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19 Dec 2025, 02:43
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Question stem: (5x+11y+7z)/x+y+z = ?
Statement 1: x=2y. Not sufficient
Statement 2: x=4k y = 2k z=2k Sufficient
B
Statement 1: x=2y. Not sufficient
Statement 2: x=4k y = 2k z=2k Sufficient
B
jkkamau
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19 Dec 2025, 02:59
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S1 tells us nothing about the boxes that weigh 7 pounds hence insufficient
S2 Since it gives us the disributiion of all the numbers in ratio form, we can use the minimum number 4;2:2 to determine the average weight per box which would be (5x4)+(7x2)+(11x2) divide by 8 hence sufficient
Ans B
S2 Since it gives us the disributiion of all the numbers in ratio form, we can use the minimum number 4;2:2 to determine the average weight per box which would be (5x4)+(7x2)+(11x2) divide by 8 hence sufficient
Ans B
Bunuel
19 Dec 2025, 03:30
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Let's say there are A, B and C boxes of each weight. Then the formula we're looking for is \(W = \frac{Total.Weight }{ Quantity} = \frac{5a + 7b + 11c }{ a+b+c}.\)
(1) We learn that \(a=2c\), but we learn nothing about B. Insufficient.
(2) Here it says that \(a:b:c=4:2:2 = 2:1:1\). Since we can define all the variables through one, we'll be able to cancel it out in the fraction and find the average weight. Sufficient.
Therefore, the answer is B.
(1) We learn that \(a=2c\), but we learn nothing about B. Insufficient.
(2) Here it says that \(a:b:c=4:2:2 = 2:1:1\). Since we can define all the variables through one, we'll be able to cancel it out in the fraction and find the average weight. Sufficient.
Therefore, the answer is B.
