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Correct option is (B) Purely Inductive circuit

Purely Inductive circuit

θ = π/2

cos π/2 = 0

Average power  = 0

Given:

No of workers = 6

6 workers take = 10 hours

Formula Used:

Total work = Efficiency × Time

Calculation:

6 workers complete some work in 10 hours.

⇒ Total work done in 6 hours = 6 × 10 = 60

From 9 am to 3 pm all 6 workers work together for 6 hours.

⇒ Amount of work done by 6 workers = 6 × 6 = 36 hours

From 3 pm to 4 pm,

⇒ Total time = 7 hours

From 4 pm to 5 pm

⇒ Total time = 8 hours

From 5 pm to 6 pm

⇒ Total time = 9 hours

Total amount of work done up to 6 pm = 36 + 7 + 8 + 9 = 60

Given:

X can do a piece of work = 25 days

Percentage increase in efficiency (a) = 25%

Calculation:

X’s one day work = 1/25

Y’s one day work = X’s one day’s work × (100 + a)/100

Y’s one day work

⇒ (1/25) × (100 + 25)/100

⇒ (1/25) × 125/100

⇒ 1/20

Y alone takes 20 days.

∴ The number of days taken by Y is 20 days.

GIVEN:

A's efficiency = 2 × B's efficiency

Time taken to complete the work together = 27 days

FORMULA USED:

Work done = efficiency × time

CALCULATION:

The ratio of efficiency (A : B) = 2 : 1

Total efficiency = 3 

Total work = 3 × 27 = 81

Time taken by B to complete the work = \(\frac{81}{1}= 81 days \)

Time taken by B to complete half the work =\(\frac{81}{2}=40\frac12days\)

Given:

A's time is double to C's time to complete a work.

C is three times as efficient as B.

Time taken by A to complete the work alone = 22 days

Concept used:

Time is inversely proportional to efficiency.

Formula used:

Total work = Efficiency × Total time

Calculation:

Ratio of time taken by A to C = 2 ∶ 1

⇒ Ratio of efficiency of A to C = 1 ∶ 2      ----(1)

Ratio of efficiency of B to C = 1 ∶ 3      (Given)      ----(2)

On comparing (1) and (2), we get

⇒ A ∶ B ∶ C = 3 ∶ 2 ∶ 6

Let the efficiency of A, B and C be 3 units, 2 units and 6 units respectively.

According to question,

Total work = Efficiency of A × Time

⇒ 3 × 22

⇒ 66 units

Efficiency of (A + B), (B + C) and (C + A) is 5 units, 8 units and 9 units respectively.

⇒ 3 days work = 22 units

    ↓_{× 3}                   ↓× 3

⇒ 9 days work = 66 units

∴ The number of days to finish the work in the given pattern is 9 days.

Given:

A is 25% more efficient than A, B takes 6 days more than A to complete a piece of work.

Concept:

Efficiency is inversely proportional to the time.

Calculation:

Ratio of efficiency of A and B = 125 ∶ 100

Ratio of time taken by A and B = 100 ∶ 125

Let A takes 100x days, and B takes 125x days to complete a piece of work.

Days taken by B – Days taken by A = 6

⇒ 125x – 100x = 6

⇒ 25x = 6

⇒ x = 6/25

Days taken by A = 100x

⇒ 100 × 6/25

⇒ 4 × 6

⇒ 24

And, days taken by B = 125x

⇒ 125 × 6/25

⇒ 5 × 6

⇒ 30

∴ B will take 30 days to complete the same work.

Given:

Time to complete work by A = 10/3 days

Time to complete work by B and C together = 20/9 days

Efficiency of B = 125% efficient than C

Concept used:

One day work = 1/Total days to complete work

Total number of days to complete work = Work/One day work

Calculation:

Time to complete work by A = 10/3 days

One day work of A = 1/(10/3)

⇒ One day work of A = 3/10

Time to complete work by B and C together = 20/9 days

⇒ One day work of B and C together = 1/(20/9)

⇒ One day work of B and C together = 9/20

One day work to complete A, B, and C together = One day work of A + One day work of B and C together

⇒ Required one day work = 3/10 + 9/20

⇒ Required one day work = (6 + 9)/20

⇒ Required one day work = 15/20

⇒ Required one day work = 3/4

Total number of days to complete work = Work/One day work

⇒ Required total days = 1/(3/4)

⇒ Required total days = 4/3 days

∴ 4/3 day to complete same work by A, B, and C.

Given :-

A can do a piece of work in 15 days

B is 25% more efficient than A and

C is 40% more efficient than B

Concept :-

Ratio of efficiency is reciprocal of time.

