Question

(A) While trekking you could climb with a speed of 2.5 km per hr for the initial distance and while 750 m per hour for the difficult part, If you could reach the peak within 5 hours; how much was the difficult part of the trek, if the peak was 9000 mtr. Away from the foot of the mountain?

Answer 1

Recall:

$\text{Time } = \dfrac{\text{Distance}}{\text{Speed}}$

Convert units:

$9000 \text{ m} = 9 \text{ km}$

$750 \text{ m/hr } = 0.75 \text{ km/hr}$

Define x:

Let x km be the distance traveled on the easy path

The distance of the difficult path = (9 - x) km

Find the time spent on the easy path:

$\text{Distance = } x \text{ km}$

$\text {Speed }= 2.5 \text{ km/h}$

$\text{Time } = \dfrac{\text{Distance}}{\text{Speed}}$

$\text{Time } = \dfrac{x}{2.5} \text { hours}$

Find the time spent on the difficult path:

$\text{Distance = } 9 - x \text{ km}$

$\text {Speed }= 0.75 \text{ km/h}$

$\text{Time } = \dfrac{\text{Distance}}{\text{Speed}}$

$\text{Time } = \dfrac{9 - x}{0.75} \text { hours}$

Given that the time spent to reach the peak is 5 hours:

$\dfrac{x}{2.5} + \dfrac{9 - x}{0.75} = 5$

$\dfrac{0.75x + 2.5(9 - x)}{2.5(0.75)} = 5$

$\dfrac{0.75x +22.5 - 2.5x}{1.875} = 5$

$\dfrac{1.75x +22.5}{1.875} = 5$

$-1.75x + 22.5 = 9.375$

$1.75x = 13.125$

$x = 7.5 \text { km}$

Find the distance of the difficult path:

$\text{Total Distance} = 9 \text{ km}$

$\text{Easy Distance} = x \text{ km}$

$\text{Easy Distance} = 7.5 \text{ km}$

$\text{Difficult Distance} = 9 - x \text{ km}$

$\text{Difficult Distance} = 9 - 7.5 \text{ km}$

$\text{Difficult Distance} = 1.5 \text{ km}$

Answer: The difficult path was 1.5 km long

Answer 2

The distance cover in difficult part of trek is 1.5 km

Step-by-step explanation:

Given as :

Total distance of the mountain = D = 9000 meters = 9 km

The total time taken to cover 9 km = 5 hours

The distance is cover in to parts ,

Let The distance cover in initial part = d km

The speed of climbing = s = 2.5 km/h

Let The time taken for initial part = t hours

Again

Let The distance cover in difficult part = (9 - d) km

The speed of climbing = S = 750 m/h = 0.75  km/h

Let The time taken for difficult part = T hours

According to question

∵  Time = $\dfrac{Distance}{Speed}$

For initial part on trek

or, t = $\dfrac{d}{s}$

Or, t = $\dfrac{d}{2.5}$    hours                      ..............1

For difficult part on trek

or, T = $\dfrac{9-d}{S}$

Or, T = $\dfrac{9-d}{0.75}$    hours                 .....................2

Now, As Total time taken to climb the trek = 5 hours

So,  From eq 1 and eq 2

        t  +  T = 5 hour

i.e     $\dfrac{d}{2.5}$  +  $\dfrac{9-d}{0.75}$  = 5

Or,   $\dfrac{10d}{25}$ + $\dfrac{900-100d}{75}$  = 5

Or,   $\dfrac{30d+900-100d}{75}$  = 5

or,  900 - 70 d = 75 × 5

Or, 900 - 70 d = 375

Or, 70 d = 900 - 375

Or, 70 d = 525

∴          d = $\dfrac{525}{70}$

i.e         d = 7.5 km

So,The distance cover in initial part = d = 7.5 km

And The distance cover in difficult part = (9 - d) = 9 - 7.5 = 1.5 km

Hence, The distance cover in difficult part of trek is 1.5 km  Answer