How Many Weeks Can The Sheep Graze?

Let the initial amount of grass be $x$ . (What? Grass are uncountable!) While, the rate be $r$ . Then, we write the amount of weeks needed for the sheep to finish consuming the grass. We have

$\frac{x+5r}{240}=5$

$\frac{x+7r}{200}=7$

Solving the simultaneous equations above yields to $x=700$ (whoa, you literally count the amount of grass, I salute you) and $r=100$ .

Again, we express the number of weeks, $n$ , using the same style as above,

$\frac{700+100n}{170}=n$

$\implies n=\boxed{10}$

The sheep in the first condition eats $240\times5 = 1200$ grass, while the sheep in the second condition eats $200\times7 = 1400$ grass

The grass in the field grows in $\frac{1400-1200}{7-5} = \frac{200}{2} = 100$ units of grass $/$ week

The initial grass is $1400 - (100\times7) = 700$ units of grass.

Let's say that the 100 sheep eats the new grass, while the other 70 eat the initial grass and it is able to last for $\frac{700}{70} =$ $\boxed{10}$ weeks

Let y be the initial grass in the field, k be the grass grown each week and z be the grass eaten by a sheep every week y + 5k = 240 x 5 x z
y + 7k = 200 x 7 x z
y + nk = 170 x n x z
just solve these three equations and u will get the answer 10.

Let F be the 'amount' of grass in the field at the start, G be the rate at which the grass replenishes per week, T be the time in weeks, and S be the number of sheep.

When the grass is depleted, then

F - ST + GT = 0

For the first example, we can see that

F - (240) (5) + (5) G = 0

F - 1200 + 5G = 0

F + 5G = 1200

Let this be equation (a)

From the second example, we can see that

F - (200) (7) + (7) G = 0

F - 1400 + 7G = 0

F + 7G = 1400

Let this be equation (b)

If we subtract equation (a) from (b) then we are left with

2G = 200

Therefore

G = 100

And so the grass replenishes at a rate of 100 (units) per week.

Now we substitute this into equation (a),

F + 5*(100) = 1200

F + 500 = 1200

F = 700

And so the field starts with 700 (units) of grass.

Now we return to our original equation

F - ST + GT = 0

And use a value of 170 for S

(700) - (170) T + (100) T = 0

700 + 100T = 170T

70T = 700

$\boxed{T = 10 weeks}$

x (initial grass) + 5 weeks * g (growth rate of grass per week) = 240 sheep * 5 weeks * s (consumption rate per sheep per week) - eq 1

x + 7 g = 200 7*s - eq 2

solving eq 1 and 2 gets: x=700s and g=100s

x + m weeks * g = 170 sheep * m weeks * s - eq 3

substituting x and g gets: m = 10 weeks

initial amt. of grass=a. let growth rate of grass =x grass/week we have : a+5x=240x5=1200 a+7x=200x7=1400 therefore, x=100 and a=700. for 170 sheep, 700+100w=170w or w=700/70=10! (the sheep can be treated as the rate at which grass reduces per week!)

240 sheeps survived for 12 weeks it means total grass available is 240 12=1200 200 sheeps survived for 7 weeks it means total grass available is 200 7=1400 by looking at total available grass its clear that 200 grass is added in 2 additional week (7 - 5) so growth rate of grass is 100 per week. No if we want survival for 170 sheeps for n weeks then available grass should de 170 n=170n, then we know that in 7 weeks 1400 is available grass , hence for n weeks available grass would be 1400+(n-7) 100 assuming that after or before 7 weeks multiplier of 100 should be added or subtracted. now comparing the equations: 1400+(n-7)*100=170n then solve it

Let x be the initial amount of grass in the field. Let gz and gr be the grazing rate per sheep and grass growth rate per week. Then

x/(240*gz - gr) = 5......(1)

x/(200*gz-gr) = 7.....(2)

Dividing (1) by (2), we get 5 (240 gz - gr) = 7 ((200 gz-gr)

1200gz - 5gr = 1400*gz - 7gr

2gr = 200*gz

1gr = 100gz.....(3)

Plugging (3) in (1), we get x/(240 gz - 100 gz) = 5

x = 140gz * 5 = 700 * gz

So the number of weeks for 170 sheep to graze the field bare would be :

700gz / (170 gz - 100 gz) = 700gz/70gz = 10

Based on: "True Basic Concept" :

$\Rightarrow$ Initial Grass + Grass Growth - Total Consumption = 0

Let: x = Initial Grass, g = Grass Growth Rate, c = Consumption per Sheep

$x + 5g - (5 \times 240 \times c) = 0$ $\Leftrightarrow$ $x + 5g -1200 \times c = 0$ ........................(I)

$x + 7g - (7 \times 200 \times c) = 0$ $\Leftrightarrow$ $x + 7g - 1400 \times c = 0$ ........................(II)

$x + n \times g - (n \times 170 \times c) = 0$ $\Leftrightarrow$ $x + n \times g - (170 \times n \times c) = 0$ .(III)

I + II

$x + 5 \times g -1200 \times c = x + 7 \times g - 1400 \times c$

$5 \times g -1200 \times c = 7g - 1400 \times c$

$2g = 200c$

$g = 100c$ $\Rightarrow$ grass growth rate = 100c ..........(IV)

I + IV

$x + 5g - 1200 \times c = 0$

$x + 5 \times 100c - 1200c = 0$

$x = 700c$ $\Rightarrow$ initial grass = 700c ..........(V)

III + IV + V

$x + n \times g - n \times 170 \times c = 0$

$700c + n \times 100c - n \times 170 \times c = 0$

$700c - 70nc =0$

$70n =700$

$n = 10$ $\Rightarrow$ the grass field becomes bare by 170 sheeps for $\boxed{10}$ $weeks$

