SOLUTION: 1. a town's old street sweeper can clean the streets in 60 h. the old sweeper together with a new sweeper can clean the streets in 15 h. how long would it take the new sweeper to d

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Question 172725This question is from textbook Algebra and Trigonomety
: 1. a town's old street sweeper can clean the streets in 60 h. the old sweeper together with a new sweeper can clean the streets in 15 h. how long would it take the new sweeper to do the job alone?
2. during 60 mi of city driving, Jenna averaged 15 mi/gal. she then drove 140 mi on an espressway and averaged 25 mi/gal for the entire 200 mi. find the average fuel consumption on the expressway.
3. members of the computer club were assessed equal amounts to raise $1200 to but some software. when 8 new members joined, the per-member assessment was reduced by $7.50. what was the new size of the club>
4.to measure the speed of the jet stream, a weather plane left its base at noon and flew 800 km directly against the stream with an air speed of 750 km/h. it then returned directly to its base, arriving at 2:24 pm. what was the speed of the jet stream?
This question is from textbook Algebra and Trigonomety

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1. a town's old street sweeper can clean the streets in 60 h. the old sweeper together with a new sweeper can clean the streets in 15 h. how long would it take the new sweeper to do the job alone?
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Old cleaner DATA:
Time = 60 hr/job ; Rate = 1/60 job/hr.
--------------------------------------
Together DATA
Time = 15 hr/job ; Rate = 1/15 job/hr.
-------------------------------------
New cleaner DATA:
Time = x hr/job ; Rate = 1/x job/hr
------------------------------------
EQUATION:
rate + rate = together rate
1/60 + 1/x = 1/15
(x+60)/60x = 1/15
----
Cross multiply to get
15(x+60) = 60x
15x -60x = -15*60
-45x = -15*60
x = 20 hr (time required for the new cleaner to do the job alone)
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2. during 60 mi of city driving, Jenna averaged 15 mi/gal. she then drove 140 mi on an espressway and averaged 25 mi/gal for the entire 200 mi. find the average fuel consumption on the expressway.
---
City DATA:
distance = 60 mi ; rate = 15 mpg
-------------
Total Drive DATA:
distance = 200 mi ; rate = 25 mpg
----------------
Expressway DATA:
distance = 140 mi ; rate = x mpg
------------------------------------
gallons + gallons = total gallons
60/15 + 140/x = 200/25
4 + 140/x = 8
140/x = 4
x = 140/4 = 35 mpg (average fuel consumption on the expressway)
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3. members of the computer club were assessed equal amounts to raise $1200 to buy some software. when 8 new members joined, the per-member assessment was reduced by $7.50. what was the new size of the club>
------
Let original size of the club be "x" members.
Then original cost per person was 1200/x dollars
-----------------------
After 8 joined there were "x+8" members.
EQUATION:
1200/(x+8) = 7.50
x+8 = 1200/7.5
x = 1200/7.5 8
x = 152 (original # of members)
-------------------------------
4.to measure the speed of the jet stream, a weather plane left its base at noon and flew 800 km directly against the stream with an air speed of 750 km/h. it then returned directly to its base, arriving at 2:24 pm. what was the speed of the jet stream?
---
Upwind DATA:
distance = 800 km ; rate = 750 km/h ; time = d/r = 800/750 = 1.067 hrs
----------------
Downwind DATA:
distance = 800 km ; rate = x km/h ; time = (2.4-1.067) hr.
----------------------------
Find downwind rate:
rate = d/t = 800/(2.4-1.067) = 600 km/h
-------------------------
Find speed of jet stream:
Upwind Equation: plane - stream = 750 km/h
Down wind Eq...: Plane + stream = 600 km/h
-----------------------------------------------
Adding the two equations you get:
2plane = 1350
plane = 675
Therefore the jec stream would be -75 km/h
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Comment: I think the data is incorrectly stated.
It takes 1.067 hrs to go the 800 km Against the wind
and a longer time (1.333 hrs) to go the same distance
With the wind. That doesn't make much sense.
===================================
Cheers,
Stan H.