calendarBack
Quant

/

Arithmetic II

/

Time & Speed
ALL MODULES

CAT 2025 Lesson : Time & Speed - Average Speed

bookmarked

2. Units of Speed and Conversions

Speed is typically expressed as kilometres per hour (km/hr) or metres per second (m/s). It, however, can be expressed in any combination of time and distance. Here are some of the common units of time and distance.

Time‾\underline{\text{Time}}
11 minute        =60\space \space \space \space \space \space \space = 60 seconds
11 hour            =60\space \space \space \space \space \space \space \space \space \space \space = 60 minutes == 36003600 seconds

Distance‾\underline{\text{Distance}}
11 metre          =100 \space \space \space \space \space \space \space \space \space = 100 centimetres == 10001000 millimetres
11 kilometre    =1000 \space \space \space = 1000 metres
11 mile             =1.609\space \space \space \space \space \space \space \space \space \space \space \space = 1.609 kilometres == 16091609 metres

Other measurements of distance‾\underline{\text{Other measurements of distance}}
11 inch             =2.54\space \space \space \space \space \space \space \space \space \space \space \space = 2.54 cm
11 foot             =12 \space \space \space \space \space \space \space \space \space \space \space \space = 12 inches        == 30.4830.48 cm
11 yard             =3\space \space \space \space \space \space \space \space \space \space \space \space = 3 feet               == 91.4491.44 cm

So, speed can be expressed as a combination of any of the aforementioned combinations, i.e., km/hr, km/s, cm/s, miles/hr, etc. The most common conversions in Time and Speed problems is km/hr to m/s and vice versa. Therefore, note the following conversions.

11 km/hr == 518 \dfrac{5}{18} m/s      and      11 m/s == 185\dfrac{18}{5} km/hr

Example 4

Express 55 km/hr as m/s.

Solution

11 kilometre == 10001000 metres and 11 hour == 36003600 seconds

55 km/hr == 5 kilometre1 hour\dfrac{5 \space \text {kilometre} }{1 \space \text {hour}} == 5×1000 metres3600 seconds\dfrac{5\times 1000 \space \text {metres}}{3600 \space \text {seconds}} == 25 metres18 seconds\dfrac{25 \space \text{metres}}{18 \space \text {seconds}} == 2518\dfrac{25}{18}

Alternatively

55 km/hr == 5×5185\times \dfrac{5}{18} m/s == 2518\dfrac{25}{18} m/s

Answer: 2518\dfrac{25}{18} m/s

Example 5

Convert 5050 m/s to km/hr.

Solution

5050 m/s = 50 metres1 second\dfrac{50 \space \text {metres}}{1 \space \text {second}} = 50×11000kilometres13600hour\dfrac{50 \times \dfrac{1}{1000}\text{kilometres}}{\dfrac{1}{3600}\text{hour}} = 180 kilometres1 hour\dfrac{180 \space \text {kilometres}}{1 \space \text {hour}} = 180180 km/hr

Alternatively

5050 m/s = 50×185 50\times \dfrac{18}{5} km/hr == 180180 km/hr

Answer: 180180 km/hr


3. Average Speed

If a distance is covered at different speeds, we cannot directly find the arithmetic mean or the average of the speeds to calculate the average speed. Instead, we have to find the total distance covered and divide it by the total time taken.

Average Speed =Total Distance CoveredTotal Time Taken \dfrac{\text{Total Distance Covered}}{\text{Total Time Taken}}

Example 6

If Kumar travels at 2525 km/hr for 22 hours and 4040 km/hr for the next 33 hours, then what is his average speed?

Solution

Total Distance covered =(25×2)+(40×3)=170= (25 \times 2) + (40 \times 3) = 170 km
Total Time taken = 2+3=52 + 3 = 5 hours

Average Speed = 1705=34\dfrac{170}{5} = 34 km/hr

Answer: 3434 km/hr

Note 1 : When Time Taken at different speeds are equal, then the average speed is the Arithmetic Mean of the Speeds.

Note 2 : When Distances covered at different speeds are equal, then the average speed is the Harmonic Mean of the Speeds.

Example 7

Aravind travels the first two hours at the speed of 2020 Km/hr, the next two hours at 3030 Km/hr and the final two hours at 7070 Km/hr. What is his average speed?

Solution

Total distance covered = (20×2)+(30×2)+(70×2)=240(20 \times 2) + (30 \times 2) + (70 \times 2) = 240 km
Total time taken =2+2+2=6= 2 + 2 + 2 = 6 hours

Average Speed =2406\dfrac{240}{6} = 40 km/hr

Alternatively (Recommended Method)

As mentioned in note 1 above, the time taken at different speeds are equal, the average speed is the arithmetic mean of the speeds.

Average Speed = 20+30+703\dfrac{20+30+70}{3} = 40 km/hr

Answer: 4040 km/hr

Example 8

John travels one-third of a distance at 2020 km/hr, one-third of the distance at 2525 km/hr and the remaining distance at 4040 km/hr. What is his average speed (in km/hr and rounded to 11 decimal point)?

Solution

Method 1

Let xx be the total distance. So, time taken at
2020 km/hr =x3×20=x60\dfrac{x}{3 \times20} = \dfrac{x}{60}

2525 km/hr =x3×25=x75\dfrac{x}{3 \times25} = \dfrac{x}{75}

4040 km/hr =x3×40=x120\dfrac{x}{3 \times40} = \dfrac{x}{120}

Total time taken =x60+x75+x120= \dfrac{x}{60} + \dfrac{x}{75} + \dfrac{x}{120} =x(10+8+5)600= \dfrac{x(10 + 8 + 5)}{600} =23x600= \dfrac{23 x}{600}

Average speed =x×60023x∼26.1= x \times \dfrac{600}{23 x} \sim 26.1 km/hr

Method 2

In this method we assume a value for the total distance, which makes it less cumbersome over using variables.

Let the total distance be 300300 km. Time taken at

2020 km/hr =10020= \dfrac{100}{20} = 55 hours

2525 km/hr =10025= \dfrac{100}{25} = 44 hours

4040 km/hr =10040= \dfrac{100}{40} = 2.52.5 hours

Average speed =3005+4+2.5=30011.5=60023∼26.1= \dfrac{300}{5 + 4 + 2.5} = \dfrac{300}{11.5} =\dfrac{600}{23} \sim 26.1 km/hr

Method 3 (Recommended Method)

As mentioned in note 2 earlier, the distances covered at different speeds are equal, the average speed is the harmonic mean of the speeds.

So, average speed = 3120+125+140\dfrac{3}{\dfrac{1}{20} + \dfrac{1}{25} + \dfrac{1}{40}} =310+8+5200= \dfrac{3}{\dfrac{10 + 8 + 5}{200}} =60023∼26.1= \dfrac{600}{23} \sim 26.1 km/hr

Answer: 26.126.1 km/hr


Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock