Informative line

Formation And Evaluation Of Expressions

Evaluation of Expression

  • Evaluate means to find the value of expressions.
  • An algebraic expression can be evaluated if values of variables are given.

For example: Evaluate \(5a^2+4b-ba\) when \(a=-3\) and \(b=1.5\)

  • Put the values of variables, \(a\) and \(b.\)

\(5×(-3)^2-4(1.5)-(1.5)×(-3)\)

  • Evaluate expression using PEMDAS rule.

\(=5×(9)-6.0-(-4.5)\)

\(=45-6+4.5\)

\(=39+4.5\)

\(=43.5\)

Illustration Questions

Evaluate \((a+2)^2\) if \(a=-1\)

A \(4\)

B \(8\)

C \(1\)

D \(0\)

×

Given: \((a+2)^2\) and \(a=-1\)

Putting the value of the variable:

\((-1+2)^2\)

Evaluating expression:

\((1)^2\)

\(=1×1\)

\(=1\)

Hence, option (C) is correct.

Evaluate \((a+2)^2\) if \(a=-1\)

A

\(4\)

.

B

\(8\)

C

\(1\)

D

\(0\)

Option C is Correct

Making Table through Expression

  • In an algebraic expression, when we give a value to its variable as an input, then it gives us a value as an output.
  • From input and output of values of an expression, we can draw an input-output table.

For example: We need to create a table of an expression \((2a+9)\) by using the first five natural numbers as input.

Input: Here variable is \('a'\) so write its values as first five natural numbers i.e. \(1,\;2,\;3,\;4,\;5\).

Output: Values of the expression after evaluation.

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline 1&2×(1)+9=11\\ 2&2×(2)+9=13\\ 3&2×(3)+9=15\\ 4&2×(4)+9=17\\ 5&2×(5)+9=19\\ \hline \end{array}\)

Illustration Questions

Create a table for the expression \(3-\dfrac{x}{2}\) by using first three negative integers as input.

A \(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&5/2\\ -2&2\\ -3&3/2\\ \hline \end{array}\)

B \(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&7/2\\ -2&4\\ -3&9/2\\ \hline \end{array}\)

C \(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&-4\\ -2&-5\\ -3&-6\\ \hline \end{array}\)

D \(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline 7/2&-1\\ 4&-2\\ 9/2&-3\\ \hline \end{array}\)

×

Given expression: \(3-\dfrac{x}{2}\)

We know that first three negative integers are \(-1,\;-2,\;-3\).

\(\therefore\) Input: Values of \(x\) i.e. \(-1,\;-2,\;-3\)

Output: Values of the expression after evaluation.

Now, form a table.

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&3-\left(\dfrac{-1}{2}\right)=3+\dfrac{1}{2}=\dfrac{7}{2}\\ -2&3-\left(\dfrac{-2}{2}\right)=3+1=4\\ -3&3-\left(\dfrac{-3}{2}\right)=3+\dfrac{3}{2}=\dfrac{9}{2}\\ \hline \end{array}\)

Thus, table is-

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&7/2\\ -2&4\\ -3&9/2\\ \hline \end{array}\)

Hence, option (B) is correct.

Create a table for the expression \(3-\dfrac{x}{2}\) by using first three negative integers as input.

A

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&5/2\\ -2&2\\ -3&3/2\\ \hline \end{array}\)

.

B

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&7/2\\ -2&4\\ -3&9/2\\ \hline \end{array}\)

C

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline -1&-4\\ -2&-5\\ -3&-6\\ \hline \end{array}\)

D

\(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline 7/2&-1\\ 4&-2\\ 9/2&-3\\ \hline \end{array}\)

Option B is Correct

Making an Expression through Table

  • The input-output table can be expressed by an expression.
  • When we need to create an expression from an input-output table, we have to analyze the relation between input and output using guess and check method.
  • Check the relation between input and output and then write the expression.
  • Let's consider an example to understand it clearly.

Example: Consider the given table-

\(\begin{array} {|c|c|} \hline \text{Input}(x)&\text{Output}\\ \hline 1&11\\ 2&13\\ 3&15\\ 4&17\\ 5&19\\ \hline \end{array}\)

Using this table we need to create an expression.

