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For the four-digit number ABCD , where A, B, C, and D all [#permalink]
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 15 May 2011, 15:01
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For the four-digit number \(ABCD\), where \(A, B, C,\) and \(D\) all represent unique digits, what is the value of \(B\)? (1) \(ABCD\) rounded to the nearest hundred is \(CD00\). (2) \(B > C + D\)
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Re: Rounding integers [#permalink]
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15 May 2011, 20:19
I think statment 1 is sufficient. Cause While rounding off ABCD to the nearest hundred digit should take it to A(B+1)00 eg: 4950 But here we see a case where it becomes CD00 ==> (A+1)000 eg: 5000 The thousand digit is also changing means B=9 and A=C+1 and D=0 Hence it can be implied that B has to be 9 Stament 2 alone does not give any information on B
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Re: Rounding integers [#permalink]
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Updated on: 16 May 2011, 00:01
Does unique digits mean different digits ? However, Considering the digits are not different. a ABCD = 1212 | 2727 | 2930 will give different values for B. not sufficient. b no values for B>C+D can be 3421 or 4741 . not sufficient. a+b gives 6970 | 5960 or so where the B = 9 always. Hence C.
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Originally posted by amit2k9 on 15 May 2011, 20:53.
Last edited by amit2k9 on 16 May 2011, 00:01, edited 1 time in total.
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Re: Rounding integers [#permalink]
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15 May 2011, 22:38
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i think answer is A. 1st statement should be sufficient. Since its given that all digits are unique, B has to be 9. ex - 2930, 3940, 4950, 5960, etc.
B is not sufficient. C & D could be 1 & 2 resp and B can be than 4,5,6 etc.
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Re: Rounding integers [#permalink]
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15 May 2011, 23:15
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amit2k9 wrote: Does unique digits mean different digits ?
However, Considering the digits are not different.
a ABCD = 1212 | 2727 | 2930 will give different values for B. not sufficient.
b no values for B>C+D can be 3421 or 4741 . not sufficient.
a+b gives 2930 | 8970 or so where the B = 9 always.
Hence C. Hi Amit, Are you sure 2930 rounds off to 3000 for nearest 100? I think it will round off to 2900. and it cant 8970 .. it has to be 8990( in your example) but its not a good number because b> c+d The numbers should be 4950, 5960, 6970 and 7980
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Re: Rounding integers [#permalink]
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16 May 2011, 18:58
agdimple333 wrote: i think answer is A. 1st statement should be sufficient. Since its given that all digits are unique, B has to be 9. ex - 2930, 3940, 4950, 5960, etc.
B is not sufficient. C & D could be 1 & 2 resp and B can be than 4,5,6 etc. I think B is required here - A -> AB = CD+1 such as- 9798, 9091, 9596, 7273 (all these rounded to nearest 100th will give CD00 but not necessarily have "9" as B). but here if you take into consideration B> C+D then only "9" fits the Bill. Such as- 8990, 7980, 1920, 3940... I hope it's clear now so C is my bet!
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Re: Rounding integers [#permalink]
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16 May 2011, 19:06
yogesh1984 wrote: agdimple333 wrote: i think answer is A. 1st statement should be sufficient. Since its given that all digits are unique, B has to be 9. ex - 2930, 3940, 4950, 5960, etc.
B is not sufficient. C & D could be 1 & 2 resp and B can be than 4,5,6 etc. I think B is required here - A -> AB = CD+1 such as- 9798, 9091, 9596, 7273 (all these rounded to nearest 100th will give CD00 but not necessarily have "9" as B). but here if you take into consideration B> C+D then only "9" fits the Bill. Such as- 8990, 7980, 1920, 3940... I hope it's clear now so C is my bet! A,B,C,D are unique digits remember!!!!
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Re: Rounding integers [#permalink]
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13 Aug 2011, 12:55
I HOPE I GOT IT RIGHT:
C MUST BE BIGGER THAN 5 AND D must be 0.(cd=50, 60, 70, 80, 90)
A MUST BE ONE NUMBER BELOW C TO MAKE THE ROUNDING C-D-0-0 FORM.
SO: 4950 IS ROUNDED TO 5-0-0-0
5960 to 6-0-0-0 6970 to 7-0-0-0 7980 to 8-0-0-0 and 8990 to 9-0-0-0. B will allways be 9
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Re: Rounding integers [#permalink]
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13 Aug 2011, 14:59
dimri10 wrote: I HOPE I GOT IT RIGHT:
C MUST BE BIGGER THAN 5 AND D must be 0.(cd=50, 60, 70, 80, 90)
A MUST BE ONE NUMBER BELOW C TO MAKE THE ROUNDING C-D-0-0 FORM.
SO: 4950 IS ROUNDED TO 5-0-0-0
5960 to 6-0-0-0 6970 to 7-0-0-0 7980 to 8-0-0-0 and 8990 to 9-0-0-0. B will allways be 9 Yes thats right.. Even i had this clarification but later got it right..
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Re: Rounding integers [#permalink]
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24 Aug 2011, 10:14
Statement 1 seems to be the solution if we take it at its face value. However, if we really read it closely, it provides no limits on direction. Most here are rounding up only. The statement is open ended, thus we can round down as well (e.g. 3110 = 3100) . Therefore multiple solutions are available,NOT Sufficient . The answer should be E….
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Re: Rounding integers [#permalink]
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25 Aug 2011, 04:43
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fluke wrote: For the four-digit number \(ABCD\), where \(A, B, C,\) and \(D\) all represent unique digits, what is the value of \(B\)?
(1) \(ABCD\) rounded to the nearest hundred is \(CD00\).
(2) \(B > C + D\) Good question! This is how I arrived at the answer: ABCD - a four digit number so A is not 0. Statement 1: ABCD is rounded to the nearest hundred. It could become AB00 - round to the lower value or (AB+1)00 - round to the upper value. It is given that it becomes CD00. Since AB is not equal to CD (distinct digits), it must have been rounded up. You added 1 to AB and both the digits changed. This means when you added 1 to B, there was a carryover to change A to C. So B must have been 9! Sufficient. Statement 2: Not enough information. Answer A
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Re: Rounding integers [#permalink]
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05 Sep 2011, 08:54
** Quote: For the four-digit number ABCD, where A, B, C, and D all represent unique digits, what is the value of B?
(1) ABCD rounded to the nearest hundred is CD00.
(2) B > C + D From statement 1Considering that when ABCD is rounded to nearest hundred, it becomes CD00... C>5 and B has to be 9 because the thousand-digit changed value as well. ---> Sufficient. From statement 2Doesn't give us any value to work with. ---> Insufficient. Answer: A
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Re: For the four-digit number ABCD , where A, B, C, and D all [#permalink]
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