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

Given:

1/6 distance at 4 km/h

1/4 distance at 6 km/h

1/3 distance at 8 km/h

Remaining 10 km at 10 km/h

Formula used:

Total time = time for each segment

Time = Distance ÷ Speed

Calculations:

Let total distance = x km

⇒ Distance at 4 km/h = x/6 ⇒ Time = (x/6) ÷ 4 = x/24

⇒ Distance at 6 km/h = x/4 ⇒ Time = (x/4) ÷ 6 = x/24

⇒ Distance at 8 km/h = x/3 ⇒ Time = (x/3) ÷ 8 = x/24

Total of above distances = x/6 + x/4 + x/3

LCM of 6, 4, 3 = 12

⇒ x/6 + x/4 + x/3 = (2x + 3x + 4x)/12 = 9x/12 = 3x/4

Remaining distance = x - 3x/4 = x/4 = 10 km

⇒ x/4 = 10 ⇒ x = 40

∴ Total distance = 40 km

Given:

Distance covered by first bus = 320 km

Time taken by first bus = 8 hours

Speed of second bus = 25% more than first bus

Distance to be covered by second bus = 200 km

Formula used:

Speed = Distance / Time

Calculations:

Speed of first bus = 320 km / 8 hours = 40 km/h

Speed of second bus = 40 km/h + 25% of 40 km/h = 40 km/h + 10 km/h = 50 km/h

Time taken by second bus to cover 200 km = Distance / Speed

Time = 200 km / 50 km/h = 4 hours

∴ The time taken by the second bus to cover 200 km is 4 hours.

Given:

Speed 1 = 3 km/h

Speed 2 = 5 km/h

Speed 3 = 6 km/h

Remaining distance = 16 km, Speed 4 = 16 km/h

Formula used:

Time = \(\frac{\text{Distance}}{\text{Speed}}\)

Calculations:

Let Total Distance = D km

Distance 1 = \(\frac{D}{10}\) km

Distance 2 = \(\frac{D}{6}\) km

Distance 3 = \(\frac{D}{5}\) km

Fraction of distance covered = \(\frac{D}{10} + \frac{D}{6} + \frac{D}{5}\)

⇒ Fraction covered = \(\frac{3D + 5D + 6D}{30}\) = \(\frac{14D}{30}\) = \(\frac{7D}{15}\)

Remaining fraction = \(D - \frac{7D}{15}\) = \(\frac{8D}{15}\)

Remaining distance =\(\frac{8D}{15}\) = 16 km

⇒ D = \(\frac{16 \times 15}{8}\) = 30 km

Time for 1st part (T1) = \(\frac{D/10}{3}\) = \(\frac{30/10}{3}\) = 1 hour

Time for 2nd part (T2) = \(\frac{D/6}{5}\) = \(\frac{30/6}{5}\) = 1 hour

Time for 3rd part (T3) = \(\frac{D/5}{6}\) = \(\frac{30/5}{6}\) = 1 hour

Time for 4th part (T4) = \(\frac{16}{16}\) = 1 hour

Total Time = T1 + T2 + T3 + T4 = 1 + 1 + 1 + 1 = 4 hours

∴ She takes 4 hours to travel the total distance.

Given:

Distance walked = 1/10 of total distance

Distance run = 1/6 of total distance

Distance (third part) = 1/5 of total distance

Remaining distance = 16 km

Formula used:

Total Distance = Sum of all individual distances

Calculations:

Let the total distance be D km.

⇒ Distance walked = D/10

⇒ Distance run = D/6

⇒ Distance (third part) = D/5

⇒ Remaining distance = 16 km

⇒ Total Distance (D) = D/10 + D/6 + D/5 + 16

⇒ To combine fractions, find the LCM of the denominators (10, 6, 5). The LCM is 30.

⇒ D = (3D/30) + (5D/30) + (6D/30) + 16

⇒ D = (3D + 5D + 6D)/30 + 16

⇒ D = 14D/30 + 16

⇒ D = 7D/15 + 16

⇒ D - 7D/15 = 16

⇒ (15D - 7D)/15 = 16

⇒ 8D/15 = 16

⇒ 8D = 16 × 15

⇒ D = (16 × 15) / 8

⇒ D = 2 × 15

⇒ D = 30 km

∴ The total distance is 30 km.

