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All the boxes in a supermarket are arranged in packs of 12 boxes each,

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All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post Updated on: 28 Apr 2016, 09:08
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All the boxes in a supermarket are arranged in packs of 12 boxes each, with no boxes left over. After new 60 additional boxes arrived and no boxes were removed, all the boxes in the supermarket were arranged in packs of 14. How many boxes were in the supermarket before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.

(2) There were fewer than 120 boxes AFTER the 60 additional boxes arrived.

Originally posted by inakihernandez on 28 Apr 2016, 08:17.
Last edited by inakihernandez on 28 Apr 2016, 09:08, edited 1 time in total.
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 28 Apr 2016, 08:46
I solved it in the following way, but I am sure there are couple of tips to save time....

(1)

12 + 60 = 72 ==> not divisible by 14
24 + 60=84/2 => 42/7 ==> divisible by 14
36 + 60= 96 ==> not divisible by 14
48 + 60= 108 ==> not divisible by 14
60 + 60= 120 ==> not divisible by 14
72 + 60= 132 ==> not divisible by 14
84 + 60= 144 ==> not divisible by 14
96 + 60= 156 ==> not divisible by 14
108 + 60= 168 = 168/2 = 84/7 ==> divisible by 14

Therefore, we can have in the beginning either 24 or 108 boxes INSUFFICIENT

(2) the only option that is divisible by 14 and has a number of boxes after the 60 new boxes below 120 is 24. Therefore 24 should be the number of boxes in the beginning. OPTION B

Has someone any better way to solve it?

Thanks!
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 28 Apr 2016, 09:01
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inakihernandez wrote:
I solved it in the following way, but I am sure there are couple of tips to save time....

(1)

12 + 60 = 72 ==> not divisible by 14
24 + 60=84/2 => 42/7 ==> divisible by 14
36 + 60= 96 ==> not divisible by 14
48 + 60= 108 ==> not divisible by 14
60 + 60= 120 ==> not divisible by 14
72 + 60= 132 ==> not divisible by 14
84 + 60= 144 ==> not divisible by 14
96 + 60= 156 ==> not divisible by 14
108 + 60= 168 = 168/2 = 84/7 ==> divisible by 14

Therefore, we can have in the beginning either 24 or 108 boxes INSUFFICIENT

(2) the only option that is divisible by 14 and has a number of boxes after the 60 new boxes below 120 is 24. Therefore 24 should be the number of boxes in the beginning. OPTION B

Has someone any better way to solve it?

Thanks!



Hi,
1) firstly there is a typo in your statement II..
Quote:
All the boxes in a supermarket are arranged in packs of 12 boxes each, with no boxes left over. After new 60 additional boxes arrived and no boxes were removed, all the boxes in the supermarket were arranged in packs of 14. How many boxes were in the supermarket before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.

(2) There were 120 boxes AFTER the 60 additional boxes arrived.


Now since you say it means <120 after 60 is added, it is OK..

Now to solve this, the procedure is same, BUT you can make your life easy by using some number properties..
your equation is --
12x+60 =14y..
12(x+5) =14y
what does this tell you about y?
THAT it should be a multiple of 6...

1) so when y=6, 12x+60 = 14*6 = 84 or x=2..
2) when y=12, 12x+60 = 14*12.. x=9..
we may not require any more..

lets see the statements--


(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.
x can be two then 12*2=24..
x CAN be 9 , then 12*9 = 108..
two solutions

Insuff

(2) There were below 120 boxes AFTER the 60 additional boxes arrived.
Only possiblity is when x=2 and y=6..
so ans =24

Suff

B
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 28 Apr 2016, 09:15
You are right. I already changed the error in the statement (2).

Thank you very much for your answer. I have a question: when you have the equation 12*(x+5)= 14y , Why Y must be a multiple of 6?
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 28 Apr 2016, 10:25
inakihernandez wrote:
You are right. I already changed the error in the statement (2).

Thank you very much for your answer. I have a question: when you have the equation 12*(x+5)= 14y , Why Y must be a multiple of 6?


