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in a pool there is a certain amount of water ( not known) . added to the pool 160 L of water, after that reduced the water amount by 10%, and finaly took out of the pool 195 L of water. at the end of the process the pool remained with 225 L of water.

1) what was the amount of water in the pool before the process?

2) did at the end of the process the water amount in the pool reduced or rised in %? if so then in how much precent %?

 

 

pls help 

 Apr 22, 2014

Best Answer 

 #1
avatar+33661 
+10

Start by letting the original volume of water in the pool be, say, w litres. 

First step is to add 160 to this: w+160

Reducing the result by 10% is to be left with 90%: (w+160)*0.9      (0.9 is 90% as a fraction).

Take out 195 litres:  (w+160)*0.9 - 195, and this equals 225, so:

(w+160)*0.9 - 195 = 225

Add 195 to both sides:  (w+160)*0.9 = 420

Multiply both sides by 10/9:  w+160 = 4200/9 = 466.67  (actually 466 and two-thirds)

Subtract 160 from both sides to get: w = 306.67

 Apr 22, 2014
 #1
avatar+33661 
+10
Best Answer

Start by letting the original volume of water in the pool be, say, w litres. 

First step is to add 160 to this: w+160

Reducing the result by 10% is to be left with 90%: (w+160)*0.9      (0.9 is 90% as a fraction).

Take out 195 litres:  (w+160)*0.9 - 195, and this equals 225, so:

(w+160)*0.9 - 195 = 225

Add 195 to both sides:  (w+160)*0.9 = 420

Multiply both sides by 10/9:  w+160 = 4200/9 = 466.67  (actually 466 and two-thirds)

Subtract 160 from both sides to get: w = 306.67

Alan Apr 22, 2014
 #2
avatar+118677 
+5

i admit i haven't the time to look properly but I doubt that there is anything wrong with Alan's answer so I am ticking it.

 Apr 22, 2014
 #3
avatar+103 
0

i understand everything in your answer alan but all untill the part that you say to multiplay it all by 10/9, why 10/9 if its 0.9? is it because i switch the sides of the 0.9?

 Apr 22, 2014
 #4
avatar+33661 
0

"i understand everything in your answer alan but all untill the part that you say to multiplay it all by 10/9, why 10/9 if its 0.9? is it because i switch the sides of the 0.9?"

 

Sorry, I should have explained more carefully!  0.9 is the same as 9/10, so by multiplying by 10/9 we eliminate it from the left-hand side of the equation.  Perhaps I should have simply said divide both sides by 0.9.

 Apr 22, 2014
 #5
avatar+103 
0
ok thanks alot but just one question, well the the second question of the assigment "by how many % did the water in the pool increased or decreased" im really sorry i cant do it alone im just not the best in math and i struglle a bit sometimes 
 Apr 22, 2014
 #6
avatar+33661 
+5

For the second part:

At the end there is 225 L  (the value given)

At the start there is 306.67 L (the value of w that we've just worked out)

So we end up with less than we started with.  

Express end as a percentage of start by dividing 225 by 306.67 and multiplying by 100:

$$\left({\frac{{\mathtt{225}}}{{\mathtt{306}}}}\right){\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{73.529\: \!411\: \!764\: \!705\: \!882\: \!4}}$$

At the end we have about 73.5% of what we started with.  Therefore we have a reduction of (100-73.5)% or 26.5%

 Apr 22, 2014
 #7
avatar+103 
0

ty so much!

 Apr 22, 2014

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