Simple way to solve this please

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Simple way to solve this please

by [email protected] » Thu Jun 27, 2013 5:29 am
133. If x, y, and z are three-digit positive integers and if
x = y + z, is the hundreds digit of x equal to the sum
of the hundreds digits of y and z ?
(1) The tens digit of x is equal to the sum of the tens
digits of y and z.
(2) The units digit of x is equal to the sum of the
units digits of y and z.
Source: — Data Sufficiency |

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by snigdha1605 » Thu Jun 27, 2013 6:27 am
IMO - OA is A

If there was a carry over from the ten's place to the hundred's place, then the ten's digit of X would not equal the sum of the ten's digit of Y and the ten's digit of Z

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by GMATGuruNY » Thu Jun 27, 2013 7:00 am
[email protected] wrote:133. If x, y, and z are three-digit positive integers and if
x = y + z, is the hundreds digit of x equal to the sum
of the hundreds digits of y and z ?
(1) The tens digit of x is equal to the sum of the tens
digits of y and z.
(2) The units digit of x is equal to the sum of the
units digits of y and z.
Let y = 2BC, z = 2EF, and x = HTU, so that the addition looks as follows:

2BC
2EF
HTU

When will it be true that H ≠ 2+2?
When we have to CARRY A 1 FROM THE TENS PLACE TO THE HUNDREDS PLACE.
To illustrate:

259
249
508

Here, because we have to carry a 1 from the tens place to the hundreds place, H = 2+2+1 = 5.

Question rephrased:

ABC
DEF
HTU

In the addition problem above, do we have to a carry a 1 from the tens place to the hundreds place?

Statement 1: The tens digit of x is equal to the sum of the tens
digits of y and z.

Since T = B+E, there is no need to carry a 1 to the hundreds place.
SUFFICIENT.

Statement 2: The units digit of x is equal to the sum of the
units digits of y and z.

Since U = C+F, we do not need to carry a 1 from the UNITS PLACE to the TENS PLACE.
But it cannot be determined whether we have to carry a 1 from the TENS PLACE to the HUNDREDS PLACE.
If T = B+E = 0+0 = 0, then there is no need to carry a 1 to the hundreds place:
If T = B+E = 9+9 = 18, then we must carry a 1 to the hundreds place.
INSUFFICIENT.

The correct answer is A.
Last edited by GMATGuruNY on Thu Jun 27, 2013 7:08 am, edited 1 time in total.
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by Brent@GMATPrepNow » Thu Jun 27, 2013 7:01 am
[email protected] wrote:133. If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?

(1) The tens digit of x is equal to the sum of the tens digits of y and z.
(2) The units digit of x is equal to the sum of the units digits of y and z.
Target question: Is the hundreds digit of x equal to the sum of the hundreds digits of y and z ?

Notice that there are essentially 3 ways for the hundreds digit of x to be different from the sum of the hundreds digits of y and z
Scenario #1: the hundreds digits of y and z add to more than 9. For example, 600 + 900 = 1500. However, we can rule out this scenario because we're told that x, y, and z are three-digit integers
Scenario #2: the tens digits of y and z add to more than 9. For example, 141 + 172 = 313.
Scenario #3: the tens digits of y and z add 9, and the units digits of y and z add to more than 9. For example, 149 + 159 = 308

Statement 1: The tens digit of x is equal to the sum of the tens digits of y and z.
This rules out scenarios 2 and 3 (plus we already ruled out scenario 1).
So, it must be the case that the hundreds digit of x equals to the sum of the hundreds digits of y and z
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The units digit of x is equal to the sum of the units digits of y and z.
This rules out scenario 3, but not scenario 2. Consider these two conflicting cases:
Case a: y = 100, z = 100 and x = 200, in which case the hundreds digit of x equals the sum of the hundreds digits of y and z
Case b: y = 160, z = 160 and x = 320, in which case the hundreds digit of x does not equal the sum of the hundreds digits of y and z
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

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