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Correct option: a) 2 mph

Explanation:

Let the speed of the stream be x hrs. Then, speed downstream=(4/x) kmph, speed upstream=(3/x) kmph; therefore(48/(4/x))+(48/(3/x))=14 or x=1/2

Speed downstream= 8 kmph and speed upstream= 6 kmph and rate of the stream=1/2*(8-6) kmph= 1 kmph

Concept:

The LCM of two or more given numbers is the least number which is exactly divisible by each of them.

Caclculation:

L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 5 = 480

Given, x should be divisible by 9.

Let (480k + 9) is divisible by 9.

x = (480k + 9)

Put k = 1, 2, 3, 4, 5…..

(i) When, k = 1, x = 489 is not divisible by 9.

(ii) When, k = 2, x = 969 is not divisible by 9.

(iii) When, k = 3, x = 1449 is divisible by 9.

⇒ x = 1449

∴ The unit digit of x is 9.

Correct option: b) 6 kmph

Explanation:

Rate downstream=(16/2) kmph= 8 kmph; rate upstream= (16/4) kmph=4 kmph speed in still water =1/2*(8+4) kmph =6 kmph

Given:

Given numbers are 18, 24 and 32

Concept Used:

LCM (Least Common Multiple): LCM of x and y is the least common multiple of x and y which is perfectly divisible by both x and y

For Example,

3 → Multiple of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, …

4 → Multiple of 4 are 4, 8, 12, 16, 20, 24, 28, 32, …

Here common are 12 and 24. But of these 12 is least

∴ LCM of 3 and 4 is 12

Given:

P does 3/4 of a piece of work = 15 days

When Q comes then both fished the work = 3 days

Concept used:

Total work = LCM

Formula used:

Efficiency = Work/Time

Calculations:

P does 3/4 of a piece of work = 15 days

Time taken by P to complete the total work = 15 × 4/3 = 20 days

Remaining work = 1 – 3/4 = 1/4

So, P + Q work 1/4 in 3 days

Now, Time taken by P and Q to complete the work total work together = 3 × 4/1 = 12 days

Total work = LCM = 60

Efficiency of Q = 5 – 3 = 2

Time taken by Q to complete the work alone = 60/2 = 30 days

∴ Time taken by Q to complete the work alone is 30 days

Given:

Boat takes time to cover a certain distance upstream = 8 hours 48 minutes

Boat takes time to cover the same distance downstream = 4 hours

Formula Used:

Speed = Distance/Time

Calculation:

Let speed of boat in still water be x km/h

Let speed of current be y km/h

Let the certain distance be D km

Upstream speed = (x – y) km/h

Downstream speed = (x + y) km/h

8 hours 48 minutes = 44/5 hours

D/(x + y) = 4           .....(i)

D/(x – y) = 44/5      .....(ii)

Dividing equation (i) by (ii)

\(⇒ \frac{{\frac{D}{{\left( {x + y} \right)}}}}{{\frac{D}{{\left( {x\;-{\rm {y}}} \right)}}}} = \;\frac{4}{{\frac{{44}}{5}}}\)

⇒ (x – y)/(x + y) = (4 × 5)/44

⇒ 44 (x – y) = 20 (x + y)

⇒ 44x – 44y = 20x + 20y

⇒ 44x – 20x = 20y + 44y

⇒ 24x = 64y

⇒ x/y = 64/24

⇒ x/y = 8/3

⇒ x : y = 8 : 3

∴ The ratio between the speed of the boat and speed of the water current is 8 : 3

Given:

Total cost = Rs. 225

Ratio of number of Rs. 1, 50 paise and 25 paise coins = 8 : 5 : 3

Calculations:

Let the number of Rs. 1, 50 paise and 25 paise coins be 8x, 5x, 3x respectively.

Total cost of Rs. 1 coins = Rs. 8x

Total cost of 50 paise coins = 0.5 × 5x = Rs. 2.5x

Total cost of 25 paise coins = 0.25 × 3x = Rs. 0.75x

Total cost = Rs. 225

⇒ 8x + 2.5x + 0.75x = 225

⇒ 11.25x = 225

⇒ x = 20

Number of Rs. 1 coins = 8x

⇒ 8 × 20

⇒ Rs. 160

∴ The number of one - rupee coins is Rs. 160.

