Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 28 Feb 2014
Posts: 267
Location: India
Concentration: General Management, International Business
GPA: 3.97
WE: Engineering (Education)

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:01
A. 43  No possible B. 44 = 20 + 6x4 C. 45 = (6x3) + (9x3) D. 46 = 20x2 + 6 E. 47 = (9x3) + 20
A is correct



Intern
Joined: 29 May 2018
Posts: 23
Location: India

A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
Updated on: 26 Jul 2019, 11:37
This question can be easily be solved by elimination.
we need to find out which of the options cannot be formed/purchased using either one or a combination of 6, 9 and 20.
A. 43  Cannot be formed using 6, 9 or 20 or the combination.
B. 44  6 x4+20 x 1 = 44. ELIMINATE
c. 45  9 x 5= 45. ELIMINATE
D 46  20 x 2+6= 46. ELIMINATE
E. 47  9 x 3= 47. ELIMINATE
ANS= A; 43.
Originally posted by Ajiteshmathur on 26 Jul 2019, 09:07.
Last edited by Ajiteshmathur on 26 Jul 2019, 11:37, edited 1 time in total.



Intern
Joined: 02 Apr 2019
Posts: 7

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:08
43 ANSWER 44 20+6*3 45 9*5 46 6*8 4720+9*3



Manager
Joined: 26 Mar 2019
Posts: 108
Concentration: Finance, Strategy

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:10
Quote: A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store? We are given that can buy apples only in packs of 6, 9 and 20. Apples cannot be bought separately. We need to identify which of the answers cannot be get using these combinations of numbers. Let us analyze each option: A. 43  This number cannot be obtained by using combinations of \(6\), \(9\) and \(20\) NOT OKB. 44  To get 44, we can use combination \(20+9+9+6=44\) or \(20+4*6=44\) OKC. 45  To get 45, we can use combination \(9*5=45\) or \(6*4+9=45\) OKD. 46  To get 46, we can use combination \(20+20+6=46\) OKE. 47  To get 47, we can use combination \(20+9*3=47\) OKAnswer: A



Director
Joined: 30 Sep 2017
Posts: 570
GMAT 1: 720 Q49 V40
GPA: 3.8

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:18
Question: What is the greatest number of apples that you cannot buy in the store, given that only packs of 6, 9, or 20 are available ?
A. \(43\) > CORRECT ANSWER. We cannot buy 43 apples, since no combination of 6, 9, and 20 adds up to 43 B. \(44 = 6 + 2*9 + 20\) > To get 44 apples, we can buy a pack of 6, two packs of 9, and a pack of 20 C. \(45 = 3*6 + 3*9\) > To get 45 apples, we can buy three packs of 6 and three packs of 9 D. \(46 = 6 + 2*20\) > To get 46 apples, we can buy a pack of 6 and two packs of 20 E. \(47 = 3*9 + 20\) > To get 47 apples, we can buy three packs of 9 and a pack of 20
Answer is (A)



Manager
Joined: 19 Apr 2017
Posts: 168
Concentration: General Management, Sustainability
GPA: 3.9
WE: Operations (Hospitality and Tourism)

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:22
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?
B. 44=6*4+20*1 C. 45=9*5 D. 46=6*1+20*2 E. 47 = 9*3+20*1
Answer A. 43



BSchool Moderator
Joined: 07 Dec 2018
Posts: 160
Location: India

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:23
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?
A. 43 B. 44 = 20*1 + 6*4 C. 45 = 9*5 D. 46 = 20*2 + 6*1 E. 47 = 20*1 + 9*3
A is the lone survivor. Ans should be (A)



Manager
Joined: 08 Jan 2018
Posts: 98
Location: India
GPA: 4
WE: Information Technology (Computer Software)

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:27
Going by option choices: 1. 6x + 9y + 20z = 43 This cannot be possible if we take any value of x, y, z the equation will not be true.
2. 6x + 9y + 20z = 44 If x = 1 , y = 2 , z =1 then the equation will be possible.
3. 6x + 9y + 20z = 45 If x = 0 , y = 5 , z =0 then the equation will be possible.
4. 6x + 9y + 20z = 46 If x = 1 , y = 0 , z =2 then the equation will be possible.
5. 6x + 9y + 20z = 47 If x = 0 , y = 3 , z =1 then the equation will be possible.
IMO the answer is A.
Please hit kudos if you like the solution.



Manager
Joined: 11 Feb 2018
Posts: 80

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:27
IMO the question is best solved using POE:
A. 43  43 apples is an impossible combo.
B. 44 9*2+6*1+20*1=44  44 apples is possible
C. 45 9*5=45 45 apples is possible
D. 46  9*0+6*1+20*2=46 46 apples is possible
E. 47  9*1+6*3+20*1=47 47 apples is possible
As all combos other than A are possible, answer is A.



