Raj and Vikram's Travels

By day $d$ , Vikram has traveled 35 yojanas already, and $5d$ more.

Raj has traveled $5+8+11+\cdots+(5+3(d-1))$ . The average distance for each day is $\dfrac{10+3(d-1)}{2}$ , and so Raj's total distance is $\dfrac{d}{2}(10+3(d-1))$ .

Then, we equate the two. $35+7d=\dfrac{d}{2}(10+3(d-1))$ Expanding, $35+7d=5d+\dfrac{3d(d-1)}{2}$ Clearing the fraction, $70+14d=10d+3d(d-1)$ Moving all terms to one side, $3d^2-7d-70=0$ By the quadratic formula, the positive solution to this is $\dfrac{7+\sqrt{7^2+4(3)(70)}}{6}=\dfrac{7+\sqrt{889}}{6}$ Since $\sqrt{889}$ is a little less than 30, $\dfrac{7+\sqrt{889}}{6}\approx\dfrac{37}{6}$ which is slightly more than 6 days. By the last assumption, the answer is $\boxed{7}$ .

Let n be the no. of Raj's day at which Raj meets Vikram. Now, Raj travels in an arithmetic progressive pattern. Sum of this arithmetic progression and equating with Vikram's no. of yojanas should give the answer. n/2 {2 5+(n-1)3}=7(n+5)

solving this, we get the value of n as 7.

(Raj-5 day delay ) # # # # # 5 , 8 , 11 ,14 , 17 , 20 =total (75)

(vikram5 dy erly) 7 7 7 7 7 7 7 7 7 7 7 = (77)

by 6th day RAJ, covered total of 75 yojana and vikram total of 77 , so on the 7th day of raj , raj will meet vikram

7th day

Lets say d is the distance travelled and t is time. Since we are in Raj´s time Domain, Vickram will have travelled 35 yojanas(Y) in t=0, so what we have to do is find both equations and find their intersection. We can do it graphically, but it will be hard. So lets do it analytically:

Vickram: $d_v=7t+35$

Rajs: it goes like this, 5+8+11+14... which is an arithmetic progression, so the sum and therefore d will be: $d=\frac{a_0+a_n}{2}\cdot n\longrightarrow d_r= \frac{5+(5+(t-1)3}{2}\cdot t=\frac{7t+3t^2}{2}$

$d_v=d_r \longrightarrow \frac{7t+3t^2}{2}=7t+35 \longrightarrow 3x^2-7x-70=0$

so we solve this equation: $t= \frac{7+\sqrt{49+840}}{6}=\frac{7}{6}+\frac{\sqrt{889}}{6}\approx 1+\frac {30}{6}\approx 6$

since $\frac{7}{6}+\frac{\sqrt{889}}{6}$ is a bit higher than 6 our t is slightly higher than 6 so it will be in the beginning of the 7 day that they will meet

7=t

Vikram=7 t+35 ;t is the days of Raj's travel. The way Raj travels gives an idea of arithmetic series Raj=(t/2)(10+3 t) equating the two equations gives t=6.1... so t=7

There will never be enough poratas for one yojana....lol http://www.youtube.com/watch?v=8_ABBgJWUPM&list=LLAQYzkrnAS1yM2aWR5W6cmg

Let Raj take $t$ days to reach the place where they meet. Therefore, Vikram takes $t + 5$ days.

Now, we know that $\text{distance} = \text{speed} \times \text{time}$ .

Thus, distance traveled by Vikram to meet at the common place is $7(t+5)$ .

We have to find the distance traveled by Raj to reach this place.

Now, Raj travels $5$ yojanas on the first day, $8$ on the second day and so on. We need to sum these up till day $t$ to get the total distance traveled till day $t$ .

First, we need a closed form of the distance traveled by Raj on the $t^{\text{th}}$ day. That is the recurrence

$p(t) = 3 + p(t-1)$ and $p(1) = 5$

Solving this, we get $p(t) = 3t + 2$ . Therefore, he has traveled

$\displaystyle\sum\limits_{t=1}^t 3t + 2$ yojanas in $t$ days. The closed form for this is $\dfrac{1}{2} t(7 + 3t)$

Thus, we equate

$\dfrac{1}{2} t(7 + 3t) = 7(t+5)$

Solving this, we get the positive root of $t$ at

$t = \dfrac{1}{6} (7 + \sqrt{889}) \approx 6.1360$ days so he reaches on the $7^{\text{th}}$ day.

The total distance in yojanas travelled by Raj on the 'n'th day of his travel is given by: \frac{n}{2} \times (2a + (n-1)d)

On the 'n'th day of Raj's travel, Vikram has travelled: 35 + 7n yojanas.

Equating the two and solving, we get: n = 6.136, -3.802

*Since the answer cannot be negative, and the 6.136th day is a part of the 7th day, the answer is * 7

Day. Vikram dis. Raj distance 1. 42 5 2. 49 13 3. 56 24 4. 63 38 5. 70 55 6. 77 75

on 7th day they would meet

As Vikram travels 7 Yojans a day ,Then we will have to find the amount of yojans that Raj has travelled which is divisible by 7,,so on 1st day Raj travels 5 Yojans..on 2nd day he travels 3 more than previous ones means 3+5=8..on 3rd day still three more means 8+3=11 and this continues and on the 7th day the equation becomes 75+23=98 which is divisible by 7 and on that day it will become 14th day of vikram's Travel..:) Hope you find it quite usefull

35+7n = 5+ 3((n-1)n/2). Got n^2-(16/3)n-20=0. Asked Wolfram alpha to simplify it and it showed me a graph which had 7.8735 as one of the roots.

use the formulas for sum of an arithmetic progression and

n=n'+4.......;where n and n' are no of days for raju and vikram resp. n'=1,2,3....