CA Foundation Exam > CA Foundation Questions > If x1/p= y1/q= z1/rand xyz =1 , then the valu...
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If x1/p = y 1/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?
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If x1/p= y1/q= z1/rand xyz =1 , then the value of p+q+r isa)1b)0c)1/2d...
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If x1/p= y1/q= z1/rand xyz =1 , then the value of p+q+r isa)1b)0c)1/2d...
Given Information:
x1/p = y1/q = z1/rand
xyz = 1
To find: Value of p, q, r
Solution:
Let's assume x1 = y1 = z1 = k
So, we have:
x1/p = k/p
y1/q = k/q
z1/rand = k/rand
Now, we know that xyz = 1
So, k^3 / (pqr) = 1
pqr = k^3
Also, we know that x1/p = y1/q = z1/rand
So, k/p = k/q = k/rand
Multiplying these equations, we get:
k^3 / (pqr) = 1/ (pqr)
pqr = 0
Therefore, the value of pqr is 0.
Hence, the correct option is B) 0.
x1/p = y1/q = z1/rand
xyz = 1
To find: Value of p, q, r
Solution:
Let's assume x1 = y1 = z1 = k
So, we have:
x1/p = k/p
y1/q = k/q
z1/rand = k/rand
Now, we know that xyz = 1
So, k^3 / (pqr) = 1
pqr = k^3
Also, we know that x1/p = y1/q = z1/rand
So, k/p = k/q = k/rand
Multiplying these equations, we get:
k^3 / (pqr) = 1/ (pqr)
pqr = 0
Therefore, the value of pqr is 0.
Hence, the correct option is B) 0.
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