Algebra word problems

Forums: Science And Math
Email this Topic • Print this Page

Hi everyone, I am in grade 11, and having some problems with solving algebraic word problems. For example:

Two planes leave airports 6400 km's apart and travel towards each other. They pass each other 2 h's later. One plane travels 200 km/h faster then the other. How far has the faster plane travelled when they pass each other?

And one more I am having trouble with.

The corner store mixes tea worth 1.40/kg and tea worth 2.00/kg to make 100 kg of tea that sells for 1.73/kg. How many kilograms of the 1.40/kg tea are needed.

I am not having problems with the elimination, or the substitution methods. But having trouble deffining my variables. First of all what would my variables be for these two questions. And also, for all other word problems, how can I help my self define the variables?

Thanks alot

Topic Stats
Top Replies
Link to this Topic

Type: Discussion • Score: 1 • Views: 961 • Replies: 3

No top replies

Re: Algebra word problems

Zion - I wrote:

Hi everyone, I am in grade 11, and having some problems with solving algebraic word problems. For example:

Two planes leave airports 6400 km's apart and travel towards each other. They pass each other 2 h's later. One plane travels 200 km/h faster then the other. How far has the faster plane travelled when they pass each other?....

Call their speeds s1 and s2. Here's what we know:

s1(2 hr) + s2(2 hr) = 6400 km
s1 = s2 + 200 km/hr

Substituting the 2nd into the first:

(s2 + 200 km/hr)(2 hr) + s2(2 hr) = 2(s2)(2 hr) + 200 km/hr(2 hr) = 6400 m

2(s2)(2 hr) + 400 km = 6400 km

2(s2)(2 hr) = 6000 km
s2 = 3000 km / 2 hr = 1500 km/hr

Now, substituting this into the 2nd equation we get:

s1 = s2 + 200 km/hr
s1 = 1500 km/hr + 200 km/hr

s1 = 1700 km/hr

So, the two planes have speeds:

s1 = 1700 km/hr
s2 = 1500 km/hr

After two hours, when the planes pass, the faster plane has travelled a distance of 1700 km/hr (2 hr) = 3400 km.

0 Replies

Re: Algebra word problems

Zion - I wrote:

...The corner store mixes tea worth 1.40/kg and tea worth 2.00/kg to make 100 kg of tea that sells for 1.73/kg. How many kilograms of the 1.40/kg tea are needed....

Kilograms of tea that sells for $1.40 = T1
Kilograms of tea that sells for $2.00 = T2

We know that T1 + T2 = 100 Kg.

The total value of the $1.40 tea is T1($1.40)
The total value of the $2.00 tea is T2($2.00)

The price per kilogram of the mixture of them is:

T1($1.40) + T2($2.00)
--------------------------
100 kg

= $1.73

Substituting into the second formula from the first:

(100 kg - T2)($1.40) + T2($2.00)
---------------------------------------
100 kg

= 100 kg($1.40)/100 kg + T2($2.00 - $1.40)/100 kg

= $1.40 + T2($.60)/100 kg = $1.73

T2($.60)/100 kg = $.33

T2 = 100 kg ($.33/$.60)

= 100kg (.55) = 55kg

Since T1 + T2 = 100kg, T1 = 45 kg

0 Replies

If you're just having trouble setting it up, don't worry about making a mess on your paper. That's what it's for.

It might help to name your variables at first (e.g., "speed of plane 1," "distance plane 1 has travelled") and make a list of what you don't have values for. Then assign symbols to the variables and set up as many equations as you can relating them to each other.

As you do more and more of this, you can keep all of the variables in you head. In the mean time, your problems don't have to look neat and concise like they do in a textbook. (Well, maybe what you turn in has to look that way, but the actual work you do doesn't.)

0 Replies

Algebra word problems

Related Topics

Quick Links

My Account

able2know