Number of days required to complete a certain work = (Work to be completed/Work efficiency)

Calculation :-

Let efficiency of A be 100 

⇒ Efficiency of B = 100 + (100 × 25%)

⇒ Efficiency of B = 100 + 25 = 125 

⇒ Efficiency of C = 125 + (125 × 40%)

⇒ Efficiency of C = 125 + 50 = 175

⇒ Ratio of efficiency of A ∶ B ∶ C = 100 ∶ 125 ∶ 175

⇒ Ratio of efficiency of A ∶ B ∶ C = 4 ∶ 5 ∶ 7

A work for 15 days with efficiency 4 

⇒ Total work done by A = 15 × 4 = 60 unit 

Now, 

(A + C) together work for 3 days.

⇒ Total efficiency of A and C together = 4 + 7 = 11 unit/day

⇒ Total work done by A and C together = 11 × 3 = 33 unit 

⇒ Remaining work = Total work - Work done by A and C together 

⇒ Remaining work = 60 - 33 

⇒ Remaining work = 27 unit 

⇒ Total efficiency of A and B together = 4 + 5 = 9 unit/day 

⇒ Required number of days = (27/9) = 3 days

⇒ Time required to complete remaining work = 3 days.

∴ Time required to complete remaining work is 3 days.

Given:

Ram = 20 days

Mohan = 30 days

Shyam = 60 days

Calculation:

Ram’s 2 days of work = (1/20) × 2

⇒ Ram’s 2 days work = 1/10

⇒ Work done by ram, mohan and shyam in 1 day = ((1/40) + (1/80) + (1/120))

⇒ Work done = 6/60

⇒ Work done = 1/10

⇒ Work done in 3 days = (1/10) + (1/10)

⇒ Work done in 3 days = 1/5

⇒ 1/5^th work is done in 3 days

⇒ Total work done = 5 × 3

⇒ Total work done = 15 days

Given:

Vandana is 40% more efficient than Diksha, Diksha takes 14 days more than Vandana to complete a piece of work.

Concept:

Efficiency is inversely proportional to the time.

Calculation:

Ratio of efficiency of Vandana and Diksha = 140 ∶ 100

Ratio of time taken by Vandana and Diksha = 100 ∶ 140

Let Vandana takes 100x days, and Diksha takes 140x days to complete a piece of work.

Days taken by Diksha – Days taken by Vandana = 9

⇒ 140x – 100x = 14

⇒ 40x = 14

⇒ x = 7/20

Days taken by Vandana = 100x

⇒ 100 × 7/20

⇒ 35

And, days taken by Diksha = 140x

⇒ 140 × 7/20

⇒ 49

∴ Vandana will take 35 days to complete the same work.

Given:

Mohini can do a piece of work in 8 days.

Rahul can do a piece of work in 12 days.

Total earning = Rs 2500

Concept:

Wages are distributed in proportion to the work done and in inverse proportion to the time taken by the individual.

Calculation:

1 day work of Mohini = 1/8

1 day work of Rahul = 1/12

Ratio of their work = 1/8 ∶ 1/12      ---- (1)

Take L.C.M. of 8, 12 and then multiply it with equation (1)

⇒ 24/8 ∶ 24/12

⇒ 3 ∶ 2

Ratio of their wages = 3 ∶ 2

Let Mohini’s share is 3x, and Rahul’s share is 2x.

Mohini’s share + Rahul’s share = 2500

⇒ 3x + 2x = 2500

⇒ 5x = 2500

⇒ x = 500

Mohini’s share = 3x

⇒ 3 × 500

⇒ 1500

Rahul’s share = 2x

⇒ 2 × 500

⇒ 1000

The difference between Mohini’s share and Rahul’s share = 1500 – 1000

⇒ 500

∴ The difference between the share of Mohini and Rahul is Rs 500.

Given:

Shaurya can do a piece of work in 15 days.

Shreya can do a piece of work in 20 days.

Total earning = Rs 2100

Concept:

Wages are distributed in proportion to the work done and in inverse proportion to the time taken by the individual.

Calculation:

1 day work of Shaurya = 1/15

1 day work of Shreya = 1/20

Ratio of their work = 1/15 ∶ 1/20      ---- (1)

Take L.C.M. of 15, 20 and then multiply it with equation (1)

⇒ 60/15 ∶ 60/20

⇒ 4 ∶ 3

Ratio of their wages = 4 ∶ 3

Let Shaurya’s share is 4x, and Shreya’s share is 3x.

Shaurya’s share + Shreya’s share = 2100

⇒ 4x + 3x = 2100

⇒ 7x = 2100

⇒ x = 300

Shreya’s share = 3x

⇒ 3 × 300

⇒ 900

∴ The share of Shreya is Rs 900.

Given:

Rahul can do a piece of work in 10 days.

Abhishek can do a piece of work in 20 days.

Total earning = Rs 1200

Concept:

Wages are distributed in proportion to the work done and in inverse proportion to the time taken by the individual.

