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

Question

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

The answer is A

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

Sayon · Answer

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

Kinshook · Answer

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

The answer is A

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 store

B. 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 store

C. 45
45 = 9*5
We can purchase 45 items from the store

D. 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 store

E. 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

freedom128 · Answer

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)

aliakberza · Answer

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.

rahaman26 · Answer

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?

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

The answer is A

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.

Sammysat · Answer

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.

himanshu14 · Answer

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

Posted from my mobile device

JonShukhrat · Answer

The question basically asks which of the following CANNOT be the results of  6x+9y+20z .  x ,  y , and  z  are non-negative 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.   43-2*20=3  out.  43-20=23  can  23  be the result of  6x+9y ? Let’s begin from the multiples of  9  because we have less calculation with it.  23-18=5  out.  23-9=14  not divisible by  6 , so out. Now let’s use multiples of  6 .  23-18=5  out.  23-12=11  out.  23-6=17  not divisible by  9  out. So  20  apple pack is not participating in  43 .

2.  Let’s start from the multiples of  9 .  43-36=7  out.  43-27=16  out.  43-18=25  out.  43-9=34  out because all the results are not divisible by  6 .

3.  Let’s use the multiples of the  6  now.  43-42=1  out.  43-36=7  out.  43-30=13  out.  43-24=19  out.  43-18=25  out.  43-12=31  out.  43-6=37  out because all the results are not divisible by  9 . Hence  6x+9y+20z
eq{43} .

B.   44 = 20 + 4*6

C.   45 = 5*9

D.   46 = 2*20 + 6

E.   47 = 20 + 3*9

Hence  A

CareerGeek · Answer

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

Posted from my mobile device

ccheryn · Answer

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

The answer is A

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

somesh86 · Answer

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.

Answer: A

Vigasimrair · Answer

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

The answer is A

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

The answer is A

unraveled · Answer

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

The answer is A

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.

Answer (A).

hiranmay · Answer

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

The answer is A

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

Abhi077 · Answer

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

globaldesi · Answer

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

The answer is A

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

hence bananas can be of the form 6x or 9x or 20 x or 6 x+9y+20z
here x y and z are >=0
Try out various combinations
 6x+9y+20z
47 = 6*0 +9*3+20*1 
46 = 6*1 +20*2
45 = 6*0+9*5 + 20*0
44 = 6*4 +20*1+ 9*0
43 .. no combination can not form it
Hence 43 A is the answer

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

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