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Given:

A boat goes upstream at a speed of 36 km/h and comes back the same distance at 60 km/h.

Formula used:

average speed = total distance travelled/total time taken

Calculations:

let the distance travelled while going upstream be x km.

distance for downstream is also the same.

so the time taken while going upstream = x/36

and the time taken while coming downstream = x/60

thus average speed = \(\frac{2x}{\frac{x}{36}\ +\ \frac{x}{60}}\)

⇒ 2x × (180/8x) 

⇒ 45 km/h

∴ The average speed in entire journey is 45 km/h

Given:

Speed while going = 18 km/hr

Speed while returning = 30 km/hr

Formula used:

Average speed for same distance = 2ab/(a + b)

Where, a = speed while going

B = speed while returning

Calculation:

Average speed = 2ab/(a + b)

⇒ 2 × 18 × 30/(18 + 30)

⇒ (36 × 30)/48

⇒ 90/4

⇒ 22.5 km/hr

∴ The average speed for the whole journey is 22.5 km/hr

Given:

Length of train A = 60 m

Length of train B = 40 m

Time taken to cross each other = 10 seconds

Formula used:

Speed = Distance/Time

Calculation:

Let the speed of both trains be x m/s

Relative speed = (x + x)km/h = 2x m/s

Now,

Time = (60 + 40)/(2x)

⇒ 10 = 100/2x

⇒ 2x = 10

⇒ x = 5 m/s

Given:

Length of train 1 = 157 m

Length of train 2 = 123 m

Time taken to cross each other in opposite direction = 7.2 s

Concept used:

Time taken by trains to cross each other in the opposite direction, t = (L₁ + L₂)/(S₁ + S₂)

Where S₁ and S₂ are speeds of trains 1 and 2 respectively.

s m/s = s × (18/5) km/hr

Calculation: 

7.2 = (157 + 123)/(S₁ + S₂)

⇒ (S₁ + S₂) = 280/7.2 m/s

⇒ 280/7.2 × (18/5) = 140 km/hr

∴ The combined speed of both trains is 140 km/hr.

Given:

Both trains running in the opposite direction

Both trains cross the man in 48 seconds and 38 seconds

Formula used:

Speed= distance/time

Relative speed = when train in opposite direction = (x+y) m/s

Calculation:

Let the speed of trains be x m/s and y m/s respectively

Length of the first train = speed × time = 48x

Length of the second train = speed × time = 38y

\(\frac{{48x + 38y}}{{x + y}} = 42\)

⇒\(48x + 38y = 42x + 42y\)

⇒ \(6x = 4y\)

⇒\(\frac{x}{y} = \frac{2}{3}\)

∴ The ratio of the speed of the trains is 2 : 3

Given:

Total time taken by two trains to cross each other = 20 seconds

Speed of both the trains = 10 m/s

Formula used:

Relative speed = Sum of lengths of both the trains/Time taken by two trains to cross each other

Calculations:

Let the sum of lengths of both the trains or the combined length of both the trains be x

∵ The speed of both the trains is equal, they can cross each other only if moving in the opposite direction

Relative speed = 10 + 10 = 20 m/s

⇒ 20 = x/20

⇒ x = 20 × 20

⇒ x = 400 m

∴ The combined length of both the trains is 400 m

Given:

The speed of two friends are 8km/hr and 10 km/hr

Person with the speed of 10 km/hr reaches 15 minutes earlier

Concept used:

Speed = Distance/Time

Detailed solution:

Let the distance be D

According to question,

(D/8) - (D/10) = 1/4

⇒ D/40 = 1/4

⇒ D = 10

∴ Distance covered to reach the destination is 10 km

Given:

Average Speed of train is 50 kmph

Distance to travel is 600 km

Resting time after every 75 km is 2 min

Concept Used:

Total time = travelling time + resting time

Time = distance/speed

Calculation:

⇒ Travelling time = 600/50 = 12 hr

Total stops = (600/75) – 1 = 8 – 1 = 7

⇒ Resting time = 7 × 2 = 14 min

∴ Total time taken to reach destination is 12 hr 14 min

Given:

Total distance to cover = 150km

Relative speeds of cars = 30 + 45 = 75km

Hours after they will meet each other =

= 150/75

= 2 hours

∴ after 2 hours they will meet.

