Introduction to Linear Polynomials — NCERT Solutions

The village starts with 750 people and gains 50 people every year.

So after $t$ years, the population is:

$P = 750 + 50t$

For the required year:

• After 6 years: $P = 750 + 50\times 6 = 750 + 300 = 1050$

Table for $t=0$ to $10$:

This is linear growth because the population increases by the same amount, 50, every year.

We use the relation $y = ax + b$.

Given points:

• When $x=10$, $y=400$
• When $x=14$, $y=500$

So,

$400 = 10a + b$

$500 = 14a + b$

Subtract the first equation from the second:

$500 - 400 = (14a + b) - (10a + b)$

$100 = 4a$

$a = 25$

Now substitute in $400 = 10a + b$:

$400 = 10(25) + b$

$400 = 250 + b$

$b = 150$

So, $a=25$ and $b=150$.

We use $y = ax + b$.

Given:

• When $x=10$, $y=800$
• When $x=15$, $y=1100$

So,

$800 = 10a + b$

$1100 = 15a + b$

Subtract the first equation from the second:

$1100 - 800 = (15a + b) - (10a + b)$

$300 = 5a$

$a = 60$

Now substitute into $800 = 10a + b$:

$800 = 10(60) + b$

$800 = 600 + b$

$b = 200$

So, $a=60$ and $b=200$.

Let the relation be

$$\,^{\circ}C = a\,^{\circ}F + b.$$

Using the two given points:

• When $^{\circ}C = 0$, $^{\circ}F = 32$:

$$0 = 32a + b$$

• When $^{\circ}C = 100$, $^{\circ}F = 212$:

$$100 = 212a + b$$

Subtract the first equation from the second:

$$100 = (212-32)a = 180a$$

So,

$$a = \frac{100}{180} = \frac{5}{9}$$

Now substitute in $0 = 32a + b$:

$$0 = 32\cdot\frac{5}{9} + b$$
$$b = -\frac{160}{9}$$

So the linear relation is

$$^{\circ}C = \frac{5}{9}\,^{\circ}F - \frac{160}{9}.$$

We use the given relation

$$^{\circ}C = a\,^{\circ}F + b.$$

From the data:

• Ice melts: $^{\circ}C = 0$, $^{\circ}F = 32$ gives

$$0 = 32a + b$$

• Water boils: $^{\circ}C = 100$, $^{\circ}F = 212$ gives

$$100 = 212a + b$$

Subtracting the first equation from the second:

$$100 = (212-32)a = 180a$$

Hence,

$$a = \frac{100}{180} = \frac{5}{9}.$$

Now substitute into $0 = 32a + b$:

$$0 = 32\cdot\frac{5}{9} + b$$
$$b = -\frac{160}{9}.$$

So, the values are $a = \frac{5}{9}$ and $b = -\frac{160}{9}$.

In the algebraic expression from Example 2,

$$200l + 160w + 50lw,$$

the terms are:

• $200l$
• $160w$
• $50lw$

The variables are:

• $l$
• $w$

The coefficients are:

• of $l$, the coefficient is $200$
• of $w$, the coefficient is $160$
• of $lw$, the coefficient is $50$

So the expression has three terms, two variables, and coefficients 200, 160, and 50.

The expression in Example 2, $200l + 160w + 50lw$, is different from the expression in Example 1, $4x + 5y + 3$.

The main difference is that Example 2 contains the product of variables in the term $50lw$, so it involves two variables multiplied together. Also, it has no constant term.

Example 1 has only terms in separate variables, $4x$ and $5y$, along with a constant term $3$.

So, Example 1 is an algebraic expression in two variables with a constant, while Example 2 has a term involving the product of two variables.

The similarity is that both expressions involve variables, coefficients, and terms.

The difference is that:

• Example 1: $4x + 5y + 3$ has two variables $x$ and $y$, and includes a constant term.
• Example 3: $10x - x^2$ has one variable $x$ only, and includes a term with power 2.

