Question

How many Split Inverter ACs did D2 sell?

• A. 75%
• B. 20%
• C. 25%
• D. 4.33%
• A. D1 and D3 sold an equal number of Split ACs.
• B. D2 sold the highest number of ACs.
• C. D1 and D3 together sold more ACs as compared to D2 and D4 together.
• D. D4 sold more Split ACs as compared to D3.
• A. 5
• B. 7
• C. 6
• D. 4

Answer 1

The Correct Option is C

Suppose, \( A \) is the total number of ACs sold. 

Given: 25% were Window variant ACs.

Window ACs = \( \frac{A}{4} \), Split ACs = \( \frac{3A}{4} \)

Let \( B \) be the total number of inverter ACs.
Among inverter ACs, 20% were Window variant.
Window Inverter ACs = \( \frac{B}{5} \), Split Inverter ACs = \( \frac{4B}{5} \)

From condition (3):

\( \frac{A}{4} - \frac{B}{5} = 6 \) and \( \frac{4B}{5} = 36 \)

Solving: \( B = 45 \), \( A = 60 \)

Distribution Table:

Total = 60
Split = 45		Window = 15
Inv = 36	Non-Inv = 9	Inv = 9	Non-Inv = 6

Assume:

• D1 and D4 sold no Window Non-Inverter ACs.
• D2 sold twice as many Window Non-Inverter ACs as D3 ⇒ D2 = 4, D3 = 2.
• D1 sold \( x \) Window Inverter ACs ⇒ Split Inverter ACs = \( 13 - x \)
• Let \( y \) be total Window ACs sold by D3 & D4 ⇒ D2 sold \( 3y \) Window ACs.
• Let \( z \) be Split Inverter ACs sold by D3 & D4 ⇒ D2 sold \( 2z \) of these.

From total Window ACs: \( x + 3y + y + y = 15 \Rightarrow x + 5y = 15 \)
Try \( x = 5, y = 2 \)

Dealer-wise Tables:

D1 Total = 15
Split = 10		Window = 5
Inv = 8	Non-Inv = 2	Inv = 5	Non-Inv = 0

D2 Total = 20
Split = 14		Window = 6
Inv = 14	Non-Inv = 0	Inv = 2	Non-Inv = 4

D3 Total = 12
Split = 10		Window = 2
Inv = 7	Non-Inv = 3	Inv = 0	Non-Inv = 2

D4 Total = 13
Split = 11		Window = 2
Inv = 7	Non-Inv = 4	Inv = 2	Non-Inv = 0

Total Non-Inverter ACs = \( 9 + 6 = 15 \)
So, percentage = \( \frac{15}{60} \times 100 = 25\% \)

Answer 2

Suppose, A is the total number of ACs sold. 

Given:

• 25% of ACs were of Window variant: \( \frac{A}{4} \)
• 75% of ACs were of Split variant: \( \frac{3A}{4} \)
• Let B be the number of Inverter ACs.
• 20% of Inverter ACs were Window: \( \frac{B}{5} \)
• So, Split Inverter ACs: \( \frac{4B}{5} \)
• From data: \( \frac{A}{4} - \frac{B}{5} = 6 \) and \( \frac{4B}{5} = 36 \)

Solving:

\[ \frac{4B}{5} = 36 \Rightarrow B = 45 \] \[ \frac{A}{4} - \frac{45}{5} = 6 \Rightarrow \frac{A}{4} = 6 + 9 = 15 \Rightarrow A = 60 \]

So, total ACs = 60, Inverter ACs = 45, hence Non-inverter ACs = \( 60 - 45 = 15 \)

Required percentage of Non-inverter ACs:

\[ \frac{15}{60} \times 100 = 25\% \]

Answer: 25%

Answer 3

Suppose, \( A \) is the total number of ACs sold. 
From the given information, 25% were Window ACs.
So, Window ACs = \( \frac{A}{4} \) and Split ACs = \( \frac{3A}{4} \)

Let \( B \) be the total number of Inverter ACs.
20% of Inverter ACs are Window ACs: \( \frac{B}{5} \), so Window Non-inverter = \( \frac{A}{4} - \frac{B}{5} = 6 \)
Also, Split Inverter ACs = 36, i.e., \( \frac{4B}{5} = 36 \Rightarrow B = 45 \)
Substituting into the first equation: \( \frac{A}{4} - 9 = 6 \Rightarrow A = 60 \)

Total ACs Sold = 60
- Split: 45
- Window: 15
- Inverter: 36 (9 Window + 27 Split)
- Non-inverter: 24 (6 Window + 18 Split)

