Problem 55 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a, b, c\) are digits, then the rational number represented by \(0 \cdot c a b a b a b \ldots\) is (A) \(\frac{99 c+a b}{990}\) (B) \(\frac{99 c+10 a+b}{99}\) (C) \(\frac{99 c+10 a+b}{990}\) (D) None of these

Step-by-Step Solution

Answer

The correct choice is (C) \(\frac{99c+10a+b}{990}\).

Step 1: Understand the given number

The number given is a repeating decimal: \(0.cab\overline{aba}\) where \(ab\) repeats indefinitely.

Step 2: Express as a fraction

To convert the repeating decimal \(0.cab\overline{aba}\) into a fraction, consider it as \(0.cababababab\ldots\). Let this number be represented by \(x\). Write two equivalent equations: \[ x = 0.cababababab\ldots \] \[ 1000x = cab.abababab\ldots \] Here, shift the decimal point three places to the right.

Step 3: Set up equation for elimination

Subtract the first equation from the second to eliminate the repeating part:\[ 1000x - x = cab.ababab\ldots - 0.cababab\ldots \]This simplifies to \[ 999x = cab - 0.cab \] which can be rewritten as \[ 999x = 100*c + 10*a + b - (0.1*a + 0.01*b) \].

Step 4: Solve for x

Simplify the equation:\[ 999x = (100c + 10a + b) - (0.1a + 0.01b) \] This equates to:\[ 999x = 100c + a - 9a/10 + b - b/100 \]Further simplify this:\[ 999x = 100c + a(1 - 0.1) + b(1 - 0.01) \]Finally, obtain:\[ 999x = 100c + (10a + b)\].

Step 5: Simplify and compare options

The final simplified version of the fraction where:\[ x = \frac{100c + 10a + b}{990} \] Compare this result with the options provided to find the correct expression.

Key Concepts

Converting Repeating Decimals to Fractions

When you see a decimal number like \(0.cabababab\ldots\), where digits repeat indefinitely, you are dealing with a repeating decimal. It's important to know how to convert this into a more manageable form - a fraction.Here's a simple way to understand it:

• Identify the non-repeating and repeating parts of the decimal. In \(0.cab\overline{aba}\), "cab" is the non-repeating part, and "aba" is the repeating part.
• Assign a variable to represent the entire decimal number. Let's say this number is \(x\).
• Multiply \(x\) by a power of 10 that aligns the repeating bits ("aba" in this case) to a new position in a second equation to help eliminate them.

By setting these equations, you can use subtraction to eliminate the repeating part and solve for \(x\) in fraction form. Always simplify the resulting expression to get the cleanest answer.

Rational Numbers

Rational numbers are numbers that can be expressed as a fraction, where both the numerator and denominator are integers, and the denominator is not zero. Repeating decimals are a special case of rational numbers.Here are a few key points about rational numbers:

• If a decimal terminates (such as 0.5) or repeats (like 0.3333...), it is a rational number.
• Rational numbers are everywhere: between any two integers or any two fractions, there exists another rational number.
• This also means that any repeating decimal can be converted into a rational number.

Understanding that repeating decimals like \(0.cab\overline{aba}\) can be altered into fractions helps us see the link between decimals and rational numbers.

Elimination Method

The elimination method is a key algebraic technique used when converting repeating decimals to fractions. It involves setting up equations to "eliminate" the repeating part of the decimal. Here's how elimination works in this context:

• Write an equation where the decimal is equal to a variable, such as \(x\).
• Multiply this equation by a power of 10, shifting the decimal point to align repeating digits in a second equation.
• Subtract the original equation from this new equation. This subtraction cancels out the parts that are repeating, allowing you to solve for \(x\).

By simplifying the resulting equation, you can express the repeating decimal as a fraction. This method is effective and systematic for handling such numbers, making complicated repeating decimals more manageable to work with.