# QUADRATIC EQUATION WORD PROBLEMS WORKSHEET WITH ANSWERS

Problem 1 :

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

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 ?

Problem 3 :

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

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.

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. Find the length of each side of the equilateral triangle. ## Solutions

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 is 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.

Then,

x  =  16

So, the required number is 16.

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  is

=  60 / x

Cost of one meter of the rod which is 2 meter shorter is

=  60 / (x - 2)

Given : If the rod was 2 meter shorter and each meter costs \$1 more.

That is, 60/(x-2) is \$1 more than 60/x.

[60 / (x - 2)]  -  [60 / x]  =  1

Simplify.

[60x - 60(x - 2)]  /  [x(x - 2)]  =  1

[60x - 60x + 120]  /  [x² - 2x]  =  1

120  /  (x2 - 2x)  =  1

120  =  x2 - 2x

0  =  x2 + 2x - 120

x2 + 2x - 120  =  0

(x + 10)(x - 12)  =  0

x  =  - 10  or  x  =  12

Because length can not be a negative number, we can ignore "- 10".

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

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).

Given : Sum of the reciprocals of the parts is 1/6.

Then, we have

1/x  +  1/(25 - x)  =  1/6

Simplify.

(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

When x  =  15,

25 - x  =  25 - 15

25 - x  =  10

When x  =  10,

25 - x  =  25 - 10

25 - x  =  15

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

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, we have

(x + 4)2 + x2  =  202

Simplify.

x2 + 8x + 16 + x2  =  400

2x2 + 8x + 16  =  400

Subtract 400 from both sides.

2x2 + 8x - 384  =  0

Divide both sides by 2.

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

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

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. Find the length of each side of the equilateral triangle.

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, we have

(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, the side represented by (x - 14) will be negative.

So, the length of each side of the equilateral triangle is 17 units. Apart from the problems given above, if you need more word problems on quadratic equations

Please click here

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

If you have any feedback about our math content, please mail us :

v4formath@gmail.com

We always appreciate your feedback.

You can also visit the 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

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation

Word problems on unit price

Word problems on unit rate

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

Double facts word problems

Trigonometry word problems

Percentage word problems

Profit and loss word problems

Markup and markdown word problems

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

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

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

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

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

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6

Featured Categories

Math Word Problems

SAT Math Worksheet

P-SAT Preparation

Math Calculators

Quantitative Aptitude

Transformations

Algebraic Identities

Trig. Identities

SOHCAHTOA

Multiplication Tricks

PEMDAS Rule

Types of Angles

Aptitude Test 