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Shortcut Trick

For half-yearly compounding, the half-yearly rate = 10% ÷ 2 = 5%.

The time period is 6 months, which is exactly one half-year cycle.

Compound Interest for 1 cycle = Principal × Rate = 8000 × 5% = 400.

∴ The correct answer is Rs. 400.

Alternate Method

Given:

Principal (P) = Rs. 8000

Annual Rate (R) = 10% per annum

Time (T) = 6 months = 1/2 year

Compounding = Half-yearly

Formula:

When interest is compounded half-yearly, New Rate (r) = R/2 and New Time (n) = 2 × T.

Calculations:

⇒ Adjusted Rate (r) = 10% ÷ 2 = 5%

⇒ Number of conversion periods (n) = 2 × (1/2) = 1 period

⇒ Amount (A) = P(1 + r/100)ⁿ

⇒ A = 8000(1 + 5/100)¹

⇒ A = 8000 × (1.05) = 8400

⇒ Compound Interest (CI) = Amount − Principal

⇒ CI = 8400 − 8000 = 400

∴ The correct answer is Rs. 400.

Additional Information

Half-Yearly Compounding Rule

Divide the annual interest rate by 2 and multiply the number of years by 2 to get the total number of conversion periods.

Quarterly Compounding Rule

Divide the annual rate by 4 and multiply the number of years by 4. Formula: A = P(1 + R/400)^4T.

CI vs SI for One Period

For the very first compounding period, the Simple Interest and Compound Interest are always equal.

Shortcut Trick

Let the Total Capacity be 48 units (LCM of 16 and 24).

Efficiency of Pipe A = +3 and Pipe B = −2.

Pipe A works for the full 30 minutes: 30 × 3 = 90 units.

Extra work to be removed by Pipe B = 90 − 48 = 42 units.

Time for Pipe B = 42 ÷ 2 = 21 minutes.

∴ The correct answer is 21 min.

Alternate Method

Given: The tank must be filled in exactly 30 minutes.

Calculation:

⇒ Let Pipe B be closed after x minutes.

⇒ Pipe A works for 30 minutes and Pipe B works for x minutes.

⇒ (Efficiency of A × 30) + (Efficiency of B × x) = Total Work

⇒ (3 × 30) + (−2 × x) = 48

⇒ 90 − 2x = 48

⇒ 2x = 90 − 48

⇒ 2x = 42

⇒ x = 21 minutes

∴ The correct answer is 21 min.

Additional Information

Inlet and Outlet Pipes

An inlet pipe fills the tank (positive efficiency), while an outlet pipe empties it (negative efficiency).

Net Efficiency

When multiple pipes operate, the net work done per unit time is the algebraic sum of their individual efficiencies.

Time and Work Relation

Total Work = Efficiency × Time. In tank problems, Total Capacity is usually the LCM of individual times.

Shortcut Trick

Speed of Train A = 36 × (5/18) = 10 m/s. Length A = 10 × 20 = 200 m.

Length B = 200 + 100 = 300 m. Platform = 500 m.

Total Distance for B = 300 + 500 = 800 m. Speed B = 18 × (5/18) = 5 m/s.

Time taken by B = 800 ÷ 5 = 160 sec.

∴ The correct answer is 160 sec.

Alternate Method

Given: Speed of Train A = 36 km/hr, Time to cross pole = 20 sec, Length of Train B = Length A + 100 m, Speed of Train B = 18 km/hr, Platform length = 500 m.

Formula Used: Distance = Speed × Time; Time = Distance ÷ Speed; 1 km/hr = 5/18 m/s.

⇒ Speed of Train A in m/s = 36 × (5/18) = 10 m/s.

⇒ Length of Train A = Speed × Time = 10 × 20 = 200 m.

⇒ Length of Train B = 200 + 100 = 300 m.

⇒ Speed of Train B in m/s = 18 × (5/18) = 5 m/s.

⇒ Total Distance to be covered by B = Length of B + Length of Platform = 300 + 500 = 800 m.

⇒ Time taken by B = 800 ÷ 5 = 160 sec.

