Comprehension:

An air conditioner (AC) company has four dealers - D1, D2, D3 and D4 in a city. It is evaluating sales performances of these dealers. The company sells two variants of ACs Window and Split. Both these variants can be either Inverter type or Non-inverter type. It is known that of the total number of ACs sold in the city, 25% were of Window variant, while the rest were of Split variant. Among the Inverter ACs sold, 20% were of Window variant.

The following information is also known:

1. Every dealer sold at least two window ACs.

2. D1 sold 13 inverter ACs, while D3 sold 5 Non-inverter ACs.

3. A total of six Window Non-inverter ACs and 36 Split Inverter ACs were sold in the city. 4. The number of Split ACs sold by D1 was twice the number of Window ACs sold by it. 5. D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.

4. D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.

5. D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.

Which of the following statements is necessarily false?

The correct answer is: D1 and D3 together sold more ACs as compared to D2 and D4 together.

Let us systematically analyze the given information and construct the complete sales data for the four dealers (D1, D2, D3, and D4) across the different categories of ACs (Window vs. Split, and Inverter vs. Non-inverter).

Let:
- Total ACs sold in the city = $T$
- Window ACs (W) = 25% of $T$ = 0.25$T$
- Split ACs (S) = 75% of $T$ = 0.75$T$
- Inverter ACs = $I$
- Non-inverter ACs = $N$

From the problem, we are given:
- Among the Inverter ACs, 20% are of Window variant (W-I). Thus, Split Inverter ACs (S-I) comprise 80% of Inverter ACs.
- We are given that the total number of Split Inverter (S-I) ACs sold is 36.
Therefore, 80% of $I=36$, which gives:
$I=\frac{36}{0.8}=45$
So, the total number of Window Inverter (W-I) ACs sold in the city is:
$20\%\text{of}45=9$

We are also given:
- The total number of Window Non-inverter (W-N) ACs sold is 6.
This allows us to find the total number of Window ACs sold:
Total Window ACs (W) = (W-I) + (W-N) = $9+6=15$.
Since Window ACs represent 25% of the total ACs sold ($T$):
$0.25T=15\Rightarrow T=60$.
So, the total Split ACs sold (S) = 75% of 60 = 45.
Since Split Inverter (S-I) = 36, the Split Non-inverter (S-N) = $45-36=9$.

Let us now determine the distribution for each dealer:
1. Window Non-inverter (W-N) ACs (Total = 6):
According to statement 4, only D2 and D3 sold W-N ACs, and the number sold by D2 was twice that of D3.
Let D3 sell $x$ W-N ACs. Then D2 sold $2x$ W-N ACs.
$x+2x=6\Rightarrow 3x=6\Rightarrow x=2$.
So, D2 sold 4 W-N ACs, D3 sold 2 W-N ACs, and D1 and D4 sold 0 W-N ACs.

2. Window ACs (W) (Total = 15):
According to statement 5, D3 and D4 sold an equal number of Window ACs, and this number was one-third of the number sold by D2.
Let the number of Window ACs sold by D3 and D4 be $y$. Then D2 sold $3y$ Window ACs.
Thus, the sum of Window ACs sold by D2, D3, and D4 is $3y+y+y=5y$.
The remaining Window ACs are sold by D1: $15-5y$.
Since every dealer sold at least two Window ACs (statement 1):
- For D3 and D4: $y\ge 2$.
- For D1: $15-5y\ge 2\Rightarrow 5y\le 13\Rightarrow y\le 2.6$.
Since the number of ACs must be an integer, we must have $y=2$.
Thus:
- D3 sold 2 Window ACs.
- D4 sold 2 Window ACs.
- D2 sold $3\times 2=6$ Window ACs.
- D1 sold $15-5\left(2\right)=5$ Window ACs.

3. Window Inverter (W-I) ACs (Total = 9):
We can find W-I for each dealer since W-I = W - W-N:
- D1: $5-0=5$ W-I ACs.
- D2: $6-4=2$ W-I ACs.
- D3: $2-2=0$ W-I ACs.
- D4: $2-0=2$ W-I ACs.

4. Split Inverter (S-I) ACs (Total = 36):
According to statement 5 (part 2), D3 and D4 sold an equal number of Split Inverter ACs, and this number was half the number of similar ACs sold by D2.
Let the number of S-I ACs sold by D3 and D4 be $z$. Then D2 sold $2z$ S-I ACs.
The total S-I ACs sold by D2, D3, and D4 is $2z+z+z=4z$.
Now we look at D1: We are given D1 sold 13 Inverter ACs (statement 2).
Since D1 sold 5 Window Inverter (W-I) ACs, the number of Split Inverter (S-I) ACs sold by D1 must be:
$13-5=8$.
Since the total S-I ACs is 36:
$8+4z=36\Rightarrow 4z=28\Rightarrow z=7$.
Thus:
- D3 sold 7 S-I ACs.
- D4 sold 7 S-I ACs.
- D2 sold $2\times 7=14$ S-I ACs.
- D1 sold 8 S-I ACs.

5. Split ACs (S) and Split Non-inverter (S-N) ACs (Total S = 45, Total S-N = 9):
- For D1: Statement 4 states that the number of Split ACs sold by D1 was twice the number of Window ACs sold by it.
Since D1 sold 5 Window ACs, D1 sold $2\times 5=10$ Split ACs.
Since D1 sold 8 S-I ACs, D1 sold $10-8=2$ S-N ACs.
- For D3: D3 sold 5 Non-inverter ACs in total (statement 2).
We know D3 sold 2 W-N ACs, so D3 sold $5-2=3$ S-N ACs.
Thus, the total Split ACs sold by D3 is S-I + S-N = $7+3=10$.
- For D2 and D4: Let D4 sell $w$ S-N ACs. Then D2 sold the remaining S-N ACs: $9-(2+3+w)=4-w$.
Thus, the total ACs sold by each dealer is:
- D1: 5 Window + 10 Split = 15 ACs.
- D2: 6 Window + (14 + 4 - w) Split = $24-w$ ACs.
- D3: 2 Window + 10 Split = 12 ACs.
- D4: 2 Window + (7 + w) Split = $9+\; w$ ACs.

Let us calculate the total ACs sold by (D1 + D3) and (D2 + D4):
- ACs sold by D1 and D3 together = $15+12=27$.
- ACs sold by D2 and D4 together = $(24-w)+(9+w)=\; 33$.
Since $27<33$, D1 and D3 together sold fewer ACs than D2 and D4 together, regardless of the value of $w$.
Therefore, the statement "D1 and D3 together sold more ACs as compared to D2 and D4 together" is necessarily false.