# Algebra

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#1
shiu planned a 210 km bike ride. However, he rode 5km per hour faster than he planned and finished his ride earlier than he planned. what is his average speed for his ride in km per hour?
0
3 years ago
#2
(Original post by nilesh2378)
shiu planned a 210 km bike ride. However, he rode 5km per hour faster than he planned and finished his ride earlier than he planned. what is his average speed for his ride in km per hour?
What have you tried?
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#3
I am really not sure?
0
3 years ago
#4
this q makes absolutely no sense.
0
3 years ago
#5
(Original post by nilesh2378)
I am really not sure?
Say the planned speed was to travel at so the speed at which he completed the journey at a speed of

So what is his average speed?
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#6
I know the answer is 35 but how?
0
3 years ago
#7
(Original post by nilesh2378)
shiu planned a 210 km bike ride. However, he rode 5km per hour faster than he planned and finished his ride earlier than he planned. what is his average speed for his ride in km per hour?
There isn't enough information to answer the question. Did you miss something out?
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#8
no i havent.
0
3 years ago
#9
The answer is in HOURS therefor you just use:
SPEED= DISTANCE/TIME
So: S= 210/6 S= 35 km/h
I have no idea how you get 6, because there are 60 minutes in an hour but there is no information to say that it should be 6.
But yeah, that's how you do it.
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3 years ago
#10
There really isn't enough information to answer the question.
0
3 years ago
#11
Firstly, I agree with others on no sufficient information.

Further, it can be interpreted in two ways. One is Shiu's speed is 5 km/h more than the planned and the other is Shiu accelerated 5 km/(h)^2 durring the ride.

Assuming planned speed S1

In first case, Shiu rode (S1 + 5). In this case, 210 = (S1+5) x (t1), where t1 is time taken

In the second case, Shiu's speed at a given time,t, is (5xt) + S1. In this case, assuming t2 is the time taken, ((S1 + 5t2 + S1) / 2) x t2 (i.e. area of trapezium under speed time graph) = 210.

In both of these cases, results are inconclusive due to lack of information.

Also, tried to see if it can be predicted from using hyperbolic functions i.e t1 = (210 / S1) and t2 = (210 / (S1 + 5)), i.e let S1 be variable as well (and form xy (i.e. t1 x S1) = C squard and use may be inverse function, domain and range for rearranging, i.e. S1 against t1, generally S against t in for both graphs) and tried to define some limits or using the area under defined limits as distance. I might have missed some thing because it has been very long time since I worked with hyberbolic functions, integration etc.

So, it may be worthwhile for some one who is hot trend on the hyberbolic functions, integration an functions etc. to try, and check if you can crack it assumong question is correct.
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