Question

Two travelers set out on a long odyssey. The first traveler starts from city X and travels north on a certain day
and covers 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3
days, a second traveler sets out from oity X in the same direction as the first traveler and on his first day he
travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the
seoond traveler be ahead of the first?
Select one:
a. From the 2nd day after the 2nd traveler starts
b. From the 3rd day after the 2nd traveler starts
0.6 days
0
d. 2 days

Answer 1

Given: Day 1: 1 km, Day 2: (1 + 2) = 3 km, Day 3: (1 + 2 + 2) = 5 km .............

To find: On how many days will the  second traveler be ahead of the first

Solution:

• Now from question, we can say that the distances covered every day by the first traveller is in AP having a=1 and d = 2 and the distances covered every day by the second traveller is also in AP with the a= 12 and the d =  1

• Now, let the number of days travelled by the second traveller be n.

• Then, the first traveller will travel for (n + 3) days.

• We have given that traveler 1 has covered a total distance of 1+ 3 + 5 + 7 + ..... + (n + 3) km.

• So, the total distance traveled by traveler 1 is :

           = n+3/2( 2 + (n+3-1)2)

           = (n+3)^2

• The distance travelled by second traveller will be 12 + 13 + 14 + .....+n

• So, the total distance traveled by traveler 2 is :

          n/2( 24 + (n-1)1)

          n(n+23)/2

• Now we have given that on second traveller is ahead of first, so:

          n(n+23)/2 >  (n+3)^2

          n^2 - 11n + 18 <0

          (n-2)(n-9) < 0

• Now either n-2>0 n-9<0 or n-2<0 n-9>0..not possible.

          So n-2>0 --> n>2

          n-9<0 --> n<9

Answer:

So second traveller was ahead of first from 3rd day to 8th day.