\(\to\) From the given table, we can observe that the output is greater than the input.

\(\to\) It means that there is a possibility of addition or multiplication or a combination of both.

If the values of output are less than the input then use subtraction or division or a combination of both.

Let's try addition,

\(1+10=11\;\)

\(2+10=12\,\,\leftarrow\)  Not similar to the table's output

Now try multiplication,

\(1×11=11\)

\(2×11=22\;\leftarrow\)  Not similar to the table's output,

\(\to\) Both the operations did not give the same output as in the table therefore, we need to use a combination of arithmetic operations.

Now use combination of addition and multiplication,

\(1×2+9=11\\ 2×2+9=13\\ 3×2+9=15\\ 4×2+9=17\\ 5×2+9=19\)

Here, the output obtained is same as in the table.

So, our expression will be

\([(\text{Input})×2]+9\)

Let input be a variable \(x.\)

\(2x+9\)

Illustration Questions

What is the expression for the given table: \(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline 9&1\\ 15&2\\ 21&3\\ \hline \end{array}\)

A \(\dfrac{x-3}{6}\)

B \(\dfrac{x+3}{6}\)

C \(\dfrac{x-6}{3}\)

D \(\dfrac{x-1}{6}\)

×

From the table, we can observe that output of the table is less than the input.

So, we can use subtraction or division or a combination of both.

Let's try subtraction:

\(9-8=1\)

\(15-8=7\)

The output values are not same.

Let's try division:

\(\dfrac{9}{9}=1\)

\(\dfrac{15}{9}=1.6\)

The output values are not same.

Let's try combination of subtraction and division:

\((9-3)\div6=1\\ (15-3)\div6=2\\ (21-3)\div6=3\)

It satisfies the values of the table.

Therefore, the expression for the given input-output table is,

\((\text{Input}-3)\div6\)

Let input be a variable \(x.\)

\(\dfrac{(x-3)}{6}\)

Hence, option (A) is correct.

What is the expression for the given table: \(\begin{array} {|c|c|} \hline \text{Input}&\text{Output}\\ \hline 9&1\\ 15&2\\ 21&3\\ \hline \end{array}\)

A

\(\dfrac{x-3}{6}\)

.

B

\(\dfrac{x+3}{6}\)

C

\(\dfrac{x-6}{3}\)

D

\(\dfrac{x-1}{6}\)

Option A is Correct

Expression Involving Tip, % and Tax

  • An expression can be written in a percentage form and vice-versa.
  • Let's understand it clearly with an example.

Example: Few years back, Mr. Jones had bought a used car for \(x\) dollars. Two years later, the value of the car was \(0.85x\) dollars. Write an expression in percentage form to show how much decline was there in the price.

Car bought for \($\,x\).

After two years, price \(=$\,0.85x\)

The expression for change in price is:

\(x-0.85x\)

\(=0.15x\)

Now we shall calculate the percentage of the above expression.

Percentage of \(0.15x\) with respect to \(x\) will be-

\(15\,\%\)

So, change in price is \(15\,\%\) decrease or \(15\,\%\) decline.

Illustration Questions

Assume current population of a city to be \(2.024\,p\) while last year it was \(2p\). Which one shows the population increment in percentage form?

A \(2.4\,\%\)

B \(0.24\,\%\)

C \(24\,\%\)

D \(1.2\,\%\)

×

Given:

Current population \(=2.024\,p\)

Last year's population \(=2p\)

Thus, change in population is:

\(2.024\,p-2p\)

\(=0.024\) (increment)

Percentage of \(0.024p\) with respect to \(2p\) will be

\(=\dfrac{002.4}{100×2}=1.2\,\%\)

So, increment in population is \(1.2\,\%\)

Hence, option (D) is correct.

Assume current population of a city to be \(2.024\,p\) while last year it was \(2p\). Which one shows the population increment in percentage form?

A

\(2.4\,\%\)

.

B

\(0.24\,\%\)

C

\(24\,\%\)

D

\(1.2\,\%\)

Option D is Correct

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