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a car consumed in its first ride 25% of her fuel. on the second ride she consumed 20% fuel of the remaining fuel in the car ( after using 25%) after the 2 rides the remaining fuel amount of was bigger by 8 than the amount of fuel used for both rides together.

what L of fuel the car had at the begining ? (before the rides)

 Apr 23, 2014

Best Answer 

 #1
avatar+128079 
+5

a car consumed in its first ride 25% of her fuel. on the second ride she consumed 20% fuel of the remaining fuel in the car ( after using 25%) after the 2 rides the remaining fuel amount of was bigger by 8 than the amount of fuel used for both rides together.

what L of fuel the car had at the begining ? (before the rides)

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

OK........let's call the amount of fuel in the tank before the rides = x

After the first trip, we used .25x, so we have [x - .25x] = .75x left

On the second trip, we used 20% of what remains = .20[.75x]

So, what we had left after the first trip less what we consumed after the second trip is "8 more" than what we used on both trips. Mathematically, this sets up as:

[.75x] - .20[.75x] - 8  = [.25x] + .20[.75x]

Notice that, if the left side is "8 more" than the right, then we have to subtract 8 to "equalize" both sides.

So we have

[.75x] - [.15x] - 8 = [.25x] + .20[.75x]

Simplifying, we have

[.60x] - 8 = [.25x] + [.15x]

[.60x] - 8 = [.40x]

Subtract [.40x] from both sides

[.20x] - 8 = 0

Add 8 to both sides

[.20x] = 8

Divide by .20 on both sides

x = 40L

Make sure that you check this answer in the original equation......I think it "works"  

 Apr 23, 2014
 #1
avatar+128079 
+5
Best Answer

a car consumed in its first ride 25% of her fuel. on the second ride she consumed 20% fuel of the remaining fuel in the car ( after using 25%) after the 2 rides the remaining fuel amount of was bigger by 8 than the amount of fuel used for both rides together.

what L of fuel the car had at the begining ? (before the rides)

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

OK........let's call the amount of fuel in the tank before the rides = x

After the first trip, we used .25x, so we have [x - .25x] = .75x left

On the second trip, we used 20% of what remains = .20[.75x]

So, what we had left after the first trip less what we consumed after the second trip is "8 more" than what we used on both trips. Mathematically, this sets up as:

[.75x] - .20[.75x] - 8  = [.25x] + .20[.75x]

Notice that, if the left side is "8 more" than the right, then we have to subtract 8 to "equalize" both sides.

So we have

[.75x] - [.15x] - 8 = [.25x] + .20[.75x]

Simplifying, we have

[.60x] - 8 = [.25x] + [.15x]

[.60x] - 8 = [.40x]

Subtract [.40x] from both sides

[.20x] - 8 = 0

Add 8 to both sides

[.20x] = 8

Divide by .20 on both sides

x = 40L

Make sure that you check this answer in the original equation......I think it "works"  

CPhill Apr 23, 2014
 #2
avatar+103 
0

TY 

it works!

 Apr 23, 2014

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