Question

What is the minimum average return Venkat would have earned during the year?

• A. 30%
• B. 31.25%
• C. 32.5%
• D. Cannot be determined
• A. I and II only
• B. II and III only
• C. I and IV only
• D. II and IV only
• A. I and II only
• B. II and III only
• C. I and III only
• D. III and IV only
• A. I only
• B. II and IV only
• C. I and III only
• D. III and IV only

Answer 1

The Correct Option is B

Step 1: Understanding the Scenario

Venkat invested in 4 companies, each from a different industry:

• Company A: Cement – Expected Return: \( 30\% \)
• Company B: IT – Expected Return: \( 40\% \)
• Company C: Auto – Expected Return: \( 10\% \)
• Company D: Steel – Expected Return: \( 20\% \)

The constraints are:

• Two companies had extraordinarily good returns.
• One company had a bad return of \( 0\% \).
• One company performed normally (i.e., met expected return).

Extraordinary is defined as:

• \( >2 \times \) expected for Cement and IT
• \( >1.5 \times \) expected for Auto

Step 2: Try Combination to Minimize Average Return

Let's test the following assignment:

• IT: Extraordinarily good → \( 2 \times 40\% = 80\% \)
• Auto: Extraordinarily good → \( 1.5 \times 10\% = 15\% \)
• Steel: Bad result → \( 0\% \)
• Cement: Normal → \( 30\% \)

Compute average: \[ \text{Average} = \frac{80 + 15 + 0 + 30}{4} = \frac{125}{4} = 31.25\% \]

 

Step 3: Try Another Valid Assignment

Now try:

• Cement: Extraordinarily good → \( 2 \times 30\% = 60\% \)
• Auto: Extraordinarily good → \( 1.5 \times 10\% = 15\% \)
• IT: Bad → \( 0\% \)
• Steel: Normal → \( 20\% \)

Compute average: \[ \text{Average} = \frac{60 + 15 + 0 + 20}{4} = \frac{95}{4} = 23.75\% \] This violates the rule since IT (allowed for \( >2 \times \)) is not used as extraordinary.

 

Step 4: Try Maxed-Out Configuration

Try maximum possible scenario:

• Cement: Extraordinarily good → \( 60\% \)
• IT: Extraordinarily good → \( 80\% \)
• Auto: Extraordinarily good → \( 15\% \)
• Steel: Bad → \( 0\% \)

Compute average: \[ \text{Average} = \frac{60 + 80 + 15 + 0}{4} = \frac{155}{4} = 38.75\% \] This exceeds our target of minimizing the average.

 

Step 5: Optimal and Minimum Valid Average

Among all valid permutations that satisfy:

• 2 extraordinary (Cement/IT > 2x, Auto > 1.5x)
• 1 bad (0%)
• 1 normal

The following yields minimum:

• IT = \( 80\% \), Auto = \( 15\% \), Steel = \( 0\% \), Cement = \( 30\% \)

\[ \text{Minimum Average} = \frac{80 + 15 + 0 + 30}{4} = \boxed{31.25\%} \]

 

Final Answer:

Option (B): \( \boxed{31.25\%} \)

Answer 2

Step 1: Setup

We are given:

• Expected Returns: 20% (Steel), 10% (Auto), 30% (Cement), 40% (IT)
• Constraints:
  • Two industries gave extraordinary returns
  • One had a bad result (0%)
  • One gave a normal expected return
• Extraordinary condition:
  • \( >2\times \) expected for Cement and IT
  • \( >1.5\times \) expected for Auto
• We need to reach an average return of 35%

Step 2: Try Assignments

Let's assign:

• Cement → 30%
• IT → 40%
• Auto → 10%
• Steel → 20%

Try one combo:

• Cement = \( 30\% \times 2 = 60\% \)
• IT = \( 0\% \)
• Auto = \( 10\% \times 1.5 = 15\% \)
• Steel = \( 20\% \)

\[ \text{Average} = \frac{60 + 0 + 15 + 20}{4} = \frac{95}{4} = 23.75\% \quad \text{(too low)} \]

Adjust:

• IT = \( 40\% \times 2 = 80\% \)
• Auto = 15%
• Steel = 0%
• Cement = 30%

\[ \text{Average} = \frac{80 + 15 + 0 + 30}{4} = \frac{125}{4} = 31.25\% \quad \text{(close)} \]

 

Step 3: Fine-Tune to Get 35%

Let’s increase Cement’s return slightly:

• Cement = approx. 42%

\[ \frac{80 + 15 + 0 + 42}{4} = \frac{137}{4} = 34.25\% \]

Almost there. Try Cement = 45%

\[ \frac{80 + 15 + 0 + 45}{4} = \frac{140}{4} = 35\% \]

Success! We have reached the exact 35% target.

