10th Maths Quarterly Exam 2024 Question Paper & Solutions - Tenkasi District

Exam Details

District: Tenkasi District
Examination: Common Quarterly Examination - 2024
Standard: 10
Subject: Mathematics
Date: 25-09-2024
Time: 3.00 Hours
Marks: 100

Part I: Answer all the following questions (14 x 1 = 14)

1) If A = {a, b, q}, B = {2, 3}, C = {p, q, r, s} then $ n[(A \cup C) \times B] $ is

a) 8
b) 20
c) 12
d) 16

Given: A = {a, b, q}, B = {2, 3}, C = {p, q, r, s}.

First, find $ A \cup C $: $ A \cup C = \{a, b, q\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\} $.

Number of elements in $ A \cup C $ is $ n(A \cup C) = 6 $.

Number of elements in B is $ n(B) = 2 $.

Then, $ n[(A \cup C) \times B] = n(A \cup C) \times n(B) = 6 \times 2 = 12 $.

Answer: c) 12

2) $ f(x) = (x+1)^3 - (x-1)^3 $ represents a function which is

a) linear
b) cubic
c) reciprocal
d) quadratic

We use the formulas: $ (a+b)^3 = a^3+3a^2b+3ab^2+b^3 $ and $ (a-b)^3 = a^3-3a^2b+3ab^2-b^3 $.

$ f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) $

$ f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 $

$ f(x) = 6x^2 + 2 $.

The highest power of x is 2, so it is a quadratic function.

Answer: d) quadratic

3) If the HCF of 65 and 117 is expressible in the form of 65m-117, then the value of m is

a) 4
b) 2
c) 1
d) 3

Find HCF of 65 and 117 using Euclid's algorithm:

$ 117 = 1 \times 65 + 52 $

$ 65 = 1 \times 52 + 13 $

$ 52 = 4 \times 13 + 0 $. The HCF is 13.

Given, $ 65m - 117 = 13 $.

$ 65m = 130 \implies m = \frac{130}{65} = 2 $.

Answer: b) 2

4) Given $ F_1 = 1, F_2 = 3 $ and $ F_n = F_{n-1} + F_{n-2} $ then $ F_5 $ is

a) 3
b) 5
c) 8
d) 11

$ F_1 = 1 $

$ F_2 = 3 $

$ F_3 = F_2 + F_1 = 3 + 1 = 4 $

$ F_4 = F_3 + F_2 = 4 + 3 = 7 $

$ F_5 = F_4 + F_3 = 7 + 4 = 11 $

Answer: d) 11

5) If $ A = 2^{65} $ and $ B = 2^{64} + 2^{63} + 2^{62} + ... + 2^0 $ which of the following is true?

a) B is $ 2^{64} $ more than A
b) A and B are equal
c) B is larger than A by 1
d) A is larger than B by 1

B is a geometric progression with first term $ a = 2^0 = 1 $, common ratio $ r = 2 $, and number of terms $ n = 65 $.

Sum of GP: $ S_n = a \frac{r^n - 1}{r - 1} $.

$ B = 1 \times \frac{2^{65} - 1}{2 - 1} = 2^{65} - 1 $.

Since $ A = 2^{65} $, we have $ B = A - 1 $.

This means A is larger than B by 1.

Answer: d) A is larger than B by 1

6) $ y^2 + \frac{1}{y^2} $ is not equal to

a) $ \frac{y^4+1}{y^2} $
b) $ \left(y + \frac{1}{y}\right)^2 - 2 $
c) $ \left(y - \frac{1}{y}\right)^2 + 2 $
d) $ \left(y - \frac{1}{y}\right)^2 - 2 $

Let's check each option:

a) $ \frac{y^4+1}{y^2} = \frac{y^4}{y^2} + \frac{1}{y^2} = y^2 + \frac{1}{y^2} $. (Equal)

b) $ \left(y + \frac{1}{y}\right)^2 - 2 = (y^2 + 2(y)(\frac{1}{y}) + \frac{1}{y^2}) - 2 = y^2 + 2 + \frac{1}{y^2} - 2 = y^2 + \frac{1}{y^2} $. (Equal)

c) $ \left(y - \frac{1}{y}\right)^2 + 2 = (y^2 - 2(y)(\frac{1}{y}) + \frac{1}{y^2}) + 2 = y^2 - 2 + \frac{1}{y^2} + 2 = y^2 + \frac{1}{y^2} $. (Equal)

d) $ \left(y - \frac{1}{y}\right)^2 - 2 = (y^2 - 2 + \frac{1}{y^2}) - 2 = y^2 + \frac{1}{y^2} - 4 $. (Not Equal)

Answer: d) $ \left(y - \frac{1}{y}\right)^2 - 2 $

7) Graph of a linear equation is a

a) straight line
b) circle
c) parabola
d) hyperbola

The graph of any linear equation in two variables (e.g., $ ax+by+c=0 $) is always a straight line.

Answer: a) straight line

8) If $ f(x) = 2x^2 $ and $ g(x) = \frac{1}{3x} $, then fog is

a) $ \frac{2x^2}{3} $
b) $ \frac{2}{3x^2} $
c) $ \frac{2}{9x^2} $
d) $ \frac{1}{6x^2} $

fog means $ f(g(x)) $.

Substitute $ g(x) $ into $ f(x) $: $ f(g(x)) = f(\frac{1}{3x}) $.

Now apply the function f: $ 2 \left( \frac{1}{3x} \right)^2 = 2 \left( \frac{1}{9x^2} \right) = \frac{2}{9x^2} $.

