Word Problem HELP!!!!!

cvandoren

New member
Joined
Nov 14, 2005
Messages
19
Tim is 100% older then his sister, but in 8 years he will be only 20% older. How old is tim and his sister now?



99 girls and 1 boy are in a school.(99% of school are girls.) How many girls lead the school if the percent of girls becomes 98%. (no more boys may enter the school)
 
problem

Now on first one, letting t be tims age, and s his sisters.
If he is 100% older then his sister now, wouldn't t = s + 100%s = 2s, so t = 2s?
ANd in 8 years, wouldnt the equation be t +8 = (s+8) + ((1/5)(s+8)) ?
 
1) Pick a variable for her current age. Write an expression for "one hundred percent of her age". Add this to her current age, to get an expression for his current age.

You now have a variable for her and an expression for him. Add "8" to each. This gives you expressions for their ages in eight years.

Take the expression for her age in eight years. Use this to find an expression for "twenty percent of her age in eight years". Then:

. . . . .(his age then) will be (her age then) plus (twenty percent of her age then)

Translate this into an equation, using the expressions you have created. Solve for the variable.

If you get stuck on either exercise, please reply showing how far you have gotten in following these steps. Thank you.

2) I'm sorry, but I don't understand what is meant by "leading" the school. Please clarify. Thank you.

Eliz.

P.S. Please post any future algebra questions to one of the "algebra" categories. Thank you.
 
cvandoren said:
99 girls and 1 boy are in a school.(99% of school are girls.) How many girls lead the school if the percent of girls becomes 98%. (no more boys may enter the school)
I'm sure you mean "leave the school".

What happens if 50 girls leave?
 
Hello, cvandoren!

The second problem has a surprising answer . . .

99 girls and 1 boy are in a school. (99% of school are girls.)
How many girls leave the school if the percent of girls becomes 98%?
(No more boys may enter the school)

Let \(\displaystyle N\) = number of girls that leave.

Then there are only \(\displaystyle 99\,-\,N\) girls
and there are only \(\displaystyle 100\,-\,N\) students.

Then the proportion of girls to students is: .\(\displaystyle \frac{99\,-\,N}{100\,-\,N}\:=\:\frac{98}{100}\)

Solve for \(\displaystyle N\) and we get: .\(\displaystyle N\,=\,50\)

Therefore, 50 girls must leave to lower the ratio to 98%.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

<u>Check</u>

Originally: 99 girls, 1 boy.
. . Percentage of girls\(\displaystyle \,=\,\frac{99}{100}\,=\,99%\)

Then: 49 girls, 1 boy.
. . Percentage of girls\(\displaystyle \,=\,\frac{49}{50}\,=\,98%\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This problem is usually presented as a "mixture problem":

There is 100 liters of a solution which is 99% acid.
How much acid must be removed to have a solution which is 98% acid?
 
Top