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All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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19 Jun 2016, 09:25

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

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes.. [#permalink]

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19 Jun 2016, 09:41

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Given: 12x + 60 = 14y

y = 12(x + 5)/14 = 6(x + 5)/7

6 is not divisible by 7. So, (x + 5) has to be divisible by 7 --> (x + 5) = 7k x = 7k - 5. Therefore x = 2, 9, 16, 23, ..........

St1: There were fewer than 110 boxes in the warehouse before the 60 additional arrived. --> 12x < 110 When x = 2 --> 12x = 24 < 110 When x = 9 --> 12x = 108 < 110 When x = 16 --> 12x = 192 > 110 There are 2 possible values --> 24 or 108 Not Sufficient

St2: There were fewer than 120 boxes in the warehouse after the 60 additional arrived. --> 12x + 60 < 120 When x = 2 --> 12x + 60 = 84 < 120 When x = 9 --> 12x + 60 = 168 > 120 There is only one possible value --> 84 Sufficient

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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20 Jun 2016, 05:44

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Vyshak wrote:

Given: 12x + 60 = 14y

y = 12(x + 5)/14 = 6(x + 5)/7

6 is not divisible by 7. So, (x + 5) has to be divisible by 7 --> (x + 5) = 7k x = 7k - 5. Therefore x = 2, 9, 16, 23, ..........

St1: There were fewer than 110 boxes in the warehouse before the 60 additional arrived. --> 12x < 110 When x = 2 --> 12x = 24 < 110 When x = 9 --> 12x = 108 < 110 When x = 16 --> 12x = 192 > 110 There are 2 possible values --> 24 or 108 Not Sufficient

St2: There were fewer than 120 boxes in the warehouse after the 60 additional arrived. --> 12x + 60 < 120 When x = 2 --> 12x + 60 = 84 < 120 When x = 9 --> 12x + 60 = 168 > 120 There is only one possible value --> 84 Sufficient

Answer: B

B for me too.

I got another way(Remainder) a bit lengthy but easy for others to understand if they don't want to follow the equations above Or if they can't interpret it

60=14*4 +4

so you need to find the difference for 12*n -14*m =4 so that when 60 is added it is divisible by 14(Remainder=0) you can follow this approach as the limitation is given till 120.If limitation was unknown then above method would suffice

with above equation you find only 2 values which agree below 120

24,28 and 108,112 if 24 is added to 60 or 108 is added to 60 and divided by 14 remainder would be 0 or divisible.(24 and 108 both divisible by 12) Till here if you have solved question is solved.

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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09 Jul 2016, 23:52

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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
_________________

You have to dig deep and find out what it takes to reshuffle the cards life dealt you

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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25 Jul 2016, 10:42

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My approach Before 60 additional boxes arrived, the box no is divisible by 12 x=12,24,36,48,60,72,84,96,108,120 After 60 additional boxes arrived, the box no is divisible by 14 y=14,28,42,56,70,84,98,112,126,140,154,168,182

Case - (1) fewer than 110 before before 60 boxes arrived x=12,24,36,48,60,72,84,96,108 Any match for y, such that y=x+60 Yes x = 24, 108 Not Sufficient

Case -(2) fewer than 120 after 60 boxes arrived x=12,24,36,48,60,72,84,96,108,120 y=14,28,42,56,70,84,98,112 x=24 Sufficient

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.

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!

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.

This is based on the application of remainder theory Here let number of boxes be x so x=12P and x+60=14Q => X=14W+10 Now as x=12P and x=14W+10 Combing the statement => x=24+84Z now sttaement 1 => x can be 24 or 108 => insuff statement 2 => x has to be 24=> suff Smash that B
_________________

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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30 Sep 2016, 19:33

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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.

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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07 Apr 2017, 20:06

1)

12n+60=14p 6n+30=7p, therefore (6n+30) must be a multiple of 7 6(n+5)=7p, therefore (n+5) is a multiple of 7 n=2 n=9 n=16 - not allowed, 12n would be more than 110. Not suff.

2) From above, if n=2 then number of boxes after addition is 84 - allowed. From above, if n=9 then number of boxes after addition is 168 - not allowed Suff B

Note n cannot equal 0 because then (n+5) from 6(n+5)=7p would not be a multiple of 7.

Kudos if you like the method! Comments please if you have improvements.

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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19 Apr 2017, 07:08

(1) LCM of 12 and 14 are 84;168 and so on... 84-60=24 which is less than 110 168-60=108 which is less than 110 Subject constrains give us 2 solutions, hence Insufficient (2) using the same reasoning as above we have: 84 is less than 120 168 is more than 120 This constrains give us only 1 solution, so initial number of boxes was 24 hence Sufficient.

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]

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28 Oct 2017, 08:23

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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