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06 Feb 2026, 00:40
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
28% (02:39) correct
72%
(02:50)
wrong
based on 47
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If x and y are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}, m = 10x + y and n = 10y + x, which is the closest to the probability that |m - n| > 10 ?
A. 0.56
B. 0.67
C. 0.69
D. 0.79
E. 0.89
A. 0.56
B. 0.67
C. 0.69
D. 0.79
E. 0.89
06 Feb 2026, 02:56
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This is a great problem that tests systematic counting with absolute values. The trap here is trying to calculate every possibility—there's a smarter way!
Step 1: Understand what m and n represent
If x and y are from {1,2,3,4,5,6,7,8,9}:
m = 10x + y creates a two-digit number with x in tens place, y in units
n = 10y + x swaps them
Example: If x=3, y=7 then m=37, n=73
Step 2: Simplify |m - n|
m - n = (10x + y) - (10y + x) = 9x - 9y = 9(x - y)
So |m - n| = 9|x - y|
We need: 9|x - y| > 10
This means: |x - y| > 10/9, which is approximately 1.11
Since x and y are integers, this simplifies to: |x - y| >= 2
Step 3: Count using the complement (much faster!)
Total outcomes: 9 x 9 = 81 (both x and y chosen from 9 values)
Instead of counting all cases where |x-y| >= 2, count where |x-y| < 2, meaning |x-y| is 0 or 1.
For |x-y| = 0 (x equals y): 9 pairs — (1,1), (2,2), ... (9,9)
For |x-y| = 1 (x and y differ by exactly 1):
(1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5), (6,7), (7,6), (7,8), (8,7), (8,9), (9,8)
That is 8 adjacent pairs x 2 directions = 16 pairs
Total with |x-y| <= 1: 9 + 16 = 25
Total with |x-y| >= 2: 81 - 25 = 56
Step 4: Calculate probability
P(|m-n| > 10) = 56/81 = 0.691...
The closest answer is C (0.72)
Common trap: Students often try to list all 81 combinations instead of using the complement. The key insight is simplifying |m-n| to 9|x-y| first, which converts a messy absolute value problem into simple integer counting. Also, don't confuse the strict inequality > 10 with >= 10 — since 9|x-y| only takes values that are multiples of 9 (0, 9, 18, 27...), the condition > 10 is equivalent to >= 18, which means |x-y| >= 2.
Takeaway: Whenever you see expressions like 10x + y on the GMAT, think about what happens when you subtract the reverse. The 9(x-y) pattern shows up frequently, and complement counting saves enormous time on probability questions.
Step 1: Understand what m and n represent
If x and y are from {1,2,3,4,5,6,7,8,9}:
m = 10x + y creates a two-digit number with x in tens place, y in units
n = 10y + x swaps them
Example: If x=3, y=7 then m=37, n=73
Step 2: Simplify |m - n|
m - n = (10x + y) - (10y + x) = 9x - 9y = 9(x - y)
So |m - n| = 9|x - y|
We need: 9|x - y| > 10
This means: |x - y| > 10/9, which is approximately 1.11
Since x and y are integers, this simplifies to: |x - y| >= 2
Step 3: Count using the complement (much faster!)
Total outcomes: 9 x 9 = 81 (both x and y chosen from 9 values)
Instead of counting all cases where |x-y| >= 2, count where |x-y| < 2, meaning |x-y| is 0 or 1.
For |x-y| = 0 (x equals y): 9 pairs — (1,1), (2,2), ... (9,9)
For |x-y| = 1 (x and y differ by exactly 1):
(1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5), (6,7), (7,6), (7,8), (8,7), (8,9), (9,8)
That is 8 adjacent pairs x 2 directions = 16 pairs
Total with |x-y| <= 1: 9 + 16 = 25
Total with |x-y| >= 2: 81 - 25 = 56
Step 4: Calculate probability
P(|m-n| > 10) = 56/81 = 0.691...
The closest answer is C (0.72)
Common trap: Students often try to list all 81 combinations instead of using the complement. The key insight is simplifying |m-n| to 9|x-y| first, which converts a messy absolute value problem into simple integer counting. Also, don't confuse the strict inequality > 10 with >= 10 — since 9|x-y| only takes values that are multiples of 9 (0, 9, 18, 27...), the condition > 10 is equivalent to >= 18, which means |x-y| >= 2.
Takeaway: Whenever you see expressions like 10x + y on the GMAT, think about what happens when you subtract the reverse. The 9(x-y) pattern shows up frequently, and complement counting saves enormous time on probability questions.
07 Feb 2026, 00:59
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kevincan I solved this question considering set elements as unique value (Can select 1 only once) but the answer is based on repeated value (Can choose 1 two times). Also, the language is saying chosen randomly. Do we need to consider elements as non unique all the time in such questions ?
