Samacheer Kalvi 10th Maths Exercise 3.15 Solutions

10th Maths Exercise 3.15 solutions (Chapter 3 Algebra) with step-by-step answers for all 12 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance12 answers · full steps below

Q1 (i)Marked price = ₹6500
Q1 (ii)Discount = ₹1250
Q2 (i) $y = 8$
Q2 (ii) $x = 4$
Q3 (i)Constant of variation $k=\dfrac{1}{2}$ ; $y = 4.5$ when $x = 9$
Q3 (ii) $x = 15$
Q4 (i)18 minutes
Q4 (ii)10 pipes
Q5 (i)Constant of variation $k = 360$ (i.e. $xy = 360$ )
Q5 (ii)Each participant gets ₹30
Q6 (i)Direct variation, $y = 15x$ ; amount = ₹90
Q6 (ii)10 hours

Q1. A garment shop announces a flat 50% discount on every purchase of …

A garment shop announces a flat 50% discount on every purchase of items for their customers. Draw the graph for the relation between the Marked Price and the Discount. Hence find

(i)

the marked price when a customer gets a discount of ₹3250 (from graph)

Show steps (6)

Let $x$ = marked price and $y$ = discount.
Discount is $50\%$ of marked price, so $y=\dfrac{50}{100}x=\dfrac{1}{2}x$ .
As $x$ increases $y$ also increases — it is a direct variation with constant of variation $k=\dfrac{1}{2}$ .
Table of points: $(1000,500),(2000,1000),(3000,1500),(4000,2000)$ . Joining them gives a straight line through the origin.
From the graph, take $y=3250$ and read the corresponding $x$ .
Check: $3250=\dfrac{1}{2}x\Rightarrow x=6500$ .

Answer

Marked price = ₹6500

(ii)

the discount when the marked price is ₹2500

From the same line $y=\dfrac{1}{2}x$ .
From the graph, at $x=2500$ read the value of $y$ .
Check: $y=\dfrac{1}{2}\times 2500=1250$ .

Answer

Discount = ₹1250

Q2. Draw the graph of $xy = 24$ , $x, y > 0$ . Using the graph find,

Draw the graph of $xy = 24$ , $x, y > 0$ . Using the graph find,

(i)

$y$ when $x = 3$ .

$xy=24\Rightarrow y=\dfrac{24}{x}$ , a constant product, so it is an inverse variation.
Table of values:
$\begin{array}{c|cccccc} x & 1 & 2 & 3 & 4 & 6 & 8\\ \hline y & 24 & 12 & 8 & 6 & 4 & 3\end{array}$
Plot $(1,24),(2,12),(3,8),(4,6),(6,4),(8,3)$ and join by a smooth curve (rectangular hyperbola).
From the graph, at $x=3$ the curve gives $y=8$ .

Answer

$y = 8$

(ii)

$x$ when $y = 6$ .

From the same curve $xy=24$ .
Draw the line $y=6$ to meet the curve and read $x$ .
Check: $x\times 6=24\Rightarrow x=4$ .

Answer

$x = 4$

Q3. Graph the following linear function $y = \dfrac{1}{2}x$ . Identify the constant of variation …

Graph the following linear function $y = \dfrac{1}{2}x$ . Identify the constant of variation and verify it with the graph. Also

(i)

find $y$ when $x = 9$ .

$y=\dfrac{1}{2}x$ is of the form $y=kx$ , so the constant of variation is $k=\dfrac{1}{2}$ .
Table of values:
$\begin{array}{c|ccccc} x & 0 & 2 & 4 & 6 & 8\\ \hline y & 0 & 1 & 2 & 3 & 4\end{array}$
Plotting these points gives a straight line through the origin; its slope $=\dfrac{1}{2}=k$ , which verifies the constant of variation.
From the graph, at $x=9$ , $y=\dfrac{1}{2}\times 9=4.5$ .

Answer

Constant of variation $k=\dfrac{1}{2}$ ; $y = 4.5$ when $x = 9$

(ii)

find $x$ when $y = 7.5$ .

From the same line $y=\dfrac{1}{2}x$ .
Put $y=7.5$ : $7.5=\dfrac{1}{2}x$ .
$x=2\times 7.5=15$ , which agrees with the graph.

