Carl decided to fence in his rectangular garden. He bought 20 fence po

Question

Carl decided to fence in his rectangular garden. He bought 20 fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly 4 yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?

(A) 256

(B) 336

(C) 384

(D) 448

(E) 512

(A) 256

(B) 336

(C) 384

(D) 448

(E) 512

PKN · Answer

Let the no of fence posts be 'x' and 'y' at one of the longer side and at one of the shorter side of the garden respectively.

The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners.
So, x+2=2(y+2)
Or, x-2y=2----------(1)

Also, total no of fence posts=20
So, 2x+2y+4=20
Or, 2x+2y=16-----------(2)
From, (1) & (2), we have x=6 and y=2
Now, length of the longer side=no of spans * gap between adjacent posts=7*4=28
Similarly. length of the shorter side=3*4=12
hence, area of the garden=28*12=336 sq. yards

Ans.(B)

garcmillan · Answer

I think this is one of the problems where doing algebra may be quite painful. Just understanding the problem and coming up with the algebraic expressions would consume more than 2 minutes.

So, what I did was to understand the problem and then get the numbers of posts in each side by applying trial and error

Since I knew one side had the double number of posts than the other side (including the corners), I started trying the next couple of numbers

2 and 4 --> I thought about this one, but I didn't even try it because clearly they don't add up 20

3 and 6 --> This would make 6*2+3*2-4=14 (I deducted 4 because the two first terms of the expression are counting the 4 corners twice so I need to deduct all the corners once --- This is similar to what we do in combinations problems!!)

4 and 8 --> This would make 4*2+8*2-4=20 BINGO!

So i know that short sides have 4 posts (including the corners) and long sides have 8 posts (including the corners)

Note that to calculate the total distance of a side in "questions of fences" when we know the spearation of each fence is always the same thing.

Distance always equals the distance between each post * number of posts (including the corners)-1 (this -1 is because we have always one space less than number of posts. i.e, 3 posts would be 2 spaces, 5 posts would be 4 spaces)

Lenght of short side (4-1)*4=12

Lenght of long side (8-1)*4=28

Area=12*28=280+2*28 ≈330 (Note I don't calculate the full multiplcation. Instead I approximate since answers are well spread out)

OPTION B

Takeaways:

(1) In foreseen complex algrebaric expressions, don't even try to come up with them. Choosing 3 groups of numbers in this kind of questions is much faster than coming up with the equations, solving them and still be sure that you got the right numbers

(2) Don't calculate final numbers (even when the multiplications is not so difficult) if you can approximate and save 2 seconds (every second counts!). In these questions if answer choices would have been 333, 334, 335, 336 and 337. Ok, fair enough, do the whole multiplication, but in all quant questions you should spend 2 sec in looking at the answer choices (after reading the question) to decide if they are sufficiently spread out and therefore to eventually do an approximation when you have come up with the equation that gives the final result

Remember all the takeaways that you get from all the quant questions and don't forget to apply them whenever it's possible so you can get used to them and apply them automatically in the exam day

Please hit the Kudos button if you liked my approach and reflection

Abhishek009 · Answer

Good Question, the question tests us the arrangement of 16 posts after arrangement of the 4 posts... IMG_20190816_200933.jpg We must keep the conditions in mind - 1. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. 2. After placing the 4 posts there are 16 posts 3. Distance between each post is 4 yards Now, afrom the above figure breeadth will be 4* 3 = 12 and Length will be 7*4 =28 Thus, Total area will be 28*12 = 336 , Answer must be (B)

Kinshook · Answer

Given:  Carl decided to fence in his rectangular garden. He bought 20 fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly 4 yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners.

Asked:  What is the area, in square yards, of Carl’s garden?

Let long side has l posts, excluding corners
and Let short side has s posts, excluding corners

He bought 20 fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly 4 yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners.

l +2 = 2 (s +2) = 2s + 4
l = 2s + 2

2l + 2s + 4 = 20
l + s = 8
3s + 2 = 8
s =2
l = 6
There are (l+1) and (s+1) 4 yards spaces on longer and shorter sides respectively

The area, in square yards, of Carl’s garden = (6+1)(2+1)*4*4 = 21 * 16 = 336 square yards

IMO B

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Carl decided to fence in his rectangular garden. He bought 20 fence po

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Carl decided to fence in his rectangular garden. He bought 20 fence po

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