Given:

15 men can finish a piece of work in 40 days.

Number of men to leave the work = 5

Calculations:

Let the number of days after which 5 men left the work be y.

15  × 40 = 15y + (15 - 5) × (45 - y)

⇒ 600 = 15y + 10(45 - y)

⇒ 600 = 15y + 450 - 10y

⇒ 600 = 5y + 450

⇒ 150 = 5y

⇒ y = 30 days

∴ After 30 days 5 men should leave the work.

Calculation∶

The series follows following pattern

147 - 126 = 21

168 - 147 = 21

189 - 168 = 21

210 - 189 = 21

231 - 210 = 21

Given

A certain number of men could do a piece of work in 60 days

With 8 more men, it could be finished in 50 days

Formula used

m₁ × d₁ = m₂ × d₂ 

Where, m = number of men and d = number of days

Calculation

Let the number of men initially was 'x'

⇒ x men can do a piece of work in 60 days 

And x + 8 men can finish the work in 50 days

⇒ x × 60 = (x + 8)× 50

⇒ 60x = 50x + 400

⇒ 10x = 400

⇒ x = 40

∴ The number of men in the beginning was 40

Given:

The number M = 147 × 84 × 126 × 39 is divisible by 7^x

Calculation:

M = 147 × 84 × 126 × 39  

⇒ 7² × 3 × 7 × 12 × 18 × 7 × 39

⇒ 7⁴ × 3 × 12 × 18 × 39

M is divisible by 7⁴.

∴ The maximum value of x is 4.

Given:

A, B and C can do a work in 10, 20 and 30 days respectively.

Formula Used:

If A and B can together finish a work in t days and alone can finish the work in a and b days respectively,

Then, 1/a + 1/b = 1/t

Work = Time × Efficiency

Calculation:

Total work = LCM of 10, 20, 30 = 60 units

Efficiency of A = 6

Efficiency of B = 3

Efficiency of C = 2

1    2    3    4    5    6    7    8    9    10 (Days)

B  BA BC  BA  B  ABC B  AB  BC  AB

3    9   5     9     3 11     3    9    5     9

57 units can be completed in 9 days.

Remaining work = (60 – 57) units = 3 units

On 10^th day 9 units will be done

Remaining 3 units will be done in = 3/9 days = 1/3 days

Total work will be done in = (9 + 1/3) days = 28/3 days

∴ The work will be finished in 28/3 days.

Concept Used-

The least number which is divisible by several numbers is the LCM of such numbers.

Calculation-

The least number which is divisible by 12, 18, 24 and 30 = LCM of 12, 18, 24 and 30

⇒ 360

Now the least number which leaves remainder as 4 when divided by 12, 18, 24 and 30 = 360 + 4

⇒ 364

Now, 364 is divisible by 7.

∴ 364 is the required number.

Given:

Numbers are 225 and 147

Concept used:

If two numbers are divided by their difference or factor of difference, leaves same remainder

Calculation:

Given numbers are 225 and 147

Difference of the numbers = 225 – 147 = 78

Calculation of factors of 78:

⇒ 1 × 78 = 78

⇒ 2 × 39 = 78

⇒ 3 × 26 = 78

⇒ 6 × 13 = 78

Factors of 78 are 1, 2, 3, 6, 13, 26, 39, 78

Only two digit factors are = 13, 26, 39, 78

∴ Only 4 numbers are possible.

Numbers are 6, 7 and 8.

Remainder = 5

Concept Used:

When we have to find the least number then we have to calculate the LCM (least common multiple)

Calculation:

We know that

If the number is divisible by the given numbers then it should be the multiple of LCM of given numbers.

∴ LCM of 6, 7 and 8 = 168

Now, lowest four digit number is 1000 and when we divide 1000 with 168 we get 160 as remainder but if we subtract the remainder we will get three digit number but we need a four digit number so the next number after 840 which is divisible by 6, 7 and 8 = 840 + 168 = 1008

So, 1008 is the number which is divisible by 6, 7 and 8 but it should leave the remainder 5 so we need to add 5 in the number to get the required number.

