Question 69076: If someone could help me out with these I'll appreciated. THank you.
Problem#6
Directions: Solve the following word problems.
Number problem. If one-half of one integer is subtracted from three-fifths of the next consecutive integer, the difference is 3. What are the two integers?
Problem# 18 same directions
Science and medicine. A passenger train can travel 325 mi in the same time a
freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi/h faster than the speed of the freight train, find the speed of each.
Problem#22 same directions
Science and medicine. Ariana took 2 h longer to drive 360 mi on the first day of a trip than she took to drive 270 mi on the second day. If her speed was the same on both days, what was the driving time each day?
Problem#26 same directions
Science and medicine. A plane flies 720 mi against a steady 30-mi/h headwind
and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?
Problem# 34. same directions
Business and finance. Kevin earned $165 interest for 1 year on an investment of
$1500. At the same rate, what amount of interest would be earned by an investment
of $2500?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Problem#6
Directions: Solve the following word problems.
Number problem. If one-half of one integer is subtracted from three-fifths of the next consecutive integer, the difference is 3. What are the two integers?
:
Let x = one integer; and (x+1) = next consecutive integer
A lot less typing if we use decimals. (3/5) = .6
:
.6(x+1) - .5x = 3
.6x + .6 - .5x = 3
.1x = 3 - .6
.1x = 2.4
x = 2.4/.1
x = 24 and 25 are the integers:
:
Check:
.6(25) - .5(24) =
15 - 12 = 3
:
:
:
Problem# 18 same directions
Science and medicine. A passenger train can travel 325 mi in the same time a
freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi/h faster than the speed of the freight train, find the speed of each.
:
Let s = speed of the freight
Then (s+25) = Pass train
:
Time = dist/speed
:
Freight time = Pass time
 = 
:
Cross multiply:
325s = 200(s+25)
:
325s = 200s + 5000
:
325s - 200s = 5000
:
125s = 5000
:
s = 5000/125
:
s = 40 mph is the speed of the freight
:
40 + 25 = 65 mph is the speed of the pass train
:
Check to see if, in fact, the times are equal:
325/65 = 200/40
5 = 5
:
:
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Problem#22 same directions
Science and medicine. Ariana took 2 h longer to drive 360 mi on the first day of a trip than she took to drive 270 mi on the second day. If her speed was the same on both days, what was the driving time each day?
:
Let t = her time on the 1st day
Then (t-2) = her time on the 2nd day
:
Speed = Dist/time
:
1st day speed = 2nd day speed
 = 
:
Cross multiply and you have:
360(t-2) = 270t
360t - 720 = 270t
360t - 270t = +720
90t = 720
t = 720/90
t = 8 hrs the 1st day
:
8 - 2 = 6 hrs the 2nd day
:
Check to see if the speeds are equal:
360/8 = 270/6
45 = 45
:
:
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Problem#26 same directions
Science and medicine. A plane flies 720 mi against a steady 30-mi/h headwind
and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?
:
Let s = speed in still air:
Then:
(s-30) = speed against the wind
(s+30) = speed with the wind
:
Time = dist/speed
:
Against time + with time = 10 hrs
 +  = 10
:
Multiply equation by (s-30)(s+30) to get rid of the denominators:
720(s+30) + 720(s-30) = 10(s-30)(s+30)
:
720s + 21600 + 720s - 21600 = 10(s^2 - 900)
:
720s + 21600 + 720s - 21600 = 10s^2 - 9000; (21600 is eliminated)
:
1440s = 10s^2 - 9000
:
10s^2 - 1440s - 9000 = 0; our old friend the quadratic equation
:
Simplify, divide by 10
s^2 - 144x - 900 = 0
:
Factor to:
(s-150)(s+6) = 0
:
S = +150 mph speed in still air, (s=-6 is ignored, obviously)
:
Check using the time:
720/120) + 720/180 =
6 + 4 = 10
:
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Problem# 34. same directions
Business and finance. Kevin earned $165 interest for 1 year on an investment of
$1500.
:
Remember P*R*T = I; substitute for P and T and I, find R
:
1500 * R * 1 = 165
1500R = 165
R = 165/1500
R = .11 or 11%
:
Check using the I = PRT formula
I = 1500*.11*1
I = $165
:
At the same rate, what amount of interest would be earned by an investment of $2500?
:
I = 2500*.11*1
I = $275 amt of interest earned on $2500 at a rate of 11%
:
:
A lot of steps here I hope you will satisfy yourself that you understand each.
If you have a question on any of these procedures, you can email me.
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