Calculation:              

1 day work of Rahul = 1/10

1 day work of Abhishek = 1/20

Ratio of their work = 1/10 ∶ 1/20      ---- (1)

Take L.C.M. of 10, 20 and then multiply it with equation (1)

⇒ 20/10 ∶ 20/20

⇒ 2 ∶ 1

Ratio of their wages = 2 ∶ 1

Let Rahul’s share is 2x, and Abhishek’s share is x.

Rahul’s share + Abhishek’s share = 1200

⇒ 2x + x = 1200

⇒ 3x = 1200

⇒ x = 400

Rahul’s share = 2x

⇒ 2 × 400

⇒ 800

∴ The share of Rahul is Rs 800.

Given:

Dolly can do a piece of work in 15 days.

Vandana can do a piece of work in 20 days.

Priyanshi can do a piece of work in 30 days.

Total earning = Rs 900

Concept:

Wages are distributed in proportion to the work done and in inverse proportion to the time taken by the individual.

Calculation:

1 day work of Dolly = 1/15

1 day work of Vandana = 1/20

1 day work of Priyanshi = 1/30

Ratio of their work = 1/15 ∶ 1/20 ∶ 1/30      ---- (1)

Take LCM of 15, 20, 30, and then multiply it with equation (1)

⇒ 60/15 ∶ 60/20 ∶ 60/30

⇒ 4 ∶ 3 ∶ 2

Ratio of their wages = 4 ∶ 3 ∶ 2

Let Dolly’s share is 4x, Vandana’s share is 3x, and Priyanshi’s share is 2x.

Dolly’s share + Vandana’s share + Priyanshi’s share = 900

⇒ 4x + 3x + 2x = 900

⇒ 9x = 900

⇒ x = 100

Dolly’s share = 4x

⇒ 4 × 100

⇒ 400

Vandana’s share = 3x

⇒ 3 × 100

⇒ 300

Priyanshi’s share = 2x

⇒ 2 × 100

⇒ 200

Difference between the share of Dolly and Priyanshi = 400 – 200

⇒ 200

∴ Difference between the share of Dolly and Priyanshi is Rs 200.  

Given:

Tej can do a piece of work in 10 days.

Aarzoo can do a piece of work in 12 days.

Nazuk can do a piece of work in 15 days.

Total earning = Rs 900

Concept:

Wages are distributed in proportion to the work done and in inverse proportion to the time taken by the individual.

Calculation:

1 day work of Tej = 1/10

1 day work of Aarzoo = 1/12

1 day work of Nazuk = 1/15

Ratio of their work = 1/10 ∶ 1/12 ∶ 1/15      ---- (1)

Take L.C.M. of 10, 12, 15, and then multiply it with equation (1)

⇒ 60/10 ∶ 60/12 ∶ 60/15

⇒ 6 ∶ 5 ∶ 4

Ratio of their wages = 6 ∶ 5 ∶ 4

Let Tej’s share is 6x, Aarzoo’s share is 5x, and Nazuk’s share is 4x.

Tej’s share + Aarzoo’s share + Nazuk’s share = 900

⇒ 6x + 5x + 4x = 900

⇒ 15x = 900

⇒ x = 60

Nazuk’s share = 4x

⇒ 4 × 60

⇒ 240

∴ The share of Nazuk is Rs 240.   

Given:

Pipes A and B can fill a tank in 18 minutes and 27 minutes, respectively.

When A, B and C are opened together, the empty tank is completely filled in 54 minutes.

Calculation:

Pipe A's 1 minute's work = 1/18

Pipe B's 1 minute's work = 1/27

Let pipe C's 1 minute's work be 1/C.

⇒ (1/18 + 1/27) - 1/C= 1/54

⇒ 1/C = 2/27

∴ Pipe C can alone empty the tank in 27/2 = 13\(\frac{1}{2}\) minutes.

When a pipe is emptying the tank then it is considered as negative work.

Given:

Time to fill the whole tank by pipe P alone = 20 minutes

Time to fill the whole tank by pipe Q alone = 30 minutes

Total capacity of the tank = 60 litres

Concept used:

In case of alternate hours, when adding two efficiencies, gives the work done for two hours.

Calculation:

Part of tank can be filled by the pipe P in 1 minute = 60/20

⇒ 3 litres

Part of tank filled by pipe Q in 1 minute = 60/30 units

⇒ 2 litres

In 2 minutes, both of the pipes will fill = (3 + 2) litres

⇒ 5 litres

Time taken to fill 60 litres = 2 × 12

⇒ 24 minutes

∴ In 24 minutes tank will get completely filled.

