Calculation

Taking 6²⁵ commons from the expression:

⇒ 6²⁵(1 + 6 + 36 + 216)

⇒ 6²⁵(259)

After simplifying the given expression we can conclude that it is compelety divisible by 259.

Given:

First pipe can fill the cistern = 3 hours

Second pipe can fill the cistern = 4 hours

Third pipe can drain the cistern = 8 hours

Calculation:

Let the total amount of work in filling a cistern be 24 units. (LCM of 3, 4 and 8)

Work done by pipe 1 in 1 hour = 24/3 = 8 units.

Work done by pipe 2 in 1 hour = 24/4 = 6 units.

Work done by pipe 3 in 1 hour = 24/ (-8) = -3 units

Total work done in 1 hour = 8 + 6 – 3 = 11 units

The time required to complete 11/12^th of the work = 11/12 × 24/ 11 = 2 hours

∴ The correct answer is 2 hours.

Given A alone can complete the work in 21 days

A and B together can complete the same work in 12 days

⇒ Total work = L.C.M of (12, 21) = 84

⇒ One day work of A = 4

⇒ One day work of (A + B) = 7

⇒ One day work of B = 3

Let A worked for x days and B worked for 16 days

⇒ 4x + 3 × 16 = 84

⇒ x = 9 days

∴ A left the work before (16 - 9 =) 7 days.

Formula Used:

Total work = Efficiency × Time taken

Calculation:

Ajay can finish the work alone in 32 days

A’s one day work = 1/32

A and B complete the whole work in = 8 days

Ajay and Bharat work on alternate days, with Bharat starting the work on the first day so, we can say B will work only 4 days A will work only:

= 8 - 4 = 4 days

If A’s 4-day work = 4/32 = 1/8

Remaining work = 1 – [1/8] = 7/8

B complete 7/8 work in = 4 days

B complete whole work in = 4 × [8/7] = 32/7 days

B alone can finish 7 times the same work in = [32/7] × 7 = 32 days

B alone can finish 7 times the same work in 32 days.

Given:

(Raja + Rocky) can complete painting work = 5 days

Formula used:

Total work = efficiency × time

Calculation:

According to the question:

⇒ (Raja + Rocky) × 2 + Raja × 4 = (Raja + Rocky) × 5

⇒ Raja × 4 = (Raja + Rocky) × 3

⇒ Raja × 1 = Rocky × 3

⇒ Raja/Rocky = 3/1

Total work = (Raja + Rocky) × 5

⇒ (3 + 1) × 5 = 4 × 5

⇒ 20 units

Time taken by Rocky alone to complete the painting work = 20/1 = 20 days

∴ The correct answer is 20 days.

Given:

An inlet pipe can fill an empty tank in  hours while an outlet pipe drains a completely filled tank in  hours.

Concept used:

Efficiency = (Total work / Total time taken)

Efficiency = work done in a single day 

Calculation:Time taken by A = 9/2 hours

Time taken by B = 36/5

Capacity of tank = LCM(9/2, 36/5) = 36 units

Efficiency of A = 36/(9/2) = 8 units

Efficiency of B = 36/(36/5) = - 5 units

tank filled in 2 hours = 8 - 5 = 3 units

tank filled in 20 hours = 30 units

and

Please note that after 20 hours, the remaining capacity = 6 units

Now in the 21st hour, pipe A will work and fill the tank so no need to add time after that.

Time taken by pipe A to fill 6 units = 6/8 = 3/4 hours

So,

Shortcut Trick

Let the total amount of work be 360 units (LCM of 40 and 18)

Let the work done by Mayukh per day be x and that by Sharan be y.

⇒ Work done by Mayukh and Sharan together in one day = x + y = 360/18 = 20 units.

⇒ Time is taken by Mayukh to complete 1/3^rd of work = 120/x

⇒ Time is taken by Sharan to complete remaining work = 240/y

the duo could complete the task in 40 days,

⇒ 120/x + 240/y = 40

⇒ 120/ (20 – y) + 240/y = 40

⇒ 120y + 240(20 – y) = 40y(20 – y)

⇒ 3y + 120 – 6y = 20y – y²

⇒ y² – 23y + 120 = 0

⇒ (y – 15)(y – 8) = 0

⇒ y = 8 (Work done per day by Sharan will be lesser than Mayukh)

∴ Time required by Sharan alone = 360/8 = 45 days.

Shortcut Trick

Option method:-  

Sharan and Mayukh, working together, can complete a task in 18 days

if both have the same efficiency Sharan and Mayukh are alone in the work for 36 days.

now, as per the question condition,

Mayukh worked faster than Sharan

It means the time required by Sharan alone must be 36 + days and

(In this condition our 2 options are eliminated 24,30)

2nd thing it can't be double either.

In this condition, we are left with only one option (1 ) 45 Days

Given:

Speed is 60 km per hour,

Train passed through a 1.5 km long tunnel in two minutes

Formula used:

Distance = Speed × Time

Calculation:

Let the length of the train be L

According to the question,

Total distance = 1500 m + L

Speed = 60(5/18)

⇒ 50/3 m/sec

Time = 2 × 60 = 120 sec

⇒ 1500 + L = (50/3)× 120

⇒ L = 2000 - 1500

⇒ L = 500 m

∴ The length of the train is 500 m.

Given:

Geeta runs 5/2 times as fast as Babita 

Geeta gives a lead of 40 m to Babita 

Formula Used:

Distance = Speed × Time 

Calculation:

Let the speed of Babita be 2x

⇒ Speed of Geeta = (5/2) × 2x = 5x

Let the distance covered by Geeta be y meters

⇒ Distance covered by Babita = (y - 40) meters

As time is constant, distance is directly proportional to speed

⇒ 2y = 5y - 200 

⇒ y = 200/3 = 66.67m 

∴ The distance from the starting point where both of them will meet is 66.67 m

Concept used:

Speed × time = distance

Calculation:

In the 1^st 20 min the thief cover distance = 4 m,

20 min in hour = 20/60 hour

Let's assume that the speed of security guard = x m/hr, where x > 12

According to the question,

⇒ (x - 12) × 20/60 = 4

⇒ x - 12 = 12

⇒ x = 24

∴ The correct answer is 24 m/h