My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 3rd, 2012, 09:29 PM   #1
Newbie
 
Joined: Mar 2012

Posts: 6
Thanks: 0

I am a little Confused.

Hello, I have a problem that I don't know how to solve. Could someone please explain to me what I am suppose to do?

"A car leaves a town traveling at 40 miles per hour. Two hours later a second car leaves the town traveling the same route and overtakes the first car in 5 hours and 20 minutes. How fast was the second car traveling?"
Siedas is offline  
 
March 3rd, 2012, 09:43 PM   #2
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: I am a little Confused.

Both cars travel the same distance, while the first car travels for 2 hr more. Let t be the time for the second car then t + 2 is the time for the first car. Let v be the speed of the second car which we assume is constant. Using distance equals rate times time, we may state:





We are told giving us:

MarkFL is offline  
March 4th, 2012, 06:47 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 20,622
Thanks: 2075

The second car made up 80 miles in hours, so its relative velocity was 15 mph. Hence the answer is 55 mph.
skipjack is offline  
March 8th, 2012, 02:39 PM   #4
Newbie
 
Joined: Mar 2012

Posts: 6
Thanks: 0

Re: I am a little Confused.

Quote:
Originally Posted by MarkFL
Both cars travel the same distance, while the first car travels for 2 hr more. Let t be the time for the second car then t + 2 is the time for the first car. Let v be the speed of the second car which we assume is constant. Using distance equals rate times time, we may state:





We are told giving us:

Could you explain to me how you got that answer?
Siedas is offline  
March 8th, 2012, 03:17 PM   #5
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: I am a little Confused.

We begin with distance equals rate times time. We let v be the rate of the second car, since we don't know the rate and this is what we're asked to find. If the second car traveled for 5 1/3 hr and the first car left 2 hours before the second then the second car traveled for 7 1/3 hr at 40 mph. Since they both traveled the same distance d we may state:



Now we solve for v:

MarkFL is offline  
March 9th, 2012, 05:52 PM   #6
Newbie
 
Joined: Mar 2012

Posts: 6
Thanks: 0

Re: I am a little Confused.

Thanks for the explanation. I have another one as well.

How much water needs to be removed from 20 gallons of 30% salt solution to change it to a 40% salt solution?
Siedas is offline  
March 9th, 2012, 06:20 PM   #7
Senior Member
 
Joined: Jan 2012

Posts: 131
Thanks: 0

Re: I am a little Confused.

30%(Salt) of 20 gallons of solution= 6 gallons
40%(Salt) of x gallons solution= 6 gallons
x=15 gallons.
H20 to be removed=5 gallons.
azizlwl is offline  
March 9th, 2012, 06:24 PM   #8
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: I am a little Confused.

I would state 30% of 20 gallons is 40% of how many (x) gallons? Writing this mathematically where 20 and x are in gallons:





Since 20 - 15 = 5, we see 5 gallons of water must be removed.
MarkFL is offline  
March 10th, 2012, 12:26 PM   #9
Newbie
 
Joined: Mar 2012

Posts: 6
Thanks: 0

Re: I am a little Confused.

Okay, I have another one.

"Suppose that the perimeter of a square equals the perimeter of a rectangle. The width of the rectangle is 9 inches less than twice the side of the square, and the length of the rectangle is 3 inches less then twice the side of the square. Find the dimensions of the square and the rectangle."
Siedas is offline  
March 10th, 2012, 01:29 PM   #10
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,934
Thanks: 1128

Math Focus: Elementary mathematics and beyond
Re: I am a little Confused.

Hi, Siedas. In future please start a new topic for a different (unrelated) question.

Let W be the width of the rectangle. Let L be the length of the rectangle. Let s be the side of the square. From the given information

W = 2s - 9 and L = 2s - 3.

Since perimeter equals 2 times length plus two times width we have

2W + 2L = 4s
2(2s - 9) + 2(2s - 3) = 4s
4s - 18 + 4s - 6 = 4s
4s - 24 = 0
4s = 24
s = 6, so the side of the square is 6, the width of the rectangle is 2(6) - 9 = 3 and the length of the rectangle is 2(6) - 3 = 9.
greg1313 is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
confused



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
So confused with this charlieboon Calculus 5 February 23rd, 2013 10:01 AM
confused. please help! ccfoose Algebra 3 April 16th, 2012 09:55 PM
Confused katfisher1208 Algebra 1 April 20th, 2010 06:59 PM
I am new and confused.... momoftwo New Users 4 August 19th, 2009 06:15 PM
So confused with this charlieboon Algebra 0 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.