CA Foundation Exam > CA Foundation Questions > If x/2=y/3=z/7, then the value of (2x-5y+4z)/...
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If x/2=y/3=z/7, then the value of (2x-5y+4z)/2y is
- a)6/23
- b)23/6
- c)3/2
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If x/2=y/3=z/7, then the value of (2x-5y+4z)/2y isa)6/23b)23/6c)3/2d)n...
x/2 = 2x/4
y/3 = (-5y) / (-15)
z/7 = 4z/28
So, the given equation becomes,
2x/4 = (-5y)/(-15) = 4z/28
= (2x-5y+4z) / (4-15+28) [Using addendo]
= (2x-5y+4z) / 17
So, (-5y) / (-15) = (2x-5y+4z) / 17
Or, y/3 = (2x-5y+4z) / 17
Or, 17/3 = (2x-5y+4z) / y [Using alternendo]
Multiply both sides with (1/2).
17/6 = (2x-5y+4z) / (2y)
The answer is 17/6.
Most Upvoted Answer
If x/2=y/3=z/7, then the value of (2x-5y+4z)/2y isa)6/23b)23/6c)3/2d)n...
Solution:
Given, x/2=y/3=z/7
Let's assume the common ratio to be k.
Therefore, x = 2k, y = 3k and z = 7k.
Required expression: (2x - 5y + 4z)/2y
Substituting the values of x, y and z, we get:
(2(2k) - 5(3k) + 4(7k))/2(3k)
Simplifying this expression, we get:
(4k - 15k + 28k)/6k
= 17k/6k
= 17/6
Hence, the value of the given expression is 17/6.
Therefore, the correct answer is option d) none of these.
Note: None of the given options match the answer we obtained.
Given, x/2=y/3=z/7
Let's assume the common ratio to be k.
Therefore, x = 2k, y = 3k and z = 7k.
Required expression: (2x - 5y + 4z)/2y
Substituting the values of x, y and z, we get:
(2(2k) - 5(3k) + 4(7k))/2(3k)
Simplifying this expression, we get:
(4k - 15k + 28k)/6k
= 17k/6k
= 17/6
Hence, the value of the given expression is 17/6.
Therefore, the correct answer is option d) none of these.
Note: None of the given options match the answer we obtained.
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Community Answer
If x/2=y/3=z/7, then the value of (2x-5y+4z)/2y isa)6/23b)23/6c)3/2d)n...
Assume each ratio to be k ,now write x,y,z in terms of k now put the values of x,y,z in expression, you will get the answer
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