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A , B , and C are digits and AB

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A , B , and C are digits and AB

by sumanr84 » Sun Jul 04, 2010 5:15 am
If A , B , and C are digits and AB is NOT= 0, what is the value of B ?

(1) AB + BA = AAC

(2) A = 1

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Got a question today on Kaplan cat similar to below PS question..interesting ..Could not do in timed mode..then learned the associated concept..try it first
https://www.beatthegmat.com/digit-issue-t60041.html
I am on a break !!
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by jube » Sun Jul 04, 2010 5:41 am
IMO C

What's the OA? Thanks!
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by selango » Sun Jul 04, 2010 5:42 am
AB NOT=0,A and B not =0 , C can be o or not

b?

From stmt1,

AB+BA=AAC

Let A=a,B=b and C=c

Let AB=10a+b

10a+b+10b+a=aac

11(a+b)=aac

Insufficient.

From stmt2,

A=1,B can be any digit.

Insufficient.

combining both,

a=1

11(1+b)=11c

11+11b=11c

b=(11c-11)/11

c=0 is the only value that satisfy the equation.

b=110-11/11=9

Hence C
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by selango » Sun Jul 04, 2010 5:57 am
what is OA?
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by sumanr84 » Sun Jul 04, 2010 6:50 am
OA : A

OE:

A digit is just one of the integers from 0 to 9. Statement (2) is much easier, so it should be handled first, in case you run out of time and need to guess.

Statement (2): INSUFFICIENT

We have no way of relating the value of A to the value of B .

Statement (1): SUFFICIENT

The two digit numbers AB and BA have a three digit sum, AAC . The largest possible sum of two two-digit numbers is 99 + 99 = 198, so A must equal 1. Rewrite the sum as 1 B + B 1 = 11 C . Since 11 C is greater than 100 and 1 B is less than 20, B 1 must be close to 100. Plug in 9 for B and you get 19 + 91 = 110. This value of B works. Are any others possible? If you try B = 8 you get 18 + 81 = 99, which is less than 100. In fact, any smaller value for B will result in a sum less than 100, so B must be 9.
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by mj78ind » Sun Jul 04, 2010 7:19 am
Another way to look at this .......

Stmt 1, ab + ba = aac, hence
10a + b + 10b + a = 100a + 10a +c
11b = 99a + c
b = 9a + c/11 ------ (1)

Now we know that b is an integer between 1 and 9 (inclusive) hence c has to be 0, as any other value of c will make b from (1) above non integer. Secondly, a has to be equal to 1 since any other value of a will make b > 10 which is not possible.

Hence, when we input a = 1, and c =0 we get b = 9. SUFFICIENT

Stmt 2 a = 1, b can be anything between 1 and 9 (inclusive) INSUFFICIENT

Hence A
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by selango » Sun Jul 04, 2010 7:21 am
sumanr84 wrote:OA : A

OE:

A digit is just one of the integers from 0 to 9. Statement (2) is much easier, so it should be handled first, in case you run out of time and need to guess.

Statement (2): INSUFFICIENT

We have no way of relating the value of A to the value of B .

Statement (1): SUFFICIENT

The two digit numbers AB and BA have a three digit sum, AAC . The largest possible sum of two two-digit numbers is 99 + 99 = 198, so A must equal 1. Rewrite the sum as 1 B + B 1 = 11 C . Since 11 C is greater than 100 and 1 B is less than 20, B 1 must be close to 100. Plug in 9 for B and you get 19 + 91 = 110. This value of B works. Are any others possible? If you try B = 8 you get 18 + 81 = 99, which is less than 100. In fact, any smaller value for B will result in a sum less than 100, so B must be 9.
From stmt1,we know AB+BA=AAC

In AAC,Hundred and tenth digit must be equal as it is A for both.

But AAC=198 means,both are not equal.

Pl correct me if I am wrong.
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by selango » Sun Jul 04, 2010 7:31 am
So from stmt1 itself,we can get A=1

11(a+b)=aac (3 digit number)

Highest value for a and b is 9.a+b=18

11*18=198.

So A=1

11(1+b)=11c

11+11b=11c

b=(11c-11)/11

c=0 is the only value that satisfy the equation.

b=110-11/11=9
--Anand--
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by san2009 » Mon Jul 05, 2010 3:57 am
some of you may find this a bit more methodical and others might not like it :D

my answer is A

First off, satement 2 is clearly insufficient as it gives us NOTHING about B

let's focus on statement 1:
AB + BA = AAC

we know that when you add 2 two-digit integers, you cannot get past 198
thus, A must be equal to 1
then you have 1B + B1 = 11C
if you write out the digits in units, tens and hundreds form, you get
10 + B + 10B + 1 = 100 + 10 + C
which can be reduced to 11B= 99+C
thus B = 9 + C/9
in order for B to be a digit, it must be a whole number/integer...the only way that is possible is if C/9 is also an integer
the only way C/9 is an integer is if C is a multiple of 9
the only DIGITs that are multiples of 9 are 0 and 9
thus C must be equal to 0...otherwise b would be 10 -- which is not possible -- see for yourself, b=9+9/9=10 --not possible
therefore, b = 9 +0/9 = 0
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