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13 Sep 2020, 03:24
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B
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:
32% (02:15) correct
68%
(02:09)
wrong
based on 118
sessions
History
Date
Time
Result
Not Attempted Yet
The number of distinct real numbers from 0 to 1(both included) that can be expressed in the form of \(\frac{a}{b}\), where a is non-negative integer and b is a positive integer less than 6, is
A. 10
B. 11
C. 12
D. 15
E. 20
A. 10
B. 11
C. 12
D. 15
E. 20
Updated on: 13 Sep 2020, 20:16
Originally posted by CrackverbalGMAT on 13 Sep 2020, 09:31.
Last edited by CrackverbalGMAT on 13 Sep 2020, 20:16, edited 2 times in total.
Last edited by CrackverbalGMAT on 13 Sep 2020, 20:16, edited 2 times in total.
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Given that \(a\) is a non negative integer. So \(a\) can take values from 0 to \(\infty\)
\(b\) is a positive integer < 6, so \(b\) can take the values 1, 2, 3, 4 and 5
Since \(\frac{a}{b}\) has to be between 0 and 1 both inclusive, \(a\) < \(b\)
Also \(\frac{a}{b}\) has to be distinct
When \(a\) = 0, \(b\) can be any of 1, 2, 3, 4 or 5 to get \(\frac{a}{b}\) = 0 = 1 outcome
When \(a\) = 1, \(b\) can be any of 1, 2, 3, 4 or 5 to get \(\frac{a}{b} = 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\) = 5 outcomes
When \(a\) = 2, \(b\) can be any of 3 or 5 to get \(\frac{2}{3}, \frac{2}{5}\) = 2 outcomes (with 4, we will get \(\frac{1}{2}\), which we have already obtained when \(a\) = 1 and \(b\) = 2
When \(a\) = 3, \(b\) can be any of 4 or 5 to get \(\frac{3}{4,}\frac{3}{5}\) = 2 outcomes
When \(a\) = 4, \(b\) can be only 5 to get \(\frac{4}{5}\) = 1 outcome
Total Possible outcomes = 1 + 5 + 2 + 2 + 1 = 11
Option B
Arun Kumar
\(b\) is a positive integer < 6, so \(b\) can take the values 1, 2, 3, 4 and 5
Since \(\frac{a}{b}\) has to be between 0 and 1 both inclusive, \(a\) < \(b\)
Also \(\frac{a}{b}\) has to be distinct
When \(a\) = 0, \(b\) can be any of 1, 2, 3, 4 or 5 to get \(\frac{a}{b}\) = 0 = 1 outcome
When \(a\) = 1, \(b\) can be any of 1, 2, 3, 4 or 5 to get \(\frac{a}{b} = 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\) = 5 outcomes
When \(a\) = 2, \(b\) can be any of 3 or 5 to get \(\frac{2}{3}, \frac{2}{5}\) = 2 outcomes (with 4, we will get \(\frac{1}{2}\), which we have already obtained when \(a\) = 1 and \(b\) = 2
When \(a\) = 3, \(b\) can be any of 4 or 5 to get \(\frac{3}{4,}\frac{3}{5}\) = 2 outcomes
When \(a\) = 4, \(b\) can be only 5 to get \(\frac{4}{5}\) = 1 outcome
Total Possible outcomes = 1 + 5 + 2 + 2 + 1 = 11
Option B
Arun Kumar
13 Sep 2020, 19:14
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arrghh, 2/4 reduces to 1/2.....so close
When you start testing the Choices with the Constraints, you quickly realize since the Fraction must be between 0 and 1 inclusive, the only Possible Numbers that Variable a can take are the following:
a can = 0 -- 1 -- 2 -- 3 -- 4 -- 5
any other Integer > 5 when it is the NUM over the DEN of b (1 or 2 or 3 or 4 or 5) will always result in a Number that is LARGER than 1
Then, you need to test all the Possibilities WATCHING FOR REPEATS and POSSIBLE SIMPLIFICIATIONS (my mistake)
if a = 0
then DEN will not matter, Value = 0 ----- 1 Way
if a = 1
1/1 -- 1/2 -- 1/3 -- 1/4 -- 1/5 ------- 5 Ways
if a = 2
2/3 ---- **2/4** ---2/5 ----- apparently 3 Ways, but since 2/4 reduces to 1/2, it is a REPEAT (my mistake again)----so 2 Ways
if a = 3
3/4 --- 3/5 ----- 2 Ways
if a = 4
4/5 ----- 1 Way
if a = 5
5/5 = 1 has already been counted ---- 0 Ways
Solution of Possible a/b Values:
1 way + 5 Ways + 2 ways + 2 ways + 1 way =
11 ways
Answer -B-
Edit: GREAT Question. Kudos + 1
When you start testing the Choices with the Constraints, you quickly realize since the Fraction must be between 0 and 1 inclusive, the only Possible Numbers that Variable a can take are the following:
a can = 0 -- 1 -- 2 -- 3 -- 4 -- 5
any other Integer > 5 when it is the NUM over the DEN of b (1 or 2 or 3 or 4 or 5) will always result in a Number that is LARGER than 1
Then, you need to test all the Possibilities WATCHING FOR REPEATS and POSSIBLE SIMPLIFICIATIONS (my mistake)
if a = 0
then DEN will not matter, Value = 0 ----- 1 Way
if a = 1
1/1 -- 1/2 -- 1/3 -- 1/4 -- 1/5 ------- 5 Ways
if a = 2
2/3 ---- **2/4** ---2/5 ----- apparently 3 Ways, but since 2/4 reduces to 1/2, it is a REPEAT (my mistake again)----so 2 Ways
if a = 3
3/4 --- 3/5 ----- 2 Ways
if a = 4
4/5 ----- 1 Way
if a = 5
5/5 = 1 has already been counted ---- 0 Ways
Solution of Possible a/b Values:
1 way + 5 Ways + 2 ways + 2 ways + 1 way =
11 ways
Answer -B-
Edit: GREAT Question. Kudos + 1
