Along a road lie an odd number of stones placed at intervals of 10m These stones have to be assemble...........

Quantitative Aptitiude (CLAT) Number System

Along a road lie an odd number of stones placed at intervals of 10m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried, out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. then, the number of stones is:

Explanation:

Let total no of stones of one side =x
total no of both sides stones = 2x
total no of stones = 2x+1
We know A.P sum = n/2(2a +( n-1)d)
Total Distance of one side, one way=
( (x/2)×((2×10 +( x-1)10))×2×2 = 4800
(x/2)×(20 + (10x - 10)= 4800/4
x × (20 +10x -10) = 1200×2
( 2x + x^2 -x) = 2400/10
x^2 +x -240 = 0
x(x-15)+ 16 (x-15) =0
(x-15) (x+16) = 9
x = 15 or x = -16 Not possible
Total no of stones = (15 × 2)+1 =31