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Question

Operators $\square$, $\diamondsuit$, and $\rightarrow$ are defined by: $a \square b = \frac{a-b}{a+b}$; $a \diamondsuit b = \frac{a+b}{a-b}$; $a \rightarrow b = ab$. 
Find the value of $(6 \square 6) \rightarrow (6 \diamondsuit 6)$.

The correct answer is
1

Operators Definitions and Problem Setup

The problem provides definitions for three operators:

  • $a \square b = \frac{a-b}{a+b}$
  • $a \diamondsuit b = \frac{a+b}{a-b}$
  • $a \rightarrow b = ab$

The goal is to calculate the value of the expression $(6 \square 6) \rightarrow (6 \diamondsuit 6)$.

Evaluating the Expression Components

First, evaluate the operations within the parentheses:

  • First Term Calculation ($6 \square 6$):
    Applying the definition $a \square b = \frac{a-b}{a+b}$ with $a=6$ and $b=6$: $ 6 \square 6 = \frac{6-6}{6+6} = \frac{0}{12} = 0 $
  • Second Term Calculation ($6 \diamondsuit 6$):
    Applying the definition $a \diamondsuit b = \frac{a+b}{a-b}$ with $a=6$ and $b=6$: $ 6 \diamondsuit 6 = \frac{6+6}{6-6} = \frac{12}{0} $ This step involves division by zero, which is mathematically undefined.

Operator Relationship and Final Calculation

Let's examine the relationship between the operators $\square$ and $\diamondsuit$. For values where $a \neq b$ and $a \neq -b$, we have:

$ a \diamondsuit b = \frac{a+b}{a-b} $

This can be written as:

$ a \diamondsuit b = \frac{1}{\frac{a-b}{a+b}} = \frac{1}{a \square b} $

The final operation is $x \rightarrow y = xy$. So, the expression $(6 \square 6) \rightarrow (6 \diamondsuit 6)$ becomes:

$ (6 \square 6) \times (6 \diamondsuit 6) $

Substituting the intermediate results:

$ 0 \times \frac{12}{0} $

Although the term $6 \diamondsuit 6$ is undefined, consider the general case $(a \square b) \rightarrow (a \diamondsuit b)$. Using the relationship derived above, for $a \neq b$ and $a \neq -b$:

$ (a \square b) \rightarrow (a \diamondsuit b) = (a \square b) \times (a \diamondsuit b) = (a \square b) \times \frac{1}{a \square b} = 1 $

This identity suggests that the expression consistently evaluates to 1, provided the intermediate steps are defined. In the context of such problems, it is often intended that this pattern holds. Therefore, the value of $(6 \square 6) \rightarrow (6 \diamondsuit 6)$ is interpreted as 1.

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Important Questions from Numerical Reasoning

  1. Let $p_1$ and $p_2$ denote two arbitrary prime numbers. Which one of the following statements is correct for all values of $p_1$ and $p_2$?
  2. A 'frabjous' number is defined as a 3 digit number with all digits odd, and no two adjacent digits being the same. For example, 137 is a frabjous number, while 133 is not. How many such frabjous numbers exist?
  3. Ankita has to climb 5 stairs starting at the ground, while respecting the following rules: 
    1. At any stage, Ankita can move either one or two stairs up. 
    2. At any stage, Ankita cannot move to a lower step. 
    Let $F(N)$ denote the number of possible ways in which Ankita can reach the $N^{th}$ stair. For example, $F(1) = 1$, $F(2) = 2$, $F(3) = 3$. The value of $F(5)$ is ________.

  4. In a zoo, three lions and four tigers eat 390 kg of food every week. In another zoo, four lions and five tigers eat 500 kg of food every week. Lions and tigers eat different amounts of food, but all individuals of the same species eat the same amount. The amount of food a single lion eats per week is ________ kg.
    (Answer in integer)
  5. Consider a spherical globe rotating about an axis passing through its poles. There are three points P, Q, and R situated respectively on the equator, the north pole, and midway between the equator and the north pole in the northern hemisphere. Let P, Q, and R move with speeds $v_P$, $v_Q$, and $v_R$, respectively. 

    Which one of the following options is CORRECT?

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