Lesson Using quadratic equations to solve word problems

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Using quadratic equations to solve word problems

In this lesson we present some typical word problems that may be solved using quadratic equations.
Solution of quadratic equations is described in the lesson Introduction into Quadratic Equations in this module.

Problem 1. Motorboat moving upstream and downstream on a river

A motorboat makes a round trip on a river 56 miles upstream and 56 miles downstream, maintaining the constant speed 15 miles per hour
relative to the water. The entire trip up and back takes 7.5 hours. What is the speed of the current?

Solution

Let

be the unknown current speed of the river in miles/hour.
When motorboat moves upstream, its speed relative to the bank of the river is 15-x

miles/hour, and the time spent moving upstream is 56%2F%2815-x%29

hours.
When motorboat moves downstream, its speed relative to the bank of the river is 15%2Bx

miles/hour, and the time spent moving downstream is 56%2F%2815%2Bx%29

hours.
So, the total time up and back is 56%2F%2815-x%29+%2B+56%2F%2815%2Bx%29

, and it is equal to 7.5 hours, according to the problem input.
This gives an equation 56%2F%2815-x%29+%2B+56%2F%2815%2Bx%29+=+7.5

.
To simplify the equation, multiply both sides by %2815-x%29%2A%2815%2Bx%29

and collect the common terms. Step by step, you get

56%2815%2Bx%29+%2B+56%2815-x%29+=+7.5%2A%2815-x%29%2815%2Bx%29

.

Answer. The speed of the current is 1 mile/hour.

For more examples of solved word problems of this type see the lesson Wind and Current problems solvable by quadratic equations
under the topic Travel and Distance of the section Word problems in this site.

Problem 2. Working together and separately to complete a job

Andrew and Bill, working together, can cover the roof of a house in 6 days.
Andrew, working alone, can complete this job in 5 days faster than Bill.
How long will it take Bill to make this job?

Solution

Let

be the number of days for Bill to cover the roof, working alone.
If Andrew works alone, he can complete this job in x-5

days.
Thus, in one single day Andrew covers 1%2F%28x-5%29

part of the roof area, while Bill covers 1%2Fx

part of the roof area.
Working together, Andrew and Bill make 1%2F%28x-5%29%2B1%2Fx

of the whole work in each single day.
Since they can cover the entire roof in 6 days working together, the equation for the unknown value

is as follows:
6%2F%28x-5%29+%2B+6%2Fx+=+1

.

To simplify this equation, multiply both sides by %28x-5%29%2Ax

, then transfer all terms from the right side to the left with the opposite signs, collect the common terms
and adjust the signs. In this way you get
6x+%2B+6%28x-5%29+=+x%28x-5%29

.
You get the quadratic equation. Apply the quadratic formula (see the lesson Introduction into Quadratic Equations) to solve this equation. You get

.
The equation has two roots: x%5B1%5D=%2817%2B13%29%2F2=15

and

.
The second root x%5B2%5D=2

does not fit the given conditions: if Bill covers the roof in two days, then Andrew has 2-5=-3 days, which has no sense.
So, the potentially correct solution is x%5B1%5D=15

: Bill covers the roof in 15 days.
Let us check it. If Bill gets the job done in 15 days, then Andrew makes it in 10 days, working separately.
Since 6%2F10%2B6%2F15=1

, this solution is correct.

Answer. Bill covers the roof in 15 days.

Problem 3. Filling the liquid reservoir

Two tubes, working together, can fill the reservoir with the liquid in 12 hours.
The larger tube, if works separately, can fill the reservoir in 18 hours faster than the smaller tube.
How long will it take to fill the reservoir using the smaller tube only?

Solution

Let

be number of hours to fill the reservoir using the smaller tube only.
Then x-18

is the time in hours to fill the reservoir using the larger tube only.
Thus, in one single hour the smaller tube will fill 1%2Fx

part of the reservoir volume, while the larger tube will fill 1%2F%28x-18%29

part of the reservoir volume in each single hour.
Working together, these two tubes will fill 1%2Fx+%2B+1%2F%28x-18%29

part of the reservoir volume in each single hour.
Since two tubes working together fill the tank in 12 hours, this gives the equation

12%2Fx+%2B+12%2F%28x-18%29+=+1

.

