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Arithmetic II

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Time & Speed

Time And Speed

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Time & Speed : Level 1
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CAT 2025 Lesson : Time & Speed - Average Speed

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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}}Time​
111 minute        =60\space \space \space \space \space \space \space = 60       =60 seconds
111 hour            =60\space \space \space \space \space \space \space \space \space \space \space = 60           =60 minutes === 360036003600 seconds

Distance‾\underline{\text{Distance}}Distance​
111 metre          =100 \space \space \space \space \space \space \space \space \space = 100         =100 centimetres === 100010001000 millimetres
111 kilometre    =1000 \space \space \space = 1000   =1000 metres
111 mile             =1.609\space \space \space \space \space \space \space \space \space \space \space \space = 1.609            =1.609 kilometres === 160916091609 metres

Other measurements of distance‾\underline{\text{Other measurements of distance}}Other measurements of distance​
111 inch             =2.54\space \space \space \space \space \space \space \space \space \space \space \space = 2.54            =2.54 cm
111 foot             =12 \space \space \space \space \space \space \space \space \space \space \space \space = 12            =12 inches        === 30.4830.4830.48 cm
111 yard             =3\space \space \space \space \space \space \space \space \space \space \space \space = 3            =3 feet               === 91.4491.4491.44 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.

111 km/hr === 518 \dfrac{5}{18}185​ m/s      and      111 m/s === 185\dfrac{18}{5}518​ km/hr

Example 4

Express 555 km/hr as m/s.

Solution

111 kilometre === 100010001000 metres and 111 hour === 360036003600 seconds

555 km/hr === 5 kilometre1 hour\dfrac{5 \space \text {kilometre} }{1 \space \text {hour}}1 hour5 kilometre​ === 5×1000 metres3600 seconds\dfrac{5\times 1000 \space \text {metres}}{3600 \space \text {seconds}}3600 seconds5×1000 metres​ === 25 metres18 seconds\dfrac{25 \space \text{metres}}{18 \space \text {seconds}}18 seconds25 metres​ === 2518\dfrac{25}{18}1825​

Alternatively

555 km/hr === 5×5185\times \dfrac{5}{18}5×185​ m/s === 2518\dfrac{25}{18}1825​ m/s

Answer: 2518\dfrac{25}{18}1825​ m/s

Example 5

Convert 505050 m/s to km/hr.

Solution

505050 m/s = 50 metres1 second\dfrac{50 \space \text {metres}}{1 \space \text {second}}1 second50 metres​ = 50×11000kilometres13600hour\dfrac{50 \times \dfrac{1}{1000}\text{kilometres}}{\dfrac{1}{3600}\text{hour}}36001​hour50×10001​kilometres​ = 180 kilometres1 hour\dfrac{180 \space \text {kilometres}}{1 \space \text {hour}}1 hour180 kilometres​ = 180180180 km/hr

Alternatively

505050 m/s = 50×185 50\times \dfrac{18}{5}50×518​ km/hr === 180180180 km/hr

Answer: 180180180 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}} Total Time TakenTotal Distance Covered​

Example 6

If Kumar travels at 252525 km/hr for 222 hours and 404040 km/hr for the next 333 hours, then what is his average speed?

Solution

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

Average Speed = 1705=34\dfrac{170}{5} = 345170​=34 km/hr

Answer: 343434 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 202020 Km/hr, the next two hours at 303030 Km/hr and the final two hours at 707070 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(20×2)+(30×2)+(70×2)=240 km
Total time taken =2+2+2=6= 2 + 2 + 2 = 6=2+2+2=6 hours

Average Speed =2406\dfrac{240}{6} 6240​ = 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}320+30+70​ = 40 km/hr

Answer: 404040 km/hr

Example 8

John travels one-third of a distance at 202020 km/hr, one-third of the distance at 252525 km/hr and the remaining distance at 404040 km/hr. What is his average speed (in km/hr and rounded to 111 decimal point)?

Solution

Method 1

Let xxx be the total distance. So, time taken at
202020 km/hr =x3×20=x60\dfrac{x}{3 \times20} = \dfrac{x}{60}3×20x​=60x​

252525 km/hr =x3×25=x75\dfrac{x}{3 \times25} = \dfrac{x}{75}3×25x​=75x​

404040 km/hr =x3×40=x120\dfrac{x}{3 \times40} = \dfrac{x}{120}3×40x​=120x​

Total time taken =x60+x75+x120= \dfrac{x}{60} + \dfrac{x}{75} + \dfrac{x}{120}=60x​+75x​+120x​ =x(10+8+5)600= \dfrac{x(10 + 8 + 5)}{600}=600x(10+8+5)​ =23x600= \dfrac{23 x}{600}=60023x​

Average speed =x×60023x∼26.1= x \times \dfrac{600}{23 x} \sim 26.1=x×23x600​∼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 300300300 km. Time taken at

202020 km/hr =10020= \dfrac{100}{20}=20100​ = 555 hours

252525 km/hr =10025= \dfrac{100}{25}=25100​ = 444 hours

404040 km/hr =10040= \dfrac{100}{40}=40100​ = 2.52.52.5 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=5+4+2.5300​=11.5300​=23600​∼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}}201​+251​+401​3​ =310+8+5200= \dfrac{3}{\dfrac{10 + 8 + 5}{200}}=20010+8+5​3​ =60023∼26.1= \dfrac{600}{23} \sim 26.1=23600​∼26.1 km/hr

Answer: 26.126.126.1 km/hr


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