The table given below shows the average number of articles (A, B and C) sold by Pesto company in three different months. It also shows the percentage of article A sold by Pesto company out of the total number of articles sold in these months. Table also reflects the total number of article C sold in these months. Read the table carefully and answer the following questions.

Months	Average number of articles sold	Percentage of article A sold	Total number of articles C sold
April	100	40%	100
May	120	30%	70
June	72	25%	80

If the price of each article at which articles A, B and C sold in the month of April are Rs. 10, Rs y and Rs. (y + 20) respectively, then the total revenue generated by selling these articles in the month of April is Rs. 4100. Find the value of 3y.

The correct answer is 15.

To find the value of $3y$, we can break down the information given for the month of April step-by-step:

Step 1: Find the total number of articles sold in April
The table shows the average number of articles (A, B, and C) sold in April is 100.
Since there are three types of articles (A, B, and C), the total number of articles sold in April is:
$\text{Total articles}=100\times 3=300$

Step 2: Find the number of each article sold in April
- Article A: The percentage of article A sold is 40% of the total articles.
$\text{Number of A sold}=40\%\text{of}300=\frac{40}{100}$
- Article C: According to the table, the total number of article C sold in April is 100.
- Article B: The remaining articles sold are of type B.
$\text{Number of B sold}=\text{Total articles}-(\text{Number of A}+\text{Number of C})$
$\text{Number of B sold}=300-(120+100)=300-220=80$

Step 3: Set up the revenue equation
The selling prices of articles A, B, and C in April are given as Rs. 10, Rs. $y$, and Rs. $(y+20)$ respectively. The total revenue generated is Rs. 4100.
Therefore:
$(\text{Number of A}\times \text{Price of A})+(\text{Number of B}\times \text{Price of B})+(\text{Number of C}\times \text{Price of C})=\text{Total Revenue}$
Substitute the values into the equation:
$(120\times 10)+(80\times y)+(100\times (y+20\left)\right)=4100$

Step 4: Solve for $y$
Simplify the terms:
$1200+80y+100y+2000=4100$
Combine the constant terms and the $y$ terms:
$3200+180y=4100$
Subtract 3200 from both sides:
$180y=4100-3200$
$180y=900$
Divide by 180:
$y=\frac{900}{180}=5$

Step 5: Find the value of $3y$
$3y=3\times 5=15$