Difficult Math Problem?

Question

A farmer owned two pieces of land. the first was eight times larger in area then the other, and the two were separated by an access road. When the farmer measured the perimeters of both sections of land, she was surprised to find that it was going to take twice as much fencing to surround the smaller area than it was to surround the larger! Both sections of land were rectangular in shape and the fences were continuous. (the gates should be counted as part of the fence). How much fencing did she need to buy?

Answer 1

There could be several answers for this question. However, if you imagine (a) and (b), the sides of the larger land, are the same, you can get what the proportion of (c) and (d), the sides of smaller land.
To make solving this problem easier for yourself, just imagine a=1 and b=1. It does not mean that a and b are really equal to 1. you can change them finally to everything you want. Just they have to be equal together.
ab=8cd & c+d=2(a+b) So 8cd=1 & c+d=4 → d=4-c → 8c(4-c)=1 → 8c^2 - 32c + 1 =0 → c=[32 ± √¯¯[(32^2) - 4*8*1] ] ÷ (2*8) → c=(32±√¯¯992) ÷ 16 → c=(32±4√¯¯62) ÷ 16 → c=2 ± (√¯¯62)/4
If c= 2 + (√¯¯62)/4, then d=4-c= 2 - (√¯¯62)/4.
If c= 2 - (√¯¯62)/4, then d=4-c= 2 + (√¯¯62)/4.
So if c be equale to every of the couple of numbers, then d would be the other one.
Now you can make a,b,c and d larger as much as you want, but they should be got larger equally. for example if you make all of them 100 times larger, then a=100, b=100, c=200 - 25√¯¯62 and d=200 + 25√¯¯62 .
Now you can test if ab=8cd & c+d=2(a+b) or not. the answer is: Yes they are.

Answer 2

This question is very hard but here's my solution:

First rectangle is 63 by 64
Area = 4032
Perimeter = 254

Second rectangle is 252 by 2
Area = 504 which is 1/8th of 4032
Perimeter = 508 which is 2 times 254

Hooray me!


** Turns out alot of other solutions exist - I used a computer program to solve

SOLUTION: 24 47 1 141
SOLUTION: 47 24 1 141
SOLUTION: 20 78 1 195
SOLUTION: 78 20 1 195
SOLUTION: 18 140 1 315
SOLUTION: 140 18 1 315
SOLUTION: 17 264 1 561
SOLUTION: 264 17 1 561
SOLUTION: 63 64 2 252
SOLUTION: 64 63 2 252
SOLUTION: 48 94 2 282
SOLUTION: 94 48 2 282
SOLUTION: 40 156 2 390
SOLUTION: 156 40 2 390
SOLUTION: 36 280 2 630
SOLUTION: 280 36 2 630
SOLUTION: 84 110 3 385
SOLUTION: 110 84 3 385
SOLUTION: 79 120 3 395
SOLUTION: 120 79 3 395
SOLUTION: 72 141 3 423
SOLUTION: 141 72 3 423
SOLUTION: 66 172 3 473
SOLUTION: 172 66 3 473
SOLUTION: 60 234 3 585
SOLUTION: 234 60 3 585
SOLUTION: 57 296 3 703
SOLUTION: 296 57 3 703
SOLUTION: 56 327 3 763
SOLUTION: 327 56 3 763
SOLUTION: 54 420 3 945
SOLUTION: 420 54 3 945
SOLUTION: 126 128 4 504
SOLUTION: 128 126 4 504
SOLUTION: 96 188 4 564
SOLUTION: 188 96 4 564
SOLUTION: 95 192 4 570
SOLUTION: 192 95 4 570
SOLUTION: 80 312 4 780
SOLUTION: 312 80 4 780
SOLUTION: 142 180 5 639
SOLUTION: 180 142 5 639
SOLUTION: 130 204 5 663
SOLUTION: 204 130 5 663
SOLUTION: 120 235 5 705
SOLUTION: 235 120 5 705
SOLUTION: 111 280 5 777
SOLUTION: 280 111 5 777
SOLUTION: 105 328 5 861
SOLUTION: 328 105 5 861
SOLUTION: 100 390 5 975
SOLUTION: 390 100 5 975
SOLUTION: 189 192 6 756
SOLUTION: 192 189 6 756
SOLUTION: 168 220 6 770
SOLUTION: 220 168 6 770
SOLUTION: 158 240 6 790
SOLUTION: 240 158 6 790
SOLUTION: 144 282 6 846
SOLUTION: 282 144 6 846
SOLUTION: 132 344 6 946
SOLUTION: 344 132 6 946
SOLUTION: 210 236 7 885
SOLUTION: 236 210 7 885
SOLUTION: 174 308 7 957
SOLUTION: 308 174 7 957
SOLUTION: 168 329 7 987
SOLUTION: 329 168 7 987

enjoy

Answer 3

Let the sides of the larger piece be R and S and of the smaller X and Y. P and Q are perimeters of each. A and B are the areas.
P = 2(R + S)
A = RS

Q = 2(X + Y)
B = XY

A = 8B ==> RS = 8XY
2P = Q ==> 4(R + S) = 2(X + Y)

Answer 4

Not enough information. You could guess that the larger piece is a square, and the smaller is a long thin rectangle, but you can't go any further than that.

Answer 5

These first two answers are correct. You've got 4 unknowns; the lengths and breadths of each paddock. But you've only got two bits of information to solve them. You need at least four clues.

Answer 6

large plot denoted by l , length l1, breadth b1
smaller plot by s,length l2, breadth b2

area of l = 8 area of s
l1 *b1 = 8 (l2 * b2)-----------@

perimeter of s = 2 perimeter of l
(l2+b2) = 2 (l1+b1) -----------@

so far so good ... u 've asked how much fencing she needed to buy and the fencing is continuous...so that would be around

like 3 times the perimeter of the larger plot....= 3(l1+b1)

Answer 7

larger land
8(d) x 8(d) = 64d = area
2[8(d) + 8(d)] = 32d + 2gates = perimeter

smaller land
32(d) x .25(d) = 8d = area
2[32(d)] + .25(d)] = 64.5d = perimeter

.5d = 4gates
d = 8gates

Answer 8

If you would specify one more piece of information, like the size of one of the plots, then this problem would have a definite solution.

Answer 9

THIS IS ONE OF THOSE TRICK PROBLEMS. THE WORD PROBLEM GIVES YOU INFO THAT YOU DONT NEED AND DOESN'T GIVE YOU THE INFO YOU DO NEED. THE ANSWER TO THIS PROBLEM IS SIMPLY "CAN NOT ANSWER TO LACK IF INFO".

THE PROBLEM DID NOT GIVE YOU ANY NUMERIC INFO. THERE FORE THERE IS NO POSSIBLE ANSWER!!!

Answer 10

yeah right