Question

A motorist leaves his home at 8 a.m. to a place 90 km away. If he drives at an average speed of (x-5)km/h, he will be 15 minutes late for his appointment. If driving at (x+10) km/h, he will be 15 minutes early. Find the value of x and the time fixed for the appointment.​

Answer 1

Solution :-

Let the original speed of motorist is x km/h & Let fixed time for the appointment is t .

Case 1 :-

→ Distance = 90km.

→ Speed = (x - 5)km/h.

→ Time = Fixed time + 15 min. = t + (15/60) = t + (1/4).

So,

→ (Distance/Speed) = Time

→ (90)/(x - 5) = t + (1/4). ----------------- Equation (1) .

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Case 2 :-

→ Distance = 90km.

→ Speed = (x + 10)km/h.

→ Time = Fixed time - 15 min. = t - (15/60) = t - (1/4).

So,

→ (Distance/Speed) = Time

→ (90)/(x + 10) = t - (1/4). ----------------- Equation (2) .

_____________________

Subtracting Equation (2) From Equation (1) , we get Now,

→ [ (90)/(x - 5) ] - [ (90)/(x + 10) ] = [ t + (1/4) ] - [ t - (1/4) ]

→ [ 90(x +10) - 90(x - 5) ] / [ (x - 5)(x + 10) ] = (1/4 + 1/4)

→ [ 90x + 900 - 90x + 450 ] / [ x² + 10x - 5x - 50 ] = (1/2)

→ x² + 5x - 50 = 2*1350

→ x² + 5x - 50 - 2700 = 0

→ x² + 5x - 2750 = 0

→ x² + 55x - 50x - 2750 = 0

→ x (x + 55) - 50(x + 55) = 0

→ (x + 55)( x - 50) = 0

→ x = (-55) & 50 [ Negative Speed ≠ ] .

So, Speed of motorist = x = 50 km/h. (Ans).

And , time fixed for the appointment = (D/S) = (90/50) = 1(4/5) Hours = 1 Hours 48 minutes.. (Ans).

Answer 2

$\Huge{\underline{\underline{\mathfrak{ Solution \colon }}}}$

The average speed is x km/hr

Let the time taken be t

If he derives at an average speed of ( x-5 ) km/ hr, he will be 15 minutes late & if he drives at an average speed of

( x + 10 ) km / hr ,he will be 15 minutes early.

A/q we have,

$⟼ \text{t }= \frac{15}{60} = \frac{90}{ \text{x} - 5} \: ............( \text{i})$

$⟼ \text{t} = \frac{15}{60} = \frac{90}{ \text{x} + 10} .............( \text{ii})$

Subtracting (i ) from (ii)

$\frac{1}{2} = \frac{90}{ \text{x} - 5} - \frac{90}{ \text{x} + 10}$

$⟹ ( \text{x} - 5)( \text{x}+ 10) = 180( \text{x} - 10 - \text{x} + 5)$

$⟹ { \text{x} }^{2} + 5 \text{x} - 50 = 180 \times 15$

$⟹ { \text{x}}^{2} + 5 \text{x} - 2750 = 0$

$⟹ \text{x} = \frac{ - 5 + \sqrt{25 + 11000} }{2} = \frac{ - 5 \sqrt{11025} }{2} \frac{ - 5 + 105}{2}$

$⟹ \text{x} = - 55$

Since,the speed can't be negative

so, the value of x = 50

Time = 90/ 50 = 1.8