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Re: In a four-digit positive integer y, the thousands digit is 2 [#permalink]
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This is a good question.

Now, the thousands digits must be a number between 1 and 9 and at the same time is 2.5 greater than the tens digit.

The question asks if the latter is inferior to 4 or bigger.

If the tens is 5 then 5*2.5= 12.5 which is impossible. The tens digit must be between 1 and 9 as well.

4*2.5=10 impossible as well.The tens digit must be 2 because is the only number that is less 2.5 times another digit the thousands 2*2.5=5

So the tens is 2 and the thousands is 5. If the tens is 2 this is < 4. So B is the answer.

Hope now is clear to you
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Re: In a four-digit positive integer y, the thousands digit is 2 [#permalink]
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Carcass wrote:
In a four-digit positive integer y, the thousands digit is 2.5 times the tens digit.

Quantity A
Quantity B
The tens digits of y
4


Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.


Let \(y\) be \(abcd\)
We are given \(a = 2.5(c)\)

Let us plug some values for \(c\):

When \(c = 1\), \(a = 2.5\) - Not Possible
When \(c = 2\), \(a = 5\) - Possible (5,b1d)
When \(c = 3\), \(a = 7.5\) - Not Possible
When \(c = 4\), \(a = 10\) - Not Possible .....

Any value of \(c\) after this will give us a value of \(a\) which is Not Possible

Col. A: \(c = 2\)
Col. B: \(4\)

Hence, option B
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Re: In a four-digit positive integer y, the thousands digit is 2 [#permalink]
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