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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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Senior Manager
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:01
1
A. 43 - No possible
B. 44 = 20 + 6x4
C. 45 = (6x3) + (9x3)
D. 46 = 20x2 + 6
E. 47 = (9x3) + 20

A is correct
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A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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Updated on: 26 Jul 2019, 11:37
1
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.
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:08
1
44- 20+6*3
45- 9*5
46- 6*8
47-20+9*3
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:10
1
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 OK
B. 44 - To get 44, we can use combination $$20+9+9+6=44$$ or $$20+4*6=44$$ OK
C. 45 - To get 45, we can use combination $$9*5=45$$ or $$6*4+9=45$$ OK
D. 46 - To get 46, we can use combination $$20+20+6=46$$ OK
E. 47 - To get 47, we can use combination $$20+9*3=47$$ OK

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

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26 Jul 2019, 09:18
1
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

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

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26 Jul 2019, 09:22
1
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

A. 43
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:23
1
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)
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:27
1
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.

Please hit kudos if you like the solution.
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:27
1
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.
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:27
1
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
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:35
1
IMO-A

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)
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:43
1
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.
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 09:44
1
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

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

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26 Jul 2019, 09:49
1
A: 43
B: 44 = 20+6+(9X2)
C: 45 = (9X5)
D: 46 = (20X2) + 6
E: 47 = 20+ (9X3)

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

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26 Jul 2019, 09:55
1
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

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

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26 Jul 2019, 10:07
1
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
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 10:08
1
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.... DISCARDED
C. 45 - 6*6 = 36; 36+9 = 45.... DISCARDED
D. 46 - 6+20+20 = 46.... DISCARDED
E. 47 - 9*3 = 27; 27+20 = 47.... DISCARDED

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

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26 Jul 2019, 10:34
1

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

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26 Jul 2019, 10:57
1
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)

Note: Strategy is to start from 1 type of pack and if not possible then add one more type of pack and so on.
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Re: A store sells apples in pack of 6, 9, or 20. By ordering 2 packs of 6  [#permalink]

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26 Jul 2019, 11:01
1
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

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