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Algebra Word Problems

Equations are frequently used to solve practical problems.

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Thursday, May 26, 2016

Equations are frequently used
to solve practical problems.

The steps involved in the method of solving
an algebra word problem are as follows.

STEP 1 :

Read the problem carefully and note down
what is given and what is required.

STEP 2 :

Select a letter or letters say x and y to represent
the unknown quantityies asked for.

STEP 3 :

Represent the word statements of the problem
in the symbolic language step by step.

STEP 4 :

Look for quantities which are equal as per
conditions given and form an equation or equations.

STEP 5 :

Solve the equations obtained in step 4.

STEP 6 :

Check the result for making sure that your answer
satisfies the requirements of the problem.

EXAMPLE 1 on Linear equations in one variable

Statement of the Problem :

One fifth of a number of butterflies in a garden are on jasmines
and one third of them are on roses. Three times the difference of
the butterflies on jasmines and roses are on lilys.If the remaining
one is flying freely find the number of butterflies in the garden.

solution to the problem :

Let x be the number of butterflies in the garden.
As per data
Number of butterflies on jasmines = x/5.
Number of butterflies on roses = x/3.
Then
difference of the butterflies on jasmines and roses
= x/3 x/5
As per data
Number of butterflies on lilys
= Three times the difference of the butterflies on jasmines and roses
= 3x/3 x/5
As per data
Number of butterflies flying freely = 1.
So number of butterflies in the garden = x
= Number of butterflies on jasmines Number of butterflies on roses
Number of butterflies on lilys Number of butterflies flying freely
= x/5 x/3 3x/3 x/5 1.
So x
= x/5 x/3 3x/3 3x/5 1.

This is the Linear Equation formed by converting
the given word statements to the symbolic language.
Now we have to solve this equation.
So x = x/5 x/3 x 3x/5 1
Cancelling x which is present on both sides we get
0 = x/5 x/3 3x/5 1
L.C.M. of the denominators 3 5 is 35 = 15.
Multiplying both sides of the equation with 15 we get
150 = 15x/5 15x/3 153x/5 151
i.e. 0 = 3x 5x 33x 15.
i.e. 0 = 8x 9x 15
i.e. 0 = x 15
i.e. 0 x = 15
i.e. x = 15.

Number of butterflies in the garden = x = 15. Ans.

Check:
Number of butterflies on jasmines = x/5 = 15/5 = 3.
Number of butterflies on roses = x/3 = 15/3 = 5.
Number of butterflies on lilys = 35 3 = 32 = 6.
Number of butterflies flying freely = 1.
Total butterflies = 3 5 6 1. = 15. Same as the Ans.verified.

EXAMPLE 2 on Linear Equations in Two Variables

Statement of the problem :

A and B each have a certain number of marbles. A says to B
" if you give 30 to me I will have twice as many as left with you."
B replies "if you give me 10 I will have thrice as many as left with you."
How many marbles does each have?

Solution to the problem :

Let x be the number of marbles A has.
And Let y be the number of marbles B has.
If B gives 30 to A then A has x 30
and B has y 30.
By data When this happens A has twice as many as left with B.
So x 30 = 2y 30
= 2y 230 = 2y 60.
i.e. x 2y = 60 30
i.e. x 2y = 90 ..........i
If A gives 10 to B then A has x 10
and B has y 10.
By data
When this happens B has thrice as many as left with A.
So y 10 = 3x 10
= 3x 310 = 3x 30
i.e. y 3x = 30 10
i.e. 3x y = 40 ...........ii

Equations i and ii are the Linear Equations formed
by converting the given word statements to the symbolic language.

Now we have to solve these simultaneous equations.
To solve i and ii Let us make y coefficients same.

ii2 gives 6x 2y = 80 ...........iii
x 2y = 90 ..........i
Subtracting 5x = 80 90 = 80 90 = 170
i.e. x = 170?5 = 34.
Using this in Equation ii we get 334 y = 40
i.e. 102 y = 40
i.e. y = 40 102 = 62
i.e. y = 62.
Thus A has 34 marbles and B has 62 marbles. Ans.

Check:
If B gives 30 to A from his 62 then A has 34 30 = 64
and B has 62 30 = 32. Twice 32 is 64. verified.
If A gives 10 to B from his 34 then A has 34 10 = 24
and B has 62 10 = 72. Thrice 24 is 72. verified.

EXAMPLE 3 on Quadratic Equations

Statement of the Problem.

A cyclist covers a distance of 60 km in a given time.
If he increases his speed by 2 kmph
he will cover the distance one hour before.
Find the original speed of the cyclist.

Solution to the Problem :

Let the original speed of the cyclist be x kmph.
Then the time cyclist takes to cover a distance of 60 km = 60/x
If he increases his speed by 2 kmph the time taken
= 60/x 2
By data the second time is less than the first by 1 hour.
So 60/x 2 = 60/x 1
Multiplying both sides with x 2x we get
60x = 60x 2 1x 2x
= 60x 120 x2 2x
i.e. x2 2x 120 = 0
Comparing this equation with
ax2 bx c = 0 we get
a = 1 b = 2 and c = 120
We know by Quadratic Formula
x
= b ?b2 4ac/2a
Applying this Quadratic Formula here we get
x = b ?b2 4ac/2a
= 2 ? 22 41 120/21
= 2 ? 4 41120/2
= 2 ?41 120/2
= 2 ?4121/2
= 2 211/2 = 111 = 111 or 111 = 10 or 12
But x can not be negative. So x = 10.
So The original speed of the cyclist = x kmph. = 10 kmph. Ans.

For more math Word problems and the relevant topics go to

http://www.mathhelpace.com/MathWordProblems.html

About the writer:nbsp;nbsp;Name : kvln
Qualifications : B.Tech. M.S. from IIT Madras
Has 14 years of teaching experience.
Love for math and love for teaching
makes me feel more than happy to help.
Has own math web site :
http://www.mathhelpace.com/

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