Given:

A takes 2 hours 30 minutes more than B to walk 40 km.

If A doubles his speed, then he can make it in 1 hour less than B.

Formula used:

Speed = Distance/Time

Calculations:

Let the initial speeds of A and B be x and y respectively and the time taken by A and B be t_A and t_B respectively.

t_A = 40/x

t_B = 40/y

t_A - t_{B =} 40/x - 40/y

⇒ 40/x - 40/y = 2.5      ----(i)

After A doubles his speed,

⇒ 40/y - 40/2x = 1      ----(ii)

On adding equation (i) and (ii),

40/x - 40/2x = 2.5 + 1

⇒ 40/x - 20/x = 3.5

⇒ 20/x = 3.5

⇒ x = 40/7 km/h

From equation (ii),

40/y - 40/2x = 1

⇒ 40/y - 40/2(40/7) = 1

⇒ 40/y - 7/2 = 1

⇒ 40/y = 1 + 7/2

⇒ 40/y = 9/2

⇒ y = 80/9 km/h

tA = 40/x

⇒ tA = 40/(40/7)

⇒ tA = 7 hours

tB = 40/y

⇒ 40/(80/9)

⇒ 9/2 hours

Average time of A and B = {7 + (9/2)}/2

⇒ 23/4 hours

⇒ 5 hours 45 minutes

∴ The average time taken by A and B to walk a 40 km distance is 5 hours 45 minutes.

Given:

Time = 10 minutes.

Formula used:

Distance = Speed × Time

Calculation:

Let the usual time be x min.

Then, Speed = 1 km/hr.

According to the question:

(1 × x) = (3/4) × (x + 10)

⇒ 4x = 3x + 30

⇒ 4x – 3x = 30

⇒ x = 30

∴ The time is 30 minutes.

Given: 

More distance covered by man = 20km 

Formula used:

Distance = speed × time 

Calculation: 

Let the actual distance cover by man be x

⇒ x + 20 = 12 × 10 

⇒ x = 100 km 

∴ The actual distance covered by man is 100 km

GIVEN :

Radha reaches company 10 min late when she travels at a speed of 20kmph

She reaches 20 min early if she travels at a speed of 30kmph.

CONCEPT :

Speed , Time & Distance

ASSUMPTION :

Let the office distance for Radha be M km.

CALCULATION :

Difference in time = 10 + 20 = 30 min = 30/60 hours

Relative speed = M/20 - M/30

⇒ M/20 - M/30 = 1/2

⇒ 10M/600 = 1/2

⇒ M = 30

The office distance for Radha is 30km.

∴ Required speed = distance/speed = 30/2 = 15kmph

Given:

Ran for 5/2 hours at 62km/hr, and for 5/4 hours at 68 km/hr 

Formula Used:

Speed = Distance/Time

Average speed = Total distance traveled/Total time taken

Calculation:

In the first case,

⇒ 62 = Distance/(5/2)

⇒ Distance = (62 × 5)/2

⇒ Distance = 155 km

In the second case,

⇒ 68 = Distance/(5/4)

⇒ Distance = (68 × 5)/4

⇒ Distance = 85 km

Using the formula for average speed,

Average speed = (155 + 85)/(5/2 + 5/4)

⇒ Average speed = 16 × 4

∴ The average speed of a train is 64 km/hr

Given :

Average speed of Arnav and Paridhi = 40

Total Distance = 180 km

Total time = 6 hours

Calculations :

Let the speed of Paridhi = SP

Let the speed of Arnav  = SA

Also , SP > SA

⇒ ( S_P + S_A )/2 = 40

⇒ (SP + SA) = 80

Time taken by Paridhi to travel, T­_P = 180/S_P

Time taken by Arnav to travel, T­_A = 180/S_A

⇒ 180/ S_A – 180/ S_P = 6      ----(1)   

Put, S_A = 80 – S_P in equation (1)

⇒ S_P² – 20S_P – 2400 = 0

⇒ S_P = 60 , –40              

⇒ S_P  = 60 km/hr 

∴ The speed of Paridhi is 60 km/hr.

