If x/3=y/4=z/7, then (x+y+z)/2z is equal to || ADRE 1.0 SLRC 2022 PAPER-III SOLVED QUESTIONS
Question:
If x/3 = y/4 = z/7
, then (x + y + z) / 2z
is equal to:
(A) 1
(B) 3
(C) 2
(D) 4
Solution:
To solve the problem, we need to find the value of (x + y + z) / 2z
given that x/3 = y/4 = z/7
.
Let's set x/3 = y/4 = z/7 = k
, where k
is a constant.
From this, we can express x
, y
, and z
in terms of k
:
x = 3k
y = 4k
z = 7k
Now, we substitute these expressions into (x + y + z)/2z
:
(x + y + z)/2z = (3k + 4k + 7k)/(2 * 7k)
Simplify the numerator and the denominator:
(3k + 4k + 7k)/(2 * 7k) = 14k/14k = 14 / 14 = 1
Thus, the expression (x + y + z)/2z
simplifies to 1.
So the correct answer is:
- (A) 1
Why this question is important?
- Tests Algebraic Skills: The problem evaluates the ability to handle and simplify algebraic expressions.
- Exam Preparation: It is relevant for various competitive exams, enhancing problem-solving techniques.
- Application of Constants: Demonstrates the use of constants to simplify complex expressions.
- Previously Asked Question: Featured in ADRE SLRC 2022 Paper-III for Grade-III posts, indicating its significance in past exams.
- Real-World Relevance: Develops skills useful for solving real-world mathematical problems.
Why are there so many math problems in ADRE 🥹?
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