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.]

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Trust this helps.

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

Cheers,

Edward