Given:

True weight = 1000 cm

False length = 800 cm

Formula Used:

Profit% = True weight - false weight / false weight × 100

Calculations:

According to the formula,

Profit% = True weight - false weight / false weight × 100

⇒ 1000 - 800 / 800 × 100

⇒ 200 / 800 × 100

⇒ 100 / 4

⇒ 25%

Given:

A diamond is broken into three parts in the ratio 1 ∶ 2 ∶ 3

The price of diamond before broken = Rs 47520

Concept:

In this type of questions we can calculate the actual weight of diamond by adding al the ratios.

Calculation:

Let the weight of 3 parts be x, 2x, and 3x

So actual weight of diamond = x + 2x + 3x = 6x

According to question

The price of diamond is in proportion to cube of its weight

The price of diamond before broken = (6x)³

⇒ 216x³

The price of diamond after broken = x³ + (2x)³ + (3x)³

⇒ x³ + 8x³ + 27x³

⇒ 36x³

The price of diamond before broken = 216x³

⇒ 216x³ = 47520

⇒ x³ = 220

Loss in the price after broken = 216x³ – 36x³

⇒ 180x³

⇒ 180 × 220

⇒ 39600

∴ The loss in the price of diamond after broken is Rs 39600.

Concept:

The product of two numbers always will be the product of their Least Common Multiple and Highest common factor

Product of 2 numbers = LCM × HCF

Given:

The product of two numbers = 192

The greatest common divisor = 4

Calculation:

192 =  4 × LCM

Given :

x is greater than or equal to 25 and y is less than or equal to 40

Calculations :

x can take any value more than or equal to 25 

y can take any value less than or equal to 40 

Now check each option to find which is correct 

Option(1) ⇒ x cannot always be greater than y.  (for example x = 26 and y = 39)

Option(2) ⇒ y - x is greater than 15 is not possible always.  (for example x = 26 and y = 39)

Option(3) ⇒ y - x is always less than or equal to 15 as their gap cannot exceed 15. (For example x = 25 and y = 40 or x = 26 and y = 39)

Option(4) ⇒ x + y greater than or equal to 65 can be correct but not always.  (For example at x = 28 and y = 30)

∴ Option 3 will be the right choice.

Calculation:

Find the prime factorization of 21

⇒ 21 = 3 × 7

Find the prime factorization of 560

⇒ 560 = 2 × 2 × 2 × 2 × 5 × 7

Multiply each factor the greater number of times it occurs above to find the LCM.

⇒ LCM = 2 × 2 × 2 × 2 × 3 × 5 × 7

⇒ LCM = 1680

∴ The least common multiple of 21 and 560 is 1680.

Calculation:

Divisibility rule of 9:

A number is divisible by 9 only when the sum of its digits is divisible by 9.

10098 → 1 + 0 + 0 + 9 + 8 = 18

10108 → 1 + 0 + 1 + 0 + 8 = 10

10089 → 1 + 0 + 0 + 8 + 9 = 18

10087 → 1 + 0 + 0 + 8 + 7 = 16

10098 and 10089 are the numbers that satisfy the condition 

Divisibility rule of 11:

A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11

In case of 10098

⇒ (1 + 0 + 8) – (0 + 9) = 9 – 9 = 0 (Satisfying the condition)

Checking second term 10089

⇒ (1 + 0 + 9) – (0 + 8) = 9 – 8 = 1 (Not satisfying the condition)

∴ The number which is divisible by both 9 and 11 is 10098.

Concept used:

Divisibility of 11 – Difference between the sum of digits at odd and even places will be zero or multiple of 11.

Divisibility of 7 – Subtract the twice of unit digit from remaining number and check if Resultant number is divisible by 7

Calculation:

Divisibility of 11 – Difference between the sum of digits at odd and even places will be zero or multiple of 11.

Divisibility of 7 – Subtract the twice of unit digit from remaining number and check if Resultant number is divisible by 7

We firstly check the numbers divisible by 11, if it satisfies then we proceed to check the divisibility of 7

16425 : (5 + 4 + 1) – (2 + 6)

⇒ 10 – 8 = 2

Not equal to zero or multiple of 11, hence not divisible by 11.

12235 : (5 + 2 +1) – (3 + 2)

⇒ 9 – 5 = 4

4 is not equal to zero or multiple of 11, hence not divisible by 11

16257 : (6 + 5) - (7 + 2 + 1)

⇒ 11 – 10 = 1

1 is not equal to zero or multiple of 11, hence not divisible by 11

16324 : (4 + 3 + 1) – (2 + 6)

⇒ 8 – 8 = 0

Difference is equal to zero, hence divisible by 11

For this only we check divisibility of 7

1632 – (2 × 4)

⇒ 1632 – 8 = 1624

162 – (2 × 4)

⇒ 162 – 8 = 154

15 – (2 × 4)

⇒ 15 – 8 = 7

This number is also divisible by 7

∴ 16324 is divisible by both 7 and 11.

