Oct 14 08:00 PM PDT  11:00 PM PDT Join a 4day FREE online boot camp to kick off your GMAT preparation and get you into your dream bschool in R2.**Limited for the first 99 registrants. Register today! Oct 15 12:00 PM PDT  01:00 PM PDT Join this live GMAT class with GMAT Ninja to learn to conquer your fears of long, kooky GMAT questions. Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

eGMAT Representative
Joined: 04 Jan 2015
Posts: 3074

Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
22 Mar 2019, 00:45
Question Stats:
36% (02:17) correct 64% (02:21) wrong based on 44 sessions
HideShow timer Statistics
eGMAT Question of the Week #40The set S is defined as {1, 2, 3, 8, 21, 30, 22, 5, 6, 11}. The chi of set is defined as the product of any 3 elements of the given set S. How many values of chi are possible such that chi is at least 7? A. 50 B. 59 C. 60 D. 61 E. 120
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4982
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
Updated on: 25 Mar 2019, 02:30
total +ve integers = 5 and ve = 5 i.e 10 integers so total chi possible no i.e product of 3 ; 10c3 = 120 now cases where no is <=7 case 1: all ve integers = 5c3 ;10 case 2: 2 +ve and 1 ve integer ; 5c2*5c1 ; 10*5=50 case 3 : 1 +ve and 2 ve ; (1,2,3 ) ; only 1 case = 1 total integers <=7 = 10+50+1 ; 61 so chi no >=7 ; 12061 ; 59 IMO B; 59 EgmatQuantExpert wrote: eGMAT Question of the Week #40The set S is defined as {1, 2, 3, 8, 21, 30, 22, 5, 6, 11}. The chi of set is defined as the product of any 3 elements of the given set S. How many values of chi are possible such that chi is at least 7? A. 50 B. 59 C. 60 D. 61 E. 120
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Originally posted by Archit3110 on 22 Mar 2019, 01:40.
Last edited by Archit3110 on 25 Mar 2019, 02:30, edited 2 times in total.



Manager
Joined: 28 Aug 2018
Posts: 226

Re: Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
22 Mar 2019, 02:01
IMo D All negative=10 2 positive and 1 negative=50 1 postive and 2 negative=1 Total 61 Posted from my mobile device
_________________
"Press +1 KUDOS"
Be very critical of my Post Ask as many questions of my post possible



Senior Manager
Joined: 25 Feb 2019
Posts: 336

Re: Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
22 Mar 2019, 02:43
What is this CHI anyways ?
Could not understand this ...
Posted from my mobile device



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3074

Re: Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
22 Mar 2019, 05:45
m1033512 wrote: What is this CHI anyways ?
Could not understand this ...
Posted from my mobile device This is just a term used to frame the question. You can also consider this as a function which gives the final output as the product of any 3 elements present in the given set S.
_________________



Manager
Joined: 05 Oct 2017
Posts: 101
Location: India
Concentration: Finance, International Business
GPA: 4
WE: Analyst (Energy and Utilities)

Re: Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
22 Mar 2019, 06:06
EgmatQuantExpert wrote: eGMAT Question of the Week #40The set S is defined as {1, 2, 3, 8, 21, 30, 22, 5, 6, 11}. The chi of set is defined as the product of any 3 elements of the given set S. How many values of chi are possible such that chi is at least 7? A. 50 B. 59 C. 60 D. 61 E. 120 Product of three elements in the set S is defined as chi We have to find the cases where chi>=7 Total negative elements = 5 Total positive elements = 5 Following are the only possibilities when product of three elements >= 7 1) 2 negative and 1 positive  2 negative out of 5 can be selected in 5C2 ways and 1 positive in 5 ways Number of possible values of chi = 5C2 * 5  1 [subtracting for the case when 2,3 and 1 will be selected] =50 1 = 49 2) three positive terms Out of 5 three positive can be chosen in 5C3 ways Number of possible values = 5C3 = 10 For every other combination such as , (3 Negative , 2 Positive + 1 negative)value of chi will be negative. Hence possible values of chi = 49 + 10 =59 IMO (B)



