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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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Joined: 25 Aug 2015
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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, 21:35
1

Try E-47. If you have 2x20s in it, you have a remainder of 7 which can't be filled with 6 or 9. However, if you have 1x20 in it, then you have a remainder of 27 which can be filled with 3 9s. Hence, E is out.

Try D- 46. If you have 2x20s in it, you have a remainder of 6 which can be filled with 6 apples. Hence, 46 is out.

Try C- 45. 45 is a multiple of 9 so it is automatically out.

Try B- 44. If you have 2x20s in it, you have a remainder of 4 which can't be filled with 6 or 9 apples. If you have 1 20 in it, you have a remainder of 24 which is a multiple of 6. So this is out.

A. 43
B. 44
C. 45
D. 46
E. 47
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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, 22:04
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 use elimination method here.

A. 43 : Correct
B. 44 : 20+24 eliminate
C. 45 : 9* 5 eliminate
D. 46 : 40 + 6 eliminate
E. 47 : 20 + 27 eliminate

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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, 22:23
1
A) 43 (no combination of 6,9 and 20 to adds up to get 43)

B) 44=1(20)+4(6)

C) 45=5(9) or 45=3(9)+3(6)

D) 46=2(20)+1(6)

E) 47=1(20)+3(9)

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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, 23:53
1
Let number of packets of 6 apples be = x
Let number of packets of 9 apples be = y
Let number of packets of 20 apples be = z

Let us go through the options now,

A. 43

No combinations of 6, 9, 20 add up to 43
=> Correct.

B. 44

6*1 + 9*3 + 20*1 = 44
=> Incorrect.

C. 45

6*0 + 9*5 + 20*0 = 45
=> Incorrect.

D. 46

6*1 + 9*0 + 20*2 = 46
=> Incorrect.

E. 47

6*3 + 9*1 + 20+1 = 47
=> Incorrect.

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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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26 Jul 2019, 23:54
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?

45 is a multiple of 9 , ---So eliminate
46 = 2*20 +6-------------Eliminate
47 = 20 + 3*9 ------------Eliminate
44 = 20 +6*4 ------------Eliminate

So left out is Option 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, 23:58
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
C. 45
D. 46
E. 47

So let's check the options:

43 -
44 = 20 + 4*6
45 = 6*6 + 9
46 = 20*2 + 6
47 = 20+3*9

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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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27 Jul 2019, 00:16
1
Lets begin from the highest option we have

A. 43
B. 44
C. 45
D. 46
E. 47

47= 9 *3 + 20*1
46= 20*2 + 6*1
45 = 9*5
44 = 6*4 +20*1
43 = Not POSSIBLE

Hence, IMO, 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  [#permalink]

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27 Jul 2019, 00:34
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
C. 45
D. 46
E. 47

Three type of packs are available i.e. 6, 9 and 20. The packs can be sold if following ways:

1. Only one type alone (3 ways – 6 only, 9 only, 20 only )
2. Two types together (3 ways – 6 & 9, 9 & 20, 20 & 6)
3. Three type together (1 ways – all three together)

Total of seven ways are there each of sum lying between 43 and 47 including. Elaborated as per table below in snapshot.

6’s and 20’s don’t result in any of the given numbers. Only option ‘A’ i.e. 43 is false here.

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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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27 Jul 2019, 00:52
To find: which option is the greatest no. of apples that can't be bought in packs of 6, 9 or 20?
Solution: A) 43- we can't buy 43 apples as its not divisible by either 6, 9,20 or combination of any
43/6=7 packs +1 apple
43/9=4 packs of 9+7 apples=4 packs of 9+1 pack of 6+1 apple
43/20=2 packs of 20+3 apples
B) 44- we can't buy 45 apples as well
44/6=7 packs of 6+2 apples
44/9=4 packs of 9 + 8 apples=4 packs of 9+1 pack of 6+2 apples
44/20=2 packs of 20+4 apples
C) 45- we can buy 45 apples as 45/9=5 packs of 9
D) 46-we can buy 46 apples as 46/20=2 packs of 20 + 6 apples=2 packs of 20+1 pack of 6
E) 47- we can't buy 47 apples as well
47/6=7 packs of 6+5 apples
47/9=5 packs of 9+2 apples
47/20=2 packs of 20+7 apples=2 packs of 20+1 pack of 6+1 apple

So, greatest no. of apples that can't be bought is 47.

Alternatively, one may start with the biggest number in the options and will find that option to be the 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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27 Jul 2019, 01:05
1
Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, ..
Multiples of 9 = 9, 18, 27, 36, 45, ..
Multiples of 20 = 20, 40, ..

(i) 43, we cannot get this number by any combination of multiples of 6, 9 or 20
(ii) 44, 6x4 + 20x1
(iii) 45, 9x5
(iv) 46, 6+20x2
(v) 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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27 Jul 2019, 02:50
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?

