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A craftsperson made 126 ornaments and put them all into boxes.

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A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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New post 26 Nov 2018, 02:28
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Difficulty:

  55% (hard)

Question Stats:

66% (02:13) correct 34% (02:04) wrong based on 123 sessions

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A craftsperson made 126 ornaments and put them all into boxes. If each box contained either 6 ornaments or 24 ornaments, how many of the boxes contained 24 ornaments?

(1) Fewer than 4 of the boxes contained 6 ornaments

(2) More than 3 of the boxes contained 24 ornaments
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Re: A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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New post 28 Nov 2018, 23:57
Can someone please explain the reason ?
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Re: A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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New post 29 Nov 2018, 00:35
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Let x = No of boxes containing 6 ornaments and y = No of boxes containing 24 ornaments
6x+24y=126

Statement 1: Fewer than 4 of the boxes contained 6 ornaments
x<4
If x=3 or 2, then y wont be an integer.
Hence x=1 is the only solution.
Statement 1 is sufficient
Statement 2: More than 3 of the boxes contained 24 ornaments
y>3
If y=4 , x= 4
If y=5 , x=1
There are 2 solutions. So B is not possible
Hence A

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Senior Manager
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Joined: 14 Feb 2017
Posts: 277
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GPA: 2.61
WE: Management Consulting (Consulting)
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A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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New post 29 Nov 2018, 15:11
Solution

From the stem we are able to form an equation:
\(6x + 24y = 126\) ornaments we are asked to solve for y

We can simplify this equation to \(x + 4y = 21\) but remember that x and y must be integers as they relate to whole boxes.

(1) Fewer than 4 boxes contain 6 ornaments

let's test values
\(x = 3
3 + 4y = 21
4y = 18\)
18 is not divisible by 4 , x=3 is not a solution

\(x=2
2 + 4y = 21
4y = 19\)
again, not a solution

\(x=1\) (the last value we can test)
\(1 + 4y = 21
4y = 20
y = 5\)
This is the only possible solution.

\((1)--> Sufficient\)

(2) More than 3 boxes contained 24 ornaments


This tells us \(y > 3\)

Let's test values for y using the original equation \(6x + 24y =126\)

\(y = 4
6x + 24(4) = 126
6x = 54
x = 9\)
one possible solution

\(y = 5
6x + 24(5) = 126
6x = 6
x= 1\)
Another possible solution

2 possible answers for x--> Inconsistent therefore \(Insufficient\)


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Re: A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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New post 30 Nov 2018, 03:36
Let x = No of boxes containing 6 ornaments
Let y = No of boxes containing 24 ornaments

So, 6x+24y=126

=> x + 4y = 21

Possible solutions to this are:
x = 1, y = 5
x = 5, y = 4
x = 9, y = 3
x= 13, y = 2
x = 17, y= 1

Statement 1

Fewer than 4 boxes contain 6 ornaments

Only 1st case, where x = 1 is the only solution.

Sufficient.

Statement 2

More than 3 boxes contained 24 ornaments

There are 2 possible solutions for this.
Hence, insufficient

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Re: A craftsperson made 126 ornaments and put them all into boxes.   [#permalink] 30 Nov 2018, 03:36
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A craftsperson made 126 ornaments and put them all into boxes.

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