Mathematics
Grade6
Easy

Question

What is the value of open square brackets open parentheses x to the power of 3 end exponent plus 4 divided by 4 close parentheses cross times 4 close square brackets plus 16

  1. 4 x to the power of 3 end exponent plus 20    
  2. x to the power of 3 end exponent    
  3. x to the power of 4 end exponent plus 20    
  4. 20    

hintHint:

This is an algebraic expression. An algebraic expression consists of two components: a variable and a constant. A variable is a quantity that doesn’t have any fixed value. It takes value according to the conditions. It is denoted by the alphabet. A constant is a quantity with a fixed value. In this question, we are given an expression. We have perform the required operations and find the final value.

The correct answer is: 4 x to the power of 3 end exponent plus 20


    The given expression is [(x3 + 4 ÷ 4) × 4 ] + 16
    There is one variable expression. The variable is: ‘x’. We have to perform the operations between them and find the right answer.
    Let’s consider the inner bracket. In inner bracket there are two terms. x^3 and (4 ÷4)
    They are unlike terms as one is variable and second is constant. Unlike terms means they have different variables.
    The whole bracket is multiplied by 4 and then to the final value 16 is added.
    Let’s solve the expression step by step
    [(x3 + 4 ÷ 4) × 4 ] + 16
    We will have to follow the BODMAS rule. We have to follow the this order.
    B stands for bracket, O stands for order, D stands for division, M stands for multiplication, A stands for addition. And S stands for subtraction.
    So we will solve (4 ÷ 4) first
    [(x3 + 4 ÷ 4) × 4 ] + 16 = [( x3 + 1) × 4 ] + 16
    Now we will use distribute law here to distribute 4 for both the terms inside the bracket.
    Distributive property - The quantity outside the bracket, or in other words, quantity in multiplication with bracket is equally distributed to the terms inside the bracket. This property is used in the above equation. And this property is distributive property
    [(x2+ 4 ÷ 4) × 4 ] + 16 = ( x3 × 4 + 1 × 4)+ 16
    [(x3 + 4 ÷ 4) × 4 ] + 16 = ( 4x3 + 4)+ 16
    No further operation is possible for the bracket term as they are unlike terms.
    So we can write
    [(x3+ 4 ÷ 4) × 4 ] + 16 = 4x3+ 4 + 16
    [(x3+ 4 ÷ 4) × 4 ] + 16 = 4x3 + 20
    No further operation is possible as both are unlike terms.
    Therefore, the value of expression is 4x+ 20
    So the option which is'4x+ 20' is the right option.

    We have to follow the proper order of operations, or we will get errors. We have to be careful about different types of variables.

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