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# need step by step instruction about how to solve this problem

from 1900 to 2000, the population rate of seattle washington grew at a rate of about 4700 people to a population of about 565,000. Baltimore maryland decreased at a rate of about 8400 people per year to a population of about 650000. If these rates continue in about how many years will seattle and baltimore have the same population

Hi Jen;
I worked on the premise that this was 1990-2000 because continuous growth over a 100-period is extremely unlikely.  You indicated 1900-2000.
Please let me know if I am mistaken.

### 4 Answers by Expert Tutors

Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (9 lesson ratings) (9)
0
Seattle increases at a rate of 4700 people per year.
Baltimore decreases at a rate of 8400 people per year.
let x= # of years until the populations are the same
565,000+4700x=650,000-8400x
4700x+8400x=650,000-565,000
13,100x=85,000
x=85,000/13,100
x=6.488549 years or about 6.5 years

Arthur,

You are the only one that I saw here that didn't read additional information into this problem.  You have Seattle with a population of 565,000 adding 4,700 people per year.  Baltimore starts with a population of 650,000 losing 8,400 people per year. You did this exactly as it was stated.  Keep it simple is exactly what you did!
Thank you Charles.
Arthur, your answer doesn't say when the 6.5 years is measured from. It's measured from 2000.
Hi Steve, I agree, putting the year down would have been more info for the student, but I only saw "in about how many years". Who knows, the problem might have asked for the date when the populations would be the same. Sometimes we don't always get all of the information in the problem.

"from [during] 1900 to 2000, the population rate of seattle washington grew at a rate of about 4700 people to a population of about 565,000 [in 2000]." "If these rates continue [from 2000] in about how many years [when] will seattle and baltimore have the same population."
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
-1
Hi Jen;
I assume "4700 people" is 4700 people/year.
I also assume this is a ten year period of 1990-2000, not 100 years from 1900-2000.  Please correct me if I am wrong.
S=initial population of Seattle
S+[(4,700 people/year)(10 years)]=565,000 people
Let's solve...
Let's note the fact that the unit of year is in both the denominator and numerator.  It cancels...
S+[(4,700 people/year)(10 years)]=565,000 people
S+[(4,700 people)(10)]=565,000 people
The only unit remaining is people.  This is what we want.
S+47,000 people=565,000 people
Let's subtract 47,000 from both sides...
S+47,000-47,000=565,000-47,000
S=518,000

B=initial population of Baltimore
B-[(8,400 people/year)(10 years)=650,000 people
B-[(8,400 people/year)(10 years)=650,000 people
B-[(8,400 people)(10)]=650,000 people
B-84,000 people=650,000 people
B-84,000 people+84,000 people=650,000 people + 84,000 people
B=734,000

x=quantity of years for both populations to be equal...
(518,000 people)+[(4,700 people/year)(x years)]=(734,000 people)-[(8,400 people/year)(x years)]
The unit of years is in the numerator and denominator of the bracketed equations.  It cancels...
(518,000 people)+[(4,700 people/year)(x years)]=(734,000 people)-[(8,400 people/year)(x years)]
(518,000 people)+[(4,700 people)(x)]=(734,000 people)-[(8,400 people)(x)]
The unit of people is in the numerator of all facets of the equation.  It cancels...
(518,000 people)+[(4,700 people)(x)]=(734,000 people)-[(8,400 people)(x)]
(518,000)+[4,700(x)]=(734,000)-[8,400(x)]
Let's subtract 518,000 from both sides...
4,700(x)=216,000-[8,400(x)]
Let's add 8,400(x) to both sides...
13,100x=216,000
Let's divide both sides by 13,100...
x=16 years

Vivan,

You read far too much into the problem. Please look at Arthur D.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
-1
It was a mistake. So it has to be done over.

S -- population of Seattle in 1900

S + 4700 ( 10 ) = 565000
S = 565,000 - 47,000 = 518,000

B -- Population of Baltimore in 1900.

B - 8400(10) = 650,000
B = 65,000 + 84,000 = 734,000

734, 000 -n ( 8400) = 518,000 + n ( 4700)

734,000 - 518,000 = n ( 4700 + 8400)

216, 000 = n ( 13,100)

n = 216,000 / 13, 100 = 16.5∼ 17
In Year 1917, both population becomes equal.

Parviz,

The whole year thing is irrelevant! Just start with Seattle having a population of 565,000 and increasing by 4,700 annually and Baltimore starting at 650,000 and decreasing at a rate of 8,400 annually. Look at Arthur D.'s explanation.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
-1
Seattle: (year 2000, 565000 people), m = 4700 people/year

Using point-slope form:

S - 565000 = 4700)(t - 2000)

S = 4700(t - 2000) + 565000

Baltimore: (2000, 650000), m = -8400

B - 650000 = -8400(t - 2000)

B = -8400(t - 2000) + 650000

When same?

S = B

4700(t-2000)+565000 = -8400(t-2000)+650000
+8400(t-2000)-565000 = +8400(t-2000)-565000
13100(t-2000) + 0 = 0 + 85000

t-2000 = 85000/13100 = 850/131

t = 850/131+2000 = 850/131+131*2000/131 = 262850/131

t ≈ 2006.48854961832061

By this model the populations were the same at 5:46:16 PM on May 26, 2006.