Given:

Speed (a) = 20 kmph

Time late (t₁) = 30 min

Increased Speed (b) = 4 kmph

Time late (t₂) = 10 min

Formula Used:

Distance = (t1 – t2) × (a + b) a/b

Calculations:

a = 20 kmph , t₁ = 30 min , b = 4 kmph , t₂ = 10 min

According to the formula,

Distance = (t₁ – t₂) × (a + b) × a/b

⇒ Distance = (30 - 10) / 60 × (20 + 4) × 20/4

⇒ Distance = 20/60 × 24 × 5

⇒ Distance = 8 × 5

⇒ Distance = 40

The distance of the police station from his home is 40 km.

Alternate method:

Let the distance = x

Here, the difference in time = 30 – 10 = 20 min = 20/60 = 1/3 hr

Speed during next journey = (20 + 4) = 24 kmph

According to the question,

⇒ x/20 – x/24 = 1/3

⇒ (6x – 5x) / 120 = 1/3

⇒ x = 40

The distance of the police station from his home is 40 km.

Given:

When the speed of man is 9/11 of its usual speed then he is late by 1 hr 45 minute.

Formula/Concept used:

Time and speed are inversely proportional to each other

Time ∝ 1/speed

Calculation:

When the speed of man is 9/11 of his actual speed then the time taken by him will be 11/9

∵ Speed and time are inversely proportional to each other

Let the usual time taken by man is t

∴ (11/9)t – t = 7/4 hours

⇒ t = 7.875 hours

∴ The usual time taken by man is 7.875 hours

GIVEN:

Ram walks at a speed of 10 km/h. Today the day was very hot so walked at 2/5^th of his average speed. He arrived at his school 18 minutes late.

FORMULA USED:

Time = Distance/Speed

CALCULATION:

Suppose the distance between his school and home = X km

Decreased speed = 10 × 2/5 = 4 kmph

According to the question:

X/4 - X/10 = 18/60 = 3/10

⇒ X = 2 km

So,

Given:

Speed 1 = 3.5 km/hr reaches 6 min late

Speed 2 = 4.5 km/h reaches 10 min early.

Formula used:

Distance = [(Speed 1 × Speed 2)/(Difference of Speeds)] × time difference

Calculation:

Distance = [(Speed 1 × Speed 2)/(Difference of Speeds)] × time difference

⇒ Distance = [(3.5 × 4.5)/1] × 16/60

⇒ 4.2 km

∴ The distance between her house and the school is 4.2 km.

Mistakes Point:

Given :

Sitesh drove 12 hours at a speed of @ 75 km/hr

Concept used :

Average speed = Total distance/Total time 

Calculations :

Let the time taken for 90 km/h be 'x' 

Total distance = 75 × 12 + 90 × x 

⇒ 900 + 90x 

Total time = (12 + x) hours 

According to the question 

The overall average speed is 81 km/h then, 

81 = (900 + 90x)/(12 + x) 

⇒ 81 × (12 + x) = 900 + 90x 

⇒ x = 8 

∴ He drove for 8 hours at the speed of 90 km/h

Given:

The man reaches the office late by 30 minutes

Formula used:

Distance = Speed × Time

Concept used:

Speed is inversely proportional to the time taken 

Calculation:

