Question

The price of the diesel is Rs. 70 per litre and the price of petrol is Rs. 40 per litre. If the profit after selling the mixture at Rs. 75 per litre by 25%. Find the ratio of diesel and petrol in the mixture.

Answer 1

Hint: In the question, we first let the ratio of diesel and petrol in the mixture to be 1:\[x\]. Determine the cost price and the selling price from the given conditions. Then, use the formula of profit percent to form an equation in \[x\]. Solve the equation to find the value of \[x\]. Lastly, substitute the value of \[x\] in the ratio 1:\[x\] to get the required ratio.

Complete step by step solution:

Let us first assume that the ratio of diesel and petrol in the mixture to be 1:\[x\].
From here it means, diesel is 1 unit and petrol is \[x\] unit in the mixture.
As it is known that the price at which an article is bought is called the cost price of that article.
Now, we can calculate the cost price of diesel and petrol by multiplying the amount of petrol by the cost of petrol and the quantity of diesel by the price of diesel.
Therefore, \[\left( {70 \times 1 + 40 \times x} \right)\] will be the cost price (C.P.).
Here, the total amount of mixture sold is \[x + 1\] units and it is sold at the rate of 75 per litre.
The price at which an item is sold is known as the selling price of that article.
So, the selling price(S.P.) of the mixture is \[75 \times \left( {x + 1} \right)\]
We know that the profit on the mixture is 25%.
The formula for profit percent is \[\dfrac{{{\text{S}}{\text{.P}} - {\text{C}}{\text{.P}}{\text{.}}}}{{{\text{C}}{\text{.P}}}} \times 100\]
We will use the given conditions to form an equation.
$
  \dfrac{{{\text{S}}{\text{.P}} - {\text{C}}{\text{.P}}{\text{.}}}}{{{\text{C}}{\text{.P}}}} \times 100 = 25 \\
  \dfrac{{75\left( {x + 1} \right) - \left( {70 \times 1 + 40 \times x} \right)}}{{\left( {70 \times 1 + 40 \times x} \right)}} \times 100 = 25 \\
$
We need to solve the brackets and cross multiply to simplify the equation. Thus, the equation is simplified as,
$
  \dfrac{{75x + 75 - \left( {70 + 40x} \right)}}{{\left( {70 + 40x} \right)}} \times 100 = 25 \\
  \dfrac{{75x + 75 - 70 - 40x}}{{\left( {70 + 40x} \right)}} \times 100 = 25 \\
  \dfrac{{35x + 5}}{{70 + 40x}} \times 100 = 25 \\
$
On dividing the equation throughout by 100, we get,
$
  \dfrac{{35x + 5}}{{70 + 40x}} = \dfrac{{25}}{{100}} \\
  \dfrac{{35x + 5}}{{70 + 40x}} = \dfrac{1}{4} \\
$
Here, we will cross multiply and then simplify the expression.
$
  4\left( {35x + 5} \right) = \left( {70 + 40x} \right) \\
  140x + 20 = 70 + 40x \\
$
Next, we solve the expression to find the value of \[x\].
$
  140x + 20 = 70 + 40x \\
  140x - 40x = 70 - 20 \\
  100x = 50 \\
  x = \dfrac{1}{2} \\
$
Substitute the value of the \[x\] in the ratio 1:\[x\] to get,
1: \[\dfrac{1}{2}\]
We avoid writing fractional values in the ratio. Therefore, the given equation can be written as 2:1
Hence, the ratio of diesel and petrol in the mixture is 2:1.

Note: Do not let the ratio as \[x:y\], because it will involve two variables which will make it difficult to solve further. Apply the correct formula and while simplifying, open the brackets first to avoid mistakes. Also, do not leave the final answer in fractional form. Write the ratio in standard form.