Calculation 

For option A, total charges for 3 nights trip and 3 day trips  = 1000 × 6 = Rs 6000

For option B, total charges for 3 nights and 3 days trip and all less than 50 km = 1200 × 3  + 700 × 3 = Rs 5700

For option C , total charges for 3 nights trip and 3 days trip and all less than 50 km = 1100 × 6 = Rs 6600

∴ Option B is best for company 

Calculation:

From the data,

In condition B day trip,

⇒ 700 per trip and night trip

⇒ 1200 per trip

In condition C more than 50 km trip

⇒ 800 per trip and less than 50 km

⇒ Rs. 1,100 per trip 

In condition B , 4 night package

⇒ 4 × 1200

⇒ Rs. 4,800

In condition C, 4 days less than 50 km trip

⇒ 4 × 1100

⇒ R.s 4,400

∴ Condition C is more preferable than B If there are 4 night trips less than 50 KM.

Given:

Required distance = 350 km, time 5 hours

Covered distance = 2 / 7 of total in 2 hours

Condition: given time = 4 hours 30 minutes

Formula used:

Speed = Distance / time

1 kmph = 5/18 m/s

Calculation:

Distance left = (1 – (2 / 7) × 350

⇒ Distance left = (5 / 7) × 350 = 250 Km

⇒ Time left = 4 hours 30 minutes – 2 hour

⇒ 2 hours 30 minutes = 2.5 hr

⇒ Required speed = 250/2.5 = 100 kmph

∴ Boy needs to travel at speed of 100 kmph in order to reach the destination 30 minutes earlier

Given:

Speed of three Motorbike racers = 10 ∶ 7 ∶ 2

Concept used:

Time = Distance/Speed

Calculation

Let, distance be x

Ratio of time taken by motorbike first, second and third = (x/10) ∶ (x/7) ∶ (x/2)

⇒ 140 × (x/10) ∶ 140 × (x/7) ∶ 140 × (x/2)

⇒ 14x ∶ 20x ∶ 70x = 7 ∶ 10 ∶ 35

∴ The ratio of time taken by motorbike first, second and third is 7 ∶ 10 ∶ 35

Let T, A and B be the average speeds(km/h) of train, auto and bike respectively.

Given, T = 4A = 2.5B

⇒ A = T/4 and B = T/2.5 = 2T/5

According to question, 135/A + 780/T + 288/B = 17

⇒ 135/(T/4) + 780/T + 288/(2T/5) = 17

⇒ 540/T + 780/T + 720/T = 17

⇒ 2040/T = 17

⇒ T = 2040/17 = 120 km/h

Thus, B = 2T/5 = 0.4 × 120 = 48 km/h

Given:

The ratio of the speed of the bike, van and lorry = 3 ∶ 7 ∶ 5 

Distance covers by lorry = 360 km 

Time taken by lorry to cover = 12 hours 

Concept used:

Speed = Distance/Time 

Average = (Total sum)/(Total terms)

Calculations:

Let the speed of the bike, van and lorry be 3x, 7x and 5x respectively 

Speed of lorry = 360/12 = 30 km/h 

⇒ 5x = 30 

⇒ x = 6 

Speed of bike = 3x = 3 × 6 = 18 

Speed of van = 7x = 7 × 6 = 42 

Average speed of bike and van = (18 + 42)/2 = 30 km/h 

∴ Average speed of bike and van is 30 km/h

Given:

The distance between A and B = 160 km

Speed of car₁ = 50 km per hour

Speed of car₂ = 30 km per hour

Concept used:

If two trains are moving in opposite directions at u m/s and v m/s, then relative speed is = (u + v) m/s

Formula used:

Time = Distance/Speed

Calculation:

Relative speed = (speed of car₁ + speed of car₂)

