Blow wind blow

This problem is best solved backwards. Let's start with the empty tree. Before it was empty, half of the leaves, and then ten more, fell. If this removes all of the leaves, then there were twenty leaves on the tree before the third gust.

To find out how many leaves were on the tree before the second gust, we work backwards again. Adding ten more leaves, we get thirty leaves. Double that to get sixty leaves. Only one gust to go!

Add ten leaves to get seventy, and then double that to get the answer, one hundred and fourty.

N is the amount of leaves at start. After the first gust of wind f leaves remain. N - blown leaves. $f=N - \frac{N}{2}-10$ $f=\frac{N}{2} - 10\hspace{2mm}(1)$ After the second gust of wind s leaves remain. f - blown leaves (2nd round). $s=f - \frac{f}{2}-10$ $s=\frac{f}{2} - 10\hspace{2mm}(2)$ After the third gust of wind 0 leaves remain. s - blown leaves (3rd round). $0=s - \frac{s}{2}-10$ $0=\frac{s}{2} - 10\hspace{2mm}(3)$ Now we work backwards rearranging equation 3 to find s. $s=20$ Rearrange equation 2 using the fact that s=20. $20=\frac{f}{2}-10$ $f=60$ Rearrange equation 1 using the fact that f=60. $60=\frac{N}{2}-10$ $N=140$ The number of leaves on the tree before the first gust of wind is therefore 140.

Let the number of leaves on the tree after the $n$ 'th wind be $l(n)$ . We known, that half the leaves will fall plus an additional 10. That means that our equation becomes $l(n+1)=\frac{1}{2} l(n)-10$ . This is a first order inhomogenous differential equation , and it can be shown that a solution to a such equation on the form $x(n+1)-ax(n)=c$ is given by: $x(n)=a^n x(0) + c \displaystyle\sum^{n-1}_{k=0} a^k$ .

In this case we get:

$l(3)=\frac{1}{2} ^3 x(0)-10(\frac{1}{2} ^0 +\frac{1}{2} ^1 +\frac{1}{2} ^2)=0$ ,

and solving this for $l(0)$ , we get $l(n)=140$

Let $x$ be the initial number of leaves. Then, after the first gust of wind, there remain $\frac{x}{2} - 10$ leaves. After the second gust of wind, there remain $\frac{\frac{x}{2} - 10}{2} - 10$ leaves. After the third gust of wind, there remain $\frac{\frac{\frac{x}{2} - 10}{2} - 10}{2} - 10 = 0$ leaves. We solve for x, the initial number of leaves: $x = 2\left(2\left(2\left(10\right) + 10\right) + 10\right) = 140.$

At the third gust of wind, half of the leaves (that remained) fell along with an additional 10. With this, there remain no more leaves in the tree. Let the half remained before the third gust be x. This implies, x-10=0(since, all the leaves have fallen now.) This implies, x=10. Thus, the total number of leaves that remained before the third gust is 10+10=20. This can also be taken in another way round. After the third gust let x leaves fall and the additional 10 leaves be left on the tree. Since x is just the half of the remainder, total no. of leaves must be 10(as exactly half i.e. 10 leaves are left behind.)

Now, before the second gust let, y be the no. of leaves remaining in the tree. This implies y/2 plus an extra 10 leaves have fallen after second gust. Also, assuming the additional 10 leaves to be in the tree we have exactly 10+20=30 leaves in the tree and 30 which have fallen(as only half the remainder have fallen). Thus total no. of leaves in the tree before the second gust is 30+30=60.

Now, let the total no. of leaves in the tree before the first gust be z. Therefore, z/2+10 leaves have fallen after the first gust. Assuming the additional 10 leaves to be in the tree then we have 60+10=70 leaves in the tree after the first gust and exactly 70 have fallen after the first gust. This, implies, total no. of leaves in the tree before the first gust was 70+70=140.

The easiest way is to start with zero and {add ten and multiply by two} three times, giving us 0, then 20, then 60, then 140.

However, I also explored this in an algebraic way as good practice, because I'm very bad at it. Let's call the starting number of leaves on the tree x , the remaining leaves after 1 gust a , the remaining leaves after 2 gusts b , and the remaining leaves after 3 gusts c . We can substitute c with 0, as we know that that is how many leaves are left after gust 3, according to the question itself. We can also form three equations form the question, which are x/2 - 10 = a ; a/2 - 10 = b ; and b/2 - 10 = c = 0 . In order to form one equation for the whole question, we place the three equations together, so that (x/2 - 10) / 2 - 10 = a/2 - 10, which in turn equals b . Furthermore, ((x/2 - 10) / 2 - 10) / 2 - 10 = b/2 - 10 = c , which in turn equals 0. So, re-written as ((x/2 - 10) / 2 - 10) / 2 - 10 = 0, we can solve for x by {adding ten and multiplying by 10} three times .