Given:

Pipe A alone can fill a tank in 12 minutes

Pipe B can fill the same tank in 15 minutes

Pipe C which is at the bottom of the tank can drain the tank at the rate of 2 litres/minute

All three pipes are kept open together when the tank is full, the tank gets emptied in one hour

Calculation:

Let C be the lime that Pipe C takes to empty the tank when Pipe A and Pipe B are closed

Pipe A + Pipe B - Pipe C kept Open together results in a full tank getting emptied in 1 hour (60 min.)

⇒ (1/12) + (/15) - (1/C) = - (1/60) 

(Negative signed indicates that the tank is getting empty)

⇒ 1/C = (1/12) + (1/15) + (1/60)

⇒ 1/C = (5 + 4 + 1)/60

⇒ 1/C = 10/60

⇒ C = 6 minutes

If Pipe C can empty the tank in 6 minutes and it drains water at the rate of 2 liters/minute

 Then the capacity of the tank 

⇒ 2 × 6

⇒ 12 liters

Given:

Manisha can complete a job in 5 days.

Formula Used:

Total work = Efficiency × Time

Calculation:

According to the question:

Manisha can do a piece of work in 5 days

Akshata can do the same work in 10 days

Nimisha can do the same work in 30 days.

So together they can do the piece of work in 1 day

⇒ (1 / 5 + 1 / 10 + 1 / 30)

⇒ (12 + 6 + 2) / 60

⇒ 20 / 60

⇒ 1 / 3

Given:

Total time by pipe A, B, and C together filled the tank = 10 hrs

A is thrice as fast as B

B is twice as fast as C

Concept used:

Time = 1/Efficiency

Calculation:

The efficiency of A = 3 × Efficiency of B

(Efficiency of A)/(Efficiency of B) = 3/1      ----(1)

The efficiency of B = 2 × Efficiency of C

(Efficiency of B)/(Efficiency of C) = 2/1      ----(2)

Combining Eqn. (1) and (2)

Efficiency ratio of A, B, and C = 6 ∶ 2 ∶ 1

Time ratio of A, b, and C = 1/6 ∶ 1/2 ∶ 1

⇒ Time ratio of A, b, and C = 1 ∶ 3 ∶ 6

Let the time taken by A, B, and C to fill the tank alone is x hrs, 3x hrs, and 6x hrs.

The volume of the tank filled in 1-hrs by A = 1/x

The volume of the tank filled in 1-hrs by B = 1/3x

The volume of the tank filled in 1-hrs by C = 1/6x

The volume of the tank filled by A, B, and C together in 1-hrs = 1-hrs volume by A + 1-hrs volume by B + 1-hrs volume by C

⇒ 1/10 = 1/x + 1/3x + 1/6x

⇒ 1/10 = (6 + 2 + 1)/6x

⇒ 1/10 = 9/6x

⇒ 6x = 90

⇒ x = 15 hrs

Total time by pipe B alone to fill the tank = 3x

⇒ Total time by pipe B alone to fill the tank = 3 × 15

⇒ Total time by pipe B alone to fill the tank = 45 hrs.

∴ The total time by pipe B alone to fill the tank is 45 hrs.

Given:

Inlet pipe = 50% in 2 hours

Drain pipe = 75% in 6 hours

Formula used:

1/T = (1/T1) - (1/T2)

Here, T1 = Tank filled in hours,  T2 = Tank emptied in hours

Calculation:

Inlet pipe = 50% in 2 hours

⇒ In order to fill full tank inlet pipe = 2 × (100/50)

⇒ Inlet pipe fills tank in 4 hours

Drain pipe = 75% in 6 hours

⇒ Time taken to drain full tank = 6 × (100/75)

⇒ Time taken to drain full tank = 6 × (4/3)

⇒ Time taken to drain full tank= 8 hours

Now,

⇒ 1/T = (1/4) – (1/8)

⇒ 1/T = (2 – 1)/8

∴ It will take 8 hours to fill the tank if both pipes are opened

Given:

Tap A can fill a tank in = 6 hours

Tap B can fill a tank in = 2 hours

Tap C can empty a tank in = 3 hours

Calculation:

Let the total capacity of cistern be LCM (6, 2, 3) = 6 litres

Now,

Tank filled by A in 1 hour = 6/6

⇒ 1 litre/hour

Tank filled by B in 1 hour = 6/2

⇒ 3 litre/hour

Tank empty by C in 1 hour = 6/3

⇒ (– 2) litre/hour

Tank filled by A, B & C in (1 + 1 +1) hour = (1 + 3 – 2)

⇒ 2 litre/3 hour

In the first 3 hour tank filled by 2 litre

In next 1 hour by tap A tank filled by 1 litre

⇒ In (3 + 1) = 4 hour tank filled by (2 + 1) = 3 litre

In next 1 hour by tap B tank filled by 3 litre

⇒ In (4 + 1) = 5 hour tank filled by (3 + 3) = 6 litre

⇒ In 5 hour the tank become fill

∴ If all three taps are open alternatively 1 hour but start with A, then the whole tank will fill in 5 hours.

