SOLUTION: I'm in algebra one is 7th grade, and i hope you can help. I'm doing these two problems and i have no idea how to solve them. Here are the directions: Set up each inequality and

Question 22002: I'm in algebra one is 7th grade, and i hope you can help. I'm doing these two problems and i have no idea how to solve them.
Here are the directions: Set up each inequality and show all work. Written explanations should be in complete sentences.
Jesse has 27 coins, some of which are dimes and the rest are quarters. Altogether the coins are worth more than $4.20. At least how many of the coins are quarters? At most how many are dimes?
Mr. A and Mrs. N are leaving school and traveling in opposite directions. If Mrs. N runs 2 miles per hour and Mr. A runs 5 miles per hour, how long would it take them to be at least 15 miles away from each other?

Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!
Let number of quarters "x".
Then number of dimes is "27-x"
Value of quarters is (25)x cents
Value of dimes is (10)(27-x) = 270-10x cents
INEQUALITY:
Value of quarters + Value of dimes >= 420 cents
25x + 270-10x >=420
15x >= 150
x >= 10 (number of quarters)
27-x <= 17 (number of dimes)
d= rt
A data: Let distance be "x"; rate is 5; time is x/5
N data: Then distance is "15-x" ; rate is 2; time is (15-x)/2
INEQUALITY(?):
Whatever the distance is between them the times are equal, so:
time of "A" = time of "N"
x/5 = (15-x)/2
2x= 5(15-x)
2x = 75-5x
7x = 75
x = (75/7) miles (A's distance"
Since "A" runs at 5 mph divide her distance by 5 to get:
75/7/5 = 15/7 = 2 (1/7) hrs.
Therefore for them to be "at least 15 miles apart" their
time must be >=2 1/7 hrs.
Cheers,
Stan H.