**Worksheet on Word Problems on Quadratic Equations :**

Worksheet given in this section will be much useful for the students who would like to practice solving word problems on quadratic equations.

**Problem 1 :**

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

**Problem 2 :**

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.

**Problem 3 :**

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.

**Problem 4 :**

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 = -16t**²** + 64t + 80.How long will the ball take to hit the ground?

**Problem 5 :**

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 ?

**Problem 1 :**

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 rectangular field respectively. **Given :** Perimeter of the rectangle is 180 m.

Then, we have

2x + 2y = 180

Divide both sides by 2.

x + y = 90

Subtract x from both sides.

y = 90 - x

Formula to find the area of a rectangle is

**Given :** Area of the rectangle is 2000 sq.m

That is,

length ⋅ width = 2000

x ⋅ y = 2000

Plug y = 90 - x.

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

y = 40

If x = 40,

y = 90 - 40

y = 50

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

**Problem 2 :**

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

Divide both sides by 2000.

p² - p - 6 = 0

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

p - 3 = 0 or p + 2 = 0

p = 3 or p = - 2

p = -2 can not be accepted. Because, price can not be negative.

Hence, price per bottle that will result zero inventory is $3.

**Problem 3 :**

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

p + 5 = 20

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

**Problem 4 :**

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 = -16t**²** + 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 = -16t**²** + 64t + 80

16t**²** - 64t - 80 = 0

t**²** - 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.

**Problem 5 :**

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 is**

**= 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 is 192 sq. inches.**

**That is, **

length ⋅ width = 192

(4x / 3) ⋅ x = 192

4x**² **/ 3 = 192

Multiply both sides by 3/4.

x**²** = 144

Take radical on both sides.

x = 12

**So, the width is 12 inches.**

**Then, the height is **

**= 4(12) / 3**

**= 16 inches**

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

Apart from the problems given above, if you need more word problems on quadratic equations

After having gone through the stuff given above, we hope that the students would have understood how to solve word problems using quadratic equations.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

You can also visit our following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Trigonometry word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**