To simplify this equation, multiply both sides by %28x-18%29%2Ax

and transfer all terms from the right side to the left with the opposite signs. Then collect the common terms
and adjust the signs. In this way you get
12x+%2B+12%28x-18%29+=+x%28x-18%29

-x%5E2+%2B+12x+%2B+12x+%2B+18x+-+218+=+0

.
You get the quadratic equation. Apply the quadratic formula (see the lesson Introduction into Quadratic Equations) to solve this equation. You get

.
The equation has two roots: x%5B1%5D=%2842%2B30%29%2F2=36

and

.
The second root x%5B2%5D=6

does not fit the given conditions: if the smaller tube fills the reservoir in 6 hours, then the larger one should make it in 6-18=-12 hours,
which has no sense.

So, the potentially correct solution is x%5B1%5D=36

: the smaller tube fills the reservoir in 36 hours.
Let us check it. If the smaller tube fills the reservoir in 36 hours, then the larger one makes it in 36-18=18 hours, working separately.
Since 12%2F36%2B12%2F18=1

, this solution is correct.

Answer. The smaller tube fills the reservoir in 36 hours.

Problem 4

If a beet-sugar factory in England processed 2000 tons less of beets per day, it would take 1 day longer to slice 60,000 tons of them.
How many tons of beets are processed per day?

Solution

Let x = "How many tons of beets are processed per day?", i.e. the value under the question.

Then the condition gives you this equation


     -  = 1,     (1)


which is direct translation of the condition to Math.


To simplify calculations, we can make an agreement that we measure the amounts not in tons, but in THOUSANDS of tons.

Then the equation (1) takes the form


     -  = 1.              (2)


You can solve the equation (2) formally by multiplying both sides by x:


60*x - 60*(x-2) = x*(x-2),

60x - 60x + 120 = x^2 - 2x

x^2 - 2x - 120 = 0

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


Of the two roots  x= 12  and  x= -10  only positive root x= 12 satisfies the condition.


Answer.  12000 tons of beets are processed per day normally.

Problem 5

A realtor bought a group of lots for $90,000. He then sells them at a gain of $3,750 per lot
and has a total profit equal to the amount he received for the last 4 lots sold.
How many lots were originally in the group ?

Solution

Let "n" be the number of lots, now unknown.


Then the bought price for each lot was  ;  the profit of selling all these lots was 3,750n dollars.


Write the equation for the profit as you read the problem

    3750n = .


Divide both sides by 3750

    n = .


Multiply both sides by n 

    n^2 = 96 + 4n


Simplfy

    n^2 - 4n - 96 = 0


Factor left side

    (n-12)*(n+8) = 0.


Of the two roots,  n= 12 and n= -8, only positive root n= 12 gives a meaningful solution.


ANSWER.  12 lots.

Problem 6

When a stone is dropped into a deep well, the number of seconds until the sound of a splash is heard is given
by the formula t = sqrt+%28x%29%2F4

, where x is the depth of the well in feet. For one particular well, the splash is heard 4 seconds
after the stone is released. How deep is the well?

Solution

4 =  + 


Let y =  be new variable.  Then the equation takes the form


4 =   + .


Multiply both sides by 1100 to get 


 = 0


 =  = .


Only positive root works  y =  = 15.16.


Then  x =  =  230 ft  (rounded)


CHECK.   +  = 4.0  seconds.   ! correct !


ANSWER.  The distance to the water surface is about 230 ft.

My other lessons on quadratic equations in this website are
    - Introduction into Quadratic Equations
    - PROOF of quadratic formula by completing the square

    - HOW TO complete the square - Learning by examples
    - HOW TO solve quadratic equation by completing the square - Learning by examples
    - Solving quadratic equations without quadratic formula
    - Who is who in quadratic equations
    - Using Vieta's theorem to solve quadratic equations and related problems

    - Find an equation of the parabola passing through given points
    - Problems on quadratic equations to solve them using discriminant
    - Relative position of a straight line and a parabola on a coordinate plane
    - Advanced minimax problems to solve them using the discriminant principle

    - Word problems on engineering constructions of parabolic shapes
    - Challenging word problems solved using quadratic equations
    - Business-related problems to solve them using quadratic equations
    - Rare beauty investment problem to solve using quadratic equation
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations
    - Entertainment problems on quadratic equations
    - Prime quadratic polynomials with real coefficients

    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

    - Problems on the area and the dimensions of a rectangle
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
    - Problems on a circular pool and a walkway around it

    - OVERVIEW of lessons on solving quadratic equations

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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