Answer

$x = 15$

Q4. The following table shows the data about the number of pipes and …

The following table shows the data about the number of pipes and the time taken to fill the same tank. Draw the graph for the above data and hence

(i)

find the time taken to fill the tank when five pipes are used.

No. of pipes ( $x$ )	2	3	6	9
Time Taken (in min) ( $y$ )	45	30	15	10

Show steps (6)

As the number of pipes $x$ increases, the time $y$ decreases — it is an inverse variation.
Let $y=\dfrac{k}{x}\Rightarrow xy=k$ .
$k=2\times 45=3\times 30=6\times 15=9\times 10=90$ .
So $xy=90$ .
Plot $(2,45),(3,30),(6,15),(9,10)$ and join by a smooth curve (rectangular hyperbola).
From the graph, when $x=5$ : $y=\dfrac{90}{5}=18$ .

Answer

18 minutes

(ii)

Find the number of pipes when the time is 9 minutes.

No. of pipes ( $x$ )	2	3	6	9
Time Taken (in min) ( $y$ )	45	30	15	10

From the same relation $xy=90$ .
Put $y=9$ : $x=\dfrac{90}{9}=10$ .
The graph also shows $x=10$ when $y=9$ .

Answer

10 pipes

Q5. A school announces that for a certain competitions, the cash price will …

A school announces that for a certain competitions, the cash price will be distributed for all the participants equally as show below.

(i)

Find the constant of variation.

No. of participants ( $x$ )	2	4	6	8	10
Amount for each participant in ₹ ( $y$ )	180	90	60	45	36

As the number of participants $x$ increases, the amount $y$ for each decreases — inverse variation.
Let $y=\dfrac{k}{x}\Rightarrow xy=k$ .
$k=2\times 180=4\times 90=6\times 60=8\times 45=10\times 36=360$ .

Answer

Constant of variation $k = 360$ (i.e. $xy = 360$ )

(ii)

Graph the above data and hence, find how much will each participant get if the number of participants are 12.

No. of participants ( $x$ )	2	4	6	8	10
Amount for each participant in ₹ ( $y$ )	180	90	60	45	36

Plot $(2,180),(4,90),(6,60),(8,45),(10,36)$ and join by a smooth curve (rectangular hyperbola).
From the graph, at $x=12$ read $y$ .
Check: $xy=360\Rightarrow y=\dfrac{360}{12}=30$ .

Answer

Each participant gets ₹30

Q6. A two wheeler parking zone near bus stand charges as below. Check …

A two wheeler parking zone near bus stand charges as below. Check if the amount charged are in direct variation or in inverse variation to the parking time. Graph the data. Also

(i)

find the amount to be paid when parking time is 6 hr.

Time (in hours) ( $x$ )	4	8	12	24
Amount ₹ ( $y$ )	60	120	180	360

As the time $x$ increases, the amount $y$ also increases, and
$\dfrac{60}{4}=\dfrac{120}{8}=\dfrac{180}{12}=\dfrac{360}{24}=15$ .
Since $\dfrac{y}{x}$ is constant, it is a direct variation with $k=15$ , so $y=15x$ .
Plot $(4,60),(8,120),(12,180),(24,360)$ — they lie on a straight line through the origin.
From the graph, at $x=6$ : $y=15\times 6=90$ .

Answer

Direct variation, $y = 15x$ ; amount = ₹90

(ii)

find the parking duration when the amount paid is ₹150.

Time (in hours) ( $x$ )	4	8	12	24
Amount ₹ ( $y$ )	60	120	180	360

From the same line $y=15x$ .
Put $y=150$ : $150=15x$ .
$x=\dfrac{150}{15}=10$ , which agrees with the graph.

