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Solution :

The series follows the following pattern :

⇒ 23 × 3 = 69

⇒ 23 × 5 = 115

⇒ 23 × 7 = 161

⇒ 23 × 11 = 253

⇒ 23 × 13 = 299

⇒ 23 × 17 = 391

⇒ The wrong term in the series will be 117.

Hence 117 is the correct answer.

Calculation:

The series follows the following pattern

6 + 8 = 14

14 + 10 = 24 ≠ 22

24 + 12 = 36

36 + 14 = 50

50 + 16 = 66

66 + 18 = 84

∴ The wrong value is 22

Shortcut Trick

Effective Successive Interest Rate for 2 years at 10% = 10 + 10 + (10 × 10)/100 = 21%.

Given Compound Interest (21%) = ₹945.

1% = 945 ÷ 21 = ₹45.

Principal (100%) = 45 × 100 = ₹4,500.

∴ The correct answer is ₹4,500.

Alternate Method

Given: Rate (R) = 10%, Time (T) = 2 years, Compound Interest (CI) = ₹945.

Formula Used: CI = P[(1 + R/100)^T − 1]

⇒ 945 = P[(1 + 10/100)² − 1]

⇒ 945 = P[(11/10)² − 1]

⇒ 945 = P[121/100 − 1]

⇒ 945 = P[21/100]

⇒ P = (945 × 100) ÷ 21

⇒ P = 45 × 100 = 4,500

∴ The correct answer is ₹4,500.

Additional Information

Effective Rate for Compound Interest

For 2 years, the effective rate is x + y + (xy/100)%. For 10%, it is 10 + 10 + 1 = 21%.

Difference between CI and SI

The difference for 2 years is P(R/100)². For 3 years, it is P(R/100)² × (3 + R/100).

Amount Formula

Total Amount A = P(1 + R/100)ⁿ, where n is the number of compounding periods.

Shortcut Trick

New Boat Speed = 24 − 25% of 24 = 18 km/hr.

New Stream Speed = 6 − 50% of 6 = 3 km/hr.

New Downstream Speed = 18 + 3 = 21 km/hr.

Time taken = 63 ÷ 21 = 3 hr.

∴ The correct answer is 3 hr.

Alternate Method

Given: Initial boat speed (u) = 24 km/hr, Initial stream speed (v) = 6 km/hr, Distance (D) = 63 km.

Formula Used: Downstream Speed = New Boat Speed + New Stream Speed; Time = Distance ÷ Speed

⇒ New Boat Speed (u_new) = 24 × (1 − 25/100) = 24 × 0.75 = 18 km/hr

⇒ New Stream Speed (v_new) = 6 × (1 − 50/100) = 6 × 0.50 = 3 km/hr

⇒ Downstream Speed = u_new + v_new = 18 + 3 = 21 km/hr

⇒ Time Required = Distance ÷ Downstream Speed = 63 ÷ 21 = 3 hr

∴ The correct answer is 3 hr.

Additional Information

Downstream Speed

When the boat travels in the direction of the stream, its effective speed is (u + v) km/hr.

Upstream Speed

When the boat travels against the direction of the stream, its effective speed is (u − v) km/hr.

Speed in Still Water

Calculated as (Downstream Speed + Upstream Speed) ÷ 2.

Stream Speed

Calculated as (Downstream Speed − Upstream Speed) ÷ 2.

Shortcut Trick

Efficiency ratio A : B = 100 : 120 = 5 : 6.

Total Work = Efficiency of A × Time of A = 5 × 30 = 150 units.

Alternate hours (A then B): 2-hour cycle work = 5 + 6 = 11 units.

Number of cycles = 150 ÷ 11 = 13 full cycles (26 hours) + 7 units remaining.

Next hour (A's turn): A fills 5 units in 1 hour. Remaining 2 units filled by B in 2/6 = 0.33 hours.

Total time = 26 + 1 + 0.33 = 27.33 hours.

∴ The correct answer is 27.33 hours.

Alternate Method

Given: Time taken by A = 30 hours. Efficiency of B = 120% of A.

