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
How many Split Inverter ACs did D2 sell?
Solution and Explanation
Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = \(\frac{A}{4}\) and Split AC's = \(\frac{3A}{4}\)
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = \(\frac{B}{5}\) and Window Non-Inverter AC's = \(\frac{4B}{5}\)
From the given condition 3, we get
\(\frac{A}{4}-\frac{B}{5}=6\) and \(\frac{4B}{5}=36\)
So , B = 46 and A = 60
| Total 60 | |||
| Split=45 | Window=15 | ||
| Inv=36 | Non-inv=9 | Inv=9 | Non-inv=6 |
So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table :
| D1 Total = | |||
| Split = | Window = x | ||
| Inv = 13-x | Non-Inv = | Inv = x | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 3y | ||
| Inv = 2z | Non-Inv = | Inv = | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = 3 | Inv = | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = | Inv = | Non-Inv = 0 |
As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
⇒ x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table :
| D1 Total = | |||
| Split = | Window = 5 | ||
| Inv = 8 | Non-Inv = | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 2z | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = | Inv = 2 | Non-Inv = 0 |
So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36.
⇒ 4z = 28
⇒ z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2×5=10
| D1 Total = 15 | |||
| Split = 10 | Window = 5 | ||
| Inv = 8 | Non-Inv = 2 | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 14 | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = 10 | Window = 2 | ||
| Inv = 7 | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = 7 | Non-Inv = | Inv = 2 | Non-Inv = 0 |
Now, by looking at the table , we get that , 14 split ACs are sold by D2.
What percentage of ACs sold were of Non-inverter type?
- 75%
- 20%
- 25%
- 4.33%
The Correct Option is C
Solution and Explanation
Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = \(\frac{A}{4}\) and Split AC's = \(\frac{3A}{4}\)
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = \(\frac{B}{5}\) and Window Non-Inverter AC's = \(\frac{4B}{5}\)
From the given condition 3, we get
\(\frac{A}{4}-\frac{B}{5}=6\) and \(\frac{4B}{5}=36\)
So , B = 46 and A = 60
| Total 60 | |||
| Split=45 | Window=15 | ||
| Inv=36 | Non-inv=9 | Inv=9 | Non-inv=6 |
So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table :
| D1 Total = | |||
| Split = | Window = x | ||
| Inv = 13-x | Non-Inv = | Inv = x | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 3y | ||
| Inv = 2z | Non-Inv = | Inv = | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = 3 | Inv = | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = | Inv = | Non-Inv = 0 |
As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
⇒ x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table :
| D1 Total = | |||
| Split = | Window = 5 | ||
| Inv = 8 | Non-Inv = | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 2z | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = | Inv = 2 | Non-Inv = 0 |
So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36.
⇒ 4z = 28
⇒ z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2×5=10
| D1 Total = 15 | |||
| Split = 10 | Window = 5 | ||
| Inv = 8 | Non-Inv = 2 | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 14 | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = 12 | |||
| Split = 10 | Window = 2 | ||
| Inv = 7 | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = 7 | Non-Inv = | Inv = 2 | Non-Inv = 0 |
| Total = 60 | |||
| Split = 45 | Window = 15 | ||
| Inv = 36 | Non-Inv = 9 | Inv = 9 | Non-Inv = 6 |
Now, by looking at the table, we get that total number of non-inverter ACs is 9 + 6 = 15.
So, needed percentage :
15 out of 60 i.e \(\frac{15}{60}\times10\) = 25%
Therefore, the correct option is (C) : 25%.
What was the total number of ACs sold by D2 and D4?
Solution and Explanation
Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = \(\frac{A}{4}\) and Split AC's = \(\frac{3A}{4}\)
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = \(\frac{B}{5}\) and Window Non-Inverter AC's = \(\frac{4B}{5}\)
From the given condition 3, we get
\(\frac{A}{4}-\frac{B}{5}=6\) and \(\frac{4B}{5}=36\)
So , B = 46 and A = 60
| Total 60 | |||
| Split=45 | Window=15 | ||
| Inv=36 | Non-inv=9 | Inv=9 | Non-inv=6 |
So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table :
| D1 Total = | |||
| Split = | Window = x | ||
| Inv = 13-x | Non-Inv = | Inv = x | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 3y | ||
| Inv = 2z | Non-Inv = | Inv = | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = 3 | Inv = | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = | Inv = | Non-Inv = 0 |
As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
⇒ x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table :
| D1 Total = | |||
| Split = | Window = 5 | ||
| Inv = 8 | Non-Inv = | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 2z | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = | Inv = 2 | Non-Inv = 0 |
So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36.
⇒ 4z = 28
⇒ z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2×5=10
| D1 Total = 15 | |||
| Split = 10 | Window = 5 | ||
| Inv = 8 | Non-Inv = 2 | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 14 | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = 10 | Window = 2 | ||
| Inv = 7 | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = 7 | Non-Inv = | Inv = 2 | Non-Inv = 0 |
Now, the total number of ACs sold by D2 and D4 :
= 60 - D1 - D3
= 60 - 15 - 12
= 33
Therefore, the correct answer is 33.
