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What fractional part of the total surface area of cube C is [#permalink]

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05 Feb 2009, 22:19

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16. What fractional part of the total surface area of cube C is red? (1) Each of 3 faces of C is exactly 1/2 red. (2) Each of 3 faces of C is entirely white.

22. Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast? (1) Candidate F received 11 of the votes. (2) Candidate H received 14 of the votes.

23. S is a set of integers such that i) if a is in S, then –a is in S, and ii) if each of a and b is in S, then ab is in S. Is –4 in S? (1) 1 is in S. (2) 2 is in S.

16. What fractional part of the total surface area of cube C is red? (1) Each of 3 faces of C is exactly 1/2 red. (2) Each of 3 faces of C is entirely white.

22. Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast? (1) Candidate F received 11 of the votes. (2) Candidate H received 14 of the votes.

23. S is a set of integers such that i) if a is in S, then –a is in S, and ii) if each of a and b is in S, then ab is in S. Is –4 in S? (1) 1 is in S. (2) 2 is in S.

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Question : 16 - IMO A Total surface area of a cube is \(6*S^2\).

From Stmt 1 - 3 sides of the cube are exactly half red. ie Surface area of the part that is colored red will be be \(3*s^2 / 2\). From this we can calculate the fraction of the surface area of the cube that is pained red. Sufficient.

From stmt2 - It is given that 3 faces are colored white. But does not say how many of the faces are colored red? nor how much part of each face is colored red. Hence insufficent.

Given 4o members voted exactly ONE of 3 candidates F, G, H ==> F + G + h = 40. Question asked is "IS F the maximum?"

From Stmt1 : F got 11 votes means, G + H = 29. ==> atleast one of G or H is more than F. So we can ans the question that F is not the maximum. Hence Sufficient.

From Stmt 2 : H got 14 votes mean, F + G = 26 ==> F can be 15 and G = 11 or F = 13 and G = 13. ==> F may be maximum. Hence it is insufficient to conclude F to be maximum making the statement Insufficient.

Question : 16 - IMO A Total surface area of a cube is \(6*S^2\).

From Stmt 1 - 3 sides of the cube are exactly half red. ie Surface area of the part that is colored red will be be \(3*s^2 / 2\). From this we can calculate the fraction of the surface area of the cube that is pained red. Sufficient.

From stmt2 - It is given that 3 faces are colored white. But does not say how many of the faces are colored red? nor how much part of each face is colored red. Hence insufficent.

IMO A too. My thoughts are the same as yours but the OA is C. Isn't stmt 1 obvious in telling us the red fraction? In this case do we need know the non red fraction? Can somebody bring some light to this?

Question : 16 - IMO A Total surface area of a cube is \(6*S^2\).

From Stmt 1 - 3 sides of the cube are exactly half red. ie Surface area of the part that is colored red will be be \(3*s^2 / 2\). From this we can calculate the fraction of the surface area of the cube that is pained red. Sufficient.

From stmt2 - It is given that 3 faces are colored white. But does not say how many of the faces are colored red? nor how much part of each face is colored red. Hence insufficent.

IMO A too. My thoughts are the same as yours but the OA is C. Isn't stmt 1 obvious in telling us the red fraction? In this case do we need know the non red fraction? Can somebody bring some light to this?

Btw, thanks for the explanation Q22 & 23!

Ok I think I understand why the ans is C.

First clue is stating that 3 sides of the cube are exactly painted half red. But didn't say anything abt the other other 3 sides of the cubes. Neither all the remaining 3 sides had red color, or atleast one them is painted red etc.

From second clue it is given that 3 sides of the cube are painted white. Now nothing to side that is painted red is given in this clue.

Combining both the clues, 3 side of the complete white. And 3 sides of the cube are painted half in red color. Hence we can find the fraction of the sides that are painted red and is sufficient.