So, Example 1 is an expression in two variables, while Example 3 is an expression in one variable with a higher power.

The perimeter of a square is $4\times$ side.

• Side $=1$ cm, perimeter $=4\times1=4$ cm
• Side $=1.5$ cm, perimeter $=4\times1.5=6$ cm
• Side $=2$ cm, perimeter $=4\times2=8$ cm
• Side $=2.5$ cm, perimeter $=4\times2.5=10$ cm
• Side $=3$ cm, perimeter $=4\times3=12$ cm

So the perimeters are 4 cm, 6 cm, 8 cm, 10 cm and 12 cm.

If the side increases by 0.5 cm, the perimeter increases by $4\times0.5=2$ cm each time.

For the chess club, the amount paid is
$$200+50m$$
where $m$ is the number of matches.

Given the player paid ₹750:
$$200+50m=750$$
$$50m=750-200=550$$
$$m=\frac{550}{50}=11$$
So the player played 11 matches.

Yes. The area expression is
$$10x-x^2=x(10-x)$$
This can be seen as an input-output process: for each value of $x$, we substitute it into the expression and get the area of the rectangle.

For $x=6$ cm,
$$10x-x^2=10(6)-6^2=60-36=24$$
So the area is 24 cm².

The pattern of square tiles is
$$1,3,5,7,9,11,13,\dots$$
Each stage increases by 2.

So the next three stages are:

• Stage 8: $15$
• Stage 9: $17$
• Stage 10: $19$

Sequence up to Stage 7:
$$1,3,5,7,9,11,13$$

The number of tiles at Stage $n$ is given by
$$2n-1$$

• For the 15th stage:

$$2(15)-1=30-1=29$$

• For the 26th stage:

$$2(26)-1=52-1=51$$

Now find the stage for 21 tiles:
$$2n-1=21 \Rightarrow 2n=22 \Rightarrow n=11$$
So Stage 11 has 21 tiles.

For 47 tiles:
$$2n-1=47 \Rightarrow 2n=48 \Rightarrow n=24$$
So Stage 24 has 47 tiles.

Bela starts with ₹100 and spends ₹5 every day.

Amount left on the 15th day:
$$100-5\times15=100-75=25$$
So ₹25 will be left.

For the entire amount to be spent:
$$100-5n=0$$
$$5n=100$$
$$n=20$$
So it will take 20 days to spend the whole amount.

For the auto-rickshaw fare, when $n\ge 2$,
$$\text{Fare}=25+15(n-2)=15n-5$$
Given fare = ₹130:
$$15n-5=130$$
$$15n=135$$
$$n=9$$
So the fare will be ₹130 for 9 km.

For travelling 15 km, use
$$C(d)=100+60d$$
So,
$$C(15)=100+60\times15=100+900=1000$$
The cost is ₹1000.

For the cost to be ₹700:
$$100+60d=700$$
$$60d=600$$
$$d=10$$
So the journey will cost ₹700 for 10 km.

The height of water is given by
$$h(t)=3-0.5t$$
At the end of 5 months:
$$h(5)=3-0.5\times5=3-2.5=0.5$$
So the height will be 0.5 m.

In the equation
$$y=20x+150,$$

• 20 is the amount added to the bill for each extra GB of data used, so it is the cost per GB.
• 150 is the fixed charge that is paid even before extra data use, so it is the fixed monthly fee.

The line is
$$y=2x+1$$
Substitute each given $x$ value:

• $x=2 \Rightarrow y=2(2)+1=5$
• $x=5 \Rightarrow y=2(5)+1=11$
• $x=7 \Rightarrow y=2(7)+1=15$ (already given)
• $x=9 \Rightarrow y=2(9)+1=19$
• $x=12 \Rightarrow y=2(12)+1=25$
• $x=20 \Rightarrow y=2(20)+1=41$

So the missing points are $(2,5)$, $(5,11)$, $(9,19)$, $(12,25)$ and $(20,41)$.

