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16 Feb 2016, 01:08
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In one can of mixed nuts, 30% is peanuts. In another can of mixed nuts that is one-half the size of the first one, 40% is peanuts. If both cans are emptied into the same bowl, what percentage of the mixed nuts in the bowl is peanuts?
A. 16 2/3%
B. 20%
C. 25%
D. 33 1/3%
E. 35%
A. 16 2/3%
B. 20%
C. 25%
D. 33 1/3%
E. 35%
16 Feb 2016, 01:46
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Let number of mixed nuts in first can = 100
Number of peanuts in first can = (30/100)*100 = 30
Since second can is one- half of first , number of mixed nuts in second can = 50
Number of peanuts in second can = (40/100)*50 = 20
Total number of peanuts in both cans = 30+20 = 50
Total number of mixed nuts in both cans = 100+50 = 150
% of peanuts if both cans are mixed = (50/150) *100% = 33.33 %
Answer D
Number of peanuts in first can = (30/100)*100 = 30
Since second can is one- half of first , number of mixed nuts in second can = 50
Number of peanuts in second can = (40/100)*50 = 20
Total number of peanuts in both cans = 30+20 = 50
Total number of mixed nuts in both cans = 100+50 = 150
% of peanuts if both cans are mixed = (50/150) *100% = 33.33 %
Answer D
08 Mar 2016, 17:08
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Wow this one is actually really reason if you see that it is a simple mixture problem:
Can 1: size 1; 30% nuts
Can 2: size 0.5 and 40% nuts
Can1+Can2= 1+0.5=0.5
(1x0.3+0.5x0.4)/1.5=33.33
Can 1: size 1; 30% nuts
Can 2: size 0.5 and 40% nuts
Can1+Can2= 1+0.5=0.5
(1x0.3+0.5x0.4)/1.5=33.33
