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Friday, November 17 2017 @ 10:16 PM CST
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By: sundarshoba (offline)  Friday, April 17 2015 @ 10:01 PM CDT (Read 1180 times)  
sundarshoba

Please assist to solve, Everyday Tan drives a distance of 25 km fro school to reach home. On a particular day, he increased his speed by 25 km/hrand reached home ten minutes earlier.What is his usual speed? Thanks

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By: echeewh (offline)  Saturday, April 18 2015 @ 06:16 AM CDT  
echeewh

Hey <sundarshoba>

Are you certain this is a P6 kind of question?

Based on the info given in the question , I have to resort to using <Quadratic> function to solve this.

Following is my worked solution:

Let the usual speed be s.
On a normal day, time taken (t) = 25/s --- (1)

when this speed was increased by 25, it becomes (s + 25) and the time taken to reach home was 10 min (1/6 hr) earlier, we can calculate the additional distance travelled in this additional (1/6) hr.

Additional dist = (s + 25) × (1/6) = (s + 25)/6

For the same time taken (t), the distance covered in the increased speed (s + 25) was [25 + (s + 25)/6].

[25 + (s + 25)/6] = (150 + s + 25)/6 = (s + 175)/6

time taken (t) = [(s + 175)/6] ÷ (s + 25)
= (s + 175)/(6s + 150) --- (2)

But (1)=(2), and we have ...

25/s = (s + 175)/(6s + 150)

Apply <Cross-multiply> method, we have ...

25 × (6s + 150) = s × (s + 175)
150s + 3750 = s^2 + 175s {s^2 = square of s}

Reorganising this into a quadratic eqn, we have ...

s^2 + 175s - 150s - 3750 = 0
s^2 + 25s - 3750 = 0

Using a calculator to set to QUAD eqn with a=1, b=25, c=-3750,

s = 50 km/h.

[Note: calculator -> press MODE --> look for EQN --> look for QUAD --> set the values for a, b, c.]

=======

Trust this helps.

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

Cheers,
Edward

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