Buying the Article:

– The merchant buys using a 125 gram weight instead of 100 grams. This means he gets 25% more of the article for the same price. So, the cost price per 100 grams is effectively \(\frac{100}{125} = 0.8\) times the original cost price.

Selling the Article:

– The merchant sells using an 80 gram weight instead of 100 grams. This means he gives 20% less of the article for the same price. So, the selling price per 100 grams is effectively \(\frac{100}{80} = 1.25\) times the original selling price.

Marking Up and Discount:

– The merchant marks up the price by 20% and then offers a 20% discount. The overall effect is calculated as follows:

\[ \text{Overall Factor} = \frac{125}{100} \times \frac{100}{80} \times \frac{120}{100} \times \frac{80}{100} = \frac{3}{2} \]

– This means the selling price is \(\frac{3}{2}\) times the cost price.

Overall Profit Percentage:

– Since the selling price is \(\frac{3}{2}\) times the cost price, the profit is \(\frac{1}{2}\) times the cost price.

– Therefore, the profit percentage is:

\[ \text{Profit Percentage} = \frac{1}{2} \times 100 = 50\% \]

Conclusion

The overall profit percentage is 50%.

Let’s analyze the data provided in the table to answer the questions:

a) In which year all the companies together produces the maximum products.

Total production in each year:
– 2000: \(35 + 20 + 40 + 31 + 31 = 157\) lakhs
– 2001: \(10 + 20 + 23 + 30 + 41 = 124\) lakhs
– 2002: \(60 + 30 + 70 + 10 + 20 = 190\) lakhs
– 2003: \(5 + 50 + 40 + 20 + 10 = 125\) lakhs

Answer: The maximum production by all companies together was in the year 2002 .

b) Production of B & C together in year 2000 is what percent of Production by D and E in year 2003?

Production of B & C in 2000: \(20 + 40 = 60\) lakhs

Production of D & E in 2003: \(20 + 10 = 30\) lakhs

Percentage: \(\frac{60}{30} \times 100\% = 200\%\)

Answer: The production of B & C together in year 2000 is 200% of the production by D and E in year 2003.

c) What is the ratio of production by A & D in 2001 to the Production by B & E in 2003?

Production by A & D in 2001: \(10 + 30 = 40\) lakhs

Production by B & E in 2003: \(50 + 10 = 60\) lakhs

Ratio: \(40:60 = 2:3\)

Answer: The ratio of production by A & D in 2001 to the production by B & E in 2003 is 2:3 .

d) If the profit generated on selling one product produced by A is Rs. 4.5, then find the total profit earned on selling all the products of A in all years together.

Total products of A in all years: \(35 + 10 + 60 + 5 = 110\) lakhs

Total profit: \(110 \times 10^5 \times 4.5 = 49,50,00,000\) Rs.

Answer: The total profit earned on selling all the products of A in all years together is Rs. 49.5 crores .

For a triangle with sides \(a\), \(b\), and \(c\), the circumradius \(R\) and the inradius \(r\) are given by:

\[ R = \frac{abc}{4K} \]

and

\[ r = \frac{K}{s} \]

where \(K\) is the area of the triangle and \(s = \frac{a + b + c}{2}\) is the semi-perimeter of the triangle.

For the given triangle with sides \(a = 40\) cm, \(b = 41\) cm, and \(c = 9\) cm:

\[ s = \frac{40 + 41 + 9}{2} = \frac{90}{2} = 45 \text{ cm} \]

Using Heron’s formula, the area \(K\) of the triangle is:

\[ K = \sqrt{s(s – a)(s – b)(s – c)} \]
\[ K = \sqrt{45(45 – 40)(45 – 41)(45 – 9)} \]
\[ K = \sqrt{45 \times 5 \times 4 \times 36} \]
\[ K = \sqrt{32400} \]
\[ K = 180 \text{ cm}^2 \]

Now, we can find the circumradius \(R\):

\[ R = \frac{abc}{4K} \]
\[ R = \frac{40 \times 41 \times 9}{4 \times 180} \]
\[ R = \frac{14760}{720} \]
\[ R = 20.5 \text{ cm} \]

And the inradius \(r\):

\[ r = \frac{K}{s} \]
\[ r = \frac{180}{45} \]
\[ r = 4 \text{ cm} \]

Finally, the value of \(2(R + r)\) is:

\[ 2(R + r) = 2(20.5 + 4) \]
\[ 2(R + r) = 2 \times 24.5 \]
\[ 2(R + r) = 49 \]

Therefore, the value of \(2(R + r)\) is 49 cm.

Given:
– \(x + y + z = 6\sqrt{3}\)
– \(x^2 + y^2 + z^2 = 36\)

1. Use the identity \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\) to expand \((x + y + z)^2\):

\[ (x + y + z)^2 = (6\sqrt{3})^2 \]
\[ x^2 + y^2 + z^2 + 2(xy + yz + zx) = 108 \]
\[ 36 + 2(xy + yz + zx) = 108 \]
\[ 2(xy + yz + zx) = 72 \]
\[ xy + yz + zx = 36 \] (Equation A)

2. Comparing Equation A with \(x^2 + y^2 + z^2 = 36\):

\[ x^2 + y^2 + z^2 = xy + yz + zx \]

This implies that \(x = y = z\), as the sum of the squares of the variables is equal to the sum of their pairwise products.

3. Therefore, the ratio of \(x : y : z\) is \(1 : 1 : 1\).

Conclusion:
The ratio \(x : y : z\) is \(1 : 1 : 1\).

Given:
– We need to find the value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\).

1. We use the identity \(\cos x \cos(60 – x) \cos(60 + x) = \frac{1}{4} \cos 3x\):

\[ \cos x \cos(60 – x) \cos(60 + x) = \frac{1}{4} \cos 3x \]

2. Applying this identity to \(\cos 50^\circ, \cos 10^\circ, \cos 110^\circ\):

\[ \cos 50^\circ \cos 10^\circ \cos 110^\circ = \frac{1}{4} \cos 150^\circ \]
\[ \cos 150^\circ = -\frac{\sqrt{3}}{2} \]
\[ \cos 50^\circ \cos 10^\circ \cos 110^\circ = -\frac{\sqrt{3}}{8} \]

3. Also, \(\cos 60^\circ = \frac{1}{2}\).

4. Substituting these values into the given expression:

\[ \frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right) \]
\[ = \frac{16}{\sqrt{3}} \times \left(-\frac{\sqrt{3}}{8}\right) \times \frac{1}{2} \]
\[ = -1 \]

Conclusion:
The value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\) is \(-1\).