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Danny can divide his herd into 5 equal parts and also to 6 equal parts

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Bunuel
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Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   05 Jul 2016, 12:24
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A
B
C
D
E

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89% (00:52) correct 11% (01:10) wrong based on 98 sessions
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Danny can divide his herd into 5 equal parts and also to 6 equal parts, but not to 9 equal parts. What could be the number of cows Danny has in his herd?

A. 120
B. 155.
C. 180
D. 336.
E. 456
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A
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FacelessMan
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   05 Jul 2016, 12:28
Danny can divide his herd into 5 equal parts and also to 6 equal parts => No of cows = Multiple of 5 and 6 = Multiple of 30.
Only option A and C qualify.
Now 2nd condition : but not to 9 equal parts : So it should not be a multiple of 9. Only A.
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   05 Jul 2016, 12:32
Bunuel wrote:
Danny can divide his herd into 5 equal parts and also to 6 equal parts, but not to 9 equal parts. What could be the number of cows Danny has in his herd?

A. 120
B. 155.
C. 180
D. 336.
E. 456


Ans.

On first look B, D & E are not divisible either by 5 or 6 - ruled out.
A & C divisible by both 5 & 6, but 180 is divisible by 9.
Hence Ans is (A).

Bunuel - can we expect kudos for immediate replies + correct answers.
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   14 Aug 2016, 09:16
Bunuel,
Should not the answer be 'A'?
155 does not divide equally into 6.
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   14 Aug 2016, 10:46
Another Vote for A
120 is not divisible by 9... but by 6 and 5.
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   14 Aug 2016, 11:27
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Bunuel wrote:
Danny can divide his herd into 5 equal parts and also to 6 equal parts, but not to 9 equal parts. What could be the number of cows Danny has in his herd?

A. 120
B. 155.
C. 180
D. 336.
E. 456


I am also with A.

Only 120 is the number which is divisible by 5 and 6 but not by 9.
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Bunuel
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   15 Aug 2016, 03:38
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abhimahna wrote:
Bunuel wrote:
Danny can divide his herd into 5 equal parts and also to 6 equal parts, but not to 9 equal parts. What could be the number of cows Danny has in his herd?

A. 120
B. 155.
C. 180
D. 336.
E. 456


I am also with A.

Only 120 is the number which is divisible by 5 and 6 but not by 9.


Edited The OA. Thank you.
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink] New post   22 Apr 2019, 22:38
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Bunuel wrote:
Danny can divide his herd into 5 equal parts and also to 6 equal parts, but not to 9 equal parts. What could be the number of cows Danny has in his herd?

A. 120
B. 155.
C. 180
D. 336.
E. 456



First, take the LCM of 5 and 6 since the total number of cows need to be divisible by both 5 & 6.

This will give us: 5*6=30

Now, we are looking for an answer which is a multiple of 30. Therefore, (B), (D), & (E) can be eliminated.

In between (A) & (B),
120 is not divisible by 90
180 is divisible by 90

Hence, choice (A) is our answer.
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Re: Danny can divide his herd into 5 equal parts and also to 6 equal parts [#permalink]
22 Apr 2019, 22:38
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