∴ The correct answer is 160 sec.

Additional Information

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Train Crossing Concepts

When a train crosses a pole or stationary person, distance = length of train. When it crosses a platform or bridge, distance = length of train + length of platform.

Relative Speed

If two trains move in the same direction, relative speed = S₁ − S₂. If they move in opposite directions, relative speed = S₁ + S₂.

Shortcut Trick

Downstream speed (u + v) = 20 km/hr

Upstream speed (u − v) = 96 ÷ 8 = 12 km/hr

Difference (u − v) represents the speed of the boat in still water minus current speed.

∴ The correct answer is 12 km/hr.

Alternate Method

Given:

Sum of speed of boat (u) and current (v) = 20 km/hr

Distance upstream = 96 km

Time taken upstream = 8 hours

Formula Used:

Upstream Speed = Distance ÷ Time = u − v

Calculations:

⇒ Downstream Speed (u + v) = 20 km/hr

⇒ Upstream Speed (u − v) = 96 ÷ 8

⇒ Upstream Speed (u − v) = 12 km/hr

⇒ The question asks for the difference between the speed of boat in still water (u) and speed of current (v).

⇒ Required difference = u − v = 12 km/hr

∴ The correct answer is 12 km/hr.

Additional Information

Still Water Speed Calculation

The speed of the boat in still water (u) is the average of downstream and upstream speeds: u = (Downstream + Upstream) ÷ 2.

Current Speed Calculation

The speed of the current (v) is half the difference between downstream and upstream speeds: v = (Downstream − Upstream) ÷ 2.

Relative Speed in Water

Downstream speed is always greater than upstream speed because the current assists the boat (u + v) vs (u − v).

Shortcut Trick

Semi-perimeter (L + B) = 248 ÷ 2 = 124 m.

Length (L) − Breadth (B) = 4 m.

Solving: L = 64 m, B = 60 m.

Area of outer path = 2w(L + B + 2w), where w = 5 m.

Area = 2 × 5 × (64 + 60 + 2 × 5) = 10 × (124 + 10) = 1340 m².

∴ The correct answer is 1340 m².

Alternate Method

Given: Perimeter = 248 m, Length (L) = Breadth (B) + 4, Path Width (w) = 5 m.

Formula Used: Perimeter = 2(L + B) | Area = L × B

⇒ 2(B + 4 + B) = 248

⇒ 2B + 4 = 124

⇒ 2B = 120 → B = 60 m

⇒ L = 60 + 4 = 64 m

⇒ Outer Length = 64 + 2 × 5 = 74 m

⇒ Outer Breadth = 60 + 2 × 5 = 70 m

⇒ Area of Path = (Outer L × Outer B) − (Inner L × Inner B)

⇒ Area = (74 × 70) − (64 × 60) = 5180 − 3840 = 1340 m²

∴ The correct answer is 1340 m².

Additional Information

Area of Outer Path

For a rectangle with length L, breadth B and path width w outside: Area = 2w(L + B + 2w).

Area of Inner Path

For a path constructed inside the rectangular field: Area = 2w(L + B − 2w).

Area of Crossroads

For two paths crossing each other at the center: Area = w(L + B − w).

Shortcut Trick

Vessel A mixture ratio 17 : 7 remains same in any quantity poured out.

Quantity poured from A to X = 40% of 120 = 48 liters.

In 48 liters: Milk = 48 × (17/24) = 34 liters, Water = 48 − 34 = 14 liters.

Vessel X initially (100 liters, equal ratio) has 50 liters Milk and 50 liters Water.

New Ratio in X = (50 + 34) : (50 + 14) = 84 : 64 = 21 : 16.

∴ The correct answer is 21 : 16.