Let the distance from the office and home be x and original speed be y

Then, the Original time is taken = x/y

Decreased speed = 2y/3

Time is taken when speed is decreased = Distance/Speed = x/(2y/3)

Now, A/Q

x/(2y/3) – x/y = 30

⇒ 3x/2y – x/y = 30

⇒ x/2y = 30

⇒ x/y = 30 × 2 = 60 minutes

∴ The actual time is taken by the man to reach the office is 60 minutes

Given:

Square of side = 75 m

And a dog runs around the square with a speed of = 6 km/hr

Formula Used:

The perimeter of a square = 4 × side of the given square

Calculation:

⇒ Speed of the dog in m/sec = 6 × 1000/3600 = 6× 5/18 = 5/3 m/sec

⇒ The perimeter of the Square = 4 × 75 m = 300 m

⇒ Time = Distance/speed

Hence, Time is taken by the dog to travel around the square

⇒ Time = (300)/(5/3) = 60 × 3 seconds or (60 × 3)/60 = 3 minutes

Given:

One-third of total journey is covered at 20 km/hr, one-fifth at 15 km/hr and rest at 10 km/hr.

Concepts used:

Time = Distance/Speed

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

Calculation:

Let the total distance covered be x.

Total distance = x/3 + x/5 + remaining distance

⇒ x = 8x/15 + remaining distance

⇒ Remaining distance = x – 8x/15 = 7x/15

Total time taken to cover the whole journey = (x/3)/20 + (x/5)/15 + (7x/15)/10 = 23x/300

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

⇒ x/(23x/300) km/hr

⇒ 300/23 km/hr

∴ Average speed for whole journey is 300/23 km/hr.

Given:

Time is taken to cover distance = 50 minutes

Speed of a bus = 48 kmph

Formula used:

Distance = speed × time

Speed = distance / time

Calculations:

Distance = 48 × (50/60) = 40 km

Speed = 40/(40/60)

⇒ 60 km/hr

∴ The bus must run to reduce the time of journey to 40 minutes it runs at 60 km/hr 

Given:

The total distance covered by Aman is 1200 km. Aman covers 60% of distance in 1 day 6 hr and remaining 40% distance in 20 hr

Formula used/Concept Used:

Speed = Distance/time

Average speed = (Total distance)/Total time

Calculation:

1 day 6 hr = 30 hr

As we know that to calculate the average speed we need total distance and total time

∴ Average speed = [1200/(30 + 20)] = 24 km/hr

∴ The average speed of Aman is 24 km/hr

Given:

The speed of the Ram in his first trip = 120 km/h

Formula used:

 \(Average\ speed =\ {Total\ distance \over Total\ time}\)

Calculation:

Let us assume the distance on the first trip is X

⇒ The total distance = 4X

⇒ The speed on first trip = 120 km/h

⇒ The speed on second trip = 60 km/h

⇒ The speed on the third trip = 30 km/h

⇒ The speed on the fourth trip = 15 km/h

⇒ The total time is taken by Ram in the whole trip = \({X\over120}\ +\ {X\over60}\ +\ {X\over30}\ +\ {X\over15}\) = \({15X\over120}\) = \(X\over8\)

⇒ The average speed = \({4X\over {X\over8}}\) = 32 km/h

∴ The required result will be 32 km/h.

Given:

Average speed of bus for the whole journey is 25 km/hr.

28 km are travelled at the speed of 22 km/hr and rest at 35 km/hr.

Concepts used:

Average speed = S1 × S2 × (d1 + d2)/(S1d2 + S2d1)

Calculation:

A bus covered 28 km at the speed of 22 km/hr and rest distance at the speed of 35 km/hr.

