In the correctly-worked addition calculation above, A, B, C, D, and E

Question

1AB
+AB
___
CDE

In the correctly-worked addition calculation above, A, B, C, D, and E each represent single-digit integers. What is the value of C?

(1) A < 5
(2) B > 4

Celerma · Answer

Statement 1: Consider A = 0,1, 2, 3, or 4 , the value of C can be determined. If value of B is more than 4 then, the carry over will be added to A+A and value of C will be one. If Value of B is less then 4 or 4, then there will be no carry over. In this case also, value of C remains 1. So Statement 1 is sufficient.

Statement 2 : B can take values such as 5, 6, 7, 8, 9. so there will a carry over.

Case 1 :
125
+25 = 150

Case 2 : 
156
+56
= 212 so the value of C is changing . Hence this statement is not sufficient.

Answer is A.

Bunuel · Answer

Check other  Addition/Subtraction/Multiplication Tables  problems from our  Special Questions Directory

kunal555 · Answer

Statement 1: 
A<5
The maximum carry over we can get from addition of two single digit numbers is 1.
Even if we get carry over of 1 from B+B, we cannot get a carry over from A+A since A is less than 5. For example, 4+4+1=9 gives no carry over.
So,there is no carry over from addition of A+A irrespective of the value of B. And c will be 1.
SUFFICIENT

Statement 2:
B>4
Doesn't provide any information about A.
INSUFFICIENT

Answer:- A

dav90 · Answer

1st statement is enough,
as at max A can have value 4, so 4+4+1(maximum possible value)=9 will not generate any carry,
and hence we can say C will be equal to 1,
hence
IMO A

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

Happy Halloween Weekend, readers (and, yes, we do find it odd that for this generation Halloween now spans several days and seems to have more relevance for 20-somethings than does pretty much any other holiday, but we’re not complaining).

As you put the final touches on your Halloween costume for the weekend – our ever-on-the-pulse-of-costume-popularity coworker, Jason (2010 – Antoine Dodson; 2011- Angry Birds. Every year he’s ahead of the curve on “most popular costume”) is going as Gangam Style – it’s not a bad time to keep your eyes one one of the next hallmarks of the fall-winter calendar, Round Two Application Deadlines. And here’s where the two coincide:

Much like innocent children dress themselves up in scary costumes for Halloween, innocent GMAT problems dress up in “scary costumes” to give you nightmares on the GMAT. If you can see through those costumes, you no longer have to fear difficult-looking problems.

Here’s what we mean. Consider the Data Sufficiency problem:

1AB
+AB
___
CDE

In the correctly-worked addition calculation above, A, B, C, D, and E each represent single-digit integers. What is the value of C?

(1) A < 5 
(2) B > 4

If your first reaction to this problem is one of fright, you’re not alone. It *looks* intimidating, much like the Haunted House employee in the ski mask with the chainsaw looks downright terrifying. But remember – he’s a volunteer for the Rotary Club earning community service credits for his college applications, and this is just an addition problem. There’s nothing to fear but fear itself (sadly, Winston Churchill isn’t a very popular Halloween costume…). So think in terms of addition – the parameters for this problem are relatively narrow. You have a three digit number that we know begins with 1 and you’re adding a two-digit number. So your min/max scenarios for this problem are:

Max: 199 + 99 = 298, in which case C = 2
Min: 100 + 00 = 100 –> But wait, then AB isn’t a two digit number. So really it’s 110 + 10 = 120, in which case C = 1. So you only have two options for C, either 1 or 2. On abstract problems, start with what you know to make them more concrete – here just establishing the parameters of the problem makes this much more solvable.

Since statement 1 tells you that A is less than 5, you now have min/max scenarios for that. The max value of A is 4 and the min value is 1 (again, it can’t be 0 because then AB isn’t a two-digit number). So your possibilities are:

Max: 149 + 49 = 198. And here you go – now the maximum value of C is 1, which means that given statement 1 it cannot be 2. Statement 1 guarantees that C = 1, so it is sufficient.

Now look at statement 2 and you should see that the units digit doesn’t matter as much. You could still have 199 + 99 (C = 2), or you could have 115 + 15 (C = 1), so this statement is not sufficient.  The answer then is A .

But more importantly, recognize this – the GMAT disguises basic concepts in “scary costumes” via abstraction, the use of large or weird numbers, or the use of technical terms (often in Reading Comp) or notation (those sequence problems tend to do this a lot). But as you know to do on Halloween, look past the scary costume to see what you’re dealing with. If the problem is abstract, make it concrete by testing it with small numbers or, as we did here, establishing parameters for the variables. If the problem uses technical language, recognize that it’s not a biology or chemistry test (and certainly not a biochem test) but rather a reasoning test, so those technical terms are there for fright’s sake almost entirely. Arguably the GMAT’s greatest weapon is that of “costume” – taking core concepts that you’ve known since you wore your Halloween costume to school and disguising it behind abstraction or distraction to get you to fear it. Don’t be intimidated by the “sheep in wolf’s clothing”, the innocent problem dressed up to look scary. This is true on Halloween just as it is on test day.

estherasare · Answer

It's well understood. Good question.

TommasoP · Answer

There's a 1 in the hundreds digit of the first addition factor, so C can either be:
1, if there's no carry over
2, if there is carry over

Take the first statement:
Because A<5,
The sum A+A can at most be 8
Whether B+B is equal to or greater than 10, meaning there's a carry over into the tens digit, the tens digit of the result would at most be 9, meaning that there can't be a carry over into the hundreds digit in any case: C must be equal to 1
SUFFICIENT

Take the second statement:
Because B>4,
A+A could be greater than or equal to 10: carry over, the hundreds digit would be 2
A+A could be 9: if there's a carry over into the tens digit, depending on the value of B, the hundreds digit would be 2
A+A could be less than 9: whether or not there's a carry over into the tens digit, the hundreds digit would be 1
NOT sufficient

The answer is A

In the correctly-worked addition calculation above, A, B, C, D, and E

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