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Bunuel
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65 = 5*13
91 = 7*13

HCF = 13
Sum : 65+91 = 156
Only 1 pair is possible

Option : A
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65 = 5*13
91 = 7*13

HCF = 13
Sum : 65+91 = 156
Only 1 pair is possible

Option : A

Copying someone else's explanation is not a very good practice.

I will encourage you to write your own explanation which will improve your understanding, accuracy and content richness of the GC community. :)
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Two numbers have the common factor 13, which means that there is no other common factor between the two numbers other than 13.
So let the two numbers be 13A, 13B

\(13(A+B) = 156\)
\(A+B = 12\)

Now, as explained above A and B will be co-primes(have no common factor other than 1)
So, the possible values for A and B are (1,11) and (5,7).

The number of pairs possible = 2

OA, B
Bunuel
The sum of two positive integers is 156 and their highest common factor is 13. How many pairs of such numbers are possible?

A. 1
B. 2
C. 3
D. 4
E. 5
 
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Explanation:
As , 13 is the highest common factor of two number and sum of 2 is 156,
Therefore, 13x+13y = 156
13(x+y) = 156 ------ 1
x+y = 12 ( where x & y has to be co-primes)

Now we have to check values for x & y.
if x = 1, then y = 11 ; put in 1, so sum is 156.
If x= 5 , then y = 7 ; put in 1, so sum is 156.

So only 2 pairs are possible.

IMO-B
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Bunuel
The sum of two positive integers is 156 and their highest common factor is 13. How many pairs of such numbers are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

156=13*12

13+13*11= hcf(13)
13*2+13*10=hcf(26)
13*3+13*9=hcf(39)
13*4+13*8=hcf(52)
13*5+13*7=hcf(13)
13*6+13*6=hcf(78)

Only those bold written has 13 hcf

Therefore only 2 pairs possible
Answer B

Posted from my mobile device
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Given: The sum of two positive integers is 156 and their highest common factor is 13.
Asked: How many pairs of such numbers are possible?

Let the two positive integers be x & y.

x + y = 156
HCF(x,y) = 13

x = 13k
y = 13m
where k & m are co-prime.

x + y = 13(k+m) = 156
k + m = 12

(k,m) = {(1,11),(5,7)}

Only 2 pairs (13, 143) & (65,91) are possible.

IMO B
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How many pairs of such numbers are possible?

why does the answer not 4 ?

(a,b) = (1, 11) (5, 7) (7, 5) (11, 1)
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How many pairs of such numbers are possible?

why does the answer not 4 ?

(a,b) = (1, 11) (5, 7) (7, 5) (11, 1)

Because (1, 11) and (11, 1) give the same pair, and (5, 7) and (7, 5) also give the same pair.
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