Average Speed MCQ Quiz - Objective Question with Answer for Average Speed - Download Free PDF

Given:

Total Distance = 600 km.

average speed was reduced = 200 km/hr 

the time of flight increased = 30 minutes

Formula Used:

∵ Average Speed = Total Distance / Time

Calculations:

Let, the original duration of flight be t hrs.

So, original average speed = 600/t

Due to bad weather speed of trip is reduced by 200 km/hr and time of flight is increased by 30 minutes i.e. 0.5 hr.

∴ Reduced average speed = (600/t) – 200

And New duration of flight = (t + 0.5) hrs

So, the new average speed = 600/(t + 0.5)

Equating,

\(\begin{array}{l} \Rightarrow \frac{{600}}{t}-200 = \frac{{600}}{{t + 0.5}}\\ \Rightarrow \frac{3}{t}-1 = \frac{3}{{t + 0.5}}\\ \Rightarrow \frac{{3-t}}{t} = \frac{3}{{t + 0.5}} \end{array}\)

⇒ t2 + 0.5t – 1.5 = 0

⇒ 2t2 + t – 3 = 0

⇒ 2t2 – 2t + 3t – 3 = 0

⇒ (t – 1)(2t + 3) = 0

⇒ (t – 1) = 0

∴ t = 1 hr

∴ The duration of the flight is 1 hr.

Given:

Q completes a journey in 99 hours.

First half of the journey is traveled at 96 kmph.

Second half of the journey is traveled at 102 kmph.

Formula used:

Total time = Time for first half + Time for second half

Time = Distance ÷ Speed

Let the total distance be D, so the first half = D/2 and second half = D/2.

Total time = (D/2 ÷ 96) + (D/2 ÷ 102)

Calculation:

Total time = 99 hours

⇒ \(\dfrac{D}{2 \times 96} + \dfrac{D}{2 \times 102} = 99\)

⇒ \(\dfrac{D}{192} + \dfrac{D}{204} = 99\)

⇒ Take LCM of 192 and 204: LCM = 3264

⇒ \(\dfrac{17D}{3264} + \dfrac{16D}{3264} = 99\)

⇒ \(\dfrac{33D}{3264} = 99\)

⇒ \(\dfrac{D}{99} = \dfrac{3264}{33}\)

⇒ D = \(\dfrac{3264 \times 99}{33}\)

⇒ D = 9792

∴ The total distance is 9792 km, and the correct answer is option (4).

Given:

Distance in 1st hour = 281 km

Distance in 2nd hour = 163 km

Total time = 2 hours

Formula used:

Average speed = Total distance ÷ Total time

Calculation:

Total distance = 281 + 163 = 444 km

Total time = 2 hours

⇒ Average speed = 444 ÷ 2 = 222 km/h

∴ The average speed is 222 km/h.

Given:

Distance of first lap = 300 m

Time for first lap = 50 sec

Distance of second lap = 300 m

Time for second lap = 150 sec

Formula used:

Average speed = Total distance / Total time

Calculation:

Total distance = 300 + 300 = 600 m

Total time = 50 + 150 = 200 sec

⇒ Average speed = 600 / 200

⇒ Average speed = 3 m/sec

∴ The correct answer is option (2).

Given:

Speed while going to school (S₁) = 10 km/h

Speed while coming back (S₂) = 40 km/h

Formula used:

Average Speed = \(\dfrac{2 \times S_1 \times S_2}{S_1 + S_2}\)

Calculation:

Average Speed = \(\dfrac{2 \times 10 \times 40}{10 + 40}\)

⇒ Average Speed = \(\dfrac{2 \times 400}{50}\)

⇒ Average Speed = \(\dfrac{800}{50}\)

⇒ Average Speed = 16 km/h

∴ The correct answer is option (3).

Given:

A man travels from A to B at a speed of 36 km/hr in 74 minutes and he travels a distance from B to C with a speed of 45 km/hr in 111 minutes. 

Formula used:

Average speed = Total distance/Total time taken

Calculation:

Time taken = 74 min : 111 min   [given]

Ratio of Time taken = 2 : 3

Average Speed = \(\frac{{36\ \times\ 2\ +\ 45\ \times\ 3}}{{2\ +\ 3}}\)

Average Speed = 207/5

Average Speed = 41.4 km/hr

∴ The average speed of whole journey is 41.4 km/h

Given:

A person travels 120 km at 80 km/h, the next 100 km at 40 km/h, and comes back to the starting point at 75 km/h

Formula used:

Average speed = total distance/total time

Calculation:

Average speed = total distance/total time

Total distance = 120 + 100 + 220

= 440 km

Total time = (120/80) + (100/40) + (220/75)

= (3/2 + 5/2 + 44/15)

= (45 + 75 + 88)/30

= 208/30

Average speed = 440/(208/30)

= (440 * 30) / 208

= 63.46 km/hr

∴ The average speed is 63.46 km/hr.

