What is the percentage profit? Coffee beans of two different qualities are mixed and sold at $20\%$ profit. If the higher quality beans are sold at the above price, then the loss is $4\%$. If the ratio of lower quality and the higher quality beans in the mixture is  $5 : 2$, then the percentage profit when the lower quality beans are sold at the same price is?
I'm confused do I need to take the variables for C.P. of different quality of seeds or presume the C.P. of the whole mixture?
If I take the individual C.P.s of both quality of seeds say l for lower quality and h for higher quality then what will be the combined C.P.?
 A: Don't worry about whether you 'need' to have variables for things or not, just write down all the relevant information and then see how you need to combine it to get the result you need. Finding a neat solution comes after finding a solution.
Let's try making some formulae:
Let $H$ be the quantity (which I'll think of as discrete units) of higher quality coffee, and $L$ the quantity of lower quality coffee. Then we know
$$L/H=5/2.$$
Let $v_H$ be the value of high quality coffee (ie the price at which selling it would make no profit or loss), and $v_L$ the value of low quality coffee. Let $v$ be the value of the mixture. Then, given the above, we see
$$v=\frac{5}{7}v_L+\frac{2}{7}v_H,$$
because every 7 units of mixture contains 5 units of low quality and 2 of high quality.
Let $P$ be the price at which the coffee is being sold. Since the mixure gives 20% profit, we get
$$P=1.2 v.$$
On the other hand, the high quality alone gives 4% loss, so
$$P=0.96 v_H.$$
Combining these gives
$$P=1.2\left(\frac{5}{7}v_L+\frac{2}{7}v_H\right)=1.2\left(\frac{5}{7}v_L+\frac{2}{7}\frac{1}{0.96}P\right),$$
so
$$\left(1-\frac{6}{5}\frac{2}{7}\frac{96}{100}\right) P=\frac{6}{5}\frac{5}{7}v_L. $$
Rearranging this gives something of the form $P=\lambda v_L$, where $\lambda$ is $1+$ profit on selling lower quality coffee, expressed as a number rather than a percentage.
