GMATinsight wrote:
doomedcat wrote:
An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?
A)30% ≤ x ≤ 40%
B)33.33% ≤ x ≤ 40%
C)32.25% ≤ x ≤ 40%
D)32.5% ≤ x ≤ 42%
E)35.5% ≤ x ≤ 42%
GMATinsight : sir could you help in giving an alternate solution to this mixture problem..
An alloy of copper and aluminum has 40% copper
i.e. Copper in first alloy = 40%
i.e. every 10 kg of alloy has 4 kg copper and 6 kg of Aluminium
An alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7
i.e. Copper in Second alloy = (2/9)*100 = 22.22%
i.e. every 10 kg of alloy has 2.22 kg copper and 7.78 kg of Zinc
As question mentions "there is more aluminum than Zinc"
for first alloy to have 7.78 kg of aluminium the total weight of first alloy = 7.78(10/6) = 12.96 kg
i.e. first alloy must be more than 12.96 kg if weight of he second alloy is 10 kg
and weight of copper in 12.96 kg of first alloy = 5.19
i.e. If total weight of two alloys = 10+12.96 = 22.96 kg
then weight of copper = 5.19+2.22 = 7.4 kg
minimum percentage of copper in final alloy = (7.4/22.96)*100 = 32.25%
Also, Maximum percentage of copper can't exceed 40% even if the second alloy becomes 0%
i.e. 32.25 < x < 40
Although I don' think that Option C should include equal to sign alongwith < sign, but that seems the best answer available hence
Answer: Option C
Archit3110Hi
GMATinsight can you pleaseprovide logical insight into this --- > why are we multiplying 7.78 by (10/6)
why are multiplying by 10/6 and not by 6/10
shouldnt concentration of aluminium be on top of fraction
thanks
i.e. 7.78 = (6/10) total weight
i.e. Total Wt. = (10/6)*7.78