Average speed = S1 × S2 × (d1 + d2)/(S1d2 + S2d1)

⇒ 25 km/hr = 22 × 35 × (28 + x)/(22 × x + 35 × 28) km/hr

⇒ 25 = 770 × (28 + x)/(22x + 980) km/hr

⇒ 25 × (22x + 980) = 770 × 28 + 770x

⇒ 25 × 22x + 25 × 980 = 21,560 + 770x

⇒ 550x + 24,500 = 21,560 + 770x

⇒ 770x - 550x = 24,500 - 21,560

⇒ 220x = 2,940

⇒ x = 2,940/220 km

⇒ x = 13.36 km

∴ Remaining distance travelled by bus is equal to 13.36 km. 

Formula used∶

Average speed = \(\frac{{{\rm{Total\;distance\;}}}}{{{\rm{Total\;time\;}}}}\)

Calculation∶

Time taken while travelling by train = 1200 / 60 = 20 hrs

Time taken while travelling by car = 900 / 45 = 20 hrs

Time taken while travelling by bus = (3000 - 1200 - 900) / 30 = 30 hr

Total time taken = 20 + 20 + 30 = 70 hr

Given∶

A car travels 1 / 3 of a distance in 2 hours, 1 / 4 of remaining in 1 hour and remaining in 3 hours.

Average speed of the car in the whole journey is 50 km / hr

Formula used∶

Average speed = Total distance Travelled / total time

Calculations∶

Let the total distance be x

And the total time = 1 + 2 + 3 = 6 hr

Average speed = \(\frac{{\rm{x}}}{6}\)         ---(1)                                          

Since average speed is 50 km / hr, on equating equation (1) with 50 km / hr, we get

Given:

Distance travelled by Bus = Distance travelled by Car = Distance travelled by Bike = 12 km

Ratio of speed by Bus, Car and Bike is 4 : 2 : 1

Total duration of journey = 7 hours

Formula used:

Speed = Distance/Time

Calculations:

Let the speed by Bus, Car and Bike be 4y, 2y, y respectively.

Total time = Time taken by Bus + Time taken by Car + Time taken by Bike

7 = (12/4y) + (12/2y) + (12/y)

⇒ 7 = (3/y) + (6/y) + (12/y)

⇒ 7 = 21/y

⇒ y = 21/7

⇒ y = 3

Time taken by car = 12/2y

⇒ 6/y

⇒ 6/3

⇒ 2 hours

∴ The time taken by Car is 2 hours.

Given:

John travels a distance by bike =\(\frac{2}{3}^{rd} \) of the total distance 

John travels a distance by foot = 6 km

Calculation:

Let us take the total distance travelled be x

John travels a distance by foot  = x - \(\frac{2}{3} \)x ⇒ \(\frac{1}{3} \) x          -----------------------(I)

John travels a distance by foot = 6 km                   -----------------------(ii)

Equating (I) and (ii)

6 = \(\frac{1}{3} \)x

x = 18 km 

∴ The total distance travelled = 18 km

Given:

Length of the train = 700 m

Length of the platform = 800 m 

Time take by Train to crosses the platform = 50 seconds 

Formula Used:

Distance = Speed × Time

Calculation:

Length of the train + length of the platform = Total distance

Total distance  = 700 + 800 = 1500 m

Total time taken by train to cross the platform = 50 seconds 

Speed of the train = Total distance / Total time = 1500/50 ⇒ 30 m/s

∴ The speed of the train is 30 m/s

Given:

Speed in 1^st half is 15 km/h

Speed in 2^nd half is 5 km/h

Total distance covered is 150 km

Formula:

Average speed for a journey when distances are equal is S = 2{xy/(x + y)}

Where,

S =average velocity

x = speed during 1^st part of journey

y = speed during 2^nd part of journey

Calculation:

Average speed for the whole journey = 2 × {(15 × 5)/(15 + 5)} km/h

⇒ Average speed = 2 × (75/20) km/h

⇒ Average speed = (75/10) km/h

Time to cover the distance = 150/(75/10) hours

⇒ Time = (150 × 10)/75 hours

⇒ Time = 20 hours

∴ Time taken to cover the distance is 20 hours

GIVEN:

⇒ Total distance = 140 kms

ASSUMPTION:

Let the upstream speed of Boat and speed of the stream be 5x and 1x.

