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Question

If $p>q=p^2+q^3$ and $p<q=p^3-q^2$, then $(1>2)<3 = $

The correct answer is
720

The problem defines two operations based on conditions:

  • If $p > q$, then the operation (let's denote it by '>') results in $p^2 + q^3$.
  • If $p < q$, then the operation (let's denote it by '<') results in $p^3 - q^2$.

We need to evaluate the expression $(1>2)<3$. The notation suggests that the symbol used directly determines the formula applied, potentially irrespective of the condition check for the evaluation itself.

Evaluating $(1>2)$

In the expression $1>2$, the operator is '>'. We apply the formula associated with this operator:

Formula for '>': $p^2 + q^3$

Substitute $p=1$ and $q=2$: $1^2 + 2^3 = 1 + 8 = 9$

Let the result be $R = 9$.

Evaluating $R<3$

Now, we need to evaluate $R<3$, which is $9<3$. The operator is '<'. We apply the formula associated with this operator:

Formula for '<': $p^3 - q^2$

Substitute $p=9$ and $q=3$: $9^3 - 3^2 = 729 - 9 = 720$

Final Result

The value of the expression $(1>2)<3$ is 720.

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Important Questions from Algebra

  1. In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

    I. x2 – 26x + 165 = 0

    II. y2 – 38y + 357 = 0

  2. Factorize the following:

    (x 2- 6xy + 9y 2) - 25

  3. If P and Q are the points on the line Joining A(-2, 5) and B(3, 1) such that

    AP = PQ = QB, then the mid point of PQ is 

  4. If a number and its reciprocal added it becomes 6, then what will be sum of its square and square of its reciprocal?

  5. If 3x + 2y = 15, and xy = 6. Find the value of (3x3/2) + (4y3/9).

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