**Question 1: **

**We will solve this question by 2 methods:**

__Solution:__Basic Concept: Let usual speed be ‘s’ and usual time taken is ‘t’. Then new speed is ‘0.8s’ and new time taken is (t + 15).

Note: In this question the unit of any parameter is not required.

Since, the distance is constant. We can apply the basic formula, distance = speed ✖ time.

i.e. S1 ✖ T1 = S2 ✖ T2.

➔ s ✖ t = 0.8s ✖ (t+15)

➔ 0.2t = 12

➔ t = 60 minutes.

➔ So, new time taken = (60+15) minutes = 75 minutes.

Shortcut method:

By taking ratio of usual and new speed i.e. s:0.8s = 5:4

Now we can take the difference of new and usual time and that will be equal to 8 minutes.

➔ 5x – 4x = 15

➔ x = 15.

So usual time taken = 4x = 60 minutes and new time taken = (60+15) = 75 minutes.

**Let the normal time taken be ‘t’. Since distance = speed ✖ time.**

__Solution:__➔ S1 ✖ T1 = S2 ✖ T2

➔ 20 ✖ (t+10) = 30 ✖ (t-20)

➔ 10t = 800. So, t = 80minutes.

So, time taken with speed 20km/hr. is 90 minutes and with speed 30 km/hr. is 60 minutes.

Now, Distance between Home and mall = 20 ✖ 90/60 = 30 ✖ 60/60 = 30km.

**Concept 4:**

Let a car goes from ‘A’ to ‘B’ with speed ‘u’ km/hr. and then it goes from ‘B’ to ‘C’ with speed ‘v’ km/hr. Then find the average speed.

Let the distance between ‘AB’ be ‘x’ and distance between ‘BC’ is ‘y’. Then:

Similarly we can also find for three speeds.

**We can directly apply the formula here i.e.**

__Solution:__

__Solution:__

__Solution:__

__Solution:__Distance travelled by train in 1 hour without stoppage = 54 km.

Distance travelled by train in 1 hour with stoppage = 45 km.

If we see here closely then we observe:

54 km distance is travelled in 60 minutes. If the train will travel 45 km without stoppage then

time taken by it is

So, for (60 – 50) minutes = 10 minutes the train stops.

**Shortcut:**

= 50 minutes. So, stoppage time = (60 – 50) = 10 minutes.

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