Solution

Given:
\[ a + b + c = 0 \]

We need to find the value of:
\[ \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \]

Since \(a + b + c = 0\), we can write \(a = -(b + c)\).

Step 1: Substitute \(a = -(b + c)\)

\[ \frac{(-b – c)^2}{(-b – c)^2 – bc} + \frac{b^2}{b^2 – c(-b – c)} + \frac{c^2}{c^2 – b(-b – c)} \]

Step 2: Simplify

\[ \frac{(b + c)^2}{(b + c)^2 – bc} + \frac{b^2}{b^2 + bc – c^2} + \frac{c^2}{c^2 + bc – b^2} \]

Step 3: Simplify further

\[ \frac{(b + c)^2}{b^2 + c^2 + 2bc – bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]

\[ \frac{(b + c)^2}{b^2 + c^2 + bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]

Step 4: Combine the fractions

\[ \frac{(b + c)^2 + b^2 + c^2}{b^2 + c^2 + bc} = \frac{b^2 + c^2 + 2bc + b^2 + c^2}{b^2 + c^2 + bc} \]
\[ = \frac{2b^2 + 2c^2 + 2bc}{b^2 + c^2 + bc} \]
\[ = 2 \frac{b^2 + c^2 + bc}{b^2 + c^2 + bc} \]
\[ = 2 \]

Conclusion

The value of \(\frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab}\) is 2.

Solution

Given:
– In \(\triangle ABC\), \(\angle A = \angle B = 60^\circ\), \(AC = \sqrt{34} \text{ cm}\).
– Lines \(AD\) and \(BD\) intersect at \(D\) with \(\angle D = 90^\circ\).
– \(DB = 3 \text{ cm}\).

Step 1: Determine the Type of Triangle

Since \(\angle A = \angle B = 60^\circ\), \(\angle C\) must also be \(60^\circ\) (as the sum of angles in a triangle is \(180^\circ\)). Therefore, \(\triangle ABC\) is an equilateral triangle.

Step 2: Find the Length of AB

Since \(\triangle ABC\) is equilateral, \(AB = AC = \sqrt{34} \text{ cm}\).

Step 3: Use Pythagoras’ Theorem

In \(\triangle ADB\), which is a right-angled triangle, we can use Pythagoras’ theorem:
\[ AB^2 = AD^2 + DB^2 \]
\[ (\sqrt{34})^2 = AD^2 + 3^2 \]
\[ 34 = AD^2 + 9 \]
\[ AD^2 = 34 – 9 \]
\[ AD^2 = 25 \]
\[ AD = \sqrt{25} \]
\[ AD = 5 \text{ cm} \]

Conclusion

The length of \(AD\) is 5 cm.

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%.

Given:
– Speed of the boat in still water = 10 km/hr
– Speed of the current = 4 km/hr
– Distance covered upstream = 12 km
– Distance covered downstream = 14 km
– Total time for the journey = 4 hrs

1. Calculate the downstream speed:
\[ \text{Downstream speed} = \text{Speed of boat in still water} + \text{Speed of current} \]
\[ \text{Downstream speed} = 10 + 4 = 14 \text{ km/hr} \]

2. Calculate the upstream speed:
\[ \text{Upstream speed} = \text{Speed of boat in still water} – \text{Speed of current} \]
\[ \text{Upstream speed} = 10 – 4 = 6 \text{ km/hr} \]

3. Calculate the time taken to cover 12 km upstream:
\[ \text{Time upstream} = \frac{\text{Distance upstream}}{\text{Upstream speed}} \]
\[ \text{Time upstream} = \frac{12}{6} = 2 \text{ hrs} \]

4. Calculate the time taken to cover 14 km downstream:
\[ \text{Time downstream} = \frac{\text{Distance downstream}}{\text{Downstream speed}} \]
\[ \text{Time downstream} = \frac{14}{14} = 1 \text{ hr} \]

5. Calculate the time for which he took rest:
\[ \text{Time for rest} = \text{Total time} – (\text{Time upstream} + \text{Time downstream}) \]
\[ \text{Time for rest} = 4 – (2 + 1) = 1 \text{ hr} \]

Conclusion:
The period of time for which he took rest is 1 hour.