Given

Speed from home to office = 20 km/h

Average speed = 24 km/h

Concept:

Let the speed while returning from the office to home be denoted by x. We can use the concept of average speed to set up an equation and solve for x.

Solution:

The formula for average speed is given by:

Average speed = 2ab/(a + b)

where a and b are the speeds of two different segments.

Using the given data, we have:

24 km/h = (2 × 20 km/h × x)/(20 km/h + x)

Simplifying this equation, we find:

480 km/h + 24x km = 40x km/h

16x = 480

x = 30 km/h

Therefore, the speed while returning is 30 km/h.

Given:

The speed of the man is 12 km/h, 24 km/h and 8 km/h

Concept Used:

Average speed = total distance/total time

The sides of an equilateral triangle are equal.

Calculation:

Let, the side of the triangle be x km.

As we know,

Time to cover x km distance at a speed of 12 km/hr = x/12 hrs

Time to cover x km distance at a speed of 24 km/hr = x/24 hrs

Time to cover x km distance at a speed of 8 km/hr = x/8 hrs

Total distance covered by the man = (x + x + x) = 3x

Total time to cover the distance (x/12 + x/24 + x/8)

∴ Average speed \( = \frac{{3x}}{{\frac{x}{{24}} + \frac{x}{{12}} + \frac{x}{8}}}\)

⇒ \(\frac{{3x}}{{\frac{{x + 2x + 3x}}{{24}}}}\)

⇒ 3x / (6x/24)

⇒ 1/2 × 24 = 12 kmph

∴ The average speed of the person is 12 km/h.

Given:

Speed₁ = 40 km/h, Distance₁ = 48 km

Speed₂ = 65 km/h, Distance₂ = 52 km

Formula used:

Time = Distance/ Speed 

Average Speed = Total Distance/ Total Time

Calculation:

Total Distance traveled = 48 km + 52 km

⇒ 100 km

Total Time = 48 km/ 40 km/h + 52 km/ 65 km/h 

⇒ 6/5 + 4/5 = 2 hours 

Average Speed = 100 km/ 2 hours 

∴ The required Speed = 50 km/h

Given:

A car travels first 33 minutes with a speed of 60 km/h,

Again 33 minutes with a speed of 70 km/h

Concept used:

Distance = velocity × time

Average speed = (Total distance)/(Total time)

Calculation:

\(\)\({\rm{Average\;speed}} = {\rm{\;}}\frac{{\frac{{33}}{{60}}\: ×\:60\: + \:\frac{{33}}{{60}}\: ×\: 70}}{{\frac{{66}}{{60}}}}\)

⇒ Average speed = (33 × 60 + 33 × 70)/66

⇒ Average speed = (60 + 70)/2 = 65 km/h

∴ The average speed of car is 65 km/h.

Shortcut Trick

Concept:

When the time interval is the same for equal distances then,

Average speed = (v₁ + v₂)/2

where v = velocity

Calculation:

Average speed = (70 + 60)/2 = 65 km/h

∴ The average speed of the car is 65 km/h.

Given:

Two trains, X and Y, travel from A to B at average speeds of 80 km/hr and 90 km/hr respectively.

Formula:

Speed = distance/time

Calculation:

Let distance be x km

According to the question

x/80 – x/90 = 1

⇒ (9x – 8x)/720 = 1

⇒ x = 720 km

Formula used:

Distance = Speed × Time

Calculation:

Let the distance be ‘D’ km

According to the question,

D/40 + D/30 = 14

⇒ (3D + 4D)/120 = 14

⇒ 7D/120 = 14

⇒ D = 240 km

∴ Distance between A & B = 240 km
Alternate Method

Formula used:

Average Speed = 2ab/(a + b)

Calculation:

Let the distance be ‘D’ km

V_av  = (2 × 40 × 30)/(40 + 30) 

⇒ V_av = 240/7 km/hr

Since person cover (D + D) i.e. 2D & it take 14 hr

⇒ 2D/14 = 240/7

⇒ D = 240 km

Confusion Points 

Average Speed = (a + b)/2

Applicable when one travels at speed a for half the time and speed b for another half of the time.

Given:

A man covers a certain distance on a scooter driving at 64 km/h.

He returns to the starting point riding on a bicycle at 16 km/h.

Formula used:

Formula:

Average speed = 2s1s2/(s1+s2)

Solution:

Average speed = 2s1s2/(s1+s2)

(2 × 64 × 16)/(64 + 16)

25.6 km/h

∴ The answer is 25.6 km/h.

GIVEN:

Ram covered 60 km in a bus in 50 min. after deboarding the bus, he took rest for 5 min. There after he took a taxi to his home 30 km away and reached in 20 min.

CONCEPT:

Basic concept of time speed and distance.

FORMULA USED:

Average speed = Total distance/Total time taken

CALCULATION:

Total distance = 60 + 30 = 90 km

Total time taken = 50 + 5 + 20 = 75 min = 5/4 hr

Hence,