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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, 11:02
Resolve by elimination, starting with the largest number in the answer options: 47: (9*3) + (20*1) > Eliminate 46: (20*2) + (6*1) > Eliminate 45: (9*5) >Eliminate 44: (6*3)+(20*1) > Eliminate 43: No combination
We can also solve by factorizing each of the answer options.
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, 11:39
For the second option we've got 44= 6×4 + 1×20 For the third option we've got 45= 9×5 For the fourth option we've got 46= 2×20 + 1×6 For the final optio we've got 47= 1×20 + 3×9 Posted from my mobile device
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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, 11:40
Best way is to check with option
A. 43 (((Correct Answer))) B. 44 (( 20 + 6*4 )) C. 45 (( 5*9 )) D. 46 (( 20 + 20 + 6)) E. 47 (( 20 + 6*3 + 9))



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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, 12:07
IMO correct answer is A  Explanation is provided as attachment 
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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, 12:29
Here store can sell apples in a pack of (6k,6k+3,6k+2)i.e (6,9,20) Now with these three nos. we can form any no.of the form
46(6m+4) : so any no of the form 6m+4 can be formed by taking two(6k+2) and rest 6k. here 20+20+6
47(6m+5) : one 6k+3 and one 6k+2 rest 6k here 9+20+6+6+6
45(6m+3) : one 6k+3 and rest 6k here 9+6+6+6+6+6+6
44(6m+2) : one 6k+2 and rest 6k here 20+ 6+6+6+6
43(6m+1) : it can be form if we take 2 (6k+2) and one (6k+3) i.e 20+20+9 which is equal to 49. so the least no. of 6m+1 form that can be formed by 6,9,20 is 49 . so 43 apples is not possible with these bags of 6,9,20 apples
hence A : 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, 12:40
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? Let's see the numbers: A. 43Any combination of 6,9, and 20 won't give us 43 B. 444 of 6 pack = 24 + 1 pack of 20 = 44 C. 455 of 9 pack = 45 D. 462 of 20 pack = 40 + 1 pack of 6 pack = 46 E. 473 of 9 pack = 27 + 1 pack of 20 = 47 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, 13:20
To rephrase, one can only sell set of 6 apples or 9 apples or 20 apples and arrange that X number of set of 6 apples, Y number of set of 9 apples, and Z number of set of 20 apples A. 43 = B. 44 = 6*4 + 20*1 C. 45 = 6*6 + 9*1 D. 46 = 6*1 + 20*2 E. 47 = 9*3 + 20*1 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, 13:52
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
As we are asked the greatest number of apples, lets start from the options in the descending way. 47 = 20 + 27 = 20 + 9*3 hence E is out 46 = 20 +20 + 6 hence D out 45 = 9*5 hence C out. 44= 20 +24 = 20 + 6* 4 Hence B out 43 = 20 +23 or 9 + 32 or 6 +37 Hence combination of 6,9 and 20 is not possible. Option 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, 14:25
the 6pack can produce packages of: 0,6,12,18,24,30,36,42,48 the 9pack can produce packages of: 0,9,18,27,36,45 the 20pack can produce packages of: 0,10,20,30,40
47 = 20 + 27 = \((20*1)+(9*3)\) 46 = 40 + 6 = \((20*2)+(6*1)\) 45 = 45 = \((9*5)\) 44 = 20 + 24 = \((20*1)+(6*4)\)
so 43 is left, by trying combinations of the possible numbers, it is not possible
A



