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.

Comments

  1. Why are there so many math problems in ADRE 🥹?

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