Divakar and Subhank set out to meet each other from Delhi and Saharanpur, 330 km apart. Divakar travels 30 km on the first day 28 km on the second day, 26 km on the third day and so on. Subhank travels 20 km ont the first day, 24 km on the second day, 28 km on the third day and so on. In how many days they meet, if they started moving towards each other at the same time?

Options

A. 4
B. 5
C. 6
D. none

Answer

To solve the problem of when Divakar and Subhank will meet, we will analyze their travel patterns and set up equations based on their distances traveled each day.

### Step-by-Step Solution:

1. **Understanding the Distance**:
- The total distance between Delhi and Saharanpur is 330 km.
- Divakar starts from Delhi and travels towards Saharanpur, while Subhank starts from Saharanpur and travels towards Delhi.

2. **Divakar's Travel Pattern**:
- Divakar travels 30 km on the first day, 28 km on the second day, 26 km on the third day, and so on. This forms an arithmetic progression (AP) where:
- First term (a) = 30 km
- Common difference (d) = -2 km
- The distance traveled by Divakar in n days can be expressed as:
\[ S_n = \frac{n}{2} \times [2a + (n-1)d] \]
Substituting the values:
\[ S_n = \frac{n}{2} \times [2 \times 30 + (n-1)(-2)] \]
\[ S_n = \frac{n}{2} \times [60 - 2n + 2] = \frac{n}{2} \times (62 - 2n) \]
\[ S_n = n(31 - n) \]

3. **Subhank's Travel Pattern**:
- Subhank travels 20 km on the first day, 24 km on the second day, 28 km on the third day, and so on. This also forms an AP where:
- First term (a) = 20 km
- Common difference (d) = 4 km
- The distance traveled by Subhank in n days can be expressed as:
\[ S_n = \frac{n}{2} \times [2a + (n-1)d] \]
Substituting the values:
\[ S_n = \frac{n}{2} \times [2 \times 20 + (n-1)(4)] \]
\[ S_n = \frac{n}{2} \times [40 + 4n - 4] = \frac{n}{2} \times (36 + 4n) \]
\[ S_n = n(18 + 2n) \]

4. **Setting Up the Equation**:
- Since they are moving towards each other, the sum of the distances they travel must equal the total distance of 330 km:
\[ n(31 - n) + n(18 + 2n) = 330 \]
- Simplifying the equation:
\[ n(31 - n + 18 + 2n) = 330 \]
\[ n(49 + n) = 330 \]
\[ n^2 + 49n - 330 = 0 \]

5. **Solving the Quadratic Equation**:
- We can factor or use the quadratic formula to solve for n:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 49\), and \(c = -330\):
\[ n = \frac{-49 \pm \sqrt{49^2 - 4 \times 1 \times (-330)}}{2 \times 1} \]
\[ n = \frac{-49 \pm \sqrt{2401 + 1320}}{2} \]
\[ n = \frac{-49 \pm \sqrt{3721}}{2} \]
\[ n = \frac{-49 \pm 61}{2} \]
- This gives two potential solutions:
1. \(n = \frac{12}{2} = 6\)
2. \(n = \frac{-110}{2} = -55\) (not valid)

6. **Conclusion**:
- The valid solution is \(n = 6\). Therefore, Divakar and Subhank will meet after **6 days**.