OG 13th Data sufficiency #133

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OG 13th Data sufficiency #133

by pchnts » Mon Nov 17, 2014 9:54 am
hello,

I have a problem with the solution of this question:

# 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.

the answer to this question is A) - 1st statement is sufficient, but what if we would have sum of numbers 359 and 549, the sum of 5 +4 is nine but then the sum of hundreds digits wouldn't be 8.

thank you
Source: — Data Sufficiency |

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by [email protected] » Mon Nov 17, 2014 10:17 am
Hi pchnts,

Unfortunately, the example that you've described does NOT fit the restriction in Fact 1.

We're told that the TENS DIGIT of X = the sum of the TENS DIGITS of Y and Z

359 + 549 = 908

The tens digit of X is 0, the sum of the tens digits of Y and Z is 9. This is not a valid example.

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by Brent@GMATPrepNow » Mon Nov 17, 2014 10:50 am
pchnts wrote: 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 to 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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by GMATGuruNY » Mon Nov 17, 2014 10:57 am
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.
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by pchnts » Mon Nov 17, 2014 11:31 am
thank you guys!

this scheme

2BC
2EF
____
2HF

really helped me out.