# @lurys

##### TANMAY GHOSH 0 karma
###### CATOfficial Quant Thread for CAT 2011 [Part 8]

Can somebody please exaplain the logic behind this qs:

A farmer has decided to build a wire fence along one straight side of his property. For this, he
planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After
he bought the posts and wire, he found that the number of posts he had bought was 5 less than
required. However, he discovered that the number of posts he had bought would be just sufficient if
he spaced them 8 m apart. What is the length of the side of his property and how many posts did he

a. 100 m, 15 b. 100 m, 16 c. 120 m, 15 d. 120 m, 16

Ans.d

No of posts=x.

6(x-1)+5*6=8(x-1).

each 6mtrs covered by two post.
so for x nos of post.distance coverded=6(x-1).
For 5 posts=6*5mtrs(as last one act as 6th post).

sum of this equal to the total distance covered by 8 posts.
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###### CATOfficial Quant Thread for CAT 2011 [Part 8]

Can somebody please exaplain the logic behind this qs:

A farmer has decided to build a wire fence along one straight side of his property. For this, he
planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After
he bought the posts and wire, he found that the number of posts he had bought was 5 less than
required. However, he discovered that the number of posts he had bought would be just sufficient if
he spaced them 8 m apart. What is the length of the side of his property and how many posts did he

a. 100 m, 15 b. 100 m, 16 c. 120 m, 15 d. 120 m, 16

Ans:d.120m,16

No. of post X.

(X-1)6+6*5=8(x-1)
For each 6mtr,2 post is required.
Then for x postTotal distance covered=6(x-1).
for another five post=6*5(last post was taken so total post is 6)
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