⇒ Speed of boat in still water = 5x + 1x = 6x

⇒ Downstream speed = 6x + 1x = 7x

CALCULATION:

According to the question.

When travelling downstream.

⇒ Let the time taken = T hours

⇒ 7x = 140/T      ----(1)

When travelling upstream.

Let the time taken = (T + 2) hours

⇒ 5x = 140/ (T + 2)      ----(2)

From eq(1) and eq(2)

⇒ T = 5 hours

⇒ x = 4

⇒ Speed of Boat in still water = 6x = 6 x 4 = 24 kmph

We have,

(cosθ + sinθ) = √2cosθ 

By squaring both sides,

(cosθ + sinθ)² = 2cos²θ 

⇒ cos²θ + sin²θ + 2sinθcosθ = 2 cos²θ 

(∵ (a + b)² = a² + b² + 2ab) 

⇒ sin²θ = cos²θ − 2sinθcosθ

⇒ 2sin²θ = cos²θ − 2sinθcosθ + sin²θ 

⇒ (cosθ − sinθ)² = 2sin²θ 

⇒ cosθ − sinθ = √2sinθ 

(By taking squaring root of both sides) 

Hence Proved.

Given,

AP is 3, 15, 27, 34, ……. 

First term of AP is a = 3 and common difference of AP is d = a₂ – a₁ 

= 15 – 3 

= 12. 

54^th term of AP is a₅₄ = a + (54 – 1)d 

(∵ a_n = a + (n – 1) d and n = 54) 

= 3 + 53 × 12 

= 3 + 636 

= 639. 

(∵ a = 3 & d = 12) 

Let n^th term of AP is 132 more than its 54^th term. 

i.e.,

a_n = 132 + a₅₄ 

= 132 + 639 = 771 

⇒ 3 + (n – 1) 12 = 771 

(∵ a = 3 & d = 12 and a_n = a + (n – 1) d)

⇒ 12(n − 1) = 771 − 3 = 768 

⇒ n − 1 = \(\frac{768 }{12}\) = 64 

⇒ n = 64 + 1 = 65.

Hence,

65^th term of AP is 132 more than its 54^th term.

Given a = 3, d = 12, n = 54.
We know that an=a+(n-1) * d   ---------- (1)

                           = 3 + (54-1) * 12
                           = 3 + 53 * 12
                          =639

                  a54 = 639.
 

Given that the term is 132 more than its 54th term = 639 + 132 = 771.
on substituting in (1), we get

771 = 3 + (n-1) * 12
768 = (n-1) * 12
768/12 = n-1
n-1 = 64
n = 65.
 

Therefore the 65th term is 132 more than its 54th term.Given a = 3, d = 12, n = 54.

We know that an=a+(n-1) * d   ---------- (1)

                            = 3 + (54-1) * 12

                            = 3 + 53 * 12

                            = 639.

                     a54 = 639.
 

Given that the term is 132 more than its 54th term = 639 + 132 = 771.

on substituting in (1), we get

771 = 3 + (n-1) * 12

768 = (n-1) * 12

768/12 = n-1

n-1 = 64

n = 65.

Therefore the 65th term is 132 more than its 54th term.

Red balls = 6 

Green balls = 8 

White balls = 8 

Total number of balls = 6 + 8 + 8 = 22 

a. The probability of getting a white or green ball = 16/20 

b. The probability of getting neither green balls nor a red ball = 8/20

correct option:

(A) 0 

(A) [Cu(H₂O)₆]SO₄ 

There is un symmetric filling of e_g subset of Cu⁺² ion, while there is symmetrical distribution in t_2g set, if the complex has same ligand there will be equal repulsion which leads to symmetrical bond length along t_2g , but due to uneven filling of electron in e_g subset, either octahedral will be elongated or compressed.