The graph of $y=3x+1$ has positive slope $3$, so it rises from left to right.

The graph of $y=-3x+1$ has negative slope $-3$, so it falls from left to right.

Both lines cut the y-axis at the same point $(0,1)$, but one rises and the other falls.

Yes. If $a$ is fixed and only $b$ varies in the equation
$$y=ax+b,$$
then the slope remains the same, so the lines are parallel. Only the y-intercept changes, so the line shifts up or down without changing its inclination.

We need any polynomial of degree 3 in $x$, and the coefficient of $x^2$ must be $-7$.

One such polynomial is
$$x^3-7x^2+2x+1$$
It has degree 3, and the coefficient of the $x^2$ term is $-7$.

Substitute $x=1$ in $5x^2-3x+7$:
$$5(1)^2-3(1)+7=5-3+7=9$$
So the value is 9.

Substitute $t=a$ into the polynomial:
$$4t^3-t^2+6=4a^3-a^2+6$$
So the value is $4a^3-a^2+6$.

Let the number be $x$.

According to the question,
$$\frac{5}{2}x+\frac{2}{3}=-\frac{7}{12}$$
Subtract $\frac{2}{3}$ from both sides:
$$\frac{5}{2}x=-\frac{7}{12}-\frac{2}{3}$$
Convert to twelfths:
$$-\frac{7}{12}-\frac{8}{12}=-\frac{15}{12}=-\frac{5}{4}$$
So,
$$\frac{5}{2}x=-\frac{5}{4}$$
Multiply both sides by $\frac{2}{5}$:
$$x=-\frac{5}{4}\times\frac{2}{5}=-\frac{1}{2}$$
The number is $-\frac{1}{2}$.

Let the smaller number be $x$. Then the other number is $5x$.

After adding 21 to both numbers, they become $x+21$ and $5x+21$.

One becomes twice the other. Since the larger number should be twice the smaller one:
$$5x+21=2(x+21)$$
$$5x+21=2x+42$$
$$3x=21$$
$$x=7$$
So the numbers are:
$$x=7, \quad 5x=35$$
Therefore, the numbers are 7 and 35.

Start with ₹800 and save ₹250 each month.

After 6 months:
$$800+250\times6=800+1500=2300$$
So the amount is ₹2300.

After 2 years = 24 months:
$$800+250\times24=800+6000=6800$$
So the amount is ₹6800.

Linear pattern for the amount after $n$ months:
$$800+250n$$

Let the tens digit be $x$ and the units digit be $y$.

Given the digits differ by 3:
$$|x-y|=3$$
The original number is $10x+y$ and the interchanged number is $10y+x$.

Their sum is 143:
$$(10x+y)+(10y+x)=143$$
$$11(x+y)=143$$
$$x+y=13$$
Now the digits differ by 3 and add to 13. The pairs are $(8,5)$ or $(5,8)$.

So the two numbers are 85 and 58.

Use the relation from the chapter:
$$y=\frac{9}{5}(x-273)+32$$
For $x=313$ K:
$$y=\frac{9}{5}(313-273)+32$$
$$y=\frac{9}{5}(40)+32=72+32=104$$
So the temperature is 104 °F.

If using exact school approximation with 273.15 K, it is about 104.6 °F, but from the chapter’s formula the answer is 104 °F.

From
$$y=\frac{9}{5}(x-273)+32,$$
we substitute $y=158$:
$$158=\frac{9}{5}(x-273)+32$$
$$126=\frac{9}{5}(x-273)$$
$$\frac{5}{9}\times126=x-273$$
$$70=x-273$$
$$x=343$$
So the temperature is 343 K.

Using the chapter’s formula, this is the computed value.

The graph passes through $(1,5)$ and $(3,11)$.

