CA Foundation Exam > CA Foundation Questions > If x/2 = y/3 = z/7, then the value of (2xR...
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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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If x/2 = y/3 = z/7, then the value of (2x–5y+4z)/2y isa)6/23b)23...
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If x/2 = y/3 = z/7, then the value of (2x–5y+4z)/2y isa)6/23b)23...
Given equation: x/2 = y/3 = z/7
To find: (2x5y 4z)/2y
Solution:
Let's first simplify the given equation,
x/2 = y/3
Cross-multiplying, we get
3x = 2y
x = (2/3)y
Using this value of x, we can simplify the equation further,
x/2 = z/7
Substituting the value of x, we get
(2/3)y / 2 = z/7
y/3 = z/7
Cross-multiplying, we get
7y = 3z
y = (3/7)z
Substituting the values of x and y in the given expression, we get
(2x5y 4z)/2y = (2(2/3)y5y4z)/(2y)
Simplifying, we get
(4y2z)/2y
Canceling out 2 from both numerator and denominator, we get
2yz/y
Canceling out y, we get the final answer
2z
Therefore, the correct option is D) none of these, and the value of (2x5y 4z)/2y is 2z.
To find: (2x5y 4z)/2y
Solution:
Let's first simplify the given equation,
x/2 = y/3
Cross-multiplying, we get
3x = 2y
x = (2/3)y
Using this value of x, we can simplify the equation further,
x/2 = z/7
Substituting the value of x, we get
(2/3)y / 2 = z/7
y/3 = z/7
Cross-multiplying, we get
7y = 3z
y = (3/7)z
Substituting the values of x and y in the given expression, we get
(2x5y 4z)/2y = (2(2/3)y5y4z)/(2y)
Simplifying, we get
(4y2z)/2y
Canceling out 2 from both numerator and denominator, we get
2yz/y
Canceling out y, we get the final answer
2z
Therefore, the correct option is D) none of these, and the value of (2x5y 4z)/2y is 2z.
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If x/2 = y/3 = z/7, then the value of (2x–5y+4z)/2y isa)6/23b)23...
2×2-5×3-4×7 ÷ 2×3 4-15+28÷6 -11+28÷6 17÷6
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If x/2 = y/3 = z/7, then the value of (2x–5y+4z)/2y isa)6/23b)23/6c)3/2d)none of theseCorrect answer is option 'D'. Can you explain this answer?
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