Alternate Method

Given:

Total mixture in Vessel A = 120 liters | Ratio (M:W) = 17 : 7

Total mixture in Vessel X = 100 liters | Ratio (M:W) = 1 : 1

Calculations:

⇒ Quantity of mixture taken from A = 40% of 120 = 0.40 × 120 = 48 liters

⇒ Milk from A = 48 × (17 / (17 + 7)) = 48 × (17/24) = 34 liters

⇒ Water from A = 48 − 34 = 14 liters

⇒ Initial Milk in X = 100 × (1/2) = 50 liters

⇒ Initial Water in X = 100 × (1/2) = 50 liters

⇒ New Milk in X = 50 + 34 = 84 liters

⇒ New Water in X = 50 + 14 = 64 liters

⇒ New Ratio = 84 : 64 = 21 : 16

∴ The correct answer is 21 : 16.

Additional Information

Mixture Property

When a portion of a mixture is removed, the ratio of components in the removed part and the remaining part remains identical to the original ratio.

Alligation Rule

It is used to find the mean value of a mixture when two ingredients are mixed: (Quantity 1 / Quantity 2) = (Rate 2 − Mean Rate) / (Mean Rate − Rate 1).

Percentage to Fraction

40% is equivalent to 2/5. Using fractions often simplifies quantitative aptitude calculations involving large numbers.

Shortcut Trick

Let the total income be X.

Successive remaining percentage: X × (100−20)% × (100−15)% × (100−30)% = 9520

X × (4/5) × (17/20) × (7/10) = 9520

X × (476/1000) = 9520 ⇒ X = 9520 × (1000/476) = 20000

∴ The correct answer is ₹20000.

Alternate Method

Given: Rent = 20%, Education = 15% of remaining, Clothes = 30% of remaining, Savings = ₹9520.

⇒ Let Income = 100 units

⇒ After Rent: 100 − 20 = 80 units

⇒ After Education: 80 − (15% of 80) = 80 − 12 = 68 units

⇒ After Clothes: 68 − (30% of 68) = 68 − 20.4 = 47.6 units

⇒ 47.6 units = 9520

⇒ 1 unit = 9520 ÷ 47.6 = 200

⇒ 100 units = 100 × 200 = 20000

∴ The correct answer is ₹20000.

Additional Information

Percentage Change (Remaining Value)

When values are subtracted from the remaining balance, it is a case of successive percentage decrease. Final Value = Initial Value × (1 − r₁/100) × (1 − r₂/100)...

Base Value Change

Always identify if the percentage is calculated on the original total income or the remaining amount to avoid calculation errors.

I. x² + 10x – 264 = 0

⇒ x² + 22x – 12x – 264 = 0

⇒ x(x + 22) – 12(x + 22) = 0

⇒ (x + 22)(x – 12) = 0

Then, x = +12 or x = -22

II. y² + 6y – 187 = 0

⇒ y² + 17y - 11y - 187 = 0

⇒ y(y + 17) - 11(y + 17) = 0

⇒ (y - 11)(y + 17) = 0

Then, y = -17 or y = +11

So, when x = +12, x > y for y = -17 and x > y for y = +11

And when x = -22, y > x for y = -17 and y > x for y = +11

∴ So, we can observe that no relationship can be determined.

I. x² - 3x – 180 = 0

⇒ x² + 12x – 15x – 180 = 0

⇒ x(x + 12) – 15(x + 12) = 0

⇒ (x + 12)(x – 15) = 0

Then, x = -12 or x = +15

II. y² + 34y + 288 = 0

⇒ y² + 18y + 16y + 288 = 0

⇒ y(y + 18) + 16(y + 18) = 0

⇒ (y + 16)(y + 18) = 0

Then, y = -16 or y = -18

So, when x = -12, x > y for y = -16 and x > y for y = -18

And when x = +15, x > y for y = -16 and x > y for y = -18

∴ So, we can observe that x > y.

x² – 19x + 84 = 0

⇒ x² – 12x – 7x + 84 = 0

⇒ x(x – 12) – 7(x – 12) = 0

⇒ (x – 12) (x – 7) = 0

⇒ x = 12 or x = 7

y² – 3y – 108 = 0

⇒ y² – 12y + 9y – 108 = 0

⇒ y(y – 12) + 9(y – 12) = 0

⇒ (y – 12) (y + 9) = 0

⇒ y = 12 or y = -9

∴ All relations are possible between x and y. So "no relationship can be established".