⇒ Relative speed = (50 + 30) = 80 km/hr

⇒ Time = Distance/Time

⇒ Time = 160/80 = 2 hours

They start at the same time 08 ∶ 10 AM

⇒ Time = 08 ∶ 10 AM + 2 = 10 : 10 AM

∴ They will meet each other at 10 : 10 AM

Given:

Speed of train = 2 times the speed of bike

Time taken by bike to cover 570 km = 15 hours

Formula used:

Speed = Distance/Time taken

Calculation:

Speed of bike = (570/15) km/hr = 38 km/hr

Speed of train = (2 × 38) = 76 km/hr

Distance travelled by train in 6 hours = (76 × 6) km = 456 km

∴ Train can cover 456 km in 6 hours

Given:

Total distance covered = 250 km

Speed in the first part of the journey = 40 km/h

Speed in the second part of the journey = 70 km/h

Speed in the third and final part of the journey = 18 km/h

Formula used:

Speed = Distance/Time

Average speed = Total distance/Total Time

Calculation:

Average speed = 250/{(20/40) + (140/70) + (90/18)}

⇒ 250/[(1/2) + 2 + 5]

⇒ 250/(15/2)

⇒ (500/15)

⇒ 33.33 km/h

∴ The average speed of the whole journey is 33.33 km/h.

Given:

speed of boat in still water = 18 km/hr

speed of the stream = 3 km/hr

Calculation:

Speed of boat in upstream = 18 – 3 = 15 km/hr

So , distance travelled by boat in upstream in 12 minutes = 15 × 12/60

= 3 km

Given:

Length of train = 141.5 m

Speed of train = 57 km/h

Time = 39 seconds

Formula used:

Speed = (distance)/(time)

Calculation:

Let the length of the platform be 'x' m.

Speed of train = 57 km/h

⇒ (57 × 1000)/(3600) m/s

⇒ 95/6 m/s

Total length = (length of train) + (length of the platform)

⇒ 141.5 + x

Speed = (distance)/(time)

⇒ distance = speed × time

⇒ 141.5 + x = (95/6) × 39

⇒ x = 617.5 – 141.5

⇒ x = 476m

∴The length of the platform is 476 m.

Given:

Length of train = 180 m

Time taken to cross platform = 6 seconds

Formula required:

Speed = Distance/Time

To convert m/s into km/h multiply by 18/5

Calculations:

Speed = 180/6

⇒ 30 m/sec

Convert m/sec in to km/hr

30 × 18/5

⇒ 108 km/hr

∴ The speed of train is 108 km/hr

Given-

Speed of a boat in still water = 30 km/h

Speed of stream = 6 km/h

Formula Used- 

Distance = Speed × Time

Concept Used- 

Relative Downstream Speed = Speed of a boat in still water + Speed of stream

Calculation-

Relative Downstream Speed = 30 + 6

⇒ 36 km/hr

Time taken = 12/36 hr

⇒ (1/3) × 60 minutes

⇒ 20 minutes

∴ Time taken by the boat to cover a distance of 12 km downstream is 20 minutes.

Formula Used:

Speed = Distance/time

Calculation:

Let assume the length of the train = x m

For platform 1

Speed = (Length of train + length of platform)/(Time)

⇒ Speed = (x + 162)/36     m/sec.                     .......................(1)

For platform 2

Speed = (Length of train + length of platform)/(Time)

⇒ Speed = (x + 120)/30  m/sec.                      ................... (2)

Speed is uniform 

So,

From equation 1 and equation 2

⇒ (x + 162)/36 = (x + 120)/30    

⇒ 5x + 810 = 6x + 720

⇒ x = 90m

Length of the train is 90m.

Speed = Distance/time

⇒ Speed = (90+ 162)/36 

⇒ Speed = 7 m/seconds

The length of the train is the same. The change in the time taken to cover the distance is because of the difference in length of the platforms.

Conclusion:

Statement A, B and D are correct

The correct option is 3 i.e. Statement A, B and D are correct.

Given:

Speed = 250 km/hr

Time₁ = 4 hours

Time₂ = 1 hour 40 minutes

⇒ (60 + 40) minutes

⇒ 100 minutes = 100/60 hours

⇒ 5/3 hours

Formula used:

Distance = Speed × Time taken

Calculation:

Distance covered = (250 × 4) km

⇒ 1000 km

Required speed = 1000 ÷ (5/3) km/hr

⇒ 1000 × (3/5) km/hr

⇒ 600 km/hr

∴ The required speed is 600 km/hr

Given-

Length of 1st train = 110 m

Speed of 1st train = 72km/hr 

Length of 1st train = 120 m

Speed of 1st train = 36 km/hr 

Formula used = 

Distance = Speed × Time

Relative Speed = Speed of Object 1 + Speed of Object 2 ( If they are moving opposite to each other)

Concept Used- 

For a train to cross another train completely it should travel a distance equal to sum of length of both the trains.

