   SEARCH HOME Math Central Quandaries & Queries  Question from Arul: Hi Please help me with this question 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 percent ryegrass, what percent of the weight of this mixture is X ? (A) 10% (B) 33 1/3% (C) 40% (D) 50% (E) 66 2/3 % Thanks Arul Arul, here's a similar problem.

Madame Campbell's class has 35% boys and 65% girls. M. Laurier's class has 50% boys and 50% girls. They all go to the track and field competition together and find that 45% of the combined group are boys. What percentage of the students at the competition are in Mme. Campbell's class?

To solve this, let C be the number of students in Mme Campbell's class and L be the number of students in M. Laurier's class.

Then the number of boys is 0.35C + 0.50L, but it is also 0.45(C+L) when they go to the track and field competition. So 0.35C + 0.50L = 0.45(C+L).

First, we can solve for L:

0.35C + 0.50L = 0.45C + 0.45L
0.05L = 0.10C
L = 2C.

The question asks what percentage of the whole group (C+L) is from Mme Campbell's class (C). So it is asking for the value of C/(C+L) as a percentage.

C / (C+L)
= C / (C + 2C)
= C / (3C)
= 1 / 3
= 33.3%.

You can solve your question the same way, Arul.

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.