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The numbers x and y are three-digit positive integers, and x

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The numbers x and y are three-digit positive integers, and x [#permalink]

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12 May 2006, 10:55
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The numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x < y, which of the following must be true?

I. The units digit of x + y is greater than the units digit of either x or y.
II. The tens digit of x + y equals 2.
III. The hundreds digit of y is at least 5.

A. II only
B. III only
C. I and II
D. I and III
E. II and III
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12 May 2006, 11:54
Think its B. (3 only)
1.Then Unit's digit can be 5 & 6, still all the conditions are fulfilled. So its not always true.
2.The sum of units digit may give carry so sum of tens digit may give 3.
3.Since we're given that y > x and addition of these numbers generate the carry, y has to be atleast 5.
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12 May 2006, 11:59
(B) III only.

I --> May or may not be true. Take 377 + 855 = 1232. Not true.
II --> May or may not be true. Same eg as above.

III-> Given x<y and x+y = 4-digit integer, let the numbers be of the form

Quote:
a7b --> x
c5d --> y
-----
efgh
-----

f = a+c+1 >= 10
=> a+c > = 9
since x <y, a <= c Hence the least value that "c" can have is 5.
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12 May 2006, 22:18
III only is th answer.

given x<y,
x = .7.
y = .5.

I. The units digit of x + y is greater than the units digit of either x or y.
Not sufficient as x could be equal 479 and y = 551, so x+y= 1030 or
x could be equal 473 and y = 553, so x+y= 1026.

II. The tens digit of x + y equals 2.
Not sufficient as tens digit can be equal 2 or 3, depends on units digits.

III. The hundreds digit of y is at least 5.
Sufficient information as y>x.
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13 May 2006, 06:57
Let x = 470 and y = 650, x+y = 1120

I. This is not true from our example above.
II. This is always true.
III. In our example above it is surely true. But lets explore this one:

If x < y and x+y => 1000, then y must be at least 501.
Hence this is true as well.

So answer is E
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13 May 2006, 06:59
M8 wrote:
III only is th answer.

given x<y,
x = .7.
y = .5.

I. The units digit of x + y is greater than the units digit of either x or y.
Not sufficient as x could be equal 479 and y = 551, so x+y= 1030 or
x could be equal 473 and y = 553, so x+y= 1026.

II. The tens digit of x + y equals 2.
Not sufficient as tens digit can be equal 2 or 3, depends on units digits.

III. The hundreds digit of y is at least 5.
Sufficient information as y>x.

M8- you said - "Not sufficient as tens digit can be equal 2 or 3, depends on units digits."

Thats a great catch, for which I was careless.

The correct answer in that case will be III only.
13 May 2006, 06:59
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The numbers x and y are three-digit positive integers, and x

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