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Efficiency ratio A : B = 100 : 125 = 4 : 5. Total work = 4 × 60 = 240 units.

Combined (A+B) work in 15 days = (4 + 5) × 15 = 135 units. Remaining = 240 − 135 = 105 units.

Efficiency of C = 105 ÷ 14 = 7.5 units/day.

Time for B+C to do 5/8^th work = (240 × 5/8) ÷ (5 + 7.5) = 150 ÷ 12.5 = 12 days.

∴ The correct answer is 12 days.

Alternate Method

Given: A completes work in 60 days. B efficiency = 125% of A.

Calculations:

⇒ Total Work = Efficiency of A × Time = 4 × 60 = 240 units.

⇒ Work done by (A + B) in 15 days = (4 + 5) × 15 = 135 units.

⇒ Remaining Work = 240 − 135 = 105 units.

⇒ Efficiency of C = Remaining Work ÷ Time = 105 ÷ 14 = 7.5 units/day.

⇒ 5/8^th of original work = 240 × (5/8) = 150 units.

⇒ Combined efficiency of (B + C) = 5 + 7.5 = 12.5 units/day.

⇒ Required Time = 150 ÷ 12.5 = 12 days.

∴ The correct answer is 12 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken. If efficiency ratio is x:y, the time ratio is y:x.

Work and Wages

Wages are always distributed in the ratio of the actual work done by individuals or the ratio of their efficiencies if they work for the same duration.

Pipes and Cisterns

Similar to Time and Work, where outlet pipes represent negative efficiency (work being destroyed).

Shortcut Trick

Ratio of Boat Speed (u) to Current Speed (v) = 4 : 3

Downstream Speed (D) = 4 + 3 = 7 units | Upstream Speed (U) = 4 − 3 = 1 unit

Total Time = Distance ÷ D + Distance ÷ U ⇒ 126/7x + 126/1x = 48

18/x + 126/x = 48 ⇒ 144/x = 48 ⇒ x = 3

Speed of boat in still water = 4 × 3 = 12 km/hr

∴ The correct answer is 12 km/hr.

Alternate Method

Given:

Ratio of still water speed (u) and current speed (v) = 4 : 3

Let u = 4x and v = 3x

Formula Used:

Downstream Speed (S_d) = u + v

Upstream Speed (S_u) = u − v

Time = Distance ÷ Speed

Calculations:

⇒ S_d = 4x + 3x = 7x

⇒ S_u = 4x − 3x = x

⇒ Total Time = (126 ÷ 7x) + (126 ÷ x) = 48

⇒ 18/x + 126/x = 48

⇒ 144/x = 48

⇒ x = 144 ÷ 48 = 3

⇒ Speed of boat in still water (u) = 4 × 3 = 12 km/hr

∴ The correct answer is 12 km/hr.

Additional Information

Relative Speed in Water

Downstream speed is the sum of boat speed and stream speed (u + v), while upstream is the difference (u − v).

Finding u and v from D and U

Speed in still water (u) = (D + U) ÷ 2 and Current speed (v) = (D − U) ÷ 2.

Average Speed for Round Trip

Average Speed = (Downstream Speed × Upstream Speed) ÷ Still Water Speed = (u² − v²) ÷ u.

Shortcut Trick

Total Simple Interest for 1 year = 3620 ÷ 2 = ₹1810.

Overall effective interest rate = (1810 ÷ 16500) × 100 = 10.9696%.

Apply Alligation between 10% and 12%: (12 − 10.9696) : (10.9696 − 10) = 1.0304 : 0.9696.

Ratio of investments = 16 : 17 ⇒ 12% part = (16500 × 16) ÷ (16+17) is not clean; try interest deviation: If all at 10%, interest = 1650 ⇒ Difference = 1810 − 1650 = 160.

2% of Scheme B = 160 ⇒ Scheme B = (160 ÷ 2) × 100 = ₹8000.

∴ The correct answer is ₹8000.

Alternate Method

Given: Total Principal = ₹16500, Total Time = 2 years, Rates = 10% and 12%, Total Interest = ₹3620.

Formula Used: Simple Interest (SI) = (P × R × T) ÷ 100

⇒ Let the amount invested at 12% be ₹x.

⇒ Amount invested at 10% = ₹(16500 − x).

⇒ Total Interest = [x × 12 × 2 / 100] + [(16500 − x) × 10 × 2 / 100]

⇒ 3620 = 0.24x + 0.20(16500 − x)

⇒ 3620 = 0.24x + 3300 − 0.20x

⇒ 3620 − 3300 = 0.04x

⇒ 320 = 0.04x

⇒ x = 320 / 0.04 = 8000

∴ The correct answer is ₹8000.

Additional Information

Simple Interest vs Compound Interest

Simple interest is calculated only on the principal, whereas compound interest is calculated on the principal plus accumulated interest. Difference for 2 years = P(R/100)².

Effective Rate of Interest

The overall rate when two different investments are combined can be found using the weighted average formula: R_avg = (P₁R₁ + P₂R₂) / (P₁ + P₂).

Shortcut Trick

Base of triangle = (2 × Area) ÷ Height = (2 × 168) ÷ 24 = 14 cm.

Given, Breadth of rectangle (b) = Base of triangle = 14 cm.

Length (l) = (Perimeter ÷ 2) − b = (84 ÷ 2) − 14 = 42 − 14 = 28 cm.

∴ The correct answer is 28 cm.

Alternate Method

Given: Height (h) = 24 cm, Area = 168 cm², Perimeter (P) = 84 cm.

Formula Used: Area of triangle = (1/2) × base × height; Perimeter of rectangle = 2(l + b).

