SOLUTION: kindly solve these problems for me: 1. Charles and Melanie are 100 km apart and are driving at constant speeds. If they drive toward each other, they will meet in 1 hr. If they

Question 155228: kindly solve these problems for me:
1. Charles and Melanie are 100 km apart and are driving at constant speeds. If they drive toward each other, they will meet in 1 hr. If they drive in the same direction, he will overtake her in 5 hr. Find their speeds.
2. Wayne invested part of his money at 8% and the rest at 6%. The income from both investments totaled $2,000. If he interchanged his investments, his income would increase by $340. How much did he have in each investment?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
1. Charles and Melanie are 100 km apart and are driving at constant speeds.
If they drive toward each other, they will meet in 1 hr. If they drive in
the same direction, he will overtake her in 5 hr. Find their speeds.
;
Let x = C's speed
Let y = M's speed
:
When they are driving towards each other, their speeds are additive:
Write a distance equation, Dist = time * speed
1(x+y) = 100
x + y = 100
or
y = (100-x)
:
Write another distance equation, when they go the same direction.
C has to travel has to travel 100 further than M in the same time (5 hrs)
5x - 100 = 5y
Replace y with (100-x)
5x - 100 = 5(100-x)
:
5x - 100 = 500 - 5x
:
5x + 5x = 500 + 100
:
10x = 600
:
x = 60 km/hr is C's speed
Find M's speed
y = 100 - 60
y = 40 km/hr is M's speed
:
Check solutions using the distance
5*60 = 300 km
5*40 = 200 km
--------------
diff = 100 km
;
:
2. Wayne invested part of his money at 8% and the rest at 6%. The income from
both investments totaled $2,000. If he interchanged his investments, his
income would increase by $340. How much did he have in each investment?
:
Originally: x = amt invested at 8% and y = amt invested at 6%
:
.08x + .06y = 2000; original income equation
.06x + .08y = 2340; switched income equation
:
Multiply the 1st equation by 8, and the 2nd equation by 6; results
.64x +.48y = 16000
.36x + 48y = 14040
----------------------subtraction eliminates y, find x
.28x + 0y = 1960
:
x =
x = $7,000 originally invested at 8%
:
Find y:
.08(7000) + .06y = 2000
560 + .06y = 2000
.06y = 2000 - 560
.06y = 1440
y =
y = $24,000
:
:
Check solution by switching the investments
.08(24000) + .06(7000) =
1920 + 420 = 2340; confirms our solutions
;
:
Do you have any questions about either of these? did you understand what we did?

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