From the conditions:
Let \( x \) be D1’s Window Inverter ACs, then Split Inverter = \( 13 - x \)
Let \( y \) be D3 and D4's Window ACs, then D2’s = \( 3y \)
Let \( z \) be Split Inverter ACs of D3 and D4, then D2 has \( 2z \)

Total Window ACs: \( x + 3y + y + y = 15 \Rightarrow x + 5y = 15 \)
Only valid integer solution: \( x = 5, y = 2 \)

Then, \( z = 7 \) from Split Inverter total: \( 8 + 2z + z + z = 36 \Rightarrow 4z = 28 \)

Dealer-wise AC Distribution:

D1:
- Window: 5 (Inverter: 5, Non-inverter: 0)
- Split: 10 (Inverter: 8, Non-inverter: 2)
- Total: 15

D2:
- Window: 6 (Inverter: 2, Non-inverter: 4)
- Split: 21 (Inverter: 14, Non-inverter: 7)
- Total: 27

D3:
- Window: 2 (Inverter: 0, Non-inverter: 2)
- Split: 10 (Inverter: 7, Non-inverter: 3)
- Total: 12

D4:
- Window: 2 (Inverter: 2, Non-inverter: 0)
- Split: 4 (Inverter: 7, Non-inverter: 0)
- Total: 6

Total ACs Sold:
D1 + D3 = 15 + 12 = 27
D2 + D4 = 60 - 27 = 33

Answer 4

Problem Analysis:
An AC company has 4 dealers: D1, D2, D3, and D4. 
ACs sold are of two types: Window and Split.
Each can be Inverter or Non-inverter.

Given Data:

• 25% of ACs are Window, 75% are Split.
• 20% of Inverter ACs are Window.
• 6 Window Non-inverter ACs and 36 Split Inverter ACs sold.
• D1 sold 13 Inverter ACs.
• D3 sold 5 Non-inverter ACs.
• D1’s Split ACs = 2 × D1’s Window ACs.
• D3 and D4 sold same number of Window ACs = ⅓ of D2’s Window ACs.
• Window Non-inverter ACs were sold only by D2 and D3; D2 sold twice as many as D3.
• D3 and D4 sold equal Split Inverter ACs = ½ of D2’s Split Inverter ACs.

Let’s Verify Each Statement:

1. D1 and D3 sold an equal number of Split ACs:
   D1 Split = 2 × D1 Window, but D3’s Split count is not directly given.
   → Cannot be determined.
2. D2 sold the highest number of ACs:
   D2 sold most Window ACs (3× of D3/D4), and the most Split Inverter ACs.
   Likely to have highest total.
   → Can be true.
3. D1 and D3 together sold more ACs than D2 and D4 together:
   From data, D2 clearly sold more than either D1 or D3.
   D4 = D3 in some categories, so D2 + D4 > D1 + D3.
   → This is false.
4. D4 sold more Split ACs compared to D3:
   D3 and D4 sold equal Split Inverter ACs; but nothing about Split Non-inverter ACs.
   → Can be true.

Final Answer: The necessarily false statement is:
"D1 and D3 together sold more ACs as compared to D2 and D4 together."

Answer 5

To determine the number of Non-inverter ACs sold by D2, we use the provided information strategically. 

1. From the problem, D3 and D4 sold equal numbers of Split Inverter and Window ACs. Let the number of Window ACs sold by D3 and D4 each be x.
2. D3 and D4 sold an equal number of Window ACs, which was one-third the number sold by D2. Therefore, D2 sold 3x Window ACs.
3. Only D2 and D3 sold Window Non-inverter ACs, with D2 selling twice the amount sold by D3. Let D3 have sold y Window Non-inverter ACs, so D2 sold 2y. We know 2y + y = 3 ACs (total 3 for D2 and D3). Therefore, y = 1, and thus D2 sold 2 ACs.
4. Six total Window Non-inverter ACs were sold. Since D1 and D4 do not sell any, D3 sells 1, and D2 sells 2. With D2 and D3 taking care of 3, we add the 6 and get the D2 and D3 numbers complete.
5. The total number of Split Inverter ACs sold was 36. As D3 and D4 sold equal quantities, each selling half the quantity of D2, summing to z for D3 and D4. Therefore, D2 sold 2z with z = 9, confirming D2 sold 18 Split Inverter ACs.

Conclusion: Given that 5 Non-inverter ACs make up the difference excluding the other categories, the number of Non-inverter ACs sold by D2 is 5.