Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 ryegrass, what percent of the weight of the mixture is X?
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Varun Reddy
It doesnt not make sense logically,
Incase i dont have time in the end and would need to guess this answer.
If the Total mixtures percentage of Ryegrass in 30% and X has 40% Ryegrass and Y has 25% Ryegrass, considering 40>25. There must be more percentage of X than of Y. How is it that mathematically its coming upto 33.3% for X and 66.67% of Y ? If i need to answer logically i would have not chosen the correct answer, i know im wrong somewhere considering the maths behind it, is not backing me up. Please help, thanks in advance.
Hi Varun, sorry for the delay in response! It's great that you are using number sense to think this through logically. Let's see if I can clear this issue up for you :-) We are dealing with a weighted average, which means that the final average of rye grass in the combined seed mixture will be most similar to that of the original mixture with more weight in the final mix. So, let's compare the percent of rye in the final mixture to that of both original mixes. Mix X is 40% rye, and mix Y is25% rye. The final mixture is 30% rye. 30 is closer to 25 than it is to 40, which means that more of the mixture must be made up of Mix Y. The greater percentage of mix Y 'weights' the final mixture and brings the final percent of rye closer to 25% than to 40%. That is why you will have a greater percent of mixture Y as compared with mixture X.
2 Explanations