Given:

A is twice efficient than B and A and B together are 3 times efficient than C 

Calculation:

Efficiency of A : Efficiency of B = 2 : 1 or 2x : x 

Efficiency of A and B : Efficiency of C = 3 : 1 

efficiency of C = x 

Now ratio of efficiencies 

A : B : C = 2 : 1 : 1 

Share of A = (2/4) × 17136 

⇒ 8568 rupees 

∴ Share of A will be Rs. 8,568

Given:

A can fill a cistern 12.5hours, B can fill a cistern 25 hours 

Calculation:   

Pipe A work in cistern = 12

⇒ Work is done by pipe A in 1 hour =2/25

Pipe  B can fill a cistern =  25 hours

⇒ work is done by pipe B in 1 hours =1/25

Work done by two pipes in 1 hour together \(\frac{2}{{25}} + \frac{1}{{25}}\)

⇒ By taking LCM 25,and 25 = 25 

⇒ two pipes compete \(\frac{3}{{25}}\)

⇒ time taking together 25/3 hours

⇒ 1 hours = 60 minute

⇒ 25/3 hours =\(\frac{{25}}{3} \times 60\) = 8 hours 20 minutes 

⇒ Due to leak 8 hours 20 minutes + 1 hour 40 minute = 10 hours 

⇒ Due to leak was done by the pipe in 1 hour =\(\frac{3}{{25}} - \frac{1}{{10}}\)

⇒ By taking LCM 25 and 10 = 250 

 ⇒ \(\frac{{30 - 25}}{{250}}\) = \(\frac{5}{{250}}\)

⇒ Time taking 250/5 hours = 50 hours 

∴ Time is taken to empty the tank by leak = 50 hours 

Given:

Time taken by pipe A to fill the tank = 10 hours

Time taken by pipe B to fill the tank = 12 hours

Time taken by pipe C to empty the tank = 6 hours

Formula used:

Total capacity = LCM

Efficiency = Total Capacity/Time

Concept used:

Efficiency of outlet pipe is always in negative

Calculation:

Total capacity = LCM(10, 12, 6) = 60

Efficiency of pipe A = 60/10 = 6 units

Efficiency of pipe B = 60/12 = 5 units

Efficiency of pipe C = 60/6 = -10 units

Now when all pipes are opened then total efficiency of all the pipes = 11 – 10 = 1

Required time = 60/1 = 60 hours

∴ The total time taken to fill the tank if all the three pipes are opened at the same time is 60 hours 

Given:

Time taken by the pipes A and B to together fill the tank = 8 hours

Time taken by the pipes B and C to together fill the tank = 12 hours

Time taken by the pipes A, B, and C to together fill the tank = 6 hours

Formula Used:

If a pipe takes x hours to fill an empty tank completely, the part of the pipe filled by the tank in one hour = 1/x

Calculation:

Part of the tank filled by A and B working together in 1 hour = 1/8

So, 1/A + 1/B = 1/8

Similarly, 1/B + 1/C = 1/12

And 1/A + 1/B + 1/C = 1/6

We know that, we can obtain the part of the work all the three complete working together for 1 hour as

(1/A + 1/B) + (1/B + 1/C) + (1/A + 1/C) = 2 (1/A + 1/B + 1/C)

⇒ 1/8 + 1/12 + (1/A + 1/C) = 2 × 1/6

⇒ (1/A + 1/C) = 1/3 – 1/8 – 1/12 

⇒ (1/A + 1/C) = 1/8

So, the time taken by A and C to complete the work working together = 1/(1/8) = 8 hours

∴ The time taken by A and C to complete the work working together is 8 hours

Given:

⇒ Pipe x's 1 hour's work = 1/15

⇒ Pipe y's 1 hour's work = 1/12

⇒ Pipe z's 1 hour's work = 1/20

Calculation:

⇒ Pipe (x + y + z)'s 1 hour's work = 1/15 + 1/12 - 1/20 = 1/10

∴ In 10 hours, hose will get completely filled if all three pipes are opened simultaneously.

Given:

Pipes A, B and C can fill an empty tank in 30/7 hours

A and B are filling pipes and C is an emptying pipe.

Time taken by A and B together to fill the tank = 15 hours 

Time taken by C to empty the tank alone = 12 hours 

Concept:

Efficiency = 1/(Total time)

Calculation: 

let time taken by B be X hr.

Work done by B in 1 hr = 1/X hr.

