Events & Promotions
|
|
GMAT 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.
Apr 2212:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
23 Sep 2020, 02:18
Kudos
Bookmarks
E
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:
47% (02:17) correct
53%
(02:43)
wrong
based on 144
sessions
History
Date
Time
Result
Not Attempted Yet
If two numbers are selected randomly from the first 100 natural numbers, what is the probability that their sum is divisible by 5?
A. \(\frac{4}{25}\)
B. \(\frac{1}{11}\)
C. \(\frac{1}{20}\)
D. \(\frac{2}{11}\)
E. \(\frac{1}{5}\)
A. \(\frac{4}{25}\)
B. \(\frac{1}{11}\)
C. \(\frac{1}{20}\)
D. \(\frac{2}{11}\)
E. \(\frac{1}{5}\)
23 Sep 2020, 03:22
Kudos
Bookmarks
[Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}[/m]
From 1 to 100, we have 10 numbers each ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0
To be divisible by 5, the number should end with a 5 or a 0
A 5 can be obtained as (0 + 5), (1 + 4), (2 + 3), (6 +9) or (7 + 8) = 5 possibilities
Each of these possibilities can have \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total possibilities for numbers ending with 5 = 5 * 100 = 500
A zero can be obtained in 2 ways
(i) (0 + 0) or (5 + 5) = 2 ways, and each of these can be obtained in \(\frac{^{10}C_1 \space * \space ^9C_1}{2}\) = 45 ways
We divide by 2 as order does not matter (to avoid double counting for e.g 5 + 15 is the same as 15 + 5)
Total ways = 2 * 45 = 90
(ii) A zero can be also obtained as the sum of (1 + 9), (2 +8), (3 + 7) or (4 + 6) = 4 ways
Each of these 4 combinations can be obtained in \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total Ways = 4 * 100 = 400
Total possibilities for numbers ending with zero = 90 + 400 = 490
Therefore total numbers divisible by 5 = 500 + 490 = 990
Total Outcomes = \(^{100}C_2 = \frac{100!}{98!2!} = \frac{100*99}{2 * 1} = 4950\)
The Required probability = \(\frac{990}{4950} = \frac{1}{5}\)
Option E
Arun Kumar
From 1 to 100, we have 10 numbers each ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0
To be divisible by 5, the number should end with a 5 or a 0
A 5 can be obtained as (0 + 5), (1 + 4), (2 + 3), (6 +9) or (7 + 8) = 5 possibilities
Each of these possibilities can have \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total possibilities for numbers ending with 5 = 5 * 100 = 500
A zero can be obtained in 2 ways
(i) (0 + 0) or (5 + 5) = 2 ways, and each of these can be obtained in \(\frac{^{10}C_1 \space * \space ^9C_1}{2}\) = 45 ways
We divide by 2 as order does not matter (to avoid double counting for e.g 5 + 15 is the same as 15 + 5)
Total ways = 2 * 45 = 90
(ii) A zero can be also obtained as the sum of (1 + 9), (2 +8), (3 + 7) or (4 + 6) = 4 ways
Each of these 4 combinations can be obtained in \(^{10}C_1 * ^{10}C_1 = 100\) ways
Total Ways = 4 * 100 = 400
Total possibilities for numbers ending with zero = 90 + 400 = 490
Therefore total numbers divisible by 5 = 500 + 490 = 990
Total Outcomes = \(^{100}C_2 = \frac{100!}{98!2!} = \frac{100*99}{2 * 1} = 4950\)
The Required probability = \(\frac{990}{4950} = \frac{1}{5}\)
Option E
Arun Kumar
23 Sep 2020, 03:23
Kudos
Bookmarks
Explanation:
Total Outcomes = 10x10 = 100
Total favourable outcome= (x,10-x) where x can be 0 too.
so 20 cases (0,10; 1,9, ........... 10,0)
Probability = 20/100 = 1/5
IMO-E
Total Outcomes = 10x10 = 100
Total favourable outcome= (x,10-x) where x can be 0 too.
so 20 cases (0,10; 1,9, ........... 10,0)
Probability = 20/100 = 1/5
IMO-E
