All boxes in a certain warehouse were arranged in stacks of 12 boxes

All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived?

(A) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
(B) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.

Looks like a LCM problem :

Let x be the # of boxes before addition (multiple of 12) and let x + 60 be after addition
LCM : of 12 and 14 = 84
84 - 60 = 12 so atleast 12 boxes were present before another 60 were added

A) x < 110, x could be anything from 12(12*1 ====> 1 stack) to 108 (12*9 ===> 9 stacks)
multiple values —> not sufficient

B) x + 60 < 120, ======> we know that the LCM of 12 and 14 ====> 84 but this is for the least value of x
take the 2nd least value of x (next multiple of 12 ie 24)
LCM of 24 and 14 is 168 > 120, hence only one ans is possible ——> sufficient

We are given that initially the boxes were arranged in 12 stacks each. => No. of boxes(Before) is a multiple of 12. Or N = 12K.

Now, after the addition of 60 boxes, they are arranged in 14 stacks each. => No. of boxes(After) is a multiple of 14. Or N + 60 = 14K'.

We can say, we have 12K + 60 = 14K' --- (1)

Option 1 : No. of boxes BEFORE < 110.

or 12K < 110, we can have K = 1,2,3,4,5,6,7,8,9

Out of these values only K = 2 and 9 ( satisfies the equation (1) above ). Hence, we have two values of K ==> INSUFFICIENT.

Option 2 : No. of boxes AFTER < 120.

or 14K < 120, we can have K = 1,2,3,4,5,6,7

Out of these values only K = 2 ( satisfies the equation (1) above ). Hence, we have only one value of K ==> SUFFICIENT.

Hence, answer is B.

Before reading the 2 options itself, I felt like the question could be solved, rendering the 2 pieces of information invalid. So, obviously I'm making a mistake. Please help me out with a solution.

Thank you!

Hi All,

Let us assume that the total number of boxes were "T".
Initially, they were stacked in R1 number of rows with 12 columns each. Therefore, 12*R1 = T
Now 60 boxes were added : T+60 or 12*R1 + 60.
Now they can be stacked in R2 number of rows with 14 columns each: 14*R2 = T

We can rewrite the equations above as: 12*R1+ 60 = 14* R2
Further: 12 ( R1 + 5) = 14 ( R2)

Now, WHENEVER you find an equation like this, know it is an LCM question. The first value when this equation will hold true will be the least common multiple of 12 & 14 = 84

************Why is it so ? When is 3*N = 4*M ?
N=4 & M=3. This is 12, which is the LCM of 3 & 4*************

So the equation above will hold true when the total number of boxes ( after adding 60) are atleast 84 or multiples of 84.
84, 168 and so on.

Now, let us find out the values of R1 and R2 for 84, 168 etc. ( i will limit to the first two numbers)

84: R1= 2 & R2= 6
168: R1= 9 & R2=12

Now since the question asks for a unique value, let us look at the statements:

Statement1: 12*R1 < 110

This means that R1 can be any value less than or equal to 9. Since R2 can then have 2 values ( 6 & 12), we dont have a unique value for the total number of boxes.

Statement2: 14*R2 < 120

This means that R2 can be any value less than or equal to 8. This further implies that R1 has a unique value of 2. This is a sufficient statement.

B is our answer.

I hope you find this useful !

The number of boxes before additional 60 boxes arrived was a multiple of 12.
The number of boxes after additional 60 boxes arrived was a multiple of 14.

So we need to find a multiple of 12, which after adding 60, becomes a multiple of 14.

Statement 1: The number of boxes before additional 60 boxes arrived is less than 110. There could be 24 or 108 boxes before additional 60 boxes arrived since adding 60 to any of these multiples of 12 yields a multiple of 14. Insufficient.

24 + 60 = 84
108 + 60 = 168

Statement 2: The number of boxes after additional 60 boxes arrived is less than 120. So the number of boxes before additional 60 boxes arrived is 24 since adding 60 to 24 gives 84, a multiple of 14. Sufficient

ANSWER: B

"6 is not divisible by 7. So, (x + 5) has to be divisible by 7"

Why does (x + 5) need to be divisible by 7?

(2*6)/12 is an integer, and neither 2 or 6 is divisible by 12...?

Hi Hermes
I would like to help you out...

"6 is not divisible by 7. So, (x + 5) has to be divisible by 7"

Why does (x + 5) need to be divisible by 7 ?
this is because since 6 & 7 share no factor other than 0, i.e. 6 & 7 are coprime. hence (x+5) has to be divisible by 7 to make y an integer.

(2*6)/12 is an integer, and neither 2 or 6 is divisible by 12...?
In the example quoted by you, both 2 & 6 share factor of 2 with 12 (not coprime with 12). hence it is possible.