Work done by  (A + B – C) in 1 hr = 7/30

According to the question:

⇒ (1/15) + (1/X) – (1/12) = 7/30

⇒ 1/X = (7/30) + (1/12) – (1/15)

⇒ 1/X = (14 + 5 – 4)/60

⇒ 1/X = 15/60

⇒ 1/X = 1/4

⇒ X = 4

∴ Time taken by B is 4 hr.

Given: There are three pipes A, B and C. A and B are filling pipes whereas C is the emptying pipe. B can fill a tank in 36 hours; C can empty the same tank in 20 hours. If all the three pipes open simultaneously, they will fill the tank in 360 hours.

Formula: If the time taken by pipes P1, P2,..Pn to fill a tank is p1, p2, ...pn and the time taken by pipes Q1, Q2, ..., Qn to empty it be q1, q2,..., qn respectively and the total time taken by all the pipes to fill the tank is t.

1/t = (1/p1 + 1/p2 + ... + 1/pn) – (1/q1 + 1/q2 + ... + 1/qn)

Calculation:

Let pipe A can fill the tank in a hours

In 1 hour A fills = 1/a part

B can fill the tank in 36 hours

In 1 hour b fills = 1/36 part

C can empty the tank in 20 hours

In 1 hour C empties = 1/20 part

A + B + C together fill the tank in 360 hours

In 1 hour they fill = 1/360 part

From the question

1/a + 1/36 – 1/20 = 1/360

⇒ 1/a = 1/360 – 1/36 + 1/20

⇒ 1/a = (1 – 10 + 18)/360

⇒ 1/a = 9/360

⇒ a = 40

∴ A will take 40 hours.

Given: Three filling pipes can fill a tank in 34, 68 and 51 minutes respectively. An emptying pipe can empty the same tank in 17 minutes.

Formula: If the time taken by pipes P1, P2,..Pn to fill a tank is p1, p2, ...pn and the time taken by pipes Q1, Q2, ..., Qn to empty it be q1, q2,..., qn respectively and the total time taken by all the pipes to fill the tank is t.

1/t = (1/p1 + 1/p2 + ... + 1/pn) – (1/q1 + 1/q2 + ... + 1/qn)

Efficiency is inversely proportional to time taken for doing work.

Total work = Efficiency × time

Calculation:

LCM of 34, 68, 51 and 17 is 204

Let the total work = 204 units

Efficiency of 1^st pipe or 1 minute work = 204/34 = 6

Efficiency of 2^nd pipe or 1 minute work = 204/68 = 3

Efficiency of 3^rd pipe or 1 minute work = 204/51 = 4

Efficiency of 4^th pipe or 1 minute work = 204/17 = -12 (emptying pipe)

Effective rate of capacity of all 4 pipes or 1 min. work by all pipe together = 6 + 3 + 4 – 12 = 1 unit

Total time taken to fill the tank = 204/1 = 204 minutes.

Given:

20 % apple juice is drawn from 60 litres of Apple juice.

Calculation:

First-time apple juice is drawn from 60 litres

20 % of 60 litres = 12 litres

12 liters of Apple juics is drawn from 60 liters

Now, 48 liters are apple juice and 12 liters of water in the jar.

In the second time, 20 % of Apple juice and 20 % of water is drawn.

20 % of 48 litres = 9.6 litres

20 % of 12 litres oof water = 2.4 litres

Apple juice left now = 48 litres – 9.6 litres

= 38.4 litres

Water left = 12 litres – 2.4 litres + 12 litres

= 21.6 litres

Ratio of milk to water : 38.4 : 21.6

= 4.8 : 2.7

= 1.6 : 0.9

Given:

Initial mixture: Tomato juice : Water = 4 : X

Final mixture: Tomato juice: Water = 2 : 1

Final mixture = 5 L mixture + 2 L water

Calculation:

2 liters of water is added to 5 liters of mixture

⇒ Final mixture = ((4/ (4 + X)) × 5 ) + ((X/(4 + X) × 5) + 2L

Hence, (((4/(4 + X)) × 5)/(((5X/(4 + X)) + 2) = 2/1

⇒ (20/(4 + X)) = (10X + 8 + 2X)/(4 + X)

⇒ 20 – 8 = 12 X

⇒ 12 = 12X

∴ Value of X = 1L

Given:

In a mixture 120 liters, the ratio of milk and water 5 : 1. If this ratio is to be 1 ∶ 4

Calculation

Quantity of milk = (120/6) × 5 = 100 liter

Quantity of water in it = (120 – 100) liters = 20 liter

New ratio = 1 ∶ 4

Let the quantity of water to be added further be x liter

According to the questions

⇒ 100/(20 + x) = 1/4

⇒ 400 = 20 + x

⇒ x = 380 liter

∴ Quantity of water to be added is 380 liter

Given:

Price of Bajra I = Rs. 45 per Kg

Price of Bajra II = Rs. 55 per Kg

Selling price of mixture = Rs. 60 per Kg

Calculation:

Cost price of a mixture = [45(1) + 55(1)]/(1 + 1)

⇒ (45 + 55)/2

⇒ 50

Selling price = Rs.60

Profit = Rs. 60 – Rs.50

⇒ Rs.10

Profit% = [(10/50) × 100]%

⇒ 20%

Given:

Initial Quantity of mixture = 60 liters

Ratio of milk and water initially in the mixture = 7 : 8

10 liters of mixture is replaced by 10 liters of milk

Calculation:

Initial amount of milk in mixture = 7/15 × 60 liters

⇒ 28 liters

Initial amount of water = 60 – 28 liters 

⇒ 32 liters

Now, Amount of milk and water in10 liters of mixture

Milk = 7/15 × 10 liters

⇒ milk = 14/3 liters

water = 8/15 × 10 liters

⇒ water = 16/3 liters

Amount of 10 liters of mixture was replaced  by 10 liters of milk 

Amount of milk in the mixture = 28 – 14/3 + 10 liters

⇒ 100/3 liters

Amount of water in the mixture = 32 – 16/3 liters 

⇒ 80/3 liters

Final ratio of milk and water = 100/3 : 80/3

⇒ 5 : 4

∴ The ratio of milk and water in the final mixture is 5 : 4

Given:

A bottle contains 20 litres of liquid A.

Calculation:

After removing 4 L of liquid A and adding 4 L of liquid B.

Mixture become this way.

16 L of A and 4 L of B

⇒ A = 16/20 and B = 4/20

Now again 4 L of mixture is taken out and 4 L of liquid B is added.

⇒ A : B = [16 – (16/20 × 4)] : [4 – (4/20 × 4) + 4]

⇒ A : B =  16× [1 – (1/5)] : 4 × [2 – (1/5)]

⇒ A : B = 16 × 4/5 : 4 × 9/5

⇒ A : B = 16 : 9

∴ The ratio of quantity of liquid A to that of liquid B in the final mixture is 16 : 9.

Given :

8 litres of water is taken out from a jar filled with mango juice. This process is done for 4 times. 

In final mixture ratio of mango juice to total solution is 16 : 81 

Formula used :

[Total amount - amount replaced each time)/total amount)ⁿ = Final concentration/Initial concentration (where n is the number of times process performed)

Calculations :

Let the total amount or mango juice be 'A' 

According to the question 

[(A - 8)/A]⁴ = 16/81 

[(A - 8)/A] = 2/3 

3A - 24 = 2A 

⇒ A = 24 

∴ Initially, there were 24 litres of mango juice in the pot

Solution:

Given:

Quantity of milk in A = 75/100 × (20) = 15 liters

Quantity of milk in B = 60/100 × (p) liters

Calculation:

⇒ 15 + (60/100) × (p) = (66/100) × (20 + p)

⇒ p = 30

∴ The required result is 30 liters.

Given:

Milkman profit = 25%

Concept:

Using ratio concept

Calculation:

Milkman profit = 25%

⇒ 25/100

⇒ 1/4

Mixture = 1 + 4 = 5 units

Water = 1 unit

Milk = 4 unit

The ratio of milk and water is 4 : 1.

∴ The ratio of mixture and milk is 5 : 4.

Given:

Total mixture = 60 litres

The ratio of milk and water = 2 ∶ 1

Calculation:

Let the quantity of milk in the mixture be 2x litres and quantity of water in the mixture is x litres

Total mixture = 2x + x = 3x = 60

⇒ x = 60/3 = 20

Milk in the mixture = 2x = 40 litres

Water in the mixture = x = 20 litres

Let the quantity of water added be y litres

After adding water the ratio of Milk ∶ Water = 1 ∶ 2

40 ∶ (20 + y) = 1 ∶ 2

⇒ 80 = 20 + y 

⇒ y = 60 litres

∴ 60 liters of water should be added

Given:

Capacity of barrel = 60 liters.

4 liters of wine is taken out and replaced by soda and 

this process is repeated two more times. (a total of 3 times)

Formula used:

x(1 - y/x)ⁿ = x(1 - y/x)ⁿ

where n = number of replacements.

