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03 Jul 2020, 23:04
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Difficulty:
55%
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Question Stats:
59% (01:45) correct
41%
(02:01)
wrong
based on 167
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HCF of positive integers x, y and z is 15, and the HCF of x and y is 30. Which of the following could be the HCF of y and z?
A. 1
B. 6
C. 14
D. 60
E. 75
A. 1
B. 6
C. 14
D. 60
E. 75
Updated on: 03 Jul 2020, 23:34
Originally posted by hsingh3031 on 03 Jul 2020, 23:15.
Last edited by hsingh3031 on 03 Jul 2020, 23:34, edited 1 time in total.
Last edited by hsingh3031 on 03 Jul 2020, 23:34, edited 1 time in total.
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E .75
HCF of x,y,z is 15
HCF of x and y is 30 ( 15 * 2)
So HCF of y and z has to be 75 ( 15*5)
Where 2 and 5 are co - prime numbers and overall HCF of all number comes out to be 15.
It HCF is 15*2*2 for the above then the HCF becomes 15*2 where all other factors are co - prime . So HCF is 75 in this case
HCF of x,y,z is 15
HCF of x and y is 30 ( 15 * 2)
So HCF of y and z has to be 75 ( 15*5)
Where 2 and 5 are co - prime numbers and overall HCF of all number comes out to be 15.
It HCF is 15*2*2 for the above then the HCF becomes 15*2 where all other factors are co - prime . So HCF is 75 in this case
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03 Jul 2020, 23:28
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Since X Y and Z have 15 as HCF.
Therefore all three are multiple of 15
X and Y have 30 as HCF
So X and Y are 30 multiples
X = 15*2*a
Y = 15*2*b
Z = 15*c
The HCF of Y and Z have to be a multiple of 15
Among options D and E are the only multiple of 15
But if 60 is the HCF than the HCF of X Y and Z would have been 30. But its not
So answer is 75.
E option
Posted from my mobile device
Therefore all three are multiple of 15
X and Y have 30 as HCF
So X and Y are 30 multiples
X = 15*2*a
Y = 15*2*b
Z = 15*c
The HCF of Y and Z have to be a multiple of 15
Among options D and E are the only multiple of 15
But if 60 is the HCF than the HCF of X Y and Z would have been 30. But its not
So answer is 75.
E option
Posted from my mobile device
