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Time for A = 10 min; Time for B = 10 + 200% of 10 = 30 min.

Efficiency ratio A : B = 30 : 10 = 3 : 1.

Total Work = 10 × 3 = 30 units.

Work done in 5 min = 5 × (3 + 1) = 20 units.

Remaining work = 30 − 20 = 10 units.

Fraction empty = 10/30 = 1/3.

∴ The correct answer is 1/3.

Alternate Method

Given: Time taken by Pipe A = 10 minutes. Pipe B takes 200% more time than A.

Formula Used: Efficiency = Total Work ÷ Time; Fraction remaining = (Total Work − Work Done) ÷ Total Work.

⇒ Time for B = 10 + (200/100 × 10) = 10 + 20 = 30 minutes.

⇒ Combined efficiency of (A + B) = 3 + 1 = 4 units/min.

⇒ Work filled in 5 minutes = 4 × 5 = 20 units.

⇒ Remaining part of the tank = 30 − 20 = 10 units.

⇒ Fraction of tank empty = 10/30 = 1/3.

∴ The correct answer is 1/3.

Additional Information

Percentage More Concept

If a value is x% more than A, the new value is A × (1 + x/100). Here, 200% more means 3 times the original value.

Pipes and Cisterns

If an inlet pipe fills in x hours and an outlet pipe empties in y hours, net work per hour = (1/x − 1/y).

Efficiency and Time

Efficiency is inversely proportional to the time taken to complete a task. Ratio of time A:B = x:y → Ratio of efficiency A:B = y:x.

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Let initial Boat speed = 5x and Current speed = x.

New Boat speed = 5x × 120% = 6x.

New Downstream speed = 6x + x = 7x.

Given: 7x = 126 ÷ 6 = 21 km/h ⇒ x = 3.

Initial Boat speed = 5 × 3 = 15 km/h.

∴ The correct answer is 15 km/h.

Alternate Method

Given:

Ratio of Boat speed (B) to Current speed (C) = 5 : 1.

Let B = 5x and C = x.

Calculations:

⇒ New Boat speed (B') = 5x + 20% of 5x = 5x + x = 6x.

⇒ New Downstream speed = B' + C = 6x + x = 7x.

⇒ Downstream speed = Distance ÷ Time = 126 ÷ 6 = 21 km/h.

⇒ 7x = 21

⇒ x = 3 km/h.

⇒ Initial speed of boat in still water = 5x = 5 × 3 = 15 km/h.

∴ The correct answer is 15 km/h.

Additional Information

Downstream and Upstream Speed

Downstream speed = Speed of boat + Speed of current. Upstream speed = Speed of boat − Speed of current.

Speed in Still Water

Speed of boat in still water = (Downstream speed + Upstream speed) ÷ 2.

Speed of Current

Speed of current = (Downstream speed − Upstream speed) ÷ 2.

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Ratio of Milk : Water = 100 : (100 − 75) = 100 : 25 = 4 : 1.

Let initial milk be 4x and water be x. Total = 5x.

After removing 10 L, remaining milk = 4x − 10 × (4/5) = 4x − 8.

Adding 3 L milk: (4x − 8) + 3 = 75 ⇒ 4x − 5 = 75 ⇒ 4x = 80.

∴ The correct answer is 80 litres.

Alternate Method

Given: Water is 75% less than milk. 10 L mixture removed. 3 L milk added. Final milk = 75 L.

Ratio Calculation:

⇒ Milk = 100 units, Water = 100 − 75 = 25 units.

⇒ Ratio Milk : Water = 4 : 1.

Step-by-step Calculation:

⇒ Let initial Milk = 4x and Water = x.

⇒ Milk removed from 10 L = 10 × [4 / (4+1)] = 10 × (4/5) = 8 litres.

⇒ Final Milk = Initial Milk − Milk removed + Milk added

⇒ 75 = 4x − 8 + 3

⇒ 75 = 4x − 5

⇒ 4x = 75 + 5 = 80

⇒ Initial Milk = 4x = 80 litres.

∴ The correct answer is 80 litres.

Additional Information

Mixture Removal Rule

When a portion of a mixture is removed, the ratio of the components in the remaining mixture remains the same as the initial ratio.

Percentage to Ratio Conversion

If A is K% less than B, then A = B × (100 − K)/100. In this case, Water = Milk × (100 − 75)/100 = Milk × 1/4.

Component Removal Formula

Quantity of component removed = (Total quantity removed) × (Component ratio / Sum of ratios).

Follow BODMAS rule to solve this question, as per the order given below,

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is,

{12.5 of (6 ÷ 3)} of {2 + (-1)}

⇒ {12.5 of (6 ÷ 3)} of {2 - 1} = {12.5 of (2)} of {1}

⇒ 25 of 1 = 25

Let ‘X’ be the required number. Then according to question,

⇒ 27 × X = 6075/15

\(\Rightarrow {\rm{;X;}} = {\rm{;}}\frac{{\left( {\frac{{6075}}{{15}}} \right)}}{{27}}\)

⇒ X = \(\frac{{6075}}{{15 \times 27}}{\rm{;}}\)

⇒ X = 15

Concept Used:

Calculation:

(560 ÷ 16) × (720 ÷ 96) = ?

⇒ 35 × 7.5 = ?

∴  ? = 262.5

Concept used:

Calculation:

Considering the following given question

[(140 ÷ 20) × 25 ÷ 7 × 250] ÷ 50 × 5 = ?

⇒ [(140 ÷ 20) × 25 ÷ 7 × 250] ÷ 50 × 5 = ?

⇒ [7 × 25 ÷ 7 × 250] ÷ 50 × 5 = ?

⇒ [25 × 250] ÷ 50 × 5 = ?

⇒ 6250 ÷ 50 × 5 = ?

⇒ 125 × 5 = ?

⇒ 625 = ?

∴ ? = 625

Concept used:

Calculation:

52% of 84% of 5/4 of 17000 =?

(52/100) × ( 84/100) × 5/4 × 17000 = ?

⇒ 26 × 21 × 17 = ?

By using the unit digit, we know, 2 must be the unit digit of the multiplication. So, option 5 will be the answer.

∴ The value of (?) is 9282

Follow BODMAS rule to solve this question, as per the order given below,

Step-1-Parts of an equation enclosed in 'Brackets' must be solved first,

Step-2-Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now, given expression:

⇒ √(7921) × 51 + 50% of 748 = (?)³

⇒ 89 × 51 + 0.5 × 748 = (?)³

⇒ 4539 + 374 = (?)³

⇒ 4913 = (?)³

⇒ ? = 4913^1/3

⇒ ? = 17

Follow BODMAS rule to solve this question, as per the order given below,

Step-1-Parts of an equation enclosed in 'Brackets' must be solved first,

Step-2-Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3-Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4-Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now the given expression:

 

\(\begin{array}{l} 6\frac{5}{6} \times 5\frac{1}{3} + 10\frac{3}{5} \times 4\frac{1}{2} =\ ?\ \Rightarrow\ ? = \frac{{41}}{6} \times \frac{{16}}{3} + \frac{{53}}{5} \times \frac{9}{2}\ \Rightarrow\ ? = \frac{{41 \times 16 \times 5 + 53 \times 9 \times 9}}{{90}}\ \Rightarrow\ ? = \frac{{3280 + 4293}}{{90}} = \frac{{7573}}{{90}} = 84\frac{{13}}{{90}} \end{array}\)