 

Step 4: Statement Validation

Let’s analyze the following statements:

• I: A (Auto or Steel) is not extraordinary → True
• II: B (e.g., IT at 0%) is not extraordinary → True in this example, but not always
• III: D (Steel at 0%) is not extraordinary → True here, but not always
• IV: C (Cement at 45%) is extraordinary → True

Only Statements I and IV are necessarily true for all possible scenarios that average to 35%.

 

Step 5: Final Answer

Correct Option: (C) I and IV only

Answer 3

Step 1: Understanding the Constraints

Given:

• Expected Returns:
  • Steel: 20%
  • Auto: 10%
  • Cement: 30%
  • IT: 40%
• Conditions:
  • Two companies gave extraordinary results:
    • Cement/IT: \( >2\times \text{expected} \)
    • Auto: \( >1.5\times \text{expected} \)
  • One company had a bad return: 0%
  • One had normal (expected) return
• Target average return = 38.75%

Step 2: Try Matching Assignment

Assignment:

• Cement: \( 30\% \times 2 = 60\% \)
• IT: \( 40\% \times 2 = 80\% \)
• Auto: \( 10\% \times 1.5 = 15\% \)
• Steel: \( 0\% \)

Average:

\[ \frac{60 + 80 + 15 + 0}{4} = \frac{155}{4} = 38.75\% \]

This matches the required average.

 

Step 3: Label and Evaluate Statements

Label companies:

• A = Auto (10%) → Returns: 15%
• B = IT (40%) → Returns: 80%
• C = Cement (30%) → Returns: 60%
• D = Steel (20%) → Returns: 0%

Evaluate statements:

• I: C (Cement 60%) belongs to Auto or Steel → False
• II: A (Auto 15%) announced extraordinary → True
• III: B (IT 80%) did not announce extraordinary → False
• IV: D (Steel 0%) announced extraordinary → False

 

Step 4: Try Different Valid Permutations

Try:

• IT = 80%
• Cement = 0%
• Auto = 15%
• Steel = 20%

\[ \text{Average} = \frac{80 + 0 + 15 + 20}{4} = \frac{115}{4} = 28.75\% \quad \text{(Too low)} \] Adjust:

• Cement = 60%
• IT = 80%
• Auto = 0%
• Steel = 15%

\[ \frac{60 + 80 + 0 + 15}{4} = \frac{155}{4} = 38.75\% \quad \text{(Valid)} \] Now check labels:

• C (Cement 60%) not Auto/Steel → False for I
• A (Auto 0%) not extraordinary → True
• B (Steel 15%) not extraordinary → True

 

Step 5: Try Another Valid Assignment

Try:

• C = 0% → Auto or Steel
• A = 80% → IT (Extraordinary)
• B = 15% → Auto (Not extraordinary)
• D = 60% → Cement (Extraordinary)

\[ \text{Average} = \frac{0 + 80 + 15 + 60}{4} = \frac{155}{4} = 38.75\% \]

Now:

• I: C = 0% → belongs to Auto or Steel → True
• III: B = 15% → Not extraordinary → True

 

Step 6: Final Answer

Correct Option: (C) I and III only

Answer 4

Step 1: Understanding C's Role

We are given:

• C belongs to Cement (30%) or IT (40%)
• Extraordinary return condition:
  • Cement or IT must give \( >2 \times \text{expected} \)
  • Auto can give \( >1.5 \times \text{expected} \)
• Other conditions:
  • One extraordinary (besides C)
  • One normal return
  • One bad (0%) return

Step 2: Case 1 – C in Cement (60%)

Other assignments:

• Auto: \( 10\% \times 1.5 = 15\% \)
• Steel: \( 0\% \)
• IT: \( 40\% \) (normal)

\[ \text{Average} = \frac{60 + 15 + 0 + 40}{4} = \frac{115}{4} = 28.75\% \]

 

Step 3: Case 2 – C in IT (80%)

Other assignments:

• Auto: \( 15\% \) (extraordinary)
• Steel: \( 0\% \)
• Cement: \( 30\% \) (normal)

\[ \text{Average} = \frac{80 + 15 + 0 + 30}{4} = \frac{125}{4} = 31.25\% \]

 

Step 4: Case 3 – C in IT (80%), Cement Bad (0%)

Other assignments:

• Cement: 0%
• Auto: 15%
• Steel: 20% (normal)

\[ \text{Average} = \frac{80 + 15 + 0 + 20}{4} = \frac{115}{4} = 28.75\% \]

 

Step 5: Final Analysis

The maximum average across all valid cases where C belongs to Cement or IT is: \[ \boxed{31.25\%} \] This is below 36.25%, so:

• I: True – maximum is strictly less than 36.25%
• II: False – 33.75% is not necessarily achievable
• III: Conditional – may or may not be true depending on other assignments
• IV: Depends on III – not universally true

Conclusion: Only Statement I is necessarily true.

 

Final Answer:

Correct Option: (A) Only I