Informative line

# 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.

$$=5×(9)-6.0-(-4.5)$$

$$=45-6+4.5$$

$$=39+4.5$$

$$=43.5$$

#### 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.

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

#### 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 $$x.$$

$$2x+9$$

#### 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.

#### 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