half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby . the rest 9 are drinking water from the pond . find the number of deer in the herd ?

rosala
May 25, 2014

#1**+8 **

OK, Rosala...let's examine these statements one-by-one and find a solution.

First. let the total number of deer = x

And half of them are grazing, so that just must be !/2 of x = (1/2)x

And "the remainder" must just be the other half = (1/2)x

And (3/4) of the remainder - (1/2)x - are playing..... "of" usually means "multiply"...so we have..... (3/4) x (1/2)x = (3/8)x ..

....and 9 others are drinking

So we have

(1/2)x + (3/4)(1/2)x + 9 = x

(1/2)x + (3/8)x + 9 = x ........ Note that..... (1/2)x = (4/8)x

(4/8)x + (3/8)x + 9 = x Add the fractions

(7/8)x + 9 = x Subtract (7/8)x from both sides

9 = x - (7/8)x Note that..... x = 1x = (8/8)x

9 = (8/8)x - (7/8)x

9 = (1/8)x Multiply by the reciprocal of (1/8) = (8/1) on both sides

(8/1)*9 = (8/1)* (1/8)x

72 = x

And there's our answer.....72 deer

CPhill
May 25, 2014

#2**+8 **

$$\frac{1}{2}+\frac{3}{8}=\frac{4}{8}+\frac{3}{8}=\frac{7}{8}$$

So the 9 deer must be $$1-\frac{7}{8}=\frac{1}{8}$$ of the total

If 1/8 of the herd is 9 then ALL of the herd must be 9*8= 72 deer.

Does that make sense?

Melody
May 25, 2014

#4**0 **

Thanks Chris,

I am eager to see what rosala thinks. She seems to have a bit of problem with understanding algebra.

Rosala, I think I just noticed you on the forum, algebra is very important so, each time you are given an algebraic solution try your best to understand. Maybe at some point the meaning will just jump out at you!

Maths really does work that way. Understanding often happens in steps.

Melody
May 25, 2014

#5**+10 **

thank u CPhill and melody for ur great answers ! CPhill after reading ur answer very carefully i could easily understand it , u ver right melody . CPhill i just wanted to say one thing that in this part 9=(1/8)x , i couldnt understand the method as i have learnt the method of cross multiplication so this method just popped up in my head and so i did it and the answer came correct and i could understand ! thank u melody for taking the initiative to make me understand and give another method ! thumbs up for bothof u ! thanks !

rosala
May 26, 2014

#8**+5 **

Hi Rosala, you called what was done 'cross multiply'.

I know this is a common term but I don't like it and I noticed that CPhill did not use it either.

Equations must be 'balanced' just like old fashioned scales had to be balanced.

If it is balanced - that means the 2 sides are equal - then if you do something to one side you must do exactly the same thing to the other side as well. Otherwise the 2 sides will not be balanced anymore!

Now let us look at this equation

$$9=\frac{1}{8}\times x$$

I need to get the $$x$$ on its own so I need to get rid of the 8.

I can do this by multiplying the RHS by 8 BUT if I multiply the right side by 8 I must also multiply the left side by 8.

$$\begin{array}{rll}

8\times 9&=&8 \times\frac{1}{8}\times x\\\\

8\times 9&=& \frac{8}{1}\times\frac{1}{8}\times x\\\\

\end{array}$$

The 8 on the top cancels with the 8 on the bottom leaving us with

$$\begin{array}{rll}

72&=&1x\\\\

x&=&72

\end{array}$$

--------------------------------------------------------------

Or ALTERNATIVELY

$$\begin{array}{rll}

9&=&\frac{1}{8}\times x\\\\

9&=&\frac{x}{8}\\\\

8\times9&=&\frac{8\times x}{8}\\\\

72&=&\frac{x}{1}\\\\

x&=&72

\end{array}$$

I've put in a couple more steps than i normally would just so that you can see what is happening more clearly.

Melody
May 26, 2014