For a linear polynomial $p(x)=ax+b$:
$$a=\frac{11-5}{3-1}=\frac{6}{2}=3$$
Now use $(1,5)$:
$$5=3(1)+b$$
$$b=2$$
Therefore,
$$p(x)=3x+2$$

From the previous part, the polynomial is
$$p(x)=3x+2$$

To find where it cuts the y-axis, put $x=0$:
$$p(0)=2$$
So it cuts the y-axis at
$$(0,2)$$

To find where it cuts the x-axis, put $p(x)=0$:
$$3x+2=0$$
$$3x=-2$$
$$x=-\frac{2}{3}$$
So it cuts the x-axis at
$$\left(-\frac{2}{3},0\right)$$

Let
$$p(x)=ax+b,\quad q(x)=cx+d$$
Given $p(0)=5$, so
$$b=5$$
Thus,
$$p(x)=ax+5$$

Also, $p(x)-q(x)$ cuts the x-axis at $(3,0)$. So at $x=3$,
$$p(3)-q(3)=0$$
Since
$$p(x)-q(x)=(a-c)x+(5-d)$$
plugging in $x=3$ gives one relation, but we also use the sum condition.

Given
$$p(x)+q(x)=6x+4$$
So comparing coefficients:
$$a+c=6$$
$$5+d=4 \Rightarrow d=-1$$
Hence
$$q(x)=cx-1$$
Now use the x-intercept condition for $p(x)-q(x)$ at $x=3$:
$$(p-q)(3)=0$$
$$( a-c)\cdot 3 + (5-(-1))=0$$
$$3(a-c)+6=0$$
$$a-c=-2$$
Now solve
$$a+c=6,\quad a-c=-2$$
Add:
$$2a=4 \Rightarrow a=2$$
Then
$$c=4$$
Therefore,
$$p(x)=2x+5,\quad q(x)=4x-1$$

For the hexagon matchstick pattern, each new stage adds one hexagon that shares one side with the previous hexagon.

• Stage 1 has 1 hexagon, so it needs 6 matchsticks.
• Every new hexagon adds 5 more matchsticks because one side is shared.

So the numbers are:

• Stage 1: $6$
• Stage 2: $6+5=11$
• Stage 3: $11+5=16$
• Stage 4: $16+5=21$
• Stage 5: $21+5=26$

Thus the completed table is 6, 11, 16, 21, 26.

If the book’s table in your question is the one shown in the chapter, these are the values that complete it.

Let the number of matchsticks in the $n^{\text{th}}$ stage be $M_n$.

• Stage 1 has 6 matchsticks.
• Each new stage adds 5 matchsticks, because the new hexagon shares one side with the previous one.

So the sequence is an arithmetic pattern:
$$6, 11, 16, 21, \dots$$
with first term $6$ and common difference $5$.

Therefore, the rule for the $n^{\text{th}}$ stage is
$$M_n = 6 + (n-1)\cdot 5 = 5n + 1.$$
So the required number of matchsticks at stage $n$ is $5n+1$.

From the pattern, the number of matchsticks in the $n^{\text{th}}$ stage is
$$M_n = 5n + 1.$$
For the 15th stage,
$$M_{15} = 5(15) + 1 = 75 + 1 = 76.$$
So, 76 matchsticks are required.

Using the rule for the pattern,
$$M_n = 5n + 1.$$
If 200 matchsticks form a stage, then
$$5n + 1 = 200.$$
So,
$$5n = 199 \quad \Rightarrow \quad n = \frac{199}{5} = 39.8.$$
Since $n$ must be a whole number stage number, 200 matchsticks cannot form a stage in this pattern.

So, 200 matchsticks cannot form a stage.

For $p(x)=ax+b$, the graph passes through $(2,3)$ and $(6,11)$.

So,
$$3=2a+b \quad \text{and} \quad 11=6a+b.$$
Subtracting,
$$11-3=6a-2a \Rightarrow 8=4a \Rightarrow a=2.$$
Then
$$3=2(2)+b \Rightarrow 3=4+b \Rightarrow b=-1.$$
Hence,
$$p(x)=2x-1.$$