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If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r is
- a)1
- b)0
- c)1/2
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1...
x1/p=y1/q=z1/r=k (say)
⇒ x = kp,y = kq,z = kr
xyz=1 (given)
kp.kq.kr=1
k(p+q+r)=1.
k(p+q+r)=k0
p+q+r = 0
Most Upvoted Answer
If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1...
Given: x1/p = y1/q = z1/r and xyz = 1
To find: the value of pqr
Solution:
Let x1/p = y1/q = z1/r = k
So, x = kp, y = kq and z = kr
Given, xyz = 1
⇒ (kp)(kq)(kr) = 1
⇒ k3pqr = 1
⇒ pqr = 1/k3
Also, x1/p = k
⇒ xp = k
⇒ (kp)p = k
⇒ k^p+1 = k
⇒ k^p = 1
Similarly, y1/q = k
⇒ yq = k
⇒ (kq)q = k
⇒ k^q+1 = k
⇒ k^q = 1
And, z1/r = k
⇒ zr = k
⇒ (kr)r = k
⇒ k^r+1 = k
⇒ k^r = 1
Now, p, q, r can be any non-zero rational number such that k^p = k^q = k^r = 1
For example, let p = 1/2, q = 1/3 and r = 1/6
Then, k^p = k^(1/2) = √k
k^q = k^(1/3) = 3√k
k^r = k^(1/6) = 6√k
Now, xyz = 1
⇒ (kp)(kq)(kr) = 1
⇒ (k^(1/2))(k^(1/3))(k^(1/6)) = 1
⇒ k^(5/6) = 1
⇒ k = 1
So, pqr = 1/k3 = 1/1 = 1
Therefore, the correct option is (b) 0.
To find: the value of pqr
Solution:
Let x1/p = y1/q = z1/r = k
So, x = kp, y = kq and z = kr
Given, xyz = 1
⇒ (kp)(kq)(kr) = 1
⇒ k3pqr = 1
⇒ pqr = 1/k3
Also, x1/p = k
⇒ xp = k
⇒ (kp)p = k
⇒ k^p+1 = k
⇒ k^p = 1
Similarly, y1/q = k
⇒ yq = k
⇒ (kq)q = k
⇒ k^q+1 = k
⇒ k^q = 1
And, z1/r = k
⇒ zr = k
⇒ (kr)r = k
⇒ k^r+1 = k
⇒ k^r = 1
Now, p, q, r can be any non-zero rational number such that k^p = k^q = k^r = 1
For example, let p = 1/2, q = 1/3 and r = 1/6
Then, k^p = k^(1/2) = √k
k^q = k^(1/3) = 3√k
k^r = k^(1/6) = 6√k
Now, xyz = 1
⇒ (kp)(kq)(kr) = 1
⇒ (k^(1/2))(k^(1/3))(k^(1/6)) = 1
⇒ k^(5/6) = 1
⇒ k = 1
So, pqr = 1/k3 = 1/1 = 1
Therefore, the correct option is (b) 0.
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If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1...
Let all the variable = with a constant k.... I. E x raise to the power 1/p=k than shift the power on k that is..... X = k raise to the power p nd do same with rest of the variables.. Than u get the value of x y and z than put it in the given eqn... Xyz=1 nd use the formulas
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If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1/2d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1/2d)none of theseCorrect answer is option 'B'. Can you explain this answer? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1/2d)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r isa)1b)0c)1/2d)none of theseCorrect answer is option 'B'. Can you explain this answer?.
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