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Saturday, September 23 2017 @ 11:17 AM CDT
 Forum Index >  Test Paper Related >  Primary 6 Matters
 Math (Speed)
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By: vivilau1112 (offline)  Thursday, August 08 2013 @ 04:27 AM CDT (Read 604 times)  
vivilau1112

Hello everyone, again! I am stuck at this question.Can someone help?Much appreciated

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By: echeewh (offline)  Friday, August 09 2013 @ 05:32 AM CDT  
echeewh

Hi <vivilau1112>,

Following is my worked solution:

*** - for alignment purpose


|----------------------------|
O ********************* H
|-> ************* |->
T ************** T1

s(T1) = s(T) - 20 , where s(T) is the usual speed and s(T1) is the slower speed.

Given that at this slower speed, T would arrive 10 mins later. So ...

t(T1H) = 10 mins = (1/6) hr
d(T1H) = [s(T) - 20] × (1/6) = [s(T) - 20] / 6
d(TT1) = 40 - [s(T) - 20] / 6
= {240 - [s(T) - 20]} / 6
= [260 - s(T)] / 6

At its usual speed,
t(TH) = 40 ÷ s(T)
= 40 / s(T)

At its slower speed, this time taken is the same as t(TT1). i.e. t(TH) = t(TT1)

t(TT1) = {[260 - s(T)] / 6} ÷ [s(T) - 20]
= [260 - s(T)] / [6s(T) - 120]

Equate this ..

40 / s(T) = [260 - s(T)] / [6s(T) - 120]
40 [6s(T) - 120] = s(T) [260 - s(T)]
240s(T) - 4800 = 260s(T) - [s(T) × s(T)]
[s(T) × s(T)] - 260s(T) + 240s(T) - 4800 = 0
[s(T) × s(T)] - 20s(T) - 4800 = 0
[s(T) + 60] [s(T) - 80] = 0

s(T) = -60 , 80

Hence, usual speed of T = 80 km/h

==========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

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By: echeewh (offline)  Friday, August 09 2013 @ 10:26 AM CDT  
echeewh

An alternative solution to this is :

At its usual speed,
t(TH) = 40 ÷ s(T)
= 40 / s(T)

At its slower speed,
t(TH) = 40 ÷ [s(T) - 20]
= 40 / [s(T) - 20]

Since time taken to reach home (H) at the slower speed would be 10 minutes (1/6 hr) longer, we have ...

40 / [s(T) - 20] - 40 / s(T) = (1/6)
{40s(T) - 40[s(T) - 20]} / s(T)[s(T) - 20] = (1/6)
[40s(T) - 40s(T) + 800] / [s(T) × s(T)] - 20s(T) = (1/6)
4800 = [s(T) × s(T)] - 20s(T)
[s(T) × s(T)] - 20s(T) - 4800 = 0
[s(T) + 60] [s(T) - 80] = 0
s(T) = -60, 80

Hence, usual speed of T = 80 km/h

=========

Cheers,
Edward

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