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by sanju09 » Sat Jul 24, 2010 2:08 am
A straight fence is to be constructed from posts 6 inches wide and separated by lengths of chain 5 feet long. If a certain fence begins and ends with a post, which of the following could not be the length of the fence in feet? (12 inches = 1 foot)
A. 17
B. 28
C. 35
D. 39
E. 50

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by Rahul@gurome » Sat Jul 24, 2010 3:23 am
Let the no. of posts used in making the fence be x and the no. of chains used be y.
Or length l of the fence in feet is( 6/12)x+5y which is (1/2)x+5y.
It is to be noted here that x and y are integers, x is greater than 1 and x has to be 1 more than y since each chain has two posts at its ends.
If x is 4 and y is 3 then l is (1/2)*4+5*3 = 17.
If x is 6 and y is 5 then l is (1/2)*6+5*5 = 28.
If x is 10 and y is 6 then l is (1/2)*10+5*6 = 35.
If x is 8 and y is 7 then l is (1/2)*8 + 5*7 = 39.
If x is 10 and y is 9 then l is (1/2)*10 + 5*9 = 50

Out of the above five choices the difference between x and y is not 1 only for l = 35.
Hence this value is rejected.

The correct answer is (C).
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by bendaniel » Sat Jul 24, 2010 4:39 am
Dear Rahul,
thanks for the xplanation. But , would this method not consume a lot of time?.. as in assuming the perfect values to fit in the derived equation?..... Is there any other method available?? ... Thank U.
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by sanju09 » Sat Jul 24, 2010 4:59 am
bendaniel wrote:Dear Rahul,
thanks for the xplanation. But , would this method not consume a lot of time?.. as in assuming the perfect values to fit in the derived equation?..... Is there any other method available?? ... Thank U.

It's true that if the fence includes x chains (a total contribution of 5 x feet to the fence), then it must include x + 1 posts {a total contribution of ½ (x + 1) feet to the fence} to complete it. The total length of the fence would then be

5 x + ½ (x + 1) feet = (11 x + 1)/2

In other words

Double the possible length of the fence must be 1 more than a multiple of 11. Now, give a quick check...
A. 17 × 2 is 1 more than a multiple of 11
B. 28 × 2 is 1 more than a multiple of 11
C. 35 × 2 is not 1 more than a multiple of 11
D. 39
E. 50
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by getso » Sat Jul 24, 2010 6:00 am
sanju09 wrote:
bendaniel wrote:Dear Rahul,
thanks for the xplanation. But , would this method not consume a lot of time?.. as in assuming the perfect values to fit in the derived equation?..... Is there any other method available?? ... Thank U.

It's true that if the fence includes x chains (a total contribution of 5 x feet to the fence), then it must include x + 1 posts {a total contribution of ½ (x + 1) feet to the fence} to complete it. The total length of the fence would then be

5 x + ½ (x + 1) feet = (11 x + 1)/2

In other words

Double the possible length of the fence must be 1 more than a multiple of 11. Now, give a quick check...
A. 17 × 2 is 1 more than a multiple of 11
B. 28 × 2 is 1 more than a multiple of 11
C. 35 × 2 is not 1 more than a multiple of 11
D. 39
E. 50
Great Explanation!!!
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