⇒ 168 = (1/2) × Base × 24

⇒ 168 = Base × 12

⇒ Base = 168 ÷ 12 = 14 cm

⇒ Breadth (b) of rectangle = 14 cm

⇒ 2(l + 14) = 84

⇒ l + 14 = 42

⇒ l = 42 − 14 = 28 cm

∴ The correct answer is 28 cm.

Additional Information

Area of Rectangle

The total region covered by a rectangle in a 2D plane is calculated as Length × Breadth (l × b).

Right-Angled Triangle Area

If the triangle is right-angled, the area is simply half the product of the two sides containing the right angle.

Semi-perimeter of Rectangle

The sum of length and breadth (l + b) is equal to half of the perimeter (P ÷ 2).

Shortcut Trick

Mixture A (40% of 250L = 100L): Water : Milk = 4 : 1 ⇒ Water = 80L, Milk = 20L.

Mixture B (Total = 60L): Water (25L) is 5 units ⇒ Milk is 7 units (35L). 80% of B = 48L.

Mixture B components in C: Water = 80% of 25 = 20L, Milk = 80% of 35 = 28L.

Mixture C Total: Milk = 20 + 28 = 48L | Water = 80 + 20 = 100L.

Ratio Milk : Water = 48 : 100 = 12 : 25.

∴ The correct answer is 12:25.

Alternate Method

Given: Mixture A = 250L (Water:Milk = 4:1). Mixture B Water = 25L (Water:Milk = 5:7). Mixture C = 40% of A + 80% of B.

Step 1: Components of Mixture A taken

⇒ Quantity of A taken = 40% of 250 = 100L

⇒ Water in A taken = (4/5) × 100 = 80L

⇒ Milk in A taken = (1/5) × 100 = 20L

Step 2: Components of Mixture B taken

⇒ In B, Water = 25L (5 units) ∴ 1 unit = 5L

⇒ Milk in B = 7 units × 5 = 35L. Total B = 60L

⇒ Quantity of B taken = 80% of 60 = 48L

⇒ Water in B taken = 80% of 25 = 20L

⇒ Milk in B taken = 80% of 35 = 28L

Step 3: Ratio in Mixture C

⇒ Total Milk in C = 20 + 28 = 48L

⇒ Total Water in C = 80 + 20 = 100L

⇒ Ratio Milk : Water = 48 : 100 = 12 : 25

∴ The correct answer is 12:25.

Additional Information

Alligation Method

Used to find the ratio in which two ingredients at given prices/concentrations are mixed to produce a mixture of a mean price/concentration.

Replacement Method

When a part of a mixture is replaced by one of the components: Final Concentration = Initial Concentration × (1 − Replacement Volume/Total Volume)ⁿ.

Compound Ratio

When two or more ratios are multiplied together (a/b × c/d), the result is the compound ratio (ac/bd).

I. x² + 18x + 77 = 0

⇒ x² + 7x + 11x + 77 = 0

⇒ x (x + 7) + 11 (x + 7) = 0

⇒ (x + 7) (x + 11) = 0

⇒ x = - 7, - 11

 

II. 4y² + 31y + 21 = 0

⇒ 4y² + 3y + 28y + 21 = 0

⇒ y (4y + 3) + 7 (4y + 3) = 0

⇒ (4y + 3) (y + 7) = 0

⇒ y = -3/4, -7  

 

 

Hence, x ≤ y

I. 2x² – 33x + 136 = 0

⇒ 2x² – 16x – 17x + 136 = 0

⇒ 2x(x – 8) – 17 (x – 8) = 0

⇒ (x – 8) (2x – 17) = 0

⇒ x = 8, 17/2

 

II. 2y² – 37y + 171 = 0

⇒ 2y² – 18y – 19y + 171 = 0

⇒ 2y (y – 9) – 19 (y – 9) = 0

⇒ (y – 9) (2y – 19) = 0,

⇒ y = 9, 19/2

 

Hence x < y

I. 5x² – 38x + 65 = 0

⇒ 5x² – 25x – 13x + 65 = 0

⇒ 5x (x – 5) – 13 (x – 5) = 0

⇒ (x – 5) (5x – 13) = 0

⇒ x = 5, 13/5

 

II. 5y² – 19y - 30 = 0

⇒ 5y² + 6y – 25y – 30 = 0

⇒ y (5y + 6) – 5 (5y + 6) = 0

⇒ (5y + 6) (y – 5) = 0

⇒ y = -6/5, 5

 

 

Hence, relationship between x and y cannot be established.

I. 12x² + 11x + 2 = 0

⇒ 12x² + 3x + 8x + 2 = 0

⇒ 3x (4x + 1) + 2 (4x + 1) = 0

⇒ (4x + 1) (3x + 2) = 0

⇒ x = -1/4, -2/3

 

II. y² + 20y + 99 = 0

⇒ y²  + 11y + 9y + 99 = 0

⇒ y (y + 11) + 9 (y + 11) = 0

⇒ (y + 11) (y + 9) = 0

⇒ y = -11, -9  

 

 

Hence, x > y

I. 2x² – 19x + 24 = 0

⇒ 2x² – 3x – 16x + 24 = 0

⇒ x (2x – 3) – 8(2x – 3) = 0

⇒ (2x – 3) (x – 8) = 0

⇒ x = 3/2, 8

 

II. y² – 25y + 144 = 0

⇒ y² – 9y – 16y + 144 = 0

⇒ y (y – 9) – 16 (y – 9) = 0

⇒ (y – 9) (y – 16) = 0

⇒ y = 9, 16  

 

 

Hence, x < y