SOLUTION: Please help me answer this question The cost of petrol rises by 2cents a litre. Last week a man bought 20 litres at the old price This week he bought 10 litres at the new pric

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Question 1209628: Please help me answer this question
The cost of petrol rises by 2cents a litre.
Last week a man bought 20 litres at the old price
This week he bought 10 litres at the new price
All together it costed $9.20
What was the old price for 1 litre
Let x be the old price
Solution
I tried saying 20 times x plus 10 times x plus two is equal to 920cents
Then I got 20x plus 10x plus 20 is equal to 920 cents
Then I said 10x plus 20x is equal to 30x = 920 cents minus 20
Then I sais 30x + 900 divided by 30
I then got the answer x = 0,30 cents
Found 2 solutions by ikleyn, johnsss:
Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.
Please help me answer this question
The cost of petrol rises by 2cents a litre.
Last week a man bought 20 litres at the old price
This week he bought 10 litres at the new price
All together it costed $9.20
What was the old price for 1 litre
Let x be the old price
Solution
I tried saying 20 times x plus 10 times x plus two is equal to 920cents
Then I got 20x plus 10x plus 20 is equal to 920 cents
Then I said 10x plus 20x is equal to 30x = 920 cents minus 20
Then I sais 30x + 900 divided by 30
I then got the answer x = 0,30 cents
~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let x be the old price per liter (x dollars per liter).

Then the new price is (x + 0.02) dollars per liter.


Write an equation as you read the problem

    20x + 10*(x+0.02) = 9.20  dollars.


Simplify and find x

    20x + 10x + 0.20 = 9.20

        30x          = 9.20 - 0.20

        30x          =    9.00

          x          =    9.00/30

          x          =    0.30


ANSWER.  The old price was 0.30 dollars per liter.

Solved.


///////////////////////////////////


The other person uses practically the same method of creating system of equations,
as people do it for centuries, but introduces his own name for it as "weighted average approach".

As far as I know, nobody and nowhere uses this term in solving/treating such problems when teaching.

In addition, this terminology "weighted average approach" is factually incorrect/irrelevant,
since in this consideration, there is no weighted average, at all: there is weighted sum, instead.

So, I do not see any reason for a reader to learn from this author,
stumbling over unnecessary and incorrect terminology that the author introduces without any need.

I'd advise to a reader to avoid such tutors and/or teachers, who earn their popularity
by introducing their own terms for well known methods and conceptions,
and then sell it as their innovations. It is a form of deceiving readers.

In addition, he uses unreadable format - to convince a reader that he is so much original ?

Web-site, which this person gives for contacts, does not work.

In short, total and fatal absurd.



Answer by johnsss(2)   (Show Source): You can put this solution on YOUR website!
To solve the problem about the price of petrol, we can use a weighted average approach (similar to the concept of nota ponderada, or "weighted average"). Here's the step-by-step solution:
Source: https://mipromedio.co/blog/como-calcular-promedio-universitario/
Let x be the old price per litre in cents.
The new price per litre is x + 2 cents.
The man bought 20 litres at the old price and 10 litres at the new price, with a total cost of 920 cents.
We can set up the equation:
20
𝑥
+
10
(
𝑥
+
2
)
=
920
20x+10(x+2)=920
Simplify the equation:
20
𝑥
+
10
𝑥
+
20
=
920
20x+10x+20=920
30
𝑥
+
20
=
920
30x+20=920
Subtract 20 from both sides:
30
𝑥
=
900
30x=900
Solve for x:
𝑥
=
900
30
=
30
 cents
x=
30
900
​
=30 cents
So, the old price per litre was 30 cents.
This approach uses the concept of a weighted average because we are calculating the total cost by weighting the quantities (20 litres at the old price and 10 litres at the new price).
Source: https://mipromedio.co/blog/como-calcular-promedio-universitario/
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