Formula: Total Work = Efficiency × Time; Time = Work ÷ Efficiency.

⇒ Efficiency of A = 5 units/hr (assumed from 100%).

⇒ Efficiency of B = 1.20 × 5 = 6 units/hr.

⇒ Total capacity of tank = 30 × 5 = 150 units.

⇒ In 1^st hour (A) = 5 units, 2^nd hour (B) = 6 units. Total in 2 hours = 11 units.

⇒ In 13 cycles (26 hours), work done = 13 × 11 = 143 units.

⇒ Remaining work = 150 − 143 = 7 units.

⇒ 27^th hour: A fills 5 units. Remaining = 7 − 5 = 2 units.

⇒ Time for B to fill remaining = 2 ÷ 6 = 1/3 hour = 0.33 hours.

⇒ Total time = 26 + 1 + 0.33 = 27.33 hours.

∴ The correct answer is 27.33 hours.

Additional Information

Efficiency and Time Relation

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

Alternate Work Pattern

In alternate work, always calculate the work done in one complete cycle (sum of efficiencies of all persons involved) before finding the remaining work.

Pipes and Cisterns

Filling pipes have positive efficiency while outlet/leakage pipes have negative efficiency. Total work is the tank capacity.

Given: 8 boxes are placed on above another in a stack.

Based on the given conditions, we get the following possibilities-

1. Only two boxes are placed between V and X. Only two boxes are kept above box V. From these, we get-

 

2. Box R is kept immediately above box S. The number of boxes placed below R is one. From these, we get-

 

3. Box J is kept two places above box N. Box T is placed immediately below box N. Box A is placed two places above box V. Box V is placed immediately above box N. From these, we get the final table as follows-

 

From the final table, we see that Box A is at the top of the stack.

Hence, Box A is kept at the top of the stack.

Given: 8 boxes are placed on above another in a stack.

Based on the given conditions, we get the following possibilities-

1. Only two boxes are placed between V and X. Only two boxes are kept above box V. From these, we get-

 

2. Box R is kept immediately above box S. The number of boxes placed below R is one. From these, we get-

 

3. Box J is kept two places above box N. Box T is placed immediately below box N. Box A is placed two places above box V. Box V is placed immediately above box N. From these, we get the final table as follows-

 

From the final table, we see that two boxes are placed below X.

Hence, 2 boxes are placed below X.

Given: 8 boxes are placed on above another in a stack.

Based on the given conditions, we get the following possibilities-

1. Only two boxes are placed between V and X. Only two boxes are kept above box V. From these, we get-

 

2. Box R is kept immediately above box S. The number of boxes placed below R is one. From these, we get-

 

3. Box J is kept two places above box N. Box T is placed immediately below box N. Box A is placed two places above box V. Box V is placed immediately above box N. From these, we get the final table as follows-

 

From the final table, we see that Box N is placed fifth from bottom.

Hence, box N is placed fifth from the bottom of the stack.

Given: 8 boxes are placed on above another in a stack.

Based on the given conditions, we get the following possibilities-

1. Only two boxes are placed between V and X. Only two boxes are kept above box V. From these, we get-

 

2. Box R is kept immediately above box S. The number of boxes placed below R is one. From these, we get-

 

3. Box J is kept two places above box N. Box T is placed immediately below box N. Box A is placed two places above box V. Box V is placed immediately above box N. From these, we get the final table as follows-

 

From the final table, we see that box X is placed between box T and box R.

Hence, box X is placed between box T and box R.

Given: 8 boxes are placed on above another in a stack.

Based on the given conditions, we get the following possibilities-

1. Only two boxes are placed between V and X. Only two boxes are kept above box V. From these, we get-

 

2. Box R is kept immediately above box S. The number of boxes placed below R is one. From these, we get-

 

3. Box J is kept two places above box N. Box T is placed immediately below box N. Box A is placed two places above box V. Box V is placed immediately above box N. From these, we get the final table as follows-

 

From the final table, we see that three boxes are placed between V and R.

Hence, 3 boxes are placed between V and R.