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Question

Find x in the following expression:

The correct answer is

\(2 \frac{1}{14}\)

To find the value of \(x\) in the given expression, we must first understand the context of the problem. Assuming it is a simplified linear equation or expression where we solve for \(x\), let's analyze the provided information to reach the correct conclusion.

The correct answer given is \(2 \frac{1}{14}\), which implies fractional or mixed number handling. Without the expression's specific details, we'll demonstrate a potential solution process for a generic mixed-fraction problem:

Suppose the problem involves solving an equation for \(x\):

Problem:

\(x = \frac{m}{n} + c\) (where \(\frac{m}{n}\) and \(c\) are based on the details of the expression)

Solution:

  • Convert any mixed numbers to improper fractions if necessary for simplification.
  • Combine like terms or solve algebraically for \(x\).
  • Simplify the resulting improper fraction back to a mixed number.

As an illustrative solution:

  1. Assume \(x = \frac{3}{2} + \frac{4}{7}\). Find a common denominator for the fractions.
  2. Convert \(\frac{3}{2}\) to \(\frac{21}{14}\) and \(\frac{4}{7}\) to \(\frac{8}{14}\).
  3. Add the fractions: \(x = \frac{21}{14} + \frac{8}{14} = \frac{29}{14}\).
  4. Convert \(\frac{29}{14}\) to a mixed number: \(x = 2 \frac{1}{14}\).

This matches the correct answer, and illustrates the typical steps of dealing with fractions and addition in solving such problems.

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Important Questions from Simplification

  1. If P = 0.3 × 0.3 + 0.03 × 0.03 - 0.6 × 0.03 and Q = 0.54, then  \(\rm \frac{P}{Q}\) is equal to:

  2. The value of \(\left(\frac{1}{2}\right)^{−2} \times\left(\frac{1}{3}\right)^{−2} \times\left(\frac{1}{4}\right)^{−2} \)  is

  3. The solution of the equation \(\frac{2}{3 x-4}+\frac{2}{2 x-6}=0 \) is:

  4. If \(\rm \sqrt{1225 \times \sqrt{32 \div x}}= 70\)  find the value of x.

  5. What will come in the place of question mark (?) in the given expression?

    \(\sqrt{21+\sqrt{49}+\sqrt{64}} \space {\%\:of\:5000}=?\)

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