Given:

Walking speed (initial) = 5 km/hr

Increased speed = 5 + 3 = 8 km/hr

Time difference = 6 hours

Formula used:

Distance = Speed × Time

Let the distance to be travelled = D km

Time taken at 5 km/hr = D / 5

Time taken at 8 km/hr = D / 8

Difference in time = 6 hours

So, (D / 5) - (D / 8) = 6

Calculation:

(D / 5) - (D / 8) = 6

⇒ (8D - 5D) / 40 = 6

⇒ 3D / 40 = 6

⇒ D = 6 × 40 / 3

⇒ D = 80

∴ The correct answer is option (1).

Let the distance be d km.

We know that,

Distance = Speed x Time

\( \Rightarrow \;\frac{d}{8} + \frac{d}{{12}} = 15\)

\( \Rightarrow \;\frac{{3d + 2d}}{{24}} = 15\)

⇒ d = 72 km 

Given data:

Total time of journey = 7 hours

Speed of car for half distance = 40 km/hr

Speed of car for remaining distance = 60 km/hr

Concept used:

Distance = Speed × Time

Calculation:

Let total distance be 2x.

Time₁ = Distance/Speed

⇒ x/40 hours

Time₂ = Distance/Speed

⇒ x/60 hours

Total time = Time₁ + Time₂

⇒ 7 = x/40 + x/60

⇒ 7 = (3x + 2x)/120

⇒ 7 = 5x/120

⇒ x = 7 × 24

⇒ x = 168 km

⇒ Total distance = 2x

⇒ 2 × 168

⇒ 336 km

∴ Total distance covered by the car is 336 km.

Alternate Method

Concept used:

Average speed = (2 × Speed₁ × Speed₂)/(Speed₁ + Speed_2­)

Calculation:

Since distance covered in both the cases is same we can apply concept of average velocity required to cover same distance.

Average speed = (2 × Speed₁ × Speed₂)/(Speed₁ + Speed_2­)

⇒ (2 × 40 × 60)/(40 + 60)

⇒ 4800/100

⇒ 48 km/hr

Distance = Speed × Time

⇒ 48 × 7

⇒ 336 km

∴ Total distance covered by the car is 336 km.

Given:

Maya goes to office at a speed of 40 km/hr, she reaches 5 minutes late.

Maya goes to office at a speed of 60 km/hr, she is 10 minutes early.

Formula Used:

Distance = Speed × Time

Calculation:

Let the original speed of Maya be x.

Let the distance to office from her home be D.

Maya goes to office at a speed of 40 km/hr, she reaches 5 minutes late.

⇒ D/40 - D/x = 5/60

⇒ D(1/40 - 1/x) = 1/12

⇒ D(x - 40/40x) = 1/12

⇒ D = 40x/12(x - 40)

Maya goes to office at a speed of 60 km/hr, she is 10 minutes early.

⇒ D/x - D/60 = 10/60

⇒D(60 - x)/60x = 1/6

⇒ 40x × (60 - x) /[12(x - 40) × 60x] = 1/6

⇒ 40x × (60 - x) × 6 = 12(x - 40) × 60x 

⇒ x = 45 km/hr

The distance = 40x/12(x - 40) = 40 × 45/12 × 5 = 30 km

∴ The distance to office from her home is 30 km.

Shortcut Trick 

Distance =  S₁ × S₂ × change in time(hr)/(S1 - S2)

Distance = 40 × 60 × (15/60)/(60 - 40) = 30 km

∴ The distance to office from her home is 30 km.

Given:

A car starts from point A towards point B, travelling at the speed of 20 km/h.

1\(\frac{1}{2}\)hours later, another car starts from point A and travelling at the speed of 30 km/h and reaches 2\(\frac{1}{2}\)hours before the first car.

Concept used:

Total distance = \(\frac{(S_1 × S_2)}{S_1-S_2} \times (T_1+T_2)\)

Here,

S₁ = speed of the faster object

S₂ = speed of the slower object

T₁ = Time1

T₂ = Time 2

Calculation:

According to the concept,

Total distance = \(\frac{(20 × 30)}{30-20} \times (1.5+2.5)\)

⇒ \(\frac{(600)}{10} \times 4\)

⇒ 240 km

∴ The distance between A and B is 240 km.

Alternate Method

Let the total distance be x km,

In 1.5 hr 1st car travel 30 km, remaining (x - 30) km

So, according to the question.

