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TANMAY GHOSH
@lurys

Posted 26 Oct '11

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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 p...

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

buy?

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.

TANMAY GHOSH
@lurys

Posted 26 Oct '11

> :: :: ::
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 p...

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

buy?

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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