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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: 27 Jul 2019, 00:27
The question basically asks which of the following CANNOT be the results of \(6x+9y+20z\). \(x\), \(y\), and \(z\) are nonnegative integers. Let’s begin analyze one by one: A. \(43\). Let’s first begin from using 20 because it is the easiest to use. 1. \(432*20=3\) out. \(4320=23\) can \(23\) be the result of \(6x+9y\)? Let’s begin from the multiples of \(9\) because we have less calculation with it. \(2318=5\) out. \(239=14\) not divisible by \(6\), so out. Now let’s use multiples of \(6\). \(2318=5\) out. \(2312=11\) out. \(236=17\) not divisible by \(9\) out. So \(20\) apple pack is not participating in \(43\). 2. Let’s start from the multiples of \(9\). \(4336=7\) out. \(4327=16\) out. \(4318=25\) out. \(439=34\) out because all the results are not divisible by \(6\). 3. Let’s use the multiples of the \(6\) now. \(4342=1\) out. \(4336=7\) out. \(4330=13\) out. \(4324=19\) out. \(4318=25\) out. \(4312=31\) out. \(436=37\) out because all the results are not divisible by \(9\). Hence \(6x+9y+20z\neq{43}\). B. \(44 = 20 + 4*6\) C. \(45 = 5*9\) D. \(46 = 2*20 + 6\) E. \(47 = 20 + 3*9\) Hence A
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Originally posted by JonShukhrat on 26 Jul 2019, 15:52.
Last edited by JonShukhrat on 27 Jul 2019, 00:27, 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, 17:17
We need to find out the answer choices which are sum of the multiples of 6,9, and 20 and eliminate them. The highest of the remaining answer choices is our Answer.
47=20(1)+9(3)=20+27. Hence E is out. 46=20(2)+6=40+6, hence D is out as well. 45 =9(5), hence C is out. 44=20+6(4)=20+24, hence B is out. The answer is therefore 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, 17:52
A. 43 —> Not Possible B. 44 = 20 + 24 = 20x1 + 6x4 —> Possible C. 45 = 36 + 9 = 6x6 + 9x1 —> Possible D. 46 = 40 + 6 = 20x2 + 6x1 —> Possible E. 47 = 20 + 27 = 20x1 + 9x3 —> Possible
IMO Option 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, 17:52
We can buy 5 packs of 9 apples = 45.........Eliminate C
We can buy 3 packs of 9 apples and 1 pack of 20 apples= 27 + 20=47.............Eliminate E
We can buy 2 packs of 20 apples and 1 pack of 6 apples = 40 + 6 =46.............Eliminate D
We can buy 1 pack of 20 apples and 4 packs of 6 apples = 20 +24 =44.............Eliminate B
Only one answer left.
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, 18:04
Since answer choices are small, it's easy to check some numbers
A) 43 not possible B) 44 = 20(1) + 6(4) C) 45 = 9(5) D) 46 = 20(2) + 6(1) E) 47 = 20(1) + 9(3)
Ans is A



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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, 19:54
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 solve one by one using combination
B. 44  4 packs of 6 apples and 1 pack of 20 apples = 44 C. 45  6 packs of 6 apples and 1 pack of 9 apples = 45 D. 46 1 pack of 6 apples and 2 packs of 20 apples = 46 E. 47 3 packs of 9 apples and 1 pack of 20 apples = 47
so i can buy above 44,45,46,and 47 no of apples as per the above combinatons explanined
but 43 apples cannot be bought by any combination, Hence that is the greatest number not possible out of the choices given
so out of the choices A. 43 cannot be bought B. 44  can be bought C. 45 can be bought D. 46 can be bought E. 47 can be bought
Hence ans 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, 20:34
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
20 = 2*2*5 6 = 2*3 9 = 3*3
So any multiple of the above combinations will be possible to buy
I tried to backsolve using the answer choices
Taking 43, 43 can't be factorized into multiples of 20, 6 or 9. So, this should be the correct answer. Other choices can be verified for confirmation. 44 = 20 + 6*4 45 = 9*5 46 = 20*2 + 6 47 = 20 + 3*9
Correct 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, 20:51
Answer:A POE B 44: 20+ (6x4)= 44 Eliminate C 45: (9x5)= 45 Eliminate D 46: (20x2)+ +6= 46 Eliminate E 47: 20+ (9x3)= 47 Eliminate
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, 20:54
A cant be bought. B 6 + 2×9 + 20 C 9x5 D 20x2 + 6 E 9x3 + 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, 21:20
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?
Eliminating options, 6*4 + 20 = 44, eliminate option B 9*5 = 45, eliminate option C 6 + 20*2 = 46, eliminate option D 9*3 + 20 = 47, eliminate option E
So, 43 apples cannot be bought.
Hence option 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, 21:22
IMO A.
Given: store sells apples in pack of 6, 9, or 20 and can be sold in whole packs only.
To find: Greatest number of apples that cannot be bought from the store.
A. 43> Cannot be summed. Correct answer. B. 44> Can be summed as = \(6 X 4+ 20 X 1.\) C. 45> Can be summed as =\(6 X 6+ 9 X 1.\) D. 46 > Can be summed as =\(20 X 2+ 6 X 1.\) E. 47 > Can be summed as = \(20 X 2+ 9 X 3.\)




Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6
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