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If m and n are positive two-digit integers, what is the value of the tens digit of m minus the tens digit of n ?

(1) m - n = 42.
(2) The units digit of m minus the units digit of n is not a multiple of 3.


 


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If \(m\) and \(n\) are positive two-digit integers, what is the value of the tens digit of \(m\) minus the tens digit of \(n\) ?

Let \(m\) be \(ab\), where \(a\) is the tens digit and \(b\) is the units digit, and \(n\) be \(cd\), where \(c\) is the tens digit and \(d\) is the units digit. The question asks to find the value of \(a - c\).

(1) \(m - n = 42\).

ab

-cd

42

If there is no borrowed 10 from the tens digit of \(m\), so if for example, we have:

54

-12

42

Then, \(a - c=4\).

But if there IS borrowed 10 from the tens digit of \(m\), so if for example, we have:

61

-19

42

Then, \(a - c=5\).

Not sufficient.

(2) The units digit of \(m\) minus the units digit of \(n\) is not a multiple of 3.

This one is clearly insufficient.

(1)+(2) Examples, we considered for the first statement (\(54 - 12\) and \(61 - 19\)) also satisfy the second statement, so even taken together the statements are not sufficient.


Answer: E
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Let m be- 10a+b
and n be- 10c+d

From statement 1, 10a+b - 10c-d= 42
---> 10(a-c) + b-d = 42
Not sufficient

From statement 2, b-d=1,2,4,5,7,8
Not sufficient

Now combining both statements, the only value that fits b-d would 2 such that 10(a-c) + (2) = 42, so 10(a-c) = 40
---> a-c = 4
Therefore C

Can someone tell me what's wrong with this approach? gmatophobia chetan2u KarishmaB Bunuel GMATPill
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SamzayWarrior
Let m be- 10a+b
and n be- 10c+d

From statement 1, 10a+b - 10c-d= 42
---> 10(a-c) + b-d = 42
Not sufficient

From statement 2, b-d=1,2,4,5,7,8
Not sufficient

Now combining both statements, the only value that fits b-d would 2 such that 10(a-c) + (2) = 42, so 10(a-c) = 40
---> a-c = 4
Therefore C

Can someone tell me what's wrong with this approach? gmatophobia chetan2u KarishmaB Bunuel GMATPill

You have taken cases where b>d, that is 62-20 or 75-33 etc. But what happens when b<d, for example 71-29.
So in those cases your equation will become.
(10a+b)-(10c+d)
Since b<d, we will take one 10 from value of a.
So, 10(a-1)+(10+b)-10c-d=10(a-1-C)+(10+b-d)
=> 10(a-1-c)+2=42
a-1-c = 4 or a-c = 5

Hope it helps
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