GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Whiz Representative
Joined: 25 May 2022
Posts: 25
Given Kudos: 1
Set S consists of all the two-digit numbers that do not have 0 in them
[#permalink]
Updated on: 06 Sep 2022, 04:04
Updated on: 06 Sep 2022, 04:04
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
94% (02:40) correct
5% (01:39) wrong based on 18 sessions
Hide Show timer Statistics
Set S consists of all the two-digit numbers that do not have 0 in them. If a number is picked at random from the set S, what is the probability that its unit digit is greater than the tens digit?
A. \(\frac{1}{9}\)
B. \(\frac{4}{9}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{9}\)
E. \(\frac{8}{9}\)
Source: GREWhiz
A. \(\frac{1}{9}\)
B. \(\frac{4}{9}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{9}\)
E. \(\frac{8}{9}\)
Source: GREWhiz
Originally posted by SaquibHGREWhiz on 06 Sep 2022, 02:39.
Last edited by SaquibHGREWhiz on 06 Sep 2022, 04:04, edited 1 time in total.
Last edited by SaquibHGREWhiz on 06 Sep 2022, 04:04, edited 1 time in total.
GRE Whiz Representative
Joined: 25 May 2022
Posts: 25
Given Kudos: 1
Set S consists of all the two-digit numbers that do not have 0 in them
[#permalink]
Updated on: 13 Sep 2022, 17:26
Updated on: 13 Sep 2022, 17:26
2
Expert Reply
Solution:
Thus, the required probability \(=\frac{36}{81}=\frac{4}{9}\)
Hence the right answer is Option B
- Total number of 2-digit numbers \(=99-10+1=90\)
- Number of 2-digit numbers with 0 in them = 10, 20, 30...90 \(= 9\)
- Therefore, total number of 2-digit numbers that do not have 0 in them \(=90-9=81\)
- An alternate way to count this would be by multiplying \(9 \times 9\)
- From 11 to 19
- Number of digit with unit digit greater than the tens digit \(=8\) (12, 13, 14...19)
- From 21 to 29
- Number of digit with unit digit greater than the tens digit \(=7\) (23, 24, 25...29)
- .
- .
- .
- From 81 to 89
- Number of digit with unit digit greater than the tens digit \(=1\) (89)
- From 91 to 99
- Number of digit with unit digit greater than the tens digit \(=0\)
- Therefore, total number of 2-digit numbers with unit digit greater than the tens digit \(=8+7+6+5+4+3+2+1=36\)
Thus, the required probability \(=\frac{36}{81}=\frac{4}{9}\)
Hence the right answer is Option B
Originally posted by SaquibHGREWhiz on 06 Sep 2022, 02:40.
Last edited by SaquibHGREWhiz on 13 Sep 2022, 17:26, edited 2 times in total.
Last edited by SaquibHGREWhiz on 13 Sep 2022, 17:26, edited 2 times in total.
Re: Set S consists of all the two-digit numbers that do not have 0 in them
[#permalink]
06 Sep 2022, 09:21
06 Sep 2022, 09:21
2
Set of Integers with no zeros = 9*9 = 81
Total Number of digits with ones place greater than tens place -->
For 11-19 there will be 8 such numbers
For 21-29 there will be 7 such numbers
…
So on
…
Up to 1 such number from 81-89
Hence the total will be = 36 {8+7+6+5+4+3+2+1}
The total probability would then be = Number of Favourable Digits / Total
i.e. =39/81 = 4/9
Total Number of digits with ones place greater than tens place -->
For 11-19 there will be 8 such numbers
For 21-29 there will be 7 such numbers
…
So on
…
Up to 1 such number from 81-89
Hence the total will be = 36 {8+7+6+5+4+3+2+1}
The total probability would then be = Number of Favourable Digits / Total
i.e. =39/81 = 4/9
