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# A lock code was product of two numbers, one taken from set A GMAT

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Intern
Joined: 01 Oct 2018
Posts: 12
A lock code was product of two numbers, one taken from set A GMAT  [#permalink]

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25 Dec 2018, 00:01
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:08) correct 34% (02:04) wrong based on 68 sessions

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A lock code was product of two numbers, one taken from set A = {2, 3, 4, 5, 9} and other taken from set B = {1, 2, 6, 7}. If all possible combinations of numbers from set A and set B are checked, what is the probability of opening the lock if entering any multiple of 6 will open the lock?

a) 1/7
b) 3/5
c) 1/2
d) 3/10
e) 7/20
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Re: A lock code was product of two numbers, one taken from set A GMAT  [#permalink]

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25 Dec 2018, 00:18
Nikhitaarorah wrote:
A lock code was product of two numbers, one taken from set A = {2, 3, 4, 5, 9} and other taken from set B = {1, 2, 6, 7}. If all possible combinations of numbers from set A and set B are checked, what is the probability of opening the lock if entering any multiple of 6 will open the lock?

a) 1/7
b) 3/5
c) 1/2
d) 3/10
e) 7/20

So total combinations possible = 20 (5 items of A multiplied by 4 items of B)

Combinations opening the locks must be multiple of 6, so they must be either 6,12,18,24,30,36 and so on

so combinations possible from given set:

2X6, 3X2, 3X6, 4X6, 5X6, 9X6,9X2, so a total of 7 combinations,

E
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Re: A lock code was product of two numbers, one taken from set A GMAT  [#permalink]

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28 Dec 2018, 11:37
Total cases : 5C1*4C1=20 & possible +ve combinations =7 [ (2,6),(3,2),(3,6),(4,6),(5,6),(9,2),(9,6),]
Hence probability= 7/20 .......Ans E
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A lock code was product of two numbers, one taken from set A GMAT  [#permalink]

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28 Dec 2018, 12:12
Nikhitaarorah wrote:
A lock code was product of two numbers, one taken from set A = {2, 3, 4, 5, 9} and other taken from set B = {1, 2, 6, 7}. If all possible combinations of numbers from set A and set B are checked, what is the probability of opening the lock if entering any multiple of 6 will open the lock?

a) 1/7
b) 3/5
c) 1/2
d) 3/10
e) 7/20

the number of possible combinations = 5 elements of A * 4 elements of B = 20 product outcomes

favorable outcomes which are multiples of 6 are (2*6=12), (3*2=6), (3*6=18), (4*6=24), (5*6=30), (9*2=18), (9*6=54) .. so only 7 outcomes

the probability of opening the lock = 7/20

which is E
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Re: A lock code was a product of 2 numbers, one taken from set A= {2,3,4,5  [#permalink]

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12 Feb 2019, 11:59
Total.no of combinations =20
Total no. Of possible conbinations =(2,3)(2,9)(6,2)(6,3)(6,4)(6,5)(6,9)

Thus 7/20, e is the correct option

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Re: A lock code was a product of 2 numbers, one taken from set A= {2,3,4,5  [#permalink]

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12 Feb 2019, 14:10
akurathi12 wrote:
A lock code was a product of 2 numbers, one taken from set A= {2,3,4,5,9} and other taken from set B={1,2,6,7}. If all possible combinations of numbers from set A and set B are checked, what is the probability of opening the lock if entering any multiple of 6 will open the lock?

A. 1/7
B. 3/5
C. 1/2
D. 3/10
E. 7/20

Set A has 5 numbers
Set B has 4 numbers

Total number of combinations = 5*4 = 20
Preferred Outcomes = all multiples of 6
=> Let's see the number of combinations wrt set B
1 --> None of the numbers form multiples of 6
2 --> 3 & 9 are eligible
6 --> All 5 numbers are eligible
7 --> None are eligible
=> Preferred outcomes = 7

Probability = Preferred/Total Outcomes = 7/20

Hence E is the correct answer.
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Re: A lock code was a product of 2 numbers, one taken from set A= {2,3,4,5   [#permalink] 12 Feb 2019, 14:10
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