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Piter
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Artemov
C
None of the statements, by themselves, are sufficient.
The combined statements give a series of nos.
60, 120, 180.. and each is divisble by 15
What is the OA?
b14kumar
Piter
ds

In my opinion, Not Sufficient.

If N is the multiple of 15 then it should be 30, 45, 60 and so on......(Should be greater than 15)

a) N is the multiple of 20.

Now 20*1 = 20 which is a multiple of 20 but is not a multiple of 15 so Insufficient.

b) N+6 is the multiple of 3.

If N=3 then N+6 = 9, which is a multiple of 3 but is not a multiple of 15 so Insufficient.


Taking a) and b) together,

If N is the multiple of 20 then unit place will always be 0.

20, 40, 60, 80, 100 and so on.......

If we club b) then (N+6) will always have 6 in the unit place. Resulting Number could be a multiple of 3 but will not be a multiple of 15 since unit digit of the multiples of 15 will be either 0 or 5.


Hence Insufficient.

- Brajesh


Hi Artemov,

How did you get a series 60, 120, 180..... be combining both statements?
Please explain.

Thanks,
Brajesh
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Stmnt 1 N could be 20 40 60 80 100 120 ….
Not sufficient

Stmnt 2So N+6 could be 3 6 9 12, 15,18 21,24,27
So N could -3 , 0 , 3 , 6 , 9 , 15 , 18 , 60…90, 120…
Not sufficient

Combining both
N could be 60 , 120 , …..which are multiples of 15 .Sufficient

So D.
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Ok, I did a mistake.....Now got it......

If we combine St1 and St2, N+6 will not be a multiple of 15 but N will be.

Hence , C should be the answer.

- Brajesh
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ashkrs
Stmnt 1 N could be 20 40 60 80 100 120 ….
Not sufficient

Stmnt 2So N+6 could be 3 6 9 12, 15,18 21,24,27
So N could -3 , 0 , 3 , 6 , 9 , 15 , 18 , 60…90, 120…
Not sufficient

Combining both
N could be 60 , 120 , …..which are multiples of 15 .Sufficient

So D.


It's not D, it's C :-)

- Brajesh
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C i believe

I) 15 is a multiple of 5 and 3
20 is a mulyiple of 5 but not 3
insuff.

II) n+6 is a multiple of 3 but this do not guarantee that n is higher than 9, because 9+6=15

both together = n+6=20x
so it means that 15 is a multiple
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Ferihere
Answer C, took about 30 seconds.


right from the start, choice was between D and E.

D is easy choise as N could be 60 , 120 , …..which are multiples of 15 .
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C. it cannot be D.

if an integer is divisible of a multiple of 3 and 20, then it is also a multiple of divisible by 60 (LCM of 3 and 20). Since 60 is a multiple/divisible of 15, so n is also a multiple or divisible of 15.

so st. 1 and 2 togather provide this information. therefore, it is C.
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ashkrs
Stmnt 1 N could be 20 40 60 80 100 120 ….
Not sufficient

Stmnt 2So N+6 could be 3 6 9 12, 15,18 21,24,27
So N could -3 , 0 , 3 , 6 , 9 , 15 , 18 , 60…90, 120…
Not sufficient

Combining both
N could be 60 , 120 , …..which are multiples of 15 .Sufficient

So D.


Sorry it was a typo on my part . Its C what I meant . Should be obvious when I said "combining both.."



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