⇒ \(\frac{(x - 30)}{20} - \frac{x}{30}\) = 2.5

⇒ x - 90 = 150

⇒ x = 240 km

Calculation:

Time taken by first man to cover journey = 2 PM – 6 AM = 8 hours

Time taken by another man to cover journey = 3 PM – 8 AM = 7 hours

Let total distance from P to Q be 56x km (LCM of 8 & 7)

⇒ Speed of first man = 7x km/hr

⇒ Speed of second man = 8x km/hr

⇒ Distance covered by first man in 2 hours = 14x km

⇒ Remaining distance = 56x – 14x = 42x km

⇒ Time taken to meet each other = 42x/ (7x + 8x) = 42/15 hrs

= 2 hrs 48 min

Given:

Driving his car at the speed of 30 km/h Vinod reaches his office 5 min late and at speed of 40 km/h, he reaches the office 3 min early.

Concept used:

Time = Distance/Speed

Calculations:

Let the time be ‘t’ minutes  to reach office

Let the distance be D.

Time for 30km/h

⇒ (t + 5)/60 = D/30      ----(1)      (1 minute = 1/60 hours)

Time for 40 km/h

⇒ (t – 3)/60 = D/40      ----(2)

Subtract equation (2) from (1)

⇒ [t + 5 - (t - 3)]/60 = D/30 - D/40

⇒ (D/30) - (D/40) = 8/60

⇒ (4D - 3D)/120 = 8/60

⇒ D/120 = 8/60

⇒ D = 16 km

∴ The correct choice is option 1.

Shortcut Trick

Difference in time = Distance/Speed

⇒ [5 - (-3)]/60 = D/30 – D/40    (8 minutes = 8/60 in hours)

⇒ 8/60 = D/30 – D/40 

⇒ D/120 = 8/60

∴ D = 16 km

Given:

The speed of a train including stoppages is 75 kmph.

The speed of a train excluding stoppages is 90 kmph

Formula used:

Speed = distance/time

Calculation:

Since,

The speed of a train is 90 km/hour excluding stoppages

And the speed of the train is 75 km/hour including stoppages.

So, due to stoppage in 1 hour it covers = (90 – 75) km = 15 km less

∴ Time taken by train in stoppage = 15/90 = 1/6 hour = 10 minutes

∴ Option 1 is the correct answer.

Given:

A car takes 50 minutes to cover a certain distance at a speed of 54 km/h

Concept used:

Time = Distance/Speed

Calculation:

Distance cover in 50 minutes = 54 × 5/6 = 45 km

Now,

New speed = 54 × 5/4 = 67.5 km/h

3/4 of 45 = 33.75 km

Now,

Time taken = (33.75/67.5) × 60 = 1/2 × 60

⇒ 30 minutes

∴ It will take 30 to cover the distance.

Shortcut TrickSpeed is inversely proportional to the time

Speed ratio =   4 : 5

Time ratio   =   5 : 4

Now, 5 = 50 minutes

So, 4 = 40 minutes

So, 40 minutes is for the total journey.

Now, for 3/4th of the journey, it takes 40 × 3/4 = 30 minutes

∴ It will take 30 minutes.

Given:

Speed of man = 14 km/h

Total distance to cover = 7 km

Formula used:

Time = distance/speed

Calculation:

If the man doesn't stop to rest, then

Time taken to complete the distance of 7 km = 7/14 = 1/2 h = 30 min

Total rest time = 7 × 6 = 42 min

Total time to cover the distance = 30 + 42 = 72 min = 1\(\frac{1}{5}\) hours

∴ The correct answer is 1\(\frac{1}{5}\) h.

Given:

Speed without stoppage = 80 km/h

Speed with stoppage = 60 km/h

Formula used:

Speed = Distance/Time

Calculation:

Distance travel by Sunil with the speed of 80 km/h in one hour,

Distance = speed × Time

⇒ Distance = 80 × 1 = 80 km

Distance travel by Sunil with the speed of 60 km/h in one hour,

Distance = speed × Time

⇒ Distance = 60 × 1 = 60 km

Now, the time is taken to cover the extra 20 km with the speed of 80 km/h

Time = Distance/Speed

⇒ Time = 20/80

⇒ Time = 1/4 hour = (1/4) × 60

⇒ Time = 15 minutes

∴ Sunil stops on an average 15 minutes per hour.Shortcut Trick

Given:

Speed excluding stoppages = 80 km/h

Speed including stoppages = 60 km/h

Formula used:

Minutes of stops per hour = [(Faster speed - Slower speed)/Faster speed] × 60

calculation:

Minutes of stoppages per hour = [(80 - 60)/80] × 60

= (20/80) × 60

= 15 min

∴ Sunil stops on an average 15 minutes per hour.