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Asked: What is the HCF of (p, q), if p and q are positive integers?

Statement 1: The product pq is 'p' more than square of p
pq = p + q^2
pq - p = q^2
p(q-1) = q^2
p = q^2/(q-1)
For p to be a positive integer (q-1) should be a factor of q^2
If q=2; p = 4; HCF(p,q) = 2
If q=even, q-1=odd; even/odd = may be even; may be feasible
But if q=odd; q-1=even; odd/even is NOT a positive integer but a fraction; NOT feasible
The only even number q = 2; where p = q^2/(q-1) is an integer;
HCF(p,q) = 2
SUFFICIENT

Statement 2: Twice the value of 'p' is one less than the sum of 'p' and 'q'
2p = p + q - 1
p = q - 1
If q = 2; p = 1; HCF(p,q) = 1
If q = 3, p = 2; HCF(p,q) = 1
For all possible combinations of p & q, HCF(p,q) = 1
SUFFICIENT

IMO D
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#1

The product pq is 'p' more than square of p
pq=p*(1+p)
Co prime
HCF ; 1
sufficient
#2
Twice the value of 'p' is one less than the sum of 'p' and 'q'
2p=p+q-1
or say
p=q-1
q-p=1
means p & q are consecutive numbers so HCF is 1
sufficient
OPTION D



What is the HCF of (p, q), if p and q are positive integers?

Statement 1: The product pq is 'p' more than square of p

Statement 2: Twice the value of 'p' is one less than the sum of 'p' and 'q'
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IMO D

What is the HCF of (p, q), if p and q are positive integers?

Statement 1: The product pq is 'p' more than square of p

pq = p^2 + p
p(p-q+1) = 0
Either , p= 0 0r (p-q+1)=0
but p = positive , so (p-q+1)=0
q-p = 1
P & q are consecutive integers.
HCF=1
Sufficient

Each statement are alone sufficient
Statement 2: Twice the value of 'p' is one less than the sum of 'p' and 'q

2p= p+q+1
p-q=1
P & q are consecutive integers.
HCF=1
Sufficient
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Find HCF(p,q) = ?

From ST(1), pq = p + p^2
pq = p(p+1)
since p is not '0', it can be concluded that q = p+1

Hence, p & q are consecutive integers
HCF(p, p+1) = 1

ST(1) is sufficient.

From ST(2), 2p = -1 + p + q
p = q-1

Hence, p & q are consecutive integers
HCF(q-1, q) = 1

ST(2) is sufficient.
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What is the HCF of (p, q), if p and q are positive integers?

Statement 1: The product pq is 'p' more than square of p

Statement 2: Twice the value of 'p' is one less than the sum of 'p' and 'q'

1) pq = p^2 + p
or, pq = p (p+1)
p and q are two consecutive integers. So their HCF is 1. Sufficient

2) 2p = p +q -1
p - q = -1
q - p = 1
same as 1. p and q are consecutive integers. HCF is 1. Sufficient

D is the answer
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Statement 1 -> q=p+1, p and q are consecutive integers. HCF = 1 - Sufficient
Statement 2 -> p=q-1, p and q are consecutive integers. HCF = 1 - Sufficient
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Answer : D.

Each statement is sufficient to say HCF = 1.

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