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A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:00
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67% (01:45) correct 33% (02:01) wrong based on 272 sessions
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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 C. 45 D. 46 E. 47
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A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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Updated on: 26 Jul 2019, 08:35
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?
A. 43 B. 44 C. 45 D. 46 E. 47 All are possible except 43 9*5=45 9*3+20=47 20+6*4=44 6*6+20=46 Hence A
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Originally posted by kitipriyanka on 26 Jul 2019, 08:13.
Last edited by kitipriyanka on 26 Jul 2019, 08:35, edited 3 times in total.



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:14
I thought of the combinations that you can buy, so disprove of the ones you can buy.
B (44) = 4(6) + 1(20) C (45) = 5(9) D (46) = 2(20) + 1(5) E (47) = 1(20) + 3(9)
I did not find anything to sum to 43



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:18
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+4*6) C. 45 ; ( 6*3+9*3) D. 46 ; ( 20*2+6*1) E. 47 ; ( 20*1+3*9)
out of options only option A ; 43 is not possible rest all are possible



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:19
44 is a possible combination 6*6+20 45 is 5 packs of (9) 47 is 20+3*(9) 46 is 2*(20)+6
leaving us with 43 So correct answer is A



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:19
we cannot have 43 and 47. 47 is the largest. E is the answer.
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A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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Updated on: 26 Jul 2019, 22:23
Let us go option by option to check which combination cannot be formed
E. 47  9 * 3 + 20 > can be formed. D. 46 – 20 * 2 + 6 > can be formed. C. 45 – 9 * 5 > can be formed. B. 44 – 6 * 4 + 20 > can be formed. A. 43 > cannot be formed.
Answer A
Originally posted by Sayon on 26 Jul 2019, 08:20.
Last edited by Sayon on 26 Jul 2019, 22:23, edited 1 time in total.



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:24
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? Given: 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. Asked: What is the greatest number of apples that you CANNOT buy in the store? Let 6x + 9y+ 20z = k where k = number of apples to be bought form the store A. 43 43 is not divisible by 6 or 9 43 = 20 + 23 but 23 is also not divisible by 6 or 9 43 = 20*2 + 3 but 3 is also not divisible by 6 or 9 We can not purchase 43 items from the storeB. 44 44 is not divisible by 6 or 9 44 = 20 + 24 but 24 is divisible by 6 44 = 20 + 6*4 We can purchase 44 items from the storeC. 45 45 = 9*5 We can purchase 45 items from the storeD. 46 46 is not divisible by 6 or 9 46 = 20 + 26 but 26 is divisible by 6 46 = 20*2 + 6 We can purchase 46 items from the storeE. 47 47 is not divisible by 6 or 9 47 = 20 + 27 but 27 is divisible by 9 47 = 20 + 9*3 We can purchase 47 items from the store
IMO A
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:24
You can solve it using the options: If I start with C 45 20+20+9+6, so C is possible.
So higher than C will be answer so,A,B,C option goes away.
From D&E, E can be eliminated by 9+9+9+20 (24+20)
So answer is D.



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:25
A
All other no of apples can be bought using the combination below:
A. 43 B. 44 : 4*6+1*20 C. 45: 5*9 D. 46: 1*6+2*20 E. 47 : 3*9+1*20



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:26
Using PoE Start from Option E, since largest is asked
E) 47 20 + 27 = 20 + 9*3 Eliminated D) 46 20*2 + 6 Eliminated C) 45 9*5 Eliminated B)44 20 + 6*4 Eliminated So, A



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:28
To solve this problem, we should just check the answers  this is the easiest way.
A. 43 The ANSWER
B. 44 20+4*6=44
C. 45 9*5=45
D. 46 2*20+6=46
E. 47 20+9*3=47
The answer is A



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:30
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?
A. 43 B. 44 C. 45 D. 46 E. 47 B. 44=20+6(4) C. 45=9(5) D. 46=20(2)+6 E. 47=20+9(3) Answer (A).



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:30
Here you must find a number that is not a multiple of the possible combinations of 6,20 and/or 9.
A. 43 , hold B. 44 = 20 + 6*4  So discard it C. 45 = 9*5  So discard it D. 46 = 20*2 + 6, so discard it E. 47 = 20 + 9*3, so dicard it
So by POE, (A) is the answer.



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:36
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?
This is an easy question. Straight away get to POE.
A. 43 This is correct it is not possible.
B. 44 This is possible 20 and 24/6.
C. 45 This is possible 45/9.
D. 46 This is possible 20*2 and 6*1
E. 47 This is possible 20*1 & 27/9
Hence the correct answer is A.



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:38
4x6 + 1x20 = 44 5x9 =45 1x6 + 2x20 = 46 3x9 + 1x20 = 47
Therefore, the answer is 43



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:44
IMO answer is A
45 = 27+18 > multiples of 9 and 6
46 = 40 + 6 > multiples of 20 and 6
44 = 6+18+20 > multiples of 6,9 and 20
47 = 27 + 20 > multiples of 9 and 20
only remaining answer is 43, which is not possible, so A



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:50
A. 43 NOT POSSIBLE
B. 44 44 = 6*4 + 20*1  POSSIBLE
C. 45 45 = 9*5  POSSIBLE
D. 46 46 = 6*1 + 20*2  POSSIBLE
E. 47 47 = 6*3 + 9*1 + 20*1  POSSIBLE
Answer is (A)



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:51
We are given that, there are pack of 6, 9 or 20 We also know that if someone buys 2 pack of 6 =2*6=12 (he gets 12 but he cant get 13) This means that the pack number is an Integer.
So we are going to use hit and trail method in this question.
1) 6*4+20=44 2) 6*6+9=45 3) 20*2+6=46 4) 9*3+20=47
This leaves us wtih ans =43=A



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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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26 Jul 2019, 08:59
Method  Eliminate possible combinations from answer choicesA. 43  No Combination exists B. 44  (1*20 Packs) + (4*6 Packs) C. 45  (5*9 Packs) D. 46  (1*6 Packs) + (2*20 Packs) E. 47  (3*9 Packs) + (1*20 Packs) IMO A
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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