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Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)

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Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 22 Mar 2019, 00:45
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Question Stats:

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e-GMAT Question of the Week #40

The 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

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Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 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 ; 120-61 ; 59
IMO B; 59


EgmatQuantExpert wrote:
e-GMAT Question of the Week #40

The 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

Image

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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.
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Re: Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 22 Mar 2019, 02:01
IMo D

All negative=10
2 positive and 1 negative=50
1 postive and 2 negative=1

Total 61

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Re: Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 22 Mar 2019, 02:43
What is this CHI anyways ?

Could not understand this ...

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Re: Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 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.
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Re: Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 22 Mar 2019, 06:06
EgmatQuantExpert wrote:
e-GMAT Question of the Week #40

The 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

Image


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)
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Re: Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 25 Mar 2019, 23:54
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shuvodip04 wrote:
EgmatQuantExpert wrote:
e-GMAT Question of the Week #40

The 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

Image


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
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Re: Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 26 Mar 2019, 04:23
foryearss wrote:
shuvodip04 wrote:
EgmatQuantExpert wrote:
e-GMAT Question of the Week #40

The 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

Image


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: (59-3) = 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?
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Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)  [#permalink]

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New post 26 Mar 2019, 05:08
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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 = 60-1=59.

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