Answer

10 hours

Quick check

Find the matrix $X$ if $2X + \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix}$
1. (1) $\begin{pmatrix} -2 & -2 \\ 2 & -1 \end{pmatrix}$
2. (2) $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
3. (3) $\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}$
4. (4) $\begin{pmatrix} 2 & 1 \\ 2 & 2 \end{pmatrix}$
Show answer
Answer: (2) $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
1. $2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix}$
2. $X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$
Question 1 of 5
The values of $a$ and $b$ if $4x^4 - 24x^3 + 76x^2 + ax + b$ is a perfect square are
1. (1) $100,120$
2. (2) $10,12$
3. (3) $-120,100$
4. (4) $12,10$
Show answer
Answer: (3) $-120,100$
1. Let $4x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2$ .
2. Expanding: $4x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2$ .
3. Compare $x^2$ terms: $36 + 4c = 76 \Rightarrow c = 10$ .
4. So $a = -12(10) = -120$ and $b = 10^2 = 100$ .
Question 2 of 5
The solution of the system $x + y - 3z = -6$ , $-7y + 7z = 7$ , $3z = 9$ is
1. (1) $x = 1,\ y = 2,\ z = 3$
2. (2) $x = -1,\ y = 2,\ z = 3$
3. (3) $x = -1,\ y = -2,\ z = 3$
4. (4) $x = 1,\ y = -2,\ z = 3$
Show answer
Answer: (1) $x = 1,\ y = 2,\ z = 3$
1. From $3z = 9$ , $z = 3$ .
2. Put $z=3$ in $-7y + 7z = 7$ : $-7y + 21 = 7 \Rightarrow y = 2$ .
3. Put $y=2,\ z=3$ in $x + y - 3z = -6$ : $x + 2 - 9 = -6 \Rightarrow x = 1$ .
Question 3 of 5
If $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 4$ matrix, how many columns does $AB$ have
1. (1) $3$
2. (2) $4$
3. (3) $2$
4. (4) $5$
Show answer
Answer: (2) $4$
1. $A_{2\times 3} B_{3\times 4}$ gives order $2 \times 4$ .
2. So $AB$ has $4$ columns.
Question 4 of 5
The square root of $\dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6}$ is equal to
1. (1) $\dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right|$
2. (2) $16\left|\dfrac{y^2}{x^2 z^4}\right|$
3. (3) $\dfrac{16}{5}\left|\dfrac{y}{xz^2}\right|$
4. (4) $\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|$
Show answer
Answer: (4) $\dfrac{16}{5}\left|\dfrac{xz^2}{y}\right|$
1. $\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}$
2. Taking square root: $\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|$
Question 5 of 5

Chapter 3 formulas

Roots of a quadratic equation
$ax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Sum of the roots
$\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}$
Product of the roots
$\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}$
Quadratic equation formed from its roots
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Discriminant
$\Delta = b^2 - 4ac$
Nature of roots: real and unequal
$\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}$
Nature of roots: real and equal
$\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}$
Nature of roots: no real roots
$\Delta < 0 \;\Rightarrow\; \text{there are no real roots}$
Cases for a system of linear equations in three variables
$\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}$
LCM of algebraic expressions
$\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}$
Quadratic polynomial and its zeroes
$p(x) = ax^2 + bx + c \;(a \neq 0) \;\Rightarrow\; p(x) \text{ has atmost two zeroes, which lie on the } x\text{-axis}$
Order of a matrix
$\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n$
Row matrix (row vector)
$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}$
Column matrix (column vector)
$A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}$
Square matrix
$\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n$
Zero matrix (null matrix)
$a_{ij} = 0 \text{ for all } i,\, j$
Transpose of a matrix
$\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})$
Diagonal matrix
$a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})$
Scalar matrix
$a_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}$
Identity matrix (unit matrix)
$a_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}$
Lower triangular matrix
$a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})$
Upper triangular matrix
$a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})$
Equality of matrices
$A = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

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Q1. A garment shop announces a flat 50% discount on every purchase of …

(i)

(ii)

Q2. Draw the graph of xy=24xy = 24xy=24 , x,y>0x, y > 0x,y>0 . Using the graph find,

(i)

(ii)

Q3. Graph the following linear function y=12xy = \dfrac{1}{2}xy=21​x . Identify the constant of variation …

(i)

(ii)

Q4. The following table shows the data about the number of pipes and …

(i)

(ii)

Q5. A school announces that for a certain competitions, the cash price will …

(i)

(ii)

Q6. A two wheeler parking zone near bus stand charges as below. Check …

(i)

(ii)

Quick check

Q2. Draw the graph of $xy = 24$ , $x, y > 0$ . Using the graph find,

Q3. Graph the following linear function $y = \dfrac{1}{2}x$ . Identify the constant of variation …