$4$ kg of rice at ' $5$ per kg is mixed with $8$ kg of rice at ` $6$ per kg. Find the average price of the mixture.

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Then, by the unitary method:

$n_1 + n_2$ corresponds to $A_2 – A_1$

$\rightarrow 1 + 2$ corresponds to $6 – 5$

That is, $3$ corresponds to $1$

$\therefore n_2$ will correspond to $\dfrac{(A_2 - A_1)n_2}{(n_1 +n_2 )} $

In this case $\frac{1}{3} \cdot 2 = 0.66$. Hence, the required answer is $5.66$.

My doubt is that I don't understand this solution given in my book, I'm stuck at how $n_2$ will correspond to $\dfrac{(A_2 - A_1)n_2}{n_1 + n_2}$.

PS: Here $A_2$ and $A_1$ are $6$ and $5$, I will attach the reference If you don't understand the terminologies, This is a basic question in Alligation and Mixture.


1 Answer 1


Let Rice A cost \$$5/$kg and Rice B cost \$$6/$kg

$4$kg of Rice A will cost you $4\times 5=\$20$

$8$kg of Rice A will cost you $8\times 6=\$48$

There will be no difference of you buying $4$kg of type A and $8$kg of type B separately compared to you buying them as a mixture. Either way, you end of spending $\$68$ for $4+8=12$kg of hybrid rice. Hence, we get

Average price $=\frac{\$68}{12kg}=\$5.67/$kg

I would suggest you refrain from using a textbook formula for such questions as this is not a general result. Also, IMO there is much more value in understanding how to solve the question and being able to derive the formula if needed.


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