A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65%

BTGmoderatorDC wrote: ↑

Fri Apr 10, 2020 4:30 pm

A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?

A) 40%

B) 45%

C) 50 %

D) 60 %

E) 55 %

OA D

Source: Manhattan Prep

Say the customer purchases x kgs of Brand A and y kgs of Brand B.

Thus,

the weight of millet = 40% of x + 65% of y = 0.4x + 0.65y
the weight of sunflower = 60% of x + 35% of y = 0.6x + 0.35y

Since millet is 50% of the mixture, i.e., (x + y) kgs.

=> 0.4x + 0.65y = 0.6x + 0.35y

x = 1.5y

Total weight of the mixture = x + y = 1.5y + y = 2.5y kgs

Thus, the proportion of Brand A = [x/(x + y)]*100% = [1.5y/(2.5y)]*100% = 60%

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorDC wrote: ↑

Fri Apr 10, 2020 4:30 pm

A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?

A) 40%

B) 45%

C) 50 %

D) 60 %

E) 55 %


OA D

Source: Manhattan Prep

We can let x = the amount of millet in Brand A and y = the amount of millet in Brand B: We can create the equation:

0.4x + 0.65y = 0.5(x + y)

40x + 65y = 50x + 50y

15y = 10x

3y = 2x

We see that we can let x = 3 and y = 2 (so that 3y = 2x). Thus, the percent of the mix that is Brand A is 3/(3 + 2) = 3/5 = 60%.

Answer: D