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01 Feb 2022, 04:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
73% (02:43) correct
27%
(02:29)
wrong
based on 278
sessions
History
Date
Time
Result
Not Attempted Yet
\(E=\{1, 5, 9, 13, 17,..., 497\}\)
What is the number of all possible ordered pairs \((a, b)\) which can be created from the members of the set E such that \(a>b\)?
A. 9750
B. 8750
C. 7750
D. 6750
E. 5750
What is the number of all possible ordered pairs \((a, b)\) which can be created from the members of the set E such that \(a>b\)?
A. 9750
B. 8750
C. 7750
D. 6750
E. 5750
aumd568
Joined: 20 Nov 2021
Last visit: 07 Nov 2022
Posts: 20
Given Kudos: 54
Location: India
Schools: Booth '24 Columbia CBS '24 Fuqua '24 Haas '24 Harvard '24 Johnson '24 Kellogg '24 Sloan '24 Stanford '24 Wharton '24
GMAT 1: 700 Q50 V35
Schools: Booth '24 Columbia CBS '24 Fuqua '24 Haas '24 Harvard '24 Johnson '24 Kellogg '24 Sloan '24 Stanford '24 Wharton '24
GMAT 1: 700 Q50 V35
Posts: 20
01 Feb 2022, 06:36
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Approach -
Step 1 : Find total number of terms in Set E -
Identified that all numbers have a difference of "4"
If we subtract the set by "1" we get,
(1-1) = 0 = 4*0
(5-1) = 4 = 4*1
(9-1) = 8 = 4*2
.
.
.
(497-1) = 496 = 4*124
Thus, number of terms = 124 + 1 (term for "0" position) = 125
Step 2 : Identify how many pairs can be formed
We can see that when we try to create pairs where first number is greater than the second the total number of combinations for n numbers is always {summation of first (n-1) numbers}
If this does not strike out to you, just use a simple case of taking 5 numbers : 1, 2, 3, 4, 5
Where we see pairs are -
(5,1), (5,2), (5,3), (5,4)
(4,1), (4,2), (4,3)
(3,1), (3,2)
(2,1)
Thus number of pairs are {summation of (5-1)} = 4 + 3 + 2 + 1 = 10
Similarly here we get,
Total number of pairs = {summation of (125-1)} = {summation of (124)}
using formula of {summation(n)} = n(n+1)/2
We get,
{summation of (124)} = 124*125/2 = 7750
Thus the answer is C
Step 1 : Find total number of terms in Set E -
Identified that all numbers have a difference of "4"
If we subtract the set by "1" we get,
(1-1) = 0 = 4*0
(5-1) = 4 = 4*1
(9-1) = 8 = 4*2
.
.
.
(497-1) = 496 = 4*124
Thus, number of terms = 124 + 1 (term for "0" position) = 125
Step 2 : Identify how many pairs can be formed
We can see that when we try to create pairs where first number is greater than the second the total number of combinations for n numbers is always {summation of first (n-1) numbers}
If this does not strike out to you, just use a simple case of taking 5 numbers : 1, 2, 3, 4, 5
Where we see pairs are -
(5,1), (5,2), (5,3), (5,4)
(4,1), (4,2), (4,3)
(3,1), (3,2)
(2,1)
Thus number of pairs are {summation of (5-1)} = 4 + 3 + 2 + 1 = 10
Similarly here we get,
Total number of pairs = {summation of (125-1)} = {summation of (124)}
using formula of {summation(n)} = n(n+1)/2
We get,
{summation of (124)} = 124*125/2 = 7750
Thus the answer is C
01 Feb 2022, 06:40
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The equation for the set is 1 + 4k, k = 124 (497 = 4*124 + 1)
for (a;b), where a>b,
- a can be 497 and b can be all the other 124 members of the set, which are less than 497
- a can be 493 and b can be all the other 123 members of set (except 497), which are less than 493
- etc...
Therefore number of ordered pairs (a;b), where a>b = 124 + 123 + 122 + ... +1
124*125/2 = 7750 (C)
for (a;b), where a>b,
- a can be 497 and b can be all the other 124 members of the set, which are less than 497
- a can be 493 and b can be all the other 123 members of set (except 497), which are less than 493
- etc...
Therefore number of ordered pairs (a;b), where a>b = 124 + 123 + 122 + ... +1
124*125/2 = 7750 (C)