Which of the following statements is necessarily false?
- D1 and D3 sold an equal number of Split ACs.
- D2 sold the highest number of ACs.
- D1 and D3 together sold more ACs as compared to D2 and D4 together.
- D4 sold more Split ACs as compared to D3.
The Correct Option is C
Solution and Explanation
Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = \(\frac{A}{4}\) and Split AC's = \(\frac{3A}{4}\)
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = \(\frac{B}{5}\) and Window Non-Inverter AC's = \(\frac{4B}{5}\)
From the given condition 3, we get
\(\frac{A}{4}-\frac{B}{5}=6\) and \(\frac{4B}{5}=36\)
So , B = 46 and A = 60
| Total 60 | |||
| Split=45 | Window=15 | ||
| Inv=36 | Non-inv=9 | Inv=9 | Non-inv=6 |
So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table :
| D1 Total = | |||
| Split = | Window = x | ||
| Inv = 13-x | Non-Inv = | Inv = x | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 3y | ||
| Inv = 2z | Non-Inv = | Inv = | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = 3 | Inv = | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = | Inv = | Non-Inv = 0 |
As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
⇒ x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table :
| D1 Total = | |||
| Split = | Window = 5 | ||
| Inv = 8 | Non-Inv = | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 2z | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = | Inv = 2 | Non-Inv = 0 |
So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36.
⇒ 4z = 28
⇒ z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2×5=10
| D1 Total = 15 | |||
| Split = 10 | Window = 5 | ||
| Inv = 8 | Non-Inv = 2 | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 14 | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = 10 | Window = 2 | ||
| Inv = 7 | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = 7 | Non-Inv = | Inv = 2 | Non-Inv = 0 |
From the table, we get that D1 and D3 sold 27 ACs together which is less than 60 - 27 = 33 which is sold by D2 and D4 together.
Therefore, the correct option is (C) :D1 and D3 together sold more ACs as compared to D2 and D4 together..
If D3 and D4 sold an equal number of ACs, then what was the number of Non-inverter ACs sold by D2?
- 5
- 7
- 6
- 4
The Correct Option is A
Solution and Explanation
Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = \(\frac{A}{4}\) and Split AC's = \(\frac{3A}{4}\)
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = \(\frac{B}{5}\) and Window Non-Inverter AC's = \(\frac{4B}{5}\)
From the given condition 3, we get
\(\frac{A}{4}-\frac{B}{5}=6\) and \(\frac{4B}{5}=36\)
So , B = 46 and A = 60
| Total 60 | |||
| Split=45 | Window=15 | ||
| Inv=36 | Non-inv=9 | Inv=9 | Non-inv=6 |
So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table :
| D1 Total = | |||
| Split = | Window = x | ||
| Inv = 13-x | Non-Inv = | Inv = x | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 3y | ||
| Inv = 2z | Non-Inv = | Inv = | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = 3 | Inv = | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = y | ||
| Inv = z | Non-Inv = | Inv = | Non-Inv = 0 |
As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
⇒ x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table :
| D1 Total = | |||
| Split = | Window = 5 | ||
| Inv = 8 | Non-Inv = | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 2z | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = z | Non-Inv = | Inv = 2 | Non-Inv = 0 |
So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36.
⇒ 4z = 28
⇒ z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2×5=10
| D1 Total = 15 | |||
| Split = 10 | Window = 5 | ||
| Inv = 8 | Non-Inv = 2 | Inv = 5 | Non-Inv = 0 |
| D2 Total = | |||
| Split = | Window = 6 | ||
| Inv = 14 | Non-Inv = | Inv = 2 | Non-Inv = 4 |
| D3 Total = | |||
| Split = 10 | Window = 2 | ||
| Inv = 7 | Non-Inv = 3 | Inv = 0 | Non-Inv = 2 |
| D4 Total = | |||
| Split = | Window = 2 | ||
| Inv = 7 | Non-Inv = | Inv = 2 | Non-Inv = 0 |
Now, If D3 and D4, sold equal number of AC's, then the table will be as given below :
| D1 Total = 15 | |||
| Split = 10 | Window = 5 | ||
| Inv = 8 | Non-Inv = 2 | Inv = 5 | Non-Inv = 0 |
| D2 Total = 21 | |||
| Split = 15 | Window = 6 | ||
| Inv = 14 | Non-Inv = 1 | 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 = 10 | Window = 2 | ||
| Inv = 7 | Non-Inv = 3 | Inv = 2 | Non-Inv = 0 |
So, by looking at the table , we can say that :
The number of non-inverter ACs sold by D2 is 4+1 = 5.
Top Questions on Data Analysis
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The following additional information is known.
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In the plots, the horizontal and vertical coordinates of point representing each firm gives their ES and PAT values respectively. The PRD values of each firm are proportional to the areas around the points representing each firm. The areas are comparable between the two plots, i.e., equal areas in the two plots represent the same PRD values for the two years.
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- The number of all positive integers up to 500 with non-repeating digits is
- CAT - 2024
- Number Systems