Hope it is clear now.

if you like my explanation, provide kudos

Solution:
\(Given:\)
Let the number of boxes before the addition of 60 boxes = ‘12x’
Let the number of boxes after the addition of 60 boxes = ‘14y’
To find:
How many boxes were in the warehouse before the 60 additional boxes arrived? We need find basically the value of “12x”.
Inferences:
From the question statement we can get;
\(14y = 12x + 60\)
Dividing by “2”; we get;
\(7y = 6x + 30\)
\(7y = 6(x + 5)\)
Here we can see that 6 is not divisible by 7; hence “(x+5)” must be divisible by “7”.
\(x + 5 = 7c\);
\(x = 7c – 5\).
Here, ‘x’ can take the values 2, 9, 16, 23…………… which will satisfy the above equation.

Analysis of statement 1: There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
This means, "12x " must be less than 110."
Let's see by substituting the values for “x”.
When x = 2; ∴\(12x=24\) which is less than 110.
When x = 9; ∴\(12x=108\) which is less than 110.
When x = 16; ∴\(12x=192\) which is greater than 110.
Here we have 2 values which satisfy the condition; i.e. x can be equal to 2 or 9.
Hence statement 1 is not sufficient to answer. We can eliminate options A and D.

Analysis of statement 2: There were fewer than 120 boxes in the warehouse after the 60 additional arrived.
This means, "12x + 60 " must be less than 120."
Let's see by substituting the values for “x”.
When x = 2; ∴\(12x+60=84\) which is less than 120.
When x = 9; ∴\(12x+60=168\) which is greater than 120.
We have a unique answer for x. Hence the number of boxes in the warehouse before the 60 additional boxes arrived = 84.
Hence statement 2 is sufficient to answer. We can eliminate the options C and E.

The correct answer option is “B”.

From my experience in practicing DS problems, I realised that we cannot assume anything. So, is it fair to assume that there are no boxes left after the boxes are arranged in 14 per group?

Given: All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over.

Asked: How many boxes were in the warehouse before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
Boxes = 12k; k is a positive integer
12k + 60 = 14m; m is a positive integer
6k + 30 = 7m
k=2; m=6
k=9; m=12
Boxes before the 60 additional arrived = {24,108}
NOT SUFFICIENT


(1) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.
Boxes = 12k; k is a positive integer
12k + 60 = 14m; m is a positive integer
6k + 30 = 7m
k=2; m=6
k=9; m=12
Boxes before the 60 additional arrived = 84
SUFFICIENT

IMO B

Let's say there are N boxes. According to the stem, N = 12a and N+60 = 14b where a and b are positive integers.

Now, consider this logic: if X = 3k + 1 and X = 4z + 3 then I can cenrtainly say that X = 12y + 10. Where k,z and y are positive. integers.

N+60 = 14b Or, N = 14b-56-4 Or, N = 14(b-4) - 4 Or, N = 14c - 4 Or, N = 14c - 14+10 Or, N = 14d +10. where d is a positive integer, meaning that N leaves a remainder of 10 when divided by 14.

N = 12a and N = 14d + 10
Hence, N = 168m + 24 using the logic above.

(1) Is sufficient now. N < 110 then N must be 24
Bunuel can you please point out flaws in my reasoning? I really don't get where I'm faltering...

Hi Bunuel,

Would you have by any chance any similar questions with which it is possible to train?

Thank you

I have a doubt here. For the first option, why do we have just 2 options for (A). Since we are checking for the number of boxes before the arrival of 60 boxes, we could have 9 options right? from 12,24,36.... 108. Only after the arrival of 60 boxes the bopxes are stacked in the ,multiples of 14 and hence the equation becomes 12x+60=14y.

Please let me know where I am going wrong

When a multiple of 12 (the original number of boxes) is increased by 60, the result is a multiple of 14 (the new number of boxes):
12x + 60 = 14y
6x + 30 = 7y
6(x+5) = 7y

Since the left side must be equal to a multiple of 7, x+5 must be equal to a multiple of 7.
Two smallest cases:
Case 1: x=2, with the result that x+5=7
Original number of boxes = 12x = 12*2 = 24
New number of boxes after the 60-box increase = 24+60 = 84

Case 2: x=9, with the result that x+5=14
Original number of boxes = 12x = 12*9 = 108
New number of boxes after the 60-box increase = 108+60 = 168

Statement 1: There were fewer than 110 boxes in the warehouse before the 60 additional arrived
Cases 1 and 2 both satisfy the condition that the original number of boxes is less than 110.
Since the original number of boxes can be different values, INSUFFICIENT.

Statement 2: There were fewer than 120 boxes in the warehouse after the 60 additional arrived.
Only Case 1 satisfies the condition that the new number of boxes is less than 120.
In Case 1, the original number of boxes = 24
SUFFICIENT.

All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
(2) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.

As per (1): 12k < 110 i.e. no. of boxes could be 12, 24, 36, 48, 60,72, 84, 96 & 108
12k+60 = 72, 84, 96, 108, 120, 132, 144, 156 & 168
12k+60 must be divisible by 14 for arranging them in a stack of 14. 84 and 168 are the two choices - not
sufficient

As per (2): 12k+60<120 i.e. no. of boxes could be 72, 84, 96, 108
12k+60 must be divisible by 14 for arranging them in a stack of 14. 84 is the only choice - sufficient

B is the right answer