Solution

Given:
– The ratio of the work done by 50 women to the work done by 25 men in the same time is 4:3.
– 18 women and 12 men can finish a piece of work in 5 days.

Step 1: Find the ratio of work done by one woman to one man

Let the work done by one woman in one day be \(W\) and the work done by one man in one day be \(M\).
\[ \frac{50W}{25M} = \frac{4}{3} \]
\[ \frac{W}{M} = \frac{4}{3} \times \frac{1}{2} = \frac{2}{3} \]

Step 2: Calculate the total work done

Total work done by 18 women and 12 men in 5 days:
\[ \text{Total work} = (18W + 12M) \times 5 \]
Using the ratio \(W = \frac{2}{3}M\):
\[ \text{Total work} = \left(18 \times \frac{2}{3}M + 12M\right) \times 5 = (12M + 12M) \times 5 = 24M \times 5 = 120M \]

Step 3: Calculate how many women can finish the work in \(20/3\) days

Let the number of women required be \(x\). They need to do the total work in \(20/3\) days:
\[ xW \times \frac{20}{3} = 120M \]
Using the ratio \(W = \frac{2}{3}M\):
\[ x \times \frac{2}{3}M \times \frac{20}{3} = 120M \]
\[ x = \frac{120 \times 3}{\frac{2}{3} \times 20} = \frac{360}{\frac{40}{3}} = \frac{360 \times 3}{40} = 27 \]

Conclusion

27 women can finish the same work in \(20/3\) days.

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 \(1 : 1 : 1\).