Manager
Joined: 09 Jun 2017
Posts: 105

Re: Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
25 Mar 2019, 23:54
shuvodip04 wrote: EgmatQuantExpert wrote: eGMAT Question of the Week #40The set S is defined as {1, 2, 3, 8, 21, 30, 22, 5, 6, 11}. The chi of set is defined as the product of any 3 elements of the given set S. How many values of chi are possible such that chi is at least 7? A. 50 B. 59 C. 60 D. 61 E. 120 Product of three elements in the set S is defined as chi We have to find the cases where chi>=7 Total negative elements = 5 Total positive elements = 5 Following are the only possibilities when product of three elements >= 7 1) 2 negative and 1 positive  2 negative out of 5 can be selected in 5C2 ways and 1 positive in 5 ways Number of possible values of chi = 5C2 * 5  1 [subtracting for the case when 2,3 and 1 will be selected] =50 1 = 49 2) three positive terms Out of 5 three positive can be chosen in 5C3 ways Number of possible values = 5C3 = 10 For every other combination such as , (3 Negative , 2 Positive + 1 negative)value of chi will be negative. Hence possible values of chi = 49 + 10 =59 IMO (B) yes , but when choosing (2 , 3 , positive number other 1 ) isn't this the same when we choose (6 , 1 , other positive number ) let's say we choose the following 2 , 3 , 30 6 , 1 , 30 both have the same result and we counted them twice , same for 2 , 3 , 8 and 6 , 1 , 8 also : 22 , 3 , negative other than 22 or 11 and 6 , 11 , negative other than 22 or 11 I'm not gonna count all possible double entrees since the only answer below 59 is is A please correct me
_________________
Hope this helps Give kudos if it does



Manager
Joined: 05 Oct 2017
Posts: 101
Location: India
Concentration: Finance, International Business
GPA: 4
WE: Analyst (Energy and Utilities)

Re: Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
26 Mar 2019, 04:23
foryearss wrote: shuvodip04 wrote: EgmatQuantExpert wrote: eGMAT Question of the Week #40The set S is defined as {1, 2, 3, 8, 21, 30, 22, 5, 6, 11}. The chi of set is defined as the product of any 3 elements of the given set S. How many values of chi are possible such that chi is at least 7? A. 50 B. 59 C. 60 D. 61 E. 120 Product of three elements in the set S is defined as chi We have to find the cases where chi>=7 Total negative elements = 5 Total positive elements = 5 Following are the only possibilities when product of three elements >= 7 1) 2 negative and 1 positive  2 negative out of 5 can be selected in 5C2 ways and 1 positive in 5 ways Number of possible values of chi = 5C2 * 5  1 [subtracting for the case when 2,3 and 1 will be selected] =50 1 = 49 2) three positive terms Out of 5 three positive can be chosen in 5C3 ways Number of possible values = 5C3 = 10 For every other combination such as , (3 Negative , 2 Positive + 1 negative)value of chi will be negative. Hence possible values of chi = 49 + 10 =59 IMO (B) yes , but when choosing (2 , 3 , positive number other 1 ) isn't this the same when we choose (6 , 1 , other positive number ) let's say we choose the following 2 , 3 , 30 6 , 1 , 30 both have the same result and we counted them twice , same for 2 , 3 , 8 and 6 , 1 , 8 also : 22 , 3 , negative other than 22 or 11 and 6 , 11 , negative other than 22 or 11 I'm not gonna count all possible double entrees since the only answer below 59 is is A please correct me I think you are right.But I could count only 3 values for which it holds true (2,3,8);(2,3,21);(2,3,30) with (6,1,8);(6,1,21);(6,1,30) So number of values: (593) = 56 Quote: also : 22 , 3 , negative other than 22 or 11 3 is not in the set. other combinations will not be possible : (22,3) with (1,8,21,30,6) and (6,11) with (2,3,22,5) no values overlap. I could not find other overlaps. Could not count it down to 50. Can you please list if you find other 6 numbers?



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3074

Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
Show Tags
26 Mar 2019, 05:08
Solution Given:• Elements of set S are {1, 2, 3, 8, 21, 30, 22, 5, 6, 11}. • chi of set is defined as the product of any 3 elements of the given set S. To find:• The number of values of Chi >=7 for set S.
Approach and Working:• The number of values of Chi being at least 7 = The number of values of Chi being greater 0  The number of values of Chi from 1 to 6.
Number of ways we can get chi >0• Chi will be greater than 0 when product of three numbers is positive.
o That can happen in two cases when we multiply 3 numbers.
1. All number positive.
• Ways to select three positive numbers= 5c3 =10 2. Two negative and one positive.
• Ways to select two negative numbers * Ways to select one positive number = 5c2 * 5c1 = 10 *5 =50 o Number of ways in which chi >0= 10 +50 =60 Number of ways we can get chi from 1 to 6.• We cannot directly know when chi can be from 1 to 6. • So, we will first calculate the least positive values of Chi and then second least and so on. o Least value of chi can be either the multiplication of least three positive integers or the multiplication of two greatest negative integers and one least positive integer.
Multiplication of three minimum positive integers = 1 * 6 *8 • This is certainly greater than 7.
Multiplication of two greatest negative integers and one least positive integer = 2 * 3 *1 =6 • So, least value of chi =6 o Second least value of chi is greater than 7. • Hence, Number of values of chi from 1 to 6 =1 The number of values of chi being at least 6 = 601=59. Correct Answer: B
_________________




Question of the Week  40 (The set S is defined as {1, 2, 3, 8,...)
[#permalink]
26 Mar 2019, 05:08






