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calreg11
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I found this question a little confusing when i solved it during the test

S-T=9(X-Y)<40
X-Y<40/9
i.e X-y<4

therefore the value of S can be 52,74...and the value of T can be 25,47

As a result the value of S-T comes to 27
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I found this question a little confusing when i solved it during the test

S-T=9(X-Y)<40
X-Y<40/9
i.e X-y<4

therefore the value of S can be 52,74...and the value of T can be 25,47

As a result the value of S-T comes to 27

That's because the math is wrong: 9(x-y)<40 --> x-y<4.4, so x-y can be 4 too, S can be 51, 62, ... and T can be 15, 26, ... S-T=36.
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I calculated it this way
9x-9y = difference.
9(x-y) = difference.
For difference to be max difference between x and y must be larger.
I calculated putting actual values
39 Difference between digits 6 : Difference between Numbers : 54 (Eliminated because its > 40)
27 Difference between digits 5 : Difference between Numbers : 45 (Eliminated because its > 40 )
30 Difference between digits 3 : Difference between Numbers : 27 ( possible answer)
36 Difference between digits 3 : Difference between Numbers : 27 (possible answer)
33 Difference between digits 0 This will never be the answer because of 0

between 36 & 30 I eliminated 30 because reverse of it will be 03, which wont be considered as two digit integers.
Is that reasoning right ?
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S-T=9(X-Y)<40
Hence the RHS should be a multiple of 9 to deduce the difference.
Now from the options we should select 36 as it is the gretaest possible option in the list.
Option D.
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calreg11
S and T are two-digit positive integers that have the same digits but in reverse order. If the positive difference between S and T is less than 40, what is the greatest possible value of S minus T?
A. 27
B. 30
C. 33
D. 36
E. 39

Two-digit integer ab can be expressed as 10a+b, for example: 45=10*4+5.

Given: S-T=(10a+b)-(10b+a)=9(a-b)<40 --> greatest multiple of 9 which is less than 40 is 36. For example S can be 51 and T can be 15.

Answer: D.

All other solutions are satisfying but I have a quick 10s Solution by PLUG IN ELIMINATION method.
1. Highest number in option is 39, so other possible number will be 93. 39+40 = 79 ---- Eliminated
2. Then Highest number in option 36, other number 63. 36+40 = 76. -- Mark the answer.
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Can't we do:

99 - 40 = 59 we know that S has the same value in reverse order: 95

95-59=36

am i right? what do you think?
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calreg11
S and T are two-digit positive integers that have the same digits but in reverse order. If the positive difference between S and T is less than 40, what is the greatest possible value of S minus T?

A. 27
B. 30
C. 33
D. 36
E. 39

all differences between reversals are multiples of 9
closest multiple of 9<40=36
36/9=difference between digits=4
so digit possibilities are 9 & 5, 8 & 4, 7 & 3, 6 & 2, 5 & 1
for all such digit possibilities, difference between reversed integers=36
D
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Two-Digit reverse order number properties:

Sum of such 2 numbers= Multiple of 11
Difference of such 2 numbers= Multiple of 9
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calreg11
S and T are two-digit positive integers that have the same digits but in reverse order. If the positive difference between S and T is less than 40, what is the greatest possible value of S minus T?

A. 27
B. 30
C. 33
D. 36
E. 39

We need 2 two-digit numbers that have the same digits but reversed e.g. 19 and 91, 26 and 62 etc.
When the gap between the digits increases, the gap between the numbers increases.
16 and 61 have a difference of 45 but this is not acceptable.
Hence 15 and 51, giving a difference of 36 would represent the largest value of S - T.

Note that diff between 15 and 51 = Diff between 26 and 62 = Diff between 37 and 73 and so on...

Answer (D)
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