Question 143667: I am having a lot of trouble with word problem questions. If someone could help me that would be great. If you could show your work so that I can understand how you got the answer would be apprectiated. Thanks so much!
*In a certain medical test designed to measure carbohydrate tolerance, and adult drinks 7 ounces of a 30% glucose solution. When the test is administered to a child, the glucose concentration must be decreased to 20%. How much 30% glucose solution and how much water should be used to prepare 7 ounces of 20% glucose solution?
*Theophylline, is prepared from an elixir with a drug concentration of 5 mg/ml and a cherry-flavored syrup that is to be added to hide the taste of the drug. How much of each must be used tp prepare 100 mililiters of solution with a drug concentratio of 2 mg/ml?
*Two children who are 224 meters apart, start walking toward each other at the same instant at rates of 1.5m/sec and 2 m/sec, respectively.
(a) When will they meet?
(b) How far will each have walked?
*A wafer cone is to hold 8 in^3 of ice cream when filled to the bottom. The diameter of the cone is 2 inches, and the top of the ice cream has the shape of a hemisphere. Find the height (h) of the cone.
*A runner starts at the beginning of a runners' path and runs at a constant rate of 6 mi/hr. Five minutes later a second runner begins at the same point, running at a rate of 8 mi/hr and following the same course. How long will it take the second runner to reach the first?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! In a certain medical test designed to measure carbohydrate tolerance, and adult drinks 7 ounces of a 30% glucose solution. When the test is administered to a child, the glucose concentration must be decreased to 20%. How much 30% glucose solution and how much water should be used to prepare 7 ounces of 20% glucose solution?
:
Let x = amt of 30% glucose solution
We want the resulting total to be 7 oz, therefore
(7-x) = amt of water required (water is a 0% solution)
:
A percent glucose equation.
:
The amt of glucose remains the same, (the percent changes by adding water)
:
.30x = .20(7)
.3x = 1.4
x = 
x = 4 oz of 30% solution
then
7 - 4 = 2 oz of water
:
:
*Theophylline, is prepared from an elixir with a drug concentration of 5 mg/ml and a cherry-flavored syrup that is to be added to hide the taste of the drug. How much of each must be used to prepare 100 milliliters of solution with a drug concentration of 2 mg/ml?
:
Let x = amt of 5 mg elixir required
then
(100-x) = amt of syrup required
:
5x = 2(100)
5x = 200
x = 
x = 40 ml of elixir
then
100-40 = 60 ml of syrup
:
;
*Two children who are 224 meters apart, start walking toward each other at the same instant at rates of 1.5m/sec and 2 m/sec, respectively.
:
The total distance of both will equal 224 meters. Write a distance equation:
Dist = speed * time
:
(a) When will they meet?
Let t = time (in sec) when they meet:
;
child 1 dist + child 2 dist = 224 meters
1.5t + 2t = 224
3.5t = 224
t = 
t = 64 seconds
:
(b) How far will each have walked?
now we know t
1.5*64 = 96 meters (child 1)
2.0*64 = 128 meters (child 2)
--------------
total = 224, confirms our solution
:
:
*A wafer cone is to hold 8 in^3 of ice cream when filled to the bottom. The diameter of the cone is 2 inches, and the top of the ice cream has the shape of a hemisphere. Find the height (h) of the cone.
:
The vol of a cone + the vol of half sphere = 8 cu in
r=1 in (half the diameter
 +  = 8
using the given values for r
 +  = 8
:
 +  = 8
Use a calc
1.0472h + 1.5708 = 8
:
1.0472h = 8 - 1.5708
:
1.0472h = 6.4292
h = 
h = 6.14 inches the height of the cone part only, the part including hemisphere would be an inch higher, 7.14"
:
:
*A runner starts at the beginning of a runners' path and runs at a constant rate of 6 mi/hr. Five minutes later a second runner begins at the same point, running at a rate of 8 mi/hr and following the same course. How long will it take the second runner to reach the first?
;
We know that when the 2nd runner catches up with the 1st, they will have run the same distance.
:
Change 5 min to hrs: 5/60 = 1/12
:
Let t = run time of the 2nd runner
then
(t+ ) = run time of the 1st runner
:
Write a dist equation: Dist = speed * time
:
Fast runner dist = slow runner dist
8t = 6(t+)
8t = 6t + 
8t - 6t -
2t =
t = *
t =  hr or we can say 15 min for the 2nd runner to catch the 1st
:
:
We can check solution by finding the distance traveled by each runner (should be equal)
1st runners time = 20 min, 2nd runner 15 min
6 * (20/60) = 2 mi
8 * (15/60) = 2 mi
:
Hope you can follow all this, you can ask me about it until 9:15 central daylight or tomorrow. ankor@dixie-net.com
|
|
|