Calculation-

Total distance to be covered = 110 + 120

= 230 m

Speed of 1st train = 72 × 5/18

= 20 m/sec

Speed of 1st train = 36 × 5/18

= 10 m/sec

Their relative Speed = 20 + 10

= 30 m/sec

Time taken to cross each other = 230/30

= 23/3 seconds

Calculation:

Rate of speed of A = 10 km/h

Rate of speed of B = 5 km/h

Let after reaching R they meet at point Y

P__________Q________________Y____________________R

10km x km 10 km 

In the time A will travel 10 + X + 10 + 10 km in the same time B will travel only x km

We know that distance = speed × time 

Here time is equal so

Time taken by A to travel 30 + x km = time taken by B to travel x km

⇒ (30 + x)/10 = x/5

⇒ x = 30

∴ the distance between P and R = 10 + 30 + 10

⇒ 50 km

The distance between P and R is 50 km.

Given:

The speed of Ram is 12 km/hr. Ram stops at every fifth kilometer for 9 min and total distance is 84 km.

Formula/Concept used:

Speed = Distance/time

Calculation:

Time taken by Ram to cover 84 km at the speed of 12 km/hr = 84/12 hr

∴ Time = 84/12 = 7 hr

Now, he stops at every fifth kilometer for 9 min

∴ Number of times he stop = 84/5 = 16.8 times and for the last kilometers he covers the full distance without any stoppage.

Total stoppage = 16 times (since for the last kilometers he covers the full distance without any stoppage.)

Time taken in stops = 9 × 16 = 144 minutes = 2 hr 24 minutes

∴ Total times = 7 hr + 2 hr 24 minutes = 9 hr 24 minutes

∴ Total time taken by Ram to cover whole distance is 9 hr 24 minutes

This traversal algorithm uses a queue to store the nodes of each level of the graph as and when they are visited. These nodes are then taken one by one and their adjacent nodes are visited and so on until all nodes have been visited. The algorithm terminates when the queue becomes empty.

Algorithm for Breadth First Traversal is as follows: 

clearq (q) 

visited (v) = TRUE 

while not empty (q) do

for all vertices w adjacent to v do 

if not visited then

{

insert (w , q) 

visited (w) = TRUE

}

delete (v, q);

Here each node of the graph is entered in the queue only once. Thus the while loop is executed n times, where n is the order of the graph. If the graph is represented by adjacency list, then only those nodes that are adjacent to the node at the front of the queue are checked therefore, the for loop is executed a total of E times, where E is the number of edges in the graph. Therefore, breadth first algorithm is O (N*E) for linked expression. If the graph is represented by an adjacency matrix, the for loop is executed once for each other node in the graph because the entire row of the adjacency matrix must be checked. Therefore, breadth first algorithm is O (N² ) for adjacency matrix representation. 

Given:

Speed of bus = 45 km/hr

Distance covered by bus in 1 hour including stoppages = 36 km

Calculation:

Time taken to cover 45 km = 60 mins

So, Time taken to cover 36 km = (60/45) × 36 mins

⇒ 48 mins

Bus stopped for = (60 – 48) mins

⇒ 12 mins

∴ The bus stopped for 12 mins in an hour

Given:

Excluding stoppages, the speed of a bus = 60 km/h

Including stoppages, the speed of a bus = 48 km/h

Formula used:

Time = Distance/Speed

Calculations:

In one hour

Distance travelled excluding stoppage = 60 km

Distance travelled including stoppage = 48 km

Difference in distance = 60 – 48 = 12 km

Because of the stoppage, 12 km less is travelled

So, stoppage time per hour = 12/60 = 1/5 h

⇒ 60/5 = 12 minutes

∴ The bus stops 12 minutes in a hours.