12(x+5)= 14y

Or, 6(x + 5) = 7y

Does it help a bit now..
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All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 02 May 2016, 18:00
Excellent Quality Question s expected from GMAC
Here using the conditions given
Let N be the Number of boxes.
N=12p
N+60 = 14q => N = 14x+10
so using the remainder combination => N = 24 + 84z (here p,q,z are any integers)
Now suing statement 1 => N= 24 or 108 => Not sufficient
Using statement 2 => N=24 is the only choice we have
Hence B is sufficient
P.S => I feel this is a 700 level Question :)

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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 01 Jun 2016, 20:17
Suppose there were initially "p" packs.

So, total boxes = 12p

Now, boxes = 12p + 60

Total packs now = (12p + 60)/14 = 12(p + 5)/14 = 6(p + 5)/7

Clearly packs have to be "integers". When will 6(p + 5)/7 be an integer?
For value of p = 2, 9, 16, 23 etc.

If p=2, total packs now = 6(p + 5)/7 = 6
If p=9, total packs now = 6(p + 5)/7 = 12
If p=16, total packs now = 6(p + 5)/7 = 18

We have to find the value of "p" (because that will give us the value of 12p, boxes in the supermarket before the 60 additional boxes arrived).

(1) says that there were fewer than 110 boxes BEFORE the 60 additional boxes arrived.

If p=2, total boxes BEFORE the 60 additional boxes arrived = 12p = 24
If p=9, total boxes BEFORE the 60 additional boxes arrived = 12p = 108

So, we don't get a unique value of 12p. Not sufficient.

(2) says that there were fewer than 120 boxes AFTER the 60 additional boxes arrived.
Now we know that p can only have a value of 2, because then, 12p = 24
AFTER the 60 additional boxes arrived, boxes = 24 + 60 = 84. So, (2) is sufficient.
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 01 Jun 2016, 22:04
inakihernandez wrote:
All the boxes in a supermarket are arranged in packs of 12 boxes each, with no boxes left over. After new 60 additional boxes arrived and no boxes were removed, all the boxes in the supermarket were arranged in packs of 14. How many boxes were in the supermarket before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.

(2) There were fewer than 120 boxes AFTER the 60 additional boxes arrived.


You can use the "integer solutions" concept to simplify the question. Check: http://www.veritasprep.com/blog/2011/06 ... -of-thumb/

12a + 60 = 14b
We need to find 12a.

14b - 12a = 60
7b - 6a = 30

First obvious solution : a = -5, b = 0
For next solution increase a by 7 and b by 6 to get a = 2, b = 6
Repeat for next solution. Increase a by 7 and b by 6 to get a = 9, b = 12
and so on...

(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.
12a is less than 110 so a is less than 9.something.
a can be 2 or 9. Not sufficient.

(2) There were fewer than 120 boxes AFTER the 60 additional boxes arrived.
14b is less than 120 so b is less than 8.something.
b must be 6. Sufficient.

Answer (B)
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,  [#permalink]

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New post 23 Dec 2017, 21:17
inakihernandez wrote:
All the boxes in a supermarket are arranged in packs of 12 boxes each, with no boxes left over. After new 60 additional boxes arrived and no boxes were removed, all the boxes in the supermarket were arranged in packs of 14. How many boxes were in the supermarket before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.

(2) There were fewer than 120 boxes AFTER the 60 additional boxes arrived.



Solution :

We can say initial no. of boxes are multiple of 12.

Then 60 boxes were added , which also are multiple of 12

So final no of boxes are multiple of 12 ( Multiple of 12 + multiple of 12 = multiple of 12)

we are also given that final no. of boxes are multiple of 14.

So we can say final no. of boxes are multiple of 12 and 14 i.e 84 , 168 ... so on

(1) There were fewer than 110 boxes BEFORE the 60 additional boxes arrived.

using this statement range of total no of boxes is 0 to 169

so we can have total no of boxes = 84 or 168 ( or initial no. of boxes 24 or 108)

Insufficient.

(2) There were fewer than 120 boxes AFTER the 60 additional boxes arrived.[/quote]

Total no boxes range from 0 to 119

So we have total no. of boxes = 84 or initial no. of boxes 24.

Sufficient.

Answer : B
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Re: All the boxes in a supermarket are arranged in packs of 12 boxes each,   [#permalink] 23 Dec 2017, 21:17
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