Manager
Joined: 12 Mar 2019
Posts: 157

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:27
IMO :A We should first see for odd values, as even values can be obtained. 43, 45, 47 : 45 : is a straight multiple of 9 so incorrect 47 : 20+9+9+9: So incorrect 43: Try and make diff permutation and combination, will not be able to make 43, So correct



Manager
Joined: 08 Jan 2018
Posts: 145
Location: India
Concentration: Operations, General Management
WE: Project Management (Manufacturing)

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:35
IMOA
Packs are of 6# 9# 20#
Let x, y, z be the number packs with 6 unit , 9 Unit, 12 Unit
6x + 9y +20z = K .....[K (Integer) = No. of apples  x, y, z  Integer ]
z=1 , x=4 , then K= 44
y=5 then K= 45
z=2 , x=1 , then K= 46
z=1 , y=3 , then K=47
So (A) 43 left out Ans. 43 (A)



Manager
Joined: 18 Sep 2018
Posts: 100

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:43
IMO A
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?
We can follow the back calculation method to solve this question. Let us test each answer choice to see whether they can be represented in the form of 6a+9b+20c (where, a, b and c are variables >=0)
A.43 There is no way to represent 43 in the form of 6a+9b+20c This is a possible answer choice, but let us test the other choices before we are 100% confident
B.44 6*4 + 20*1 = 24+20 = 44 => wrong
C.45 9*5 = 45 => wrong
D.46 6*1 + 20*2 = 6+40 = 46 => wrong
E.47 9*3 + 20 = 27+20 = 47 => wrong
Hence, 43 is the correct answer.



Manager
Joined: 30 Nov 2017
Posts: 194
WE: Consulting (Consulting)

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:44
Given that: The apples are sold in pack of 6,9 or 20 To find the greatest number of apples that cannot be bought Starting from Option C. Given 45: therefore, 45 = 9 pack x 5 (PROVEN it can be bought) Option B. Given 44: therefore, 44 = 20 pack x 1 + 6 pack x 4 = 20 + 24 (PROVEN it can be bought) Option D. Given 46: therefore, 46 = 20 pack x 2 + 6 pack x 1 = 40 + 6 (PROVEN it can be bought) Option E. Given 47: therefore, 47 = 20 pack x 1 + 9 pack x 3 = 20 + 27 (PROVEN it can be bought) Option A. Given 43: therefore, 43 is not equal to any combination of 6,9 or 20 Final answer = 43
Hence answer choice A.
_________________
Be Braver, you cannot cross a chasm in two small jumps...



Intern
Joined: 09 Feb 2017
Posts: 16
Location: India
Concentration: Marketing, Operations
GMAT 1: 660 Q49 V31 GMAT 2: 700 Q48 V37
GPA: 2.87
WE: Engineering (Manufacturing)

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:49
A: 43 B: 44 = 20+6+(9X2) C: 45 = (9X5) D: 46 = (20X2) + 6 E: 47 = 20+ (9X3)
Answer is A



Manager
Joined: 06 Aug 2018
Posts: 97

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 09:55
The most easy to elimunate is 45 as 9*5=45
20+18+6=44
Then
20+9*3=47
Then
20+9*2+6*2=46
We only can't get 43
It is A
Posted from my mobile device



Senior Manager
Joined: 13 Feb 2018
Posts: 496

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 10:07
As we are dealing with small numbers POE is the best IMO
A) 43; you cant make equation 6a+9b+20c=43 a, b and c are non negative integers B) 44; 20+4*6=44 possible C) 45; 9*5=45 possible D) 46; 2*20+6=46 Possible E) 47; 20+9+3*6 Possible
IMO Ans: A



Manager
Joined: 24 Jun 2019
Posts: 113

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 10:08
Quote: A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?
I found it easiest to try and eliminate all options untill one remains. Calculating it directly seems extremely difficult A. 43  LAST MAN STANDING  It is impossible to buy 43... since it is the only option left, it must also be the greatest number you cant buy. B. 44  6*4 = 24; 24 + 20 = 44.... DISCARDEDC. 45  6*6 = 36; 36+9 = 45.... DISCARDEDD. 46  6+20+20 = 46.... DISCARDEDE. 47  9*3 = 27; 27+20 = 47.... DISCARDEDANSWER: ABONUS Numberphile video! https://www.youtube.com/watch?v=vNTSugyS038



Intern
Joined: 11 Jun 2014
Posts: 26

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 10:34
Answer:A
combination should be 6x+9y+20z x=3,y=1,z=1 get 47 simll x=1 z=2 get 46 x=3 y=3 get 45 x=1 y=2 z=1 get 44 hence answer 43



Manager
Joined: 24 Jan 2019
Posts: 107
Location: India
Concentration: Strategy, Finance
GPA: 4

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 10:57
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?
A. 43 (no combination possible) B. 44 = (20+6*4) C. 45 = (9*5) D. 46 = (20*2+6) E. 47 = (20+9*3)
ANSWER: A
Note: Strategy is to start from 1 type of pack and if not possible then add one more type of pack and so on.



Intern
Joined: 20 Nov 2018
Posts: 19

Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
Show Tags
26 Jul 2019, 11:01
A store sells apples in pack of 6, 9, or 20. By buying 2 packs of 6, you can get 12. But you cannot buy 13 apples, since no combination of 6, 9, and 20 adds up to 13. What is the greatest number of apples that you CANNOT buy in the store?
Lets go through the options and see if we can find a combination of 6,9, and 20 to result in the number given as our answer choices. A. 43  I can not find a possible combination of 6, 9, and 20 to result as 43. just keep it. B. 44  20+(6*4)= 20+24 = 44 C. 45  9*5=45
D. 46  6+(20*2)= 6+40= 46
E. 47 20+(6*3)+9= 20+18+9= 47
Bingo..!! All the other number of apples we CAN buy at the store through some combination of 6, 9, and 20 but not 43.
So, IMO A is the answer.




Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
[#permalink]
26 Jul 2019, 11:01



Go to page
Previous
1 2 3 4 5
Next
[ 85 posts ]



