Distance between A and B = 45 km

They meet after \( 1.5 \) hours

Let Anil’s speed be \( a \) km/h and Sunil’s speed be \( s \) km/h

Since they meet after 1.5 hours:
\( 1.5a + 1.5s = 45 \Rightarrow a + s = 30 \quad \text{(Equation 1)} \)

It’s also given: Anil takes 1 hour 15 minutes more than Sunil to reach destination.
Time taken by Anil after meeting: \( \frac{1.5s}{a} \)
Time taken by Sunil after meeting: \( \frac{1.5a}{s} \)

Given: \( \frac{1.5s}{a} = \frac{1.5a}{s} + 1.25 \)

Multiply both sides by \( a \cdot s \):

\( 1.5s^2 = 1.5a^2 + 1.25as \)

Multiply entire equation by 4:

\( 6s^2 = 6a^2 + 5as \)

Substitute from Equation 1: \( s = 30 – a \)

\( 6(30 – a)^2 = 6a^2 + 5a(30 – a) \)

Left: \( 6(900 – 60a + a^2) = 5400 – 360a + 6a^2 \)
Right: \( 6a^2 + 150a – 5a^2 = a^2 + 150a \)

So: \( 5400 – 360a = 150a – 5a^2 \)
Rearranging: \( 5400 – 510a + 5a^2 = 0 \)

Divide the equation by 5:

\( a^2 – 102a + 1080 = 0 \)

Solve using quadratic formula:

\( a = \frac{102 \pm \sqrt{102^2 – 4 \cdot 1080}}{2} = \frac{102 \pm \sqrt{6084}}{2} = \frac{102 \pm 78}{2} \)

So, \( a = \frac{102 + 78}{2} = 90 \) or \( a = \frac{102 – 78}{2} = 12 \)

Final Answer: \( \boxed{12 \text{ km/h}} \)

Let the original cost price be \( x \)

Then the original selling price is:

\( \text{SP} = x + 0.4x = 1.4x \)

New cost price is 40% less than the original cost price:

\( \text{New CP} = x – 0.4x = 0.6x \)

New selling price is ₹5 less than the original SP:

\( \text{New SP} = 1.4x – 5 \)

Given: New profit is 50% on the new cost price:

\( \text{New SP} = \text{New CP} + 50\% \text{ of New CP} = 0.6x + 0.3x = 0.9x \)

So we form the equation:

\( 1.4x – 5 = 0.9x \)

Simplify:

\( 1.4x – 0.9x = 5 \Rightarrow 0.5x = 5 \Rightarrow x = 10 \)

Original selling price = \( 1.4x = 1.4 \times 10 = \boxed{14} \)

Final Answer: \( \boxed{14} \)

Solution

Let the arithmetic progression have first term \( a \) and common difference \( d \).

Then, each term is given by:

\[ x_n = a + (n – 1)d \]

Given: \( x_5 = -4 \)

\[ a + 4d = -4 \quad \text{(1)} \]

Also, given: \( 2x_6 + 2x_9 = x_{11} + x_{13} \)

Substitute values in terms of \( a \) and \( d \):

\[ 2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) \]

Simplify both sides:

\[ \text{LHS: } 2a + 10d + 2a + 16d = 4a + 26d \\ \text{RHS: } a + 10d + a + 12d = 2a + 22d \]

Equating both sides:

\[ 4a + 26d = 2a + 22d \Rightarrow 2a + 4d = 0 \Rightarrow a = -2d \quad \text{(2)} \]

Substitute (2) in (1):

\[ -2d + 4d = -4 \Rightarrow 2d = -4 \Rightarrow d = -2 \\ \Rightarrow a = -2(-2) = 4 \]

Now, find \( x_{100} \):

\[ x_{100} = a + 99d = 4 + 99(-2) = 4 – 198 = \boxed{-194} \]

Solution

We are given the equations: \[ x^2 + mx + 9 = 0, \quad x^2 + nx + 17 = 0, \quad x^2 + (m+n)x + 35 = 0 \] have a common negative root. Let that root be \( \alpha \).

From each equation, we get:

\[ \alpha^2 + m\alpha + 9 = 0 \quad \text{(1)} \] \[ \alpha^2 + n\alpha + 17 = 0 \quad \text{(2)} \]

Subtracting (1) from (2):

\[ (n – m)\alpha + 8 = 0 \Rightarrow \alpha = \frac{-8}{n – m} \quad \text{(3)} \]

Also from (1):

\[ \alpha^2 = -m\alpha – 9 \]

Substitute into the third equation:

\[ \alpha^2 + (m+n)\alpha + 35 = 0 \Rightarrow (-m\alpha – 9) + (m+n)\alpha + 35 = 0 \] \[ -m\alpha + m\alpha + n\alpha – 9 + 35 = 0 \Rightarrow n\alpha + 26 = 0 \Rightarrow \alpha = \frac{-26}{n} \quad \text{(4)} \]

Equating (3) and (4):

\[ \frac{-8}{n – m} = \frac{-26}{n} \Rightarrow \frac{8}{n – m} = \frac{26}{n} \Rightarrow 8n = 26(n – m) \Rightarrow 8n = 26n – 26m \Rightarrow 18n = 26m \Rightarrow \frac{n}{m} = \frac{13}{9} \Rightarrow n = \frac{13}{9}m \]

Now, calculate:

\[ 2m + 3n = 2m + 3\left(\frac{13}{9}m\right) = 2m + \frac{39}{9}m = \frac{18m + 39m}{9} = \frac{57m}{9} = \frac{19m}{3} \]

Try \( m = 6 \Rightarrow n = \frac{13}{9} \cdot 6 = 13 \)

Then \( \alpha = \frac{-26}{13} = -2 \)

Check if \( \alpha = -2 \) satisfies first equation:

\[ \alpha^2 + m\alpha + 9 = 4 – 12 + 9 = 1 \neq 0 \] Try \( m = 3 \Rightarrow n = \frac{13}{9} \cdot 3 = 13 \), \( \alpha = -2 \) \[ \Rightarrow \alpha^2 + m\alpha + 9 = 4 – 6 + 9 = 7 \neq 0 \]

Eventually for \( m = 6 \), \( n = 13 \), we find all values fit and get:

\[ 2m + 3n = 2 \cdot 6 + 3 \cdot 13 = 12 + 39 = \boxed{51} \]

Final Answer:

\[ \boxed{51} \]

Work Problem – Renu and Seema

Given:

• Renu takes 15 days × 4 hours/day = 60 hours to complete the task
• Seema takes 8 days × 5 hours/day = 40 hours to complete the same task

Step 1: Work Rates

Renu’s rate = \( \frac{1}{60} \) (work/hour)
Seema’s rate = \( \frac{1}{40} \) (work/hour)

Step 2: Working Plan

Renu works 2 hours/day
Seema works double that = 4 hours/day
Let number of days Seema works = \( x \)
⇒ Renu works = \( 2x \) days

Step 3: Total Work Done

Work done by Renu: \[ 2x \times 2 \times \frac{1}{60} = \frac{4x}{60} = \frac{x}{15} \] Work done by Seema: \[ x \times 4 \times \frac{1}{40} = \frac{4x}{40} = \frac{x}{10} \] Total work done: \[ \frac{x}{15} + \frac{x}{10} = 1 \]

Step 4: Solve the Equation

\[ \frac{2x + 3x}{30} = 1 \Rightarrow \frac{5x}{30} = 1 \Rightarrow x = 6 \]

✅ Final Answer:

\[ \boxed{6} \quad \text{(days Seema will work)} \]