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

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Manager
Joined: 14 Feb 2017
Posts: 122
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GPA: 2.61
WE: Management Consulting (Consulting)
A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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26 Nov 2018, 01:28
2
00:00

Difficulty:

35% (medium)

Question Stats:

70% (01:43) correct 30% (01:31) wrong based on 73 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
Intern
Joined: 16 Jun 2018
Posts: 23
Location: India
Schools: Stern '21
GMAT 1: 700 Q49 V41
GPA: 4
Re: A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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28 Nov 2018, 22:57
Can someone please explain the reason ?
Intern
Joined: 11 Dec 2013
Posts: 25
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)
Re: A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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28 Nov 2018, 23:35
1
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

Need KUDOS pls
Manager
Joined: 14 Feb 2017
Posts: 122
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GPA: 2.61
WE: Management Consulting (Consulting)
A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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29 Nov 2018, 14:11
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 $$Insufficient$$

Intern
Status: Attempting to cross the 750 barrier
Joined: 15 Apr 2017
Posts: 6
GMAT 1: 530 Q39 V24
Re: A craftsperson made 126 ornaments and put them all into boxes.  [#permalink]

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30 Nov 2018, 02:36
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

Do give a Kudos if you found the solution good and easy to understand.
Re: A craftsperson made 126 ornaments and put them all into boxes. &nbs [#permalink] 30 Nov 2018, 02:36
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