Thread: Word problems Homework Help Urgent

1. Word problems Homework Help Urgent

I am extremely bad at word problems and I have been sitting and staring at these 3 problems and I still have no idea on how to set them up. Please is there anyone with either good advice as to how figure out word problems or just how to solve these ones. Thanks!

1. a woman earns 15% more than her husband. Together they make $69,875 per year? What is the husbands annual salary? 2. a movie star, unwilling to give his age, posed the following riddle. "seven years a go I was eleven times as old as my daughter. Now I am four times as old as she is. What is the star's age? 3. Mary has$3.00 in nickles, dimes, and quarters. If she has twice as many dimes as quarters and five more nickles than dimes. How many of each coin does she have?

2. Originally Posted by lucy2284
I am extremely bad at word problems and I have been sitting and staring at these 3 problems and I still have no idea on how to set them up. Please is there anyone with either good advice as to how figure out word problems or just how to solve these ones. Thanks!

1. a woman earns 15% more than her husband. Together they make $69,875 per year? What is the husbands annual salary? 2. a movie star, unwilling to give his age, posed the following riddle. "seven years a go I was eleven times as old as my daughter. Now I am four times as old as she is. What is the star's age? 3. Mary has$3.00 in nickles, dimes, and quarters. If she has twice as many dimes as quarters and five more nickles than dimes. How many of each coin does she have?
Hi Lucy,

Problem 1:

Let x = husband's salary
Let x + .15x = woman's salary

$x+x+.15x=69875$

Solve for x

Problem 2:

Let x = daughter's age now
Let x - 7 = daughter's age 7 years ago

11(x - 7) = 4x - 7

Find x and then multiply by 4 to get star's age

Problem 3:

Let q = # quarters
Let 2q = # dimes
Let 2q + 5 = # nickels

25q + 10(2q) + 5(2q + 5) = 300

Solve for q, then 2q, then 2q + 5

3. Hello, Lucy!

2. A movie star, unwilling to give his age, posed the following riddle.
"Seven years ago I was eleven times as old as my daughter.
Now I am four times as old as she is."
What is the star's age?

For an Age Problem, I usually set up a chart.

Make a row for each person; make a column for each time period.

. . $\begin{array}{c||c|c|}
& \text{Now} & \text{7 yrs.ago} \\ \hline \hline
\text{Father}& & \\ \hline \text{Daughter} & & \\ \hline\end{array}$

Let $x$ = daughter's age now.
Then $4x$ = father's age now.
. . Write those in the first column.

. . $\begin{array}{c||c|c|}
& \text{Now} & \text{7 yrs.ago} \\ \hline \hline
\text{Father}& 4x & \\ \hline \text{Daughter} & x & \\ \hline\end{array}$

Seven years ago, they were both seven years younger.
The father was was $4x-7$ years old, his daughter was $x-7.$
. . Write those in the second column.

. . $\begin{array}{c||c|c|}
& \text{Now} & \text{7 yrs.ago} \\ \hline \hline
\text{Father}& 4x & 4x-7 \\ \hline \text{Daughter} & x & x-7 \\ \hline\end{array}$

The equaton comes from the second column.
What infomation were we given?
"Seven years ago, he was eleven times as old as his daugher."

There! . . . $4x-7 \;=\;11(x-7)$

Solve for $x$ and we get: . $x = 10$

But they asked for the father's age . . . This is: . $4x \,=\,\boxed{40}$