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01 Sep 2009, 15:15
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ABCD are all different digits. Is 0.AB + 0.CD>1?
1) The least of the four digits is 4
2) 0.AB x 0.CD > 1/2
1) The least of the four digits is 4
2) 0.AB x 0.CD > 1/2
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01 Sep 2009, 16:12
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Ans should be D
1 If minimum value is 4 and all are different numbers then the lowest possible number
0.45 + 0.67 > 1
2 0.ab*0.cd >1/2 will also work for any combination you try.
1 If minimum value is 4 and all are different numbers then the lowest possible number
0.45 + 0.67 > 1
2 0.ab*0.cd >1/2 will also work for any combination you try.
01 Sep 2009, 16:44
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0.AB + 0.CD > 1 ??
1.) Least value of A,B,C and D is 4 -- i hope everyone agrees that A,B,C and D can only be integers digits 0-9.
So, Least value of 0.AB + 0.CD = when A and C are least = 0.4B + 0.5D -- All are different..
Also, if B is 6 then d should be 7 or vice-versa..
in both cases; 0.AB + 0.CD = 0.46 + 0.57> 1 and 0.47 + 0.56 > 1 ...hence, sufficient.
2.) 0.AB * 0.CD > 0.50
=> when 0.AB = 0.70 and 0.CD = 0.70 then 0.AB * 0.CD = 0.49 which is less than 0.50
=> 0.AB and 0.Cd should be atleast greater than 0.70..Hence, sufficient.
Answer is D.
1.) Least value of A,B,C and D is 4 -- i hope everyone agrees that A,B,C and D can only be integers digits 0-9.
So, Least value of 0.AB + 0.CD = when A and C are least = 0.4B + 0.5D -- All are different..
Also, if B is 6 then d should be 7 or vice-versa..
in both cases; 0.AB + 0.CD = 0.46 + 0.57> 1 and 0.47 + 0.56 > 1 ...hence, sufficient.
2.) 0.AB * 0.CD > 0.50
=> when 0.AB = 0.70 and 0.CD = 0.70 then 0.AB * 0.CD = 0.49 which is less than 0.50
=> 0.AB and 0.Cd should be atleast greater than 0.70..Hence, sufficient.
Answer is D.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
