Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 07 Jun 2015
Posts: 3
Location: United States
Concentration: Entrepreneurship, Leadership

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
19 Jun 2016, 09:25
8
This post received KUDOS
20
This post was BOOKMARKED
Question Stats:
57% (01:43) correct 43% (01:53) wrong based on 658 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.



SC Moderator
Joined: 13 Apr 2015
Posts: 1678
Location: India
Concentration: Strategy, General Management
WE: Analyst (Retail)

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes.. [#permalink]
Show Tags
19 Jun 2016, 09:41
18
This post received KUDOS
16
This post was BOOKMARKED
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



Senior Manager
Joined: 02 Mar 2012
Posts: 339

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
20 Jun 2016, 05:44
1
This post received KUDOS
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. Answer will be B



Manager
Joined: 17 Sep 2015
Posts: 95

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
09 Jul 2016, 23:52
5
This post received KUDOS
3
This post was BOOKMARKED
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



Intern
Joined: 29 Apr 2011
Posts: 42

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
25 Jul 2016, 10:42
8
This post received KUDOS
1
This post was BOOKMARKED
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
B is the answer



Intern
Joined: 03 Jun 2016
Posts: 3

Test 5 GMATPrep problem [#permalink]
Show Tags
20 Aug 2016, 05:22
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!
Attachments
Capture1.PNG [ 14.04 KiB  Viewed 13072 times ]



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3454

Re: Test 5 GMATPrep problem [#permalink]
Show Tags
20 Aug 2016, 05:50
3
This post received KUDOS
3
This post was BOOKMARKED
Srishti_15 wrote: 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. Hence, answer is B.
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub Verbal Resources: All SC Resources at one place  All CR Resources at one place
Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2602
GRE 1: 323 Q169 V154

Re: Test 5 GMATPrep problem [#permalink]
Show Tags
21 Aug 2016, 02:48
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
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Intern
Joined: 09 Oct 2015
Posts: 44

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
30 Sep 2016, 19:33
2
This post received KUDOS
2
This post was BOOKMARKED
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 !



Manager
Joined: 13 Dec 2013
Posts: 163
Location: United States (NY)
Concentration: Nonprofit, International Business
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
07 Apr 2017, 20:06
1
This post received KUDOS
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.



Senior Manager
Joined: 05 Dec 2016
Posts: 260
Concentration: Strategy, Finance

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
19 Apr 2017, 07:08
(1) LCM of 12 and 14 are 84;168 and so on... 8460=24 which is less than 110 16860=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.
Answer B



Intern
Joined: 26 Jul 2014
Posts: 13

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
18 Oct 2017, 02:38
Bunuel: I tried to solve this problem by using remainder concept. a = 12n and a = 14m  60 => a = 84q + 2 (1) a < 110 => 84q + 2 < 110 => q only can be 1 (sufficient) (2) a + 60 < 120 => 84q + 2 < 60 => 84q < 58 (illogical?) Can you point out what is wrong with my approach? Thank you so much!



Intern
Joined: 24 Oct 2016
Posts: 25

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
28 Oct 2017, 08:23
4
This post received KUDOS
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



Intern
Joined: 24 Dec 2017
Posts: 3

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
24 Dec 2017, 04:39
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 "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...?



DS Forum Moderator
Joined: 27 Oct 2017
Posts: 450
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
24 Dec 2017, 19:20
1
This post received KUDOS
Hi HermesI 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 Hermes wrote: 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 "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...?
_________________
SC: Confusable words All you need for Quant, GMAT PS Question Directory,GMAT DS Question Directory Error log/Key Concepts Combination Concept: Division into groups



Intern
Joined: 06 Dec 2016
Posts: 12

Re: All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
04 Mar 2018, 01:09
Quick solution: Initial boxes = 12X (X is no. of stacks) New boxes =14Y = 12X+60 (Y is no of stacks) <=> 6(X+5)=7Y => Prethink: 6x7=7x6 or X=2, Y=6 sastified (1) 12X<110 <=> 1=<X< 9 => insufficient (2) 12X+60< 20 <=> 1=<X< 5 14Y < 120 <=> 1=<Y<8 ONLY Pair x=2, y=6 sastified => B



Intern
Joined: 27 Sep 2017
Posts: 11

All boxes in a certain warehouse were arranged in stacks of 12 boxes [#permalink]
Show Tags
07 Mar 2018, 11:46
Stat 1 clearly insuff.
Stat 2 : 
14x<120 So, X >=1 and X<=8
14X is the number of boxes after 60 boxes arrived. 14X60 is the number of boxes before 60 boxes arrive. So we are looking for a value of X between "1 and 8" that satisfies : 14x60/12 = Integer.
the only value that satisfies this condition is when x = 6
stat.2 Suff.




All boxes in a certain warehouse were arranged in stacks of 12 boxes
[#permalink]
07 Mar 2018, 11:46






