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01 Oct 2025, 01:20
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B
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Difficulty:
65%
(hard)
Question Stats:
60% (01:49) correct
40%
(02:12)
wrong
based on 147
sessions
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All the two-digit positive integer whose unit digit is greater than their ten’s digit are selected. If all these numbers are written one after the other in a sequence, how many digits are there in the resulting number?
(A) 90
(B) 72
(D) 60
(D) 54
(E) 36
(A) 90
(B) 72
(D) 60
(D) 54
(E) 36
01 Oct 2025, 01:29
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We need to count how many such 2-digit numbers exist and then multiply by 2 (since each number has 2 digits).
Condition: unit digit > tens digit.
Tens digit can be 1–9. For each tens digit `t`, units digit can be any of (t+1 ... 9). That gives (9 – t) possibilities.
So total numbers = (9 – 1) + (9 – 2) + ... + (9 – 8) + (9 – 9).
= 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 numbers.
Each number has 2 digits → 36 × 2 = 72 digits.
Answer: B
Condition: unit digit > tens digit.
Tens digit can be 1–9. For each tens digit `t`, units digit can be any of (t+1 ... 9). That gives (9 – t) possibilities.
So total numbers = (9 – 1) + (9 – 2) + ... + (9 – 8) + (9 – 9).
= 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 numbers.
Each number has 2 digits → 36 × 2 = 72 digits.
Answer: B
01 Oct 2025, 04:14
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unit digit > tens digit
Total cases = 8*9/2 = 36
Therefore, total digits = 36*2 = 72
GMAT-Club-Forum-vd0j0cmp.png [ 3.09 KiB | Viewed 926 times ]
Total cases = 8*9/2 = 36
Therefore, total digits = 36*2 = 72
Attachment:
GMAT-Club-Forum-vd0j0cmp.png [ 3.09 KiB | Viewed 926 times ]
