# Eddie had some engine oil. He sold 2/3 of his engine oil a Monday. He then sold 4/9 of...

#### ydubrovensky4

##### New member
Eddie had some engine oil. He sold 2/3 of his engine oil a Monday. He then sold 4/9 of the remainder and an extra 340 litres on Tuesday. On Wednesday, he sold 1/2 of the remainder and an extra 200 litres. Eddie then kept the remaining 1260 litres of engine oil for himself. How much engine oil did he have at first?

The first equation I started with was:
x - 2/3*x = 1/3*x

I then used this equation to solve for the amount of engine oil Eddie had left after selling 4/9 of the remainder and an extra 340 litres on Tuesday:

1/3*x - 4/9*(1/3*x) - 340 = 1/9*x - 340

I then used this equation to solve for the amount of engine oil Eddie had left after selling 1/2 of the remainder and an extra 200 litres on Wednesday:

1/9*x - 340 - 1/2*(1/9*x) - 200 = 1/18*x - 540

#### limiTS

##### New member
You have the right idea, but your second line is wrong: [imath]\displaystyle\frac{x}{3}-\frac{4}{9}\cdot\frac{x}{3}-340=\frac{5x}{27}-340[/imath]

• blamocur and Steven G

#### Steven G

##### Elite Member
1/3*x - 4/9*(1/3*x) - 340 = 1/9*x - 340

I then used this equation to solve for the amount of engine oil Eddie had left after selling 1/2 of the remainder and an extra 200 litres on Wednesday:

1/9*x - 340 - 1/2*(1/9*x) - 200 = 1/18*x - 540
As pointed out the 1/9x is wrong.
On Wednesday, he sold 1/2 of the remainder and an extra 200 litres
After he sold some oil on Tues he had (using you error) 1/9*x - 340
For Wed, why are you not subtracting 1/2[1/9*x - 340] from 1/9*x - 340? For the record, if you sell half of something, then you have the other half remaining. That is, [1/9*x - 340] - 1/2[1/9*x - 340] - 200= 1/2[1/9*x - 340] - 200

Why didn't you finish solving for x? Did you have trouble doing that? What exactly was your trouble?

• blamocur and limiTS

#### khansaheb

##### Full Member
• stapel

#### ydubrovensky4

##### New member
As pointed out the 1/9x is wrong.
On Wednesday, he sold 1/2 of the remainder and an extra 200 litres
After he sold some oil on Tues he had (using you error) 1/9*x - 340
For Wed, why are you not subtracting 1/2[1/9*x - 340] from 1/9*x - 340? For the record, if you sell half of something, then you have the other half remaining. That is, [1/9*x - 340] - 1/2[1/9*x - 340] - 200= 1/2[1/9*x - 340] - 200

Why didn't you finish solving for x? Did you have trouble doing that? What exactly was your trouble?
I can’t solve for x.

#### blamocur

##### Senior Member
I can’t solve for x.
You have there a linear equation with one variable. It can be simplified to [imath]ax=b[/imath], which will be easy to solve.

#### stapel

##### Super Moderator
Staff member
The first equation I started with was:
x - 2/3*x = 1/3*x

This is, technically, an equation, because it has an "equals" sign in it. But it's actually just a simplification:

This will be true for any value of [imath]x[/imath]. This is not something that can be "solved" for anything.

You were given the following (VERY good) advice:
What is 'x'?

Have you defined your variables yet?
Eddie had some engine oil. He sold 2/3 of his engine oil a Monday. He then sold 4/9 of the remainder and an extra 340 litres on Tuesday. On Wednesday, he sold 1/2 of the remainder and an extra 200 litres. Eddie then kept the remaining 1260 litres of engine oil for himself. How much engine oil did he have at first?

Reading the entirety of the exercise, you are asked to find the amount of oil that he'd started with. So perhaps do your definition as:

[imath]\qquad \textrm{the amount of oil in the begining: } x[/imath]

Then work day-by-day. For instance, at the end of Monday:

[imath]\qquad \textrm{two-thirds sold, so one-third left: } \frac{x}{3}[/imath]

Then look at end-of-business on Tuesday:

Continue, simplifying as you go. Set your final expression equal to the amount that he kept for himself. Solve that equation.

#### Steven G

##### Elite Member
I can’t solve for x.
Why didn't you finish solving for x? Did you have trouble doing that? What exactly was your trouble?

• stapel and blamocur

#### ydubrovensky4

##### New member
This is, technically, an equation, because it has an "equals" sign in it. But it's actually just a simplification:

This will be true for any value of [imath]x[/imath]. This is not something that can be "solved" for anything.

You were given the following (VERY good) advice:

Have you defined your variables yet?