Let the normal train be train A and the express train be train B 

From 10 am to 2 pm distance travelled by train A = 4 × 60 = 240 km

Let the train A moves a distance of x km after 2 pm before they meet ,

Now the express train B have to travel (240 + x) km in the same time as the time taken by train A to travel x km ;

⇒ (240 + x)/120 = x/60

⇒ 240 + x = 2x

⇒ x = 240 km

V time taken to travel 240 km by train A = 240 /60 = 4hr

⇒ They meet 4 hr after 2pm i.e. 6pm 

Calculation:

Let the time for which he traveled in snow be ‘x’ hr and time for which he travelled when it stopped snowing be ‘y’ hr.

Given, Mitul travels at a speed of a km/hr when it’s snowing and at a speed of (a – 1) km/hr when it stops snowing.

Average speed = 6.4 km/hr

⇒ Thus, a > 6.4 > a – 1

⇒ a > 6.4 or a < 7.4

∴ As n is an integer, a = 7 km/hr.

Now, Mitul travels at a speed of 7 km/hr when it’s snowing and at a speed of 6 km/hr when it’s not snowing.

∴ Total distance traveled = 7x + 6y

Total time taken = x + y

⇒ (7x + 6y)/x + y = 6.4

⇒ 7x + 6y = 6.4x + 6.4y

⇒ 0.6x = 0.4y

⇒ y = 1.5x

∴ Total distance traveled = 7x + 9x = 16x

∴ Fraction of distance traveled in the snow = 7x/16x

⇒ 7/16

The fraction of the distance which he covered while it was snowing is 7/16.

The Quick Sort technique exhibits its worst-case behavior when the input data is “Already Completely Sorted”.

The given list is 142 543 123 65 453 879 572

 As we know that insertion sort is based on the idea of inserting records into an existing sorted file. To insert a record, we must find the proper place for the record to be inserted and this requires searching the list. 

An insertion sort will have the worst-case when the list is reversely sorted i.e. in a decreasing order. So for the above given list the insertion sort exhibits its worse case when the list is in following order

879 572 543 453 434 142 123 65

Again since it is the property of the insertion sort that it searches for the correct position in the list for an element, so this technique exhibits its best case computing when the list is already sorted in ascending order as given below.

65 123 142 434 453 543 572 879 

Algorithm for Binary Search:- 

Assuming that a [] is the array of items to be searched, n is the number of items in the array a, and target is the value of the item that is to be searched. 

int binsearch( int target, int a[], int n)

{

int low=0; 

int high=1; 

int mid; 

while (low<=high) 

{

mid=(low+high)/2;

if (target==a[mid])

return (mid);

if (target<a[mid])

high=mid-1;

else

low=mid+1;

} 

return (-1);

}

Sequential search is preferred over binary search in the following conditions:- 

i). If the list is short sequential search is easy to write and efficient than binary search. 

ii) If the list is unsorted then binary search cannot be used, in that case we have to use sequential search. 

iii) If the list is unordered and haphazardly constructed, the linear search may be the only way to find anything in it. 

Given:

Speed of train = 60 kmph

Time taken to travel in train = 1 hour 20 mins

Speed of bus = 50 kmph

Time is taken to travel in bus = 54 mins

Speed of boat = 15 kmph

Time is taken to travel in boat = 1 hour 36 mins

Calculations:

Jhon traveled distance by train = 1 hour 20 mins × 60 kmph 

⇒ (80/60) × 60 = 80 km

Jhon travelled distance by bus = 54 minutes × 50 kmph

⇒ (54/60) × 50 = 45 km

Jhon travelled by boat = 1 hour 36 mins × 15 km

⇒ (96/60) × 15 = 24 km

Total distance travelled by Jhon = distance travelled by train + distance travelled by bus + distance travelled by boat

⇒ 80 + 45 + 24 = 149 km

∴ The total distance travelled by Jhon = 149 km  

Given:

CD = 540 km

Speed of the train = 36 kmph

The train leaves C for D at 9:00 am

Their is a stop in the journey for 1 hour 20 minutes

Formula used:

Speed = Distance/Time

Calculation:

Time taken by the train if it does not stop = 540/36 = 15 hours

Hence, total time for the journey = 15 hours + 1 hour 20 minutes = 16 hour 20 minutes

12 hour will complete at 9:00 pm, further 4 hour 20 minutes added and journey will get completed at 1:20 am

∴ The train will reach D at 1:20 am