let x = number of leaves in the tree.. y = number of leaves after the first blow.. z = number of leaves after the second blow

after the first wind's blow,

x - [ $\frac{1}{2}$ x + 10 ] = y => x - $\frac{1}{2}$ x - 10 = y

after the second wind's blow,

y - [ $\frac{1}{2}$ y + 10 ] = z => y - $\frac{1}{2}$ y - 10 = z

after the third wind's blow,

z - [ $\frac{1}{2}$ z + 10 ] = 0 => z - $\frac{1}{2}$ z - 10 = 0

we can use the backward solution here..

so,

                z =  \\\( \\frac\{1\}\{2\} \\\)z + 10;

=> $\frac{1}{2}$ z = 10; => z = 20

and

                 y =  \\\( \\frac\{1\}\{2\} \\\)y + 10 + (20);

=> $\frac{1}{2}$ y = 30; => y = 60

and

                 x =  \\\( \\frac\{1\}\{2\} \\\)x + 10 + (60);

=> $\frac{1}{2}$ x = 70; => x = 140

Let x = initial number of leaves

At first wind blows the remaining number of

leaves = x-20 / 2

At the second blow the number of leaves

remaining = x-60 / 4

At the third blow where there were no more

leaves left = x-200+60= 0

Therefore x = 140

let the no. of leaves be 'x'. When the first gust of wind blows, 1/2x+10 are fall. remaining leaves are 1/2x-10. When the second gust of wind blows, 1/2(1/2x-10)+10 are fall after the second gust. remaining leaves after second gust are 1/4x-15. When the third gust of wind blows, 1/2(1/4x-15)+10 are fall after third gust. remaining leaves are 1/8x-35/2 . remaining leaves = 0 (given). 1/8x-35/2=0. 1/8x=35/2. x=140.........

Assume total no. of leaves on tree = x

   No. of leaves on trees              No.of leaves fallen

After 1st gust of wind (x/2)-10 (x/2)+10

After 2nd gust of wind (x/4)-15 (x/4)+5

After 3rd gust of wind (x/8)-(35/2)
(x/8)+(35/2)

Now there are two ways available

M1: After 3 rd gust of wind there are no leaves on tree so (x/8)-(35/2)=0
x=140

M2: Take the summation of leaves fallen which is equal to the total no. of leaves on tree
so (x/2)+10+(x/4)+5+(x/8)+(5/2)=x
x=140

Take n as the number of leaves . after the first wind leaves remains = n - (n/2 +10) = n/2 - 10 leaves falling after second wind = n/4 - 5 + 10 = n/4 +5 after second wind , leaves remaining = (n/2-10) - (n/4 +5) = n/4 - 15 Leaves falling after third wind = n/8 - 15/2 + 10 = n/8 + 5/2 leaves remaining after third wind = (n/4-15) -(n/8 + 5/2) = n/8 - 35/2 Set this to 0 , n/8 - 35/2 = 0 so that n = 140

$x_1 = \frac{1}{2} x_0 - 10 \\ x_2 = \frac{1}{2} x_1 - 10 \\ x_3 = \frac{1}{2} x_2 - 10 \\$ Therefore, $x_0 = 2 (x_1 + 10) \\ x_1 = 2 (x_2 + 10) \\ x_2 = 2 (x_3 + 10) \\$ Therefore, $x_3 = 0 \\ x_2 = 20 \\ x_1 = 60 \\ x_0 = 140$

Let the total number of leaves on the tree be x. Then, after the first gust of wind, the number of leaves remaining is x/2 -10.

After the second gust of wind, the number of leaves remaining is (x/2 - 10)/2 -10 = x/4 - 5 -10 = x/4 -15.

After the third gust, the number of leaves remaining is (x/4 - 15)/2 -10 = x/8 - 15/2 -10.

According to the problem, x/8 - 15/2 - 10 = 0.

=> x/8 = 10 + 15/2 = 35/2

=> x = (35X8)/2 = 35X4 = 140.

If we draw the situation of tree, we can get the last part of tree leafs is 10.

the last-first wind falling 20 leafs down (a part of tree and +10 [last 10 leafs])

the last-second wind falling 50 leafs down (a part of tree[20 leafs] + 10 leafs before)

the last-third wind (The first wind) falling 70 leafs down (a part of tree [without plus 10 leafs])

and the total of leaf is 20 + 50 + 70 = 140

let us assume total no of leaves in tree=x. For first gust:x-(x/2)-10=y So no of leaves remain on tree after first gust=y. Repeat the procedure for second gust:y-(y/2)-10=z. So no of leaves remain on tree after first gust=z. Repeat the procedure for third gust:z-(z/2)-10=0{given}. So find z,y,x

Lets say the total is x

1st fall = $\frac {1}{2} x + 10$ , and the remain was $\frac {1}{2} x - 10$

2nd fall is half and -10 from remain so it will be = $\frac {1}{2} ( \frac {1}{2} x - 10) - 10$

3rd will be the half and -10 from that = $\frac {1}{2} ( \frac {1}{2} ( \frac {1}{2} x - 10) - 10) - 10 = 0$

just do the math and it will get 140 ^^