Reading the entirety of the exercise, you are asked to find the amount of oil that he'd started with. So perhaps do your definition as:

[imath]\qquad \textrm{the amount of oil in the begining: } x[/imath]

Then work day-by-day. For instance, at the end of Monday:

[imath]\qquad \textrm{two-thirds sold, so one-third left: } \frac{x}{3}[/imath]

Then look at end-of-business on Tuesday:

Continue, simplifying as you go. Set your final expression equal to the amount that he kept for himself. Solve that equation.
Monday: Eddie sold 2/3 of his oil, so he had $x - 2/3x = x/3$ left.

Tuesday: 4/9 of the remainder, which is 4/9 * x/3 = 4x/27. He also sold an extra 340 liters, so he had $x/3 - 4x/27 - 340 = 7x/81 - 340$ left.

Wednesday: 1/2 of the remainder, which is 1/2 * (7x/81 - 340) = 7x/162 - 170. He also sold an extra 200 liters, so he had $7x/162 - 170 - 200 = 7x/162 - 370$ left.

Thursday: kept the remaining 1260 liters of oil, so we have the equation:

7x/162 - 370 = 1260

are you sure that is how you solve it? The answer I’m getting is 3503.33

Last edited:

#### stapel

##### Super Moderator
Staff member
Reading the entirety of the exercise, you are asked to find the amount of oil that he'd started with. So perhaps do your definition as:

[imath]\qquad \textrm{the amount of oil in the begining: } x[/imath]

Then work day-by-day. For instance, at the end of Monday:

[imath]\qquad \textrm{two-thirds sold, so one-third left: } \frac{x}{3}[/imath]

Then look at end-of-business on Tuesday:

Continue, simplifying as you go. Set your final expression equal to the amount that he kept for himself. Solve that equation.
Monday: Eddie sold 2/3 of his oil, so he had $x - 2/3x = x/3$ left.

Tuesday: 4/9 of the remainder, which is 4/9 * x/3 = 4x/27. He also sold an extra 340 liters, so he had $x/3 - 4x/27 - 340 = 7x/81 - 340$ left.

Wednesday: 1/2 of the remainder, which is 1/2 * (7x/81 - 340) = 7x/162 - 170. He also sold an extra 200 liters, so he had $7x/162 - 170 - 200 = 7x/162 - 370$ left.

Thursday: kept the remaining 1260 liters of oil, so we have the equation:

7x/162 - 370 = 1260

are you sure that is how you solve it? The answer I’m getting is 3503.33

I wasn't paying attention, and switched my meaning on Tuesday. Sorry!

Yes, at the end of Monday, there remains [imath]\frac{x}{3}[/imath] of the original amount of oil.

On Tuesday, he sells [imath]\frac{4}{9}[/imath] of what was left at the end of Monday, so he still has [imath]\frac{5}{9}[/imath] of what he had at the end of Monday, or [imath]\frac{5x}{27}[/imath]. Then he sells an additional [imath]340[/imath] units so, at the end of Tuesday, he has [imath]\frac{5x}{27} - 340[/imath] units available.

On Wednesday, he sells half of what was left, so he still has half:

[imath]\qquad \frac{1}{2}\left(\frac{5x}{27} - 340\right) = \frac{5x}{54} - 170[/imath]

Then he sells another [imath]200[/imath] units so, at the end of the day, he has the following amount left:

[imath]\qquad \frac{5x}{54} - 170 - 200 = \frac{5x}{54} - 370[/imath]

At this point, he has [imath]1260[/imath] units left, so:

[imath]\qquad \frac{5x}{54} - 370 = 1260[/imath]

Checking:

On Monday, he sold [imath]\frac{2}{3}[/imath] of what he had, so he kept [imath]\frac{1}{3}[/imath]:

On Tuesday, he sold [imath]\frac{4}{9}[/imath] of the [imath]5868[/imath] units, so he had [imath]\frac{5}{9}[/imath] left:

Then he sold another [imath]340[/imath] units:

[imath]\qquad 3260 - 340 = 2920[/imath]

On Wednesday, he sold [imath]\frac{1}{2}[/imath] of what was left, which means that he still had half of what was left:

[imath]\frac{1}{2}(2920) = 1460[/imath]

Then he sold another [imath]200[/imath] units, so he was left with [imath]1260[/imath] units, as required.

(P.S. Normally, a solution would not / should not be posted so soon, but I steered you wrong, so I felt obligated. Again, apologies for the confusion.)

#### Steven G

##### Elite Member
$x/3 - 4x/27 - 340 = 7x/81 - 340$ left.
1/3 - 4/27 = 7/81 is not correct. Although the lcm of 3 and 27 is 27 you can use 81 but then I guarantee you that your result will be able to be reduced. Since 7/81 is not reducible, I am sure that your computation is incorrect.