For which of the following 6a+9b+20c cannot equal to an integer (a,b, and c are all positive integers)

A. 43 = 20+23 - can 6a+9b equal to 23. No, because 6*4 is 24, 2*9+6=24, 20+6=26, 6*3+9=27, 3*9=27 OR 23-6=17 (prime number cannot be divided by composite number), 23-9=14(not divisible by 6 or 9), 23-12=11(prime number cannot be divided by a composite number), 23-18=5(prime number cannot be divided by a coposite number). As we see, from 6a+9b+20c we cannot get 43. Keep A

B. 44 = 20+6+2*9 - works
C. 45 = 5*9 - works
D. 46 = 20*2+6 - works
E. 47=20+3*9 - works

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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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27 Jul 2019, 03:11
2
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--> correct :can't be bought: 43 can be make w/ any combination of 6,9 & 20
B. 44--> can be bought: 44=20+4*6
C. 45--> can be bought: 45 = 5*9
D. 46--> can be bought: 46=2*20+6
E. 47--> can be bought: 47=20 + 3*9
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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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27 Jul 2019, 03:50
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?

By looking at answer choices we need to find the greatest number which does not satisfy the eq.
N = 6a+9b+20c (where a, b, c >=0).

Now,
44 = 6*4 + 20*1
45 = 9*5
47 = 9*3 + 20*1

Both, 43 and 46 doesn't fit the combination of 6, 9 and 20.
and greatest of both is 46.

A. 43
B. 44
C. 45
D. 46
E. 47

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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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27 Jul 2019, 04:08
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?

Here, the combination is 6x+9y+20z

A. 43 it is not possible in any combination
B. 44 is possible when x= 4 and z= 1
C. 45 is possible when x= 6 and y= 1
D. 46 is possible when x= 1 and z= 2
E. 47 is possible when y = 3 and z = 1

so, the correct answer choice 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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27 Jul 2019, 04:44
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?

So, the question is the greatest:
!= not equal
6*x+9*y+20*z != a

Let's input variants (start from the greatest):
E. 47
6*x+9*y+20*z = 47
x - y - z - 47
any - any - 2 - not 47
0 - 3 - 1 - 47
D. 46
6*x+9*y+20*z = 46
x - y - z - 46
1 - 0 - 2 - 46
C. 45
6*x+9*y+20*z = 45
x - y - z - 45
0 - 5 - 0 - 45
B. 44
6*x+9*y+20*z = 44
x - y - z - 44
4 - 0 - 1 - 44
A. 43
6*x+9*y+20*z = 43
x - y - z - 43
any - any - any != 43

A. 43
B. 44
C. 45
D. 46
E. 47
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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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27 Jul 2019, 05:16
1
Lets start from the greatest number from the option
47 = (20 + 9*3i.e 27 =47) hence NO
46 = (20*2 + 6) hence No
45 = (9*5) hence No
44 = (20+6*4)hence No
43 Not possible
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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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27 Jul 2019, 05:23
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
C. 45
D. 46
E. 47

Solution:
I tried this question by hit and trial.
By picking up the various answer options i tried to see if the combination of various packs of apples is possible or not .Also we need to find the greatest number of apples which cannot be bought.

44 is possible to buy with 1 pack of 6+2 pack of 9+1 pack of 20=6+18+20=44
45 is possible to buy 5 pack of 9=45
46 is possible to buy with 1 pack of 6 and 2 pack of 20=6+40=46
47 is possible to buy with 3 pack of 9 and 1 pack of 20=27+20=47
No combination would buy us 43

Hence A IMO
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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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27 Jul 2019, 05:36
A. 43 --
B. 44--6*4 +20
C. 45--9*5
D. 46
E. 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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27 Jul 2019, 05:44
1
Solution:

In this question, it is easier to start with the answer choices and try out various combinations to fit into that particular answer choice.
According to the Stem, we can use any combinations of 6,,9, 20 or maybe just a single number

A. 43 = Now 43 cannot be obtained by any combinations we try. Hence lets keep this one.
B .44 = 6 + (9 X 2) + 20 Here we use 1 pack of 6, 2 packs of 9 & one pack of 20. Hence we can buy apples in this combination so this option is out.
C .45 = 5 X 9 . we can straightaway use 5 packs of 9 and get 45 apples. Hence this option is out.
D. 46 = we can obtain this by using 2 packs of 20 and one pack of 6. Hence this is out.
E. 47 = If we use a larger pack of 20. we have 27 remaining for which we can use 3 packs of 9 apples and hence this combination is possible. Rule out this choice

So the answer must 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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27 Jul 2019, 06:04
1
Lets check each of the choices in descending order of their value
E. 47; 47 can be expressed as 9+9+9+20 =>Incorrect
D. 46; 46 can be expressed as 6+20+20 =>Incorrect
C.45; 45 can be expressed as 9+9+9+9+9 =>Incorrect
B.44; 44 can be expressed as 6+6+6+6+20 =>Incorrect
A. 43; Must be correct

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] 27 Jul 2019, 06:04

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