Latest Bank Exam Aptitude Question SOLUTION: The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: A. 1 B. 2 C. 3 D. 4

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The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A. 1
B. 2
C. 3
D. 4

Asked In Bank Exam jai Narayan Mishra (12 years ago)

Solved Tej Raj Read Solution (8)

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Read Solution (8) :

Ashwin Dinesh  (9 years ago)

why not (6,2) ?

Is this solution Helpfull? Yes (15) | No (3)  

Tej Raj   (12 years ago) Best Solution

let no. are 13x and 13y
13x(13y)=2028
xy=12
thus no of such pair is (1,12) and (3,4)

Is this solution Helpfull? Yes (14) | No (26)  

Debraj Das  (9 years ago)

Let no.s are 13a and 13b.

so,
13a x 13b = 2028
or,
ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Ans: there are 2 such pairs.

Is this solution Helpfull? Yes (12) | No   

elavarasan  (9 years ago)

let be the numbers are 13x and 13 y
here 13x(13y)=2028
here x and y are 12.
then the number of pair is (1,12) and (3,4).

Is this solution Helpfull? Yes (3) | No (4)  

narender singh  (8 years ago)

(6,2) is not considered because they are not co-prime number

Is this solution Helpfull? Yes (2) | No   

Nirmalrajsanjeev  (8 years ago)

given that product of numbers=2028 and hcf=13
let 13a*13b=2028
ab=2028/13*13
ab=12
if a=1, b=12 than 13,156
if a=3, b=4 than 39,52
two pairs

Is this solution Helpfull? Yes (1) | No (1)  

roopavani  (8 years ago)

yeah why not(6,2)??

Is this solution Helpfull? Yes | No   

MUKKAVALLI BHARGAVA KRISHNA CHAITANYA  (8 years ago)

(6,2) is not a valid pair cause 13*6=78, 13*2=26 , for which the HCF is not 13, here they asked about the possible pairs whose HCF is 13. so (1,12),(3,4) is correct answer

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