Question

Alex and his sister Gina were asked to mow the lawn. Each of them are to mow half of the lawn. The lawn is rectangular in shape and measure 120'x58'. If Alex started in one corner and went around the edges of the lawn how many times would he have to mow around the edges to mow to his half.

Answer 1

i dont think its stated well

Answer 2

depends on the width of the mower

Answer 3

the last sentence, makes no sense

Answer 4

no but the way I am thinking about it is they have a big rectangle representing the whole thing , then you need to calculate the width from the edge he must mow, to split it into two equal sections

Answer 5

Let the mower width be x.

Answer 6

oh Im so sorry the mower cuts a 2foot wide path

Answer 7

ohh well that makes it bout 1000times easier
:P

Answer 8

a 24inch cut eh .. is it a john deere ?

Answer 9

do we make a perfect cut so that each pass is exactly on the cut line of the one before? or do we have to 1/4 step it over to make sure we get it all?

Answer 10

how sharp is the blade, cause i always get peices left over that defy all cutting conventions :)

Answer 11

@amistre64;
i dont know the question did not say

Answer 12

There is 6960 square feet for the entire lawn so Alex has 1/2 of that to mow or 3480 sq ft. First orbit would be 356x square feet. If the mower is 9.8 feet wide that would do it lol.

Answer 13

Oh, it is 2 ft. wide That makes the problem interesting as each orbit is smaller. So much for the 9 ft mower!

Answer 14

3480 is the area we wanna cut; each pass on a perimeter takes off l*2'

2(120-0)+ 2(58-2)
+ 2(120-2) + 2(58-4)
+ 2(120-4) + 2(58-6)
+ 2(120-6) + 2(58-8) ..etc

Answer 15

2(178) - 2(1)
+ 2(178) - 2(3)
+ 2(178) - 2(5)
+ 2(178) - 2(7) ......

Answer 16

356n - 2(1+3+5+7+9+11+13 ...)

Answer 17

B{n} = 356n + 2(2n -1) ... right?

Answer 18

+2 means -2 :)

Answer 19

@mrain do u know the answer ? my answer is way to big.......

Answer 20

is it something abt 387

Answer 21

Looks like he would do more than his half on the completion of his 6th orbit.

Answer 22

myne answer for the orbit was abt something 9.87

Answer 23

i do not know the answer, very sorry, wish i did

Answer 24

whats the area for the first orbit

Answer 25

@radar; thats what i got

Answer 26

myne was 384= 120*2 + (58-4)*2

Answer 27

it would take his about 5.5 orbits to equal almost half but it wont equal the exact half, thats what i am understanding

Answer 28

i used AP........ 384 is the first term and every time area gets reduced by 16 then by formula .... 3840 = 384 +(n-1)(-16)...where n should be number of trip.....pls do tell me if i am wrong....

Answer 29

sounds right but i truly dont know what is actually right

Answer 30

i dont think that works because you come out to a negative

Answer 31

352 + (352-2n)(n-1)

352-2n
n-1
--------
352n -2n^2
-352 +2n
------------
354n -2n^2

that should do it :)

2380 = 354n - 2n^2

2n^2 -354n +2380 = 0

354/4 +- sqrt(354^2 -4(2380)(2))/4

354/4 +- sqrt(106276)/4

354 + 326
---------- = 680/4 = 170
4

354 - 326
---------- = 28/4 = 7 ... id say 7 passes
4

Answer 32

23
352
+348
+344
+340
+336
+332
+328
-----
2380 ; 7 passes

Answer 33

whats 352?

Answer 34

2(120-0+58-2) = 352 is the area of going half way around

Answer 35

2(120-2+58-4) = 2(120+58-6)=344 .... bummer forgot how to add :)

Answer 36

dont u think it should be -4 not -2

Answer 37

area cut each time cumulative area
--------------- -------------
2(120-0+58-2) = 352, 352
2(120-2+58-4) = 344, 696
2(120-4+58-6) = 336, 1032
2(120-6+58-8) = 328, 1360
2(120-8+58-10) = 320, 1680
2(120-10+58-12) = 312, 1992
2(120-12+58-14) = 304, 2296
....................... = 296, 2592

8 times

Answer 38

8 times is really twice as much since i am counting 2 times per cycle

4 times around a complete perimeter?

Answer 39

i'm saying length should be deducted by -4 after a complete cycle ....don't u think so?

Answer 40

Area1 = 120*2 + 56*2 + 118*2 + 54*2
= 2(120+56+118+54)
= 2(348)
= 696

Area2 = 2(116+52+114+50)
= 2(332)
= 664

Area3 = 2(112+48+110+46)
= 2(316)
= 632

Total area after 3 passes around = 1992

108(2) = 216; +1992 = 2208
44(2) = 88; +2208 = 2296
42 more feet gives 84, 2296+84 = 2380 :)

Answer 41

area is reduced by 32 each pass

Answer 42

-2 is for one side ..hence, -4 for both sides.......as it must traverse both parallel sides of the lawn for a cycle...and 2ft is the movers length..

Answer 43

ackk ... reduced by 16 i mean ;)

Answer 44

ok i get it ... i was counting for only one side ....thanks amistre..:)

Answer 45

so the first area should be 2(120*2 +(58-4)*2)= 348*2

Answer 46

am i right?

Answer 47

that is correct; you formed it differently, but its still correct :)

Answer 48

now how do we determine what it means by "how many times to mow around the edges"?

Answer 49

does one side = 1 pass? or do 4 sides = 1pass?

Answer 50

the second one

Answer 51

corresponding areas 696 664 632 600 568 536....it makes 6 passes

Answer 52

(600)4 = 2400 which is already too much :)

Answer 53

or is that in total yard area?

Answer 54

area total equals 3696 while we need 3480 so divide the last area by 4 as there are 4 sides ... we get 134 and it makes 5.5 passes..........5*4 + .5*4 = 22 corners

Answer 55

mrain what do u think about the solution

Answer 56

i agree thats what i got but i wasn't sure if it was correct

Answer 57

...... i got my numbers for area convoluted ... :) the good thing about my age is in 3 days ill have forgotten all about it

Answer 58

what grade level would you guess this question is?