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

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)

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.

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.

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.

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

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 \(6x+9y+20z\neq{43}\).

B. \(44 = 20 + 4*6\)

C. \(45 = 5*9\)

D. \(46 = 2*20 + 6\)

E. \(47 = 20 + 3*9\)

Hence A

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

Pls Hit kudos if you like the solution

Posted from my mobile device

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

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

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 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.

Answer (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--> 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

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

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

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

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