Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 13)
13.
| Reduce | 128352 | to its lowest terms. |
| 238368 |
Answer: Option
Explanation:
128352) 238368 ( 1
128352
---------------
110016 ) 128352 ( 1
110016
------------------
18336 ) 110016 ( 6
110016
-------
x
-------
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 ÷ 18336 7
Therefore, ------ = -------------- = --
238368 238368 ÷ 18336 13
Discussion:
59 comments Page 4 of 6.
Yash b said:
1 year ago
Simple divisibility rule;
128352 is divisible by 7, 3
238368 is divisible by 13.
So, option is 7/13.
128352 is divisible by 7, 3
238368 is divisible by 13.
So, option is 7/13.
(20)
Sanket shinde said:
7 years ago
7*6=42.
13*6=78.
Just see unit digit number it's 2 and 8 respectively so my answer is 7/13 only.
13*6=78.
Just see unit digit number it's 2 and 8 respectively so my answer is 7/13 only.
Shahul said:
1 decade ago
Hi sona,
Why you didn't subtract 110016-18336?
Why you stop'd subtracting the above values?
Why you didn't subtract 110016-18336?
Why you stop'd subtracting the above values?
Barath s said:
4 years ago
Reduce the no to 12/23.
it is approx 1/2.
So, find an option that is close to 1/2(i.e) 7/13.
it is approx 1/2.
So, find an option that is close to 1/2(i.e) 7/13.
(3)
Saloni said:
8 years ago
Is there any easy and quick method to solve this? Please explain me a simple method.
Rushikesh said:
4 years ago
I think option A is also divisible and also lowest. So why can't it be opton A?
(14)
Kavita said:
4 years ago
@Punni.
How can we take out factors of such a big number? Explain, please.
How can we take out factors of such a big number? Explain, please.
(1)
Pankaj said:
1 decade ago
Please somebody answer why after we have to stop subtracting after.
18336?
18336?
Sachin said:
2 years ago
(2*2*2*2*2*3*7*191)/(2*2*2*2*2*3*2483)
Gives;
(7*191)/(2483) = 7/13.
Gives;
(7*191)/(2483) = 7/13.
(7)
Bhargav said:
7 years ago
Can anyone explain the solution in another method to get it easily?
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