10 weeks

Let the amount of the grass in the field be 'G' sq m to begin with. Assume it grows at the rate of 'x' sq m per week and the sheep grazes at the rate of 's' sq m per day. Then setting up the equations

G -240*5s +5x=0

G-200*7s +7x =0

Solve the two equations by equating the value of G -1200s+5x = -1400 + 7x which gives x = 100s and G = 700s

If the grass lasts n weeks when 170 sheep graze then the equation is G - 170 ns +nx =0, substituting the values 700s -170ns + 100ns=0 700 = 70 n which gives n = 10 weeks

rate of increase of gross per week is x 240 sheep covered an area of 240 5=1200 200 sheep covered an area of 200 7=1400 in 2 weeks rate of increase of gross is 2x then 1200+2x=1400 x=100 now in y weeks rate of increase of gross is 100y 1400+y 100=170 (7+y) y=3 no of weeks to become field will be bear is 7+3=10 weeks

Let initial grass quantity= a, Let grass grows at a constant rate of= y per week, Let each sheep grazes x quantity of grass in 1 week. According to first statement, a+5y=240 5x .....(i) According to second statement, a+7y=200 7x .....(ii) Then, subtracting (ii) from (i), we have: 2y=200x
so, y=100x and a=700x Let the no. of weeks required for 170 sheep to graze the field be z.

Then, a+zy=170 z x .....(iii)

Subtracting (iii) from (ii), we have (7-z)y=1400x--170zx

Put y=100x and solve for z

Let F, c, N, s and T be the amount of grass on the F ield at the present, the C onstant speed at which the grass grows, the N umber of sheep on the field and the S peed at which sheep eat and the T otal time it takes for the field to be bare.

Field is bare when: Amount of Grass = Amount of grass eaten F + T c = N s*T

Field is bare when s=240 and T=5 F + 5c = 240s*5 F + 5c = 1200s ... (1)

Field is bare when s=200 and T=7 F + 7c = 200s*7 F + 7c = 1400s ... (2)

(2) - (1) 2c = 200s c = 100s ... (3)

Substitute (3) into (1) F + 500s = 1200s F = 700s ... (4)

Substitute (3) and (4) into original equation 700s + 100Ts = N s T T(N-100) = 700 ...(5)

Substitute N = 170 into (5) T(170 - 100) = 700 T = 10

It would take 10 weeks for the field to be bare.

This is a very step by step process for those who have trouble.

240=5 weeks 200=7 weeks 170=10 weeks.because of the diifference b/w 5&7 is 2,7&10 is 3.

I=Inicial grass field ,T=Number of weeks,C=Grows Tax rate,S=Number of sheeps R=Grass field

I+TC-S.T=R ....................................................................................................................................................................................................................................... If 240 sheep were to graze on it, it will become bare in 5 weeks........................... I+5C-240.5=0............... I+5C-1200=0 ........................................................................................................................................................................................................................................... If 200 sheep were to graze on it, it will become bare in 7 weeks............................. I+7C-200.7=0............ I+7C-1400=0 ..................................................................................................................................................................................................................................... How many weeks would it take 170 sheep to graze, before the field becomes bare?.............. I+TC-170.T=0.......... I+TC-170T=0 ........................................................................................................................................................................................................................................

I+5C-1200=I+7C-1400............. 5C-1200=7C-1400........... 200=2C............. C=100 .................................................................................................................................................................................................................................... I+500-1200=0.......... I=700 ........................................................................................................................................................................................................................................ 700+100T-170T=0........ 700-70T=0......... T=10 ...............................................................................................................................................................................................................................

let, the amount of grass already existing when the sheep start to graze be y; the rate of growth of the grass be = x units/week; the rate of consumption of grass by the ONE sheep be = z units/week;

Therefore Case 1(240 sheep) The amount of grass existing + the amount of grass grown = amount of grass consumed by 240 sheep.; y + xt = znt where, t is number of weeks(time) and n is number of sheep.; In this case, y + x 5 = z 240*5 => y + 5x = 1200z .............................(1);

Case 2(200 sheep) y + xt = znt y + 7x = z* 200*7 => y + 7x = 1400z...............................(2); By solving Equations (1) and (2) we get: y/7 = x;............(3); y/700 = z;...........(4);

Case 3 t = ?; n = 170; y +xt = znt => y+ (y/7)t = (y/700) 170 t.......................subsituting (3) and (4); => 1+t/7 = (17/70)t => 1 = (7/70)t => t = 10 Hence, ANSWER = 10 weeks.;

suppose the amount of grass at the start is G suppose the amount of grass grown each day is g suppose the amount of grass one sheep eats each day is k

then G+5g = 240k * 5 G+7g = 200k * 7

or dividing both sides by k G/k+5g/k=1200 G/k+7g/k=1400

Solving above equations for G/k and g/k we get g/k = 100 G/K = 700

Hence for given scenario we have 700 + 100 * Weeks = 170 * Weeks So Weeks = 700/70 = 10

let g be the amount of grass (measured in what a sheep eat in the unit time) and s be number of sheep the amount of grass increases regularly per time thus dg = k (where k is constant) thus g = integration of k due to time = kt + c (where c is constant) but what sheep eat is ts then kt + c - ts = 0 thus t(k-s)+c=0 thus t(s-k) = c thus t1(s1-k)=t2(s2-k) = t3(s3-k) thus 5(240-k)=7(200-k)=t3(170-k) thus 1400 - 5k = 1200 - 7k thus 2 k = 200 ==> k = 100 thus 70t3=700 ==> t = 10 weeks