Solution

Let’s denote the marks obtained by Anil, Barun, and Chandan as \(A\), \(B\), and \(C\) respectively.

Given:
– Anil’s marks are 28.57% less than Barun’s marks: \(A = B – 0.2857B = 0.7143B\).
– Barun’s marks are 11.11% less than Chandan’s marks: \(B = C – 0.1111C = 0.8889C\).
– The difference between the marks obtained by Anil and Chandan is 80.5: \(C – A = 80.5\).

Step 1: Express Anil’s marks in terms of Chandan’s marks

Using the relation between Anil’s and Barun’s marks:
\[ A = 0.7143B \]
Using the relation between Barun’s and Chandan’s marks:
\[ B = 0.8889C \]

So, Anil’s marks in terms of Chandan’s marks:
\[ A = 0.7143 \times 0.8889C = 0.6351C \]

Step 2: Use the difference between Anil’s and Chandan’s marks

\[ C – A = 80.5 \]
\[ C – 0.6351C = 80.5 \]
\[ 0.3649C = 80.5 \]
\[ C = \frac{80.5}{0.3649} \]
\[ C = 220.5 \]

Step 3: Find Barun’s marks

Using the relation between Barun’s and Chandan’s marks:
\[ B = 0.8889C \]
\[ B = 0.8889 \times 220.5 \]
\[ B = 196 \]

Conclusion

The marks obtained by Barun are 196.

Solution

In a regular hexagon, all sides are equal, and all internal angles are 120°. The diagonals of a regular hexagon divide it into six equilateral triangles.

Let’s denote the side length of the hexagon as \(a\).

Area of the Hexagon PQRSTU:

The area of an equilateral triangle with side length \(a\) is given by:
\[ \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4}a^2 \]

Since the hexagon is made up of six equilateral triangles, the area of the hexagon is:
\[ \text{Area of hexagon} = 6 \times \frac{\sqrt{3}}{4}a^2 = \frac{3\sqrt{3}}{2}a^2 \]

Area of Quadrilateral PQOU:

Quadrilateral PQOU is made up of two equilateral triangles, POQ and UOQ. Therefore, the area of quadrilateral PQOU is:
\[ \text{Area of quadrilateral PQOU} = 2 \times \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{2}a^2 \]

Ratio of Areas:

The ratio of the area of quadrilateral PQOU to the area of hexagon PQRSTU is:
\[ \text{Ratio} = \frac{\text{Area of quadrilateral PQOU}}{\text{Area of hexagon}} = \frac{\frac{\sqrt{3}}{2}a^2}{\frac{3\sqrt{3}}{2}a^2} = \frac{1}{3} \]

Conclusion

The ratio of the area of quadrilateral PQOU to the area of hexagon PQRSTU is 1:3.

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 .

Given:
– \(x^4 + \frac{1}{x^4} = 322\) and \(x > 1\).

1. We know that \((a + b)^2 = a^2 + b^2 + 2ab\). Therefore, \((a + b)^2 – 2ab = a^2 + b^2\).

2. Applying this to \(x^2 + \frac{1}{x^2}\):
\[ \left(x^2 + \frac{1}{x^2}\right)^2 – 2 \times x^2 \times \frac{1}{x^2} = x^4 + \frac{1}{x^4} \]
\[ \left(x^2 + \frac{1}{x^2}\right)^2 = 322 + 2 \]
\[ \left(x^2 + \frac{1}{x^2}\right)^2 = 324 \]
\[ x^2 + \frac{1}{x^2} = \pm 18 \]
Since \(x > 1\), we take the positive root:
\[ x^2 + \frac{1}{x^2} = 18 \]

3. Similarly, for \(x^3 – \frac{1}{x^3}\), we use the identity \((a – b)^2 + 2ab = a^2 + b^2\):
\[ \left(x^2 – \frac{1}{x^2}\right)^2 + 2 = 18 \]
\[ \left(x^2 – \frac{1}{x^2}\right)^2 = 16 \]
\[ x^2 – \frac{1}{x^2} = \pm 4 \]
Since \(x > 1\), we take the positive root:
\[ x^2 – \frac{1}{x^2} = 4 \]

4. Now, cubing both sides:
\[ \left(x^3 – \frac{1}{x^3}\right) – 3 \times x \times \frac{1}{x} \left(x^2 – \frac{1}{x^2}\right) = 64 \]
\[ x^3 – \frac{1}{x^3} – 3 \times 4 = 64 \]
\[ x^3 – \frac{1}{x^3} = 64 + 12 \]
\[ x^3 – \frac{1}{x^3} = 76 \]

Conclusion:
The value of \(x^3 – \frac{1}{x^3}\) is 76.

Let’s first find the work done by each group in one day, which we can call their work rate. We’ll denote the total work to develop the app as 1 unit of work.

– 4 men can develop the app in 3 days, so their work rate is \( \frac{1}{3 \times 4} = \frac{1}{12} \) of the work per day per man.
– 3 women can develop the app in 6 days, so their work rate is \( \frac{1}{6 \times 3} = \frac{1}{18} \) of the work per day per woman.
– 6 boys can develop the app in 4 days, so their work rate is \( \frac{1}{4 \times 6} = \frac{1}{24} \) of the work per day per boy.

Now, 3 men and 6 boys worked together for 1 day. The work done by them in 1 day is:

\[ 3 \times \frac{1}{12} + 6 \times \frac{1}{24} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \text{ of the work} \]

So, half of the work is remaining.

If only women are to finish the remaining half of the work in 1 day, the number of women required is:

\[ \text{Number of women} = \frac{\text{Remaining work}}{\text{Work rate of one woman}} = \frac{1/2}{1/18} = 9 \text{ women} \]

Therefore, 9 women would be required to finish the remaining work in 1 day.