root is 12. Find the number.

Solution :

Let "x" be the required number.

Its positive square root is √x

Given : Difference between x and √x = 12

x - √x = 12

x - 12 = √x

(x-12)² = x

x² -24x + 144 = x

x² -25x + 144 = 0

(x - 9)(x-16) = 0

x = 9 or x = 16

x = 9 does not satisfy the condition given in the question.

Hence the required number is 16.

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Problem 2 :

A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod?

Solution :

Let "x" be the length of the given rod.

Then the length of the rod 2 meter shorter is (x-2) and the total cost of both the rods is $60 (because cost would remain unchanged)

Cost of one meter of the given rod = 60/x

Cost of one meter of the rod which is 2 meter shorter = 60/(x-2)

From the question, 60/(x-2) = (60/x) + 1

60/(x-2) = (60+x)/x

60x = (x-2)(60+x)

60x = x² + 58x - 120

x² -2x -120 = 0

(x+10)(x-12) = 0

x = -10 or x = 12

x = -10 can not be accepted. Because length can not be negative.

Hence, the length of the given rod is 12 m.

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Problem 3 :

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

Solution :

Let "x" be one of the parts of 25. Then the other part is (25 - x).

Sum of the reciprocals of the parts = 1/6

1/x + 1/(25 - x) = 1/6

(25-x+x) / x(25 - x) = 1/6

25 / (25x - x²) = 1/6

6(25) = 25x - x²

150 = 25x - x²

x² - 25x + 150 = 0

(x - 15)(x - 10) = 0

x = 15 or x = 10

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

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Problem 4 :

The hypotenuse of a right angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Solution :

Let "x" and "x+4" be the lengths of other two sides.

Using Pythagorean theorem, (x+4)² + x² = 202

x² + 8x + 16 + x² - 400 = 0

2x² + 8x - 384 = 0

x² + 4x - 192 = 0

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

x = -16 or x = 12

x = -16 can not be accepted. Because length can not be negative.

If x = 12,

x + 4 = 12 + 4 = 16

Hence, the other two sides of the triangle are 12 cm and 16 cm.

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Problem 5 :

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

Solution :

Let "x" be the length of each side of the equilateral triangle.

Then, the sides of the right angle triangle are (x-12), (x-13) and (x-14)

In the above three sides, the side represented by (x -12) is hypotenuse (because that is the longest side).

Using Pythagorean theorem, (x-12)² = (x-13)² + (x-14)2

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

x² - 30x + 221 = 0

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

x = 13 or x = 17.

x = 13 can not be accepted. Because, if x = 13, one of the sides of the right angle triangle would be negative.

Hence, the side of the equilateral triangle is 17 units.

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Problem 6 :

The area of a rectangular field is 2000 sq.m and its perimeter is 180 m. Find the length and width of the field.

Solution :

Let "x" and "y" be the length and width of the rectangle respectively.

Perimeter = 180 m (given)

2x + 2y = 180 ---> x + y = 90

So, y = 90 - x

Area = 2000 sq.m

xy = 2000

x(90-x) = 2000

90x - x² = 2000

x² - 90x + 2000 = 0

(x-50)(x-40) = 0

x = 50 or x = 40

If x = 50, y = 90 - 50 = 40

If x = 40, y = 90 - 40 = 50

Hence, the length and breadth of the rectangle are 40 m and 50 m respectively.

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Problem 7 :

A distributor of apple juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D is given by D = -2000p² + 2000p + 17000. Find the price per bottle that will result zero inventory.

Solution :

Stock in the store is 5000 bottles. If the inventory be zero, the demand has to be 5000 bottles.

To know the price for zero inventory, we have to plug D = 5000 in the quadratic equation D = -2000p2 + 2000p + 17000

5000 = -2000p² + 2000p + 17000

2000p² - 2000p - 12000 = 0

p² - p - 6 = 0

(p + 3)(p - 2) = 0

p = -3 or p = 2

p = -3 can not be accepted. Because, price can not be negative.

Hence, price per bottle that will result zero inventory is $2.

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Problem 8 :

Two squares have sides p cm and (p+5) cm. The sum of their squares is 625 sq.cm. Find the sides of the squares.

Solution :

According to the question, we have

p² + (p+5)² = 625

p² + p² + 10p + 25 = 625

2p² + 10p - 600 = 0

p² + 5p - 300 = 0

(p -15)(p + 20) = 0

p = 15 or p = -20

p = -20 can not be accepted. Because, the side of the square can not be negative.

If p = 15, p + 5 = 15 + 5 = 20

Hence, the sides of the square are 15 cm and 20 cm.

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Problem 9 :

A ball is thrown upwards from a rooftop which is above from the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball "h" from the ground at time "t" seconds is given by, h = -16t² + 64t + 80.How long will the ball take to hit the ground?

Solution :

When the ball hits the ground, height "h" = 0.

So, we have

0 = -16t² + 64t + 80

16t² - 64t - 80 = 0

t² - 4t - 5 = 0

(t - 5)(t+1) = 0

t = 5 or t = -1

t = -1 can not be accepted. Because time can never be a negative value.

Hence, the ball will take 5 seconds to hit the ground.

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Problem 10 :

A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

Solution :

After enlargement, let "x" be the width of picture.

Then, the height of the picture = 4x / 3

(From the given information, it is very clear that the picture is in the shape of rectangle. So, the given height can be considered as length of the rectangle)

Given : Area after enlargement = 192 square inches.

So, we have

Length X width = 192 sq. inches

(4x/3)(x) = 192

4x²/3 = 192

x² = 192(3/4)

x² = 144

x = 12

Therefore the width = 12 inches.

The height = 4(12) / 3 = 16 inches

Hence, the dimensions of the picture are 12 inches and 16 inches.

After having gone through the step by step solutions for all the problems on "Quadratic equation word problems worksheet with answers", we hope that the students would have understood how to solve quadratic equation word problems.

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