Calculation:

Suppose a solution contains x units of wine

from which y units are taken out and replaced by soda

After n repeated operations, quantity of pure wine

remaining in solution:

⇒ x(1 - y/x)n = x(1 - y/x)n

So, wine in the barrel now:

⇒ 60(1 - 4/60)³ = 60(1 - 1/15)³

⇒ 48.78 ≈  49

Given: 

Quantity of mixture = 8 liters

Calculation: 

Let the percentage of quantity of mixture be 100x 

                   Milk            Water

Initial           95%             5% = 19 : 1 ----(1)

Final            96%             4% = 24 : 1   ---(2) 

In both condition the ratio of water in mixture is same 

So, 20x = 8 

⇒ x = 2/5 

Change in Milk = 5x = 5 × 2/5 = 2 liters 

∴ MIlk added in mixture to reduce 4% of water is 2 liters 

Given:

Ratio of water and acid in A = 4 ∶ 5

Ratio of water and acid in B = 1 ∶ 2

Calculations:

Taking the ratio of water and acid in A and B in the equal unit,

Ratio of water and acid in A = 4 ∶ 5

Ratio of water and acid in B = 3 ∶ 6

Ratio of water and acid in the mixture = (4x + 3y)/(5x + 6y)

⇒ (4x + 3y)/(5x + 6y) = 8/13

⇒ 52x + 39y = 40x + 48y

⇒ 12x = 9y

⇒ x/y = 9/12

⇒ x/y = 3/4

∴ The ratio of x and y is 3 ∶ 4

Analytic Method: The word “analytic” is derived from the word “analysis” which means “breaking up” or resolving a thing into its constituent elements. The original meaning of the word analysis is to unloose or to separate things that are together. In this method, we break up the unknown problem into simpler parts and then see how these can be recombined to find the solution. So, we start with what is to be found out and then think of further steps or possibilities the may connect the unknown built the known and find out the desired result. 
Merits

• It develops the power of thinking and reasoning
• It develops originality and creativity amongst the students.
• It helps in a clear understanding of the subject because the students have to go through the whole process themselves.
• It is a psychological method.
• No cramming is required in this method.
• Teaching by this method, the teacher carries the class with him.
• It develops self-confidence and self-reliant in the pupil. Knowledge gained by this method is more solid and durable.
• It is based on the heuristic method.

Demerits

• It is a time-consuming and lengthy method, so it is uneconomical.
• In it, facts are not presented in neat and systematic order.
• This method is not suitable for all the topics in mathematics.
• This does not find favour with all the students because below-average students fail to follow this method.
• Every teacher cannot use this method successfully

Hence, we conclude that the above statement is of analysis method.

• Inductive Method: The inductive approach is based on the process of induction. It is a method of constructing a formula with the help of a sufficient number of concrete examples. Inductive approach proceeds from particular cases to general rules of formulae, concrete instance to abstract rules, known to unknown, simple to complex
• Deductive method: This method is based on deduction. In this approach, we proceed from general to particular and from abstract and concrete. The deductive approach proceeds form general rule to specific instances, unknown to known, abstract rule to a concrete instance, complex to simple 
• Synthetic Method: The word "synthetic" is derived from the word ‘synthesis which means to combine. In this method, we combine several facts, perform cer­tain mathematical operations, and arrive at the solution. In this method, we start with the known data and connect it with the unknown part.

Problem-solving skills are skills that enable people to handle unexpected situations or difficult challenges at work. Problem-solving is an essential skill in the workplace and personal situations. Problem-solving and decision making are closely related skills, and making a decision, solving may sound obvious but often requires more thought and analysis.

Problem-solving abilities are skills that allow individuals to efficiently and effectively find solutions to issues. These skills include the following:

• Problem-solving is the ability to use appropriate methods to tackle unexpected challenges in an organized manner.
• Problem-solving is a mental process and is related to the analytical power of the brain that concludes the part of the larger problem process that includes problem finding and problem shaping.
• It is based on the conviction that an analytic mind and perseverance can help in overcoming problems. Therefore, it reposes faith in the ability of distant learners to analyze the problematic situation and identify the resources at their disposal with which they can overcome the problem. 
• Analytic skills allow us to evaluate problems, both simple and complex.
• Effective problem solving requires a controlled mixture of analytical and creative thinking.
• The act of defining a problem, determining the cause of the problem, identifying and selecting alternatives for a solution, and implementing a solution.
• Problem-solving is all about using logic as well as imagination to make sense of a situation and come up with an intelligent solution.

Hence, we can conclude that problem-solving skill is related to analytical power of the brain.

Given:

Sum of 3 numbers = 136

First number : Second number = 2 : 3

Second number : Third number = 5 : 3

Concept used:

If the ratio of A : B and B : C is given 

Then, to find ratio of A : B : C

To make ratio of B equal we have to apply the formula below.

A : B₁ and B₂ : C

A : B : C = A × B₂ : B₁ × B₂ : B₂ × C

Calculation:

Let the first, second and third numbers be A,B and C respectively.

⇒ A + B + C = 136

A : B = 2 : 3

B : C = 5 : 3

⇒ A : B : C = 10 : 15 : 9

⇒ A + B + C = 10x + 15x + 9x

⇒ A + B + C = 34x

⇒ 34x = 136

⇒ x = 4

⇒ B = 15x 

⇒ B = 60

∴ The second number is 60.