Answer: c) $ \frac{2}{9x^2} $

9) In $ \Delta LMN $, $ \angle L = 60^\circ, \angle M = 50^\circ $. If $ \Delta LMN \sim \Delta PQR $ then value of $ \angle R $ is

a) $ 40^\circ $
b) $ 70^\circ $
c) $ 30^\circ $
d) $ 110^\circ $

In $ \Delta LMN $, the sum of angles is $ 180^\circ $.

$ \angle N = 180^\circ - (\angle L + \angle M) = 180^\circ - (60^\circ + 50^\circ) = 180^\circ - 110^\circ = 70^\circ $.

Since $ \Delta LMN \sim \Delta PQR $, the corresponding angles are equal.

$ \angle L = \angle P, \angle M = \angle Q, \angle N = \angle R $.

Therefore, $ \angle R = \angle N = 70^\circ $.

Answer: b) $ 70^\circ $

10) In a $ \Delta ABC $, AD is the bisector of $ \angle BAC $. If AB=8cm, BD=6cm and DC=3cm. The length of the side AC is

a) 6 cm
b) 4 cm
c) 3 cm
d) 8 cm

By Angle Bisector Theorem, the ratio of the sides adjacent to the angle is equal to the ratio of the segments the bisector divides the opposite side into.

$ \frac{AB}{AC} = \frac{BD}{DC} $

$ \frac{8}{AC} = \frac{6}{3} \implies \frac{8}{AC} = 2 $

$ AC = \frac{8}{2} = 4 $ cm.

Answer: b) 4 cm

11) The straight line given by the equation x=11 is

a) Parallel to x-axis
b) Parallel to y-axis
c) Passing through the origin
d) Passing through the point (0, 11)

The equation $ x = c $ (where c is a constant) represents a vertical line.

A vertical line is always parallel to the y-axis.

Answer: b) Parallel to y-axis

12) The slope of the line joining (12, 3) and (4, a) is $ \frac{1}{8} $. The value of 'a' is

a) 1
b) 4
c) -5
d) 2

Slope $ m = \frac{y_2 - y_1}{x_2 - x_1} $.

$ \frac{1}{8} = \frac{a - 3}{4 - 12} = \frac{a - 3}{-8} $

Multiply both sides by -8: $ -1 = a - 3 $.

$ a = 3 - 1 = 2 $.

Answer: d) 2

13) When proving that a quadrilateral is a Parallelogram by using slopes you must find

a) The slopes of two sides
b) The slopes of two pair of opposite sides
c) The lengths of all sides
d) Both the lengths and slopes of two sides.

A key property of a parallelogram is that its opposite sides are parallel.

In coordinate geometry, parallel lines have equal slopes.

Therefore, to prove a quadrilateral is a parallelogram using slopes, one must show that both pairs of opposite sides have equal slopes.

Answer: b) The slopes of two pair of opposite sides

14) If $ (\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = K + \tan^2\alpha + \cot^2\alpha $, then the value of K is equal to

a) 9
b) 7
c) 5
d) 3

Expand the LHS:

$ (\sin^2 \alpha + \csc^2 \alpha + 2\sin \alpha \csc \alpha) + (\cos^2 \alpha + \sec^2 \alpha + 2\cos \alpha \sec \alpha) $

Since $ \sin \alpha \csc \alpha = 1 $ and $ \cos \alpha \sec \alpha = 1 $:

$ (\sin^2 \alpha + \cos^2 \alpha) + \csc^2 \alpha + \sec^2 \alpha + 2 + 2 $

Using identities $ \sin^2 \alpha + \cos^2 \alpha = 1 $, $ \csc^2 \alpha = 1 + \cot^2 \alpha $, $ \sec^2 \alpha = 1 + \tan^2 \alpha $:

$ 1 + (1 + \cot^2 \alpha) + (1 + \tan^2 \alpha) + 4 $

$ 1 + 1 + 1 + 4 + \tan^2 \alpha + \cot^2 \alpha = 7 + \tan^2 \alpha + \cot^2 \alpha $

Comparing with $ K + \tan^2\alpha + \cot^2\alpha $, we get $ K=7 $.

Answer: b) 7

Part II: Answer any 10 questions. 28th question is compulsory (10 x 2 = 20)

15) If A={m, n}; B = $ \phi $, Find i) A×B ii) A×A

i) $ A \times B $: The cartesian product with an empty set is always an empty set.

$ A \times B = \{m, n\} \times \phi = \phi $

ii) $ A \times A $:

$ A \times A = \{m, n\} \times \{m, n\} = \{(m,m), (m,n), (n,m), (n,n)\} $

16) A function f is defined by $ f(x)=3-2x $. Find x such that $ f(x^2) = [f(x)]^2 $

Given $ f(x)=3-2x $.

LHS: $ f(x^2) = 3 - 2(x^2) = 3 - 2x^2 $

RHS: $ [f(x)]^2 = (3-2x)^2 = 9 - 12x + 4x^2 $

Equating LHS and RHS: $ 3 - 2x^2 = 9 - 12x + 4x^2 $

$ 6x^2 - 12x + 6 = 0 $

Divide by 6: $ x^2 - 2x + 1 = 0 $

$ (x-1)^2 = 0 \implies x-1=0 \implies x=1 $

The value of x is 1.

17) Find the value of k, such that fog=gof if $ f(x)=3x+2 $ and $ g(x)=6x-k $

LHS: fog = $ f(g(x)) = f(6x-k) = 3(6x-k) + 2 = 18x - 3k + 2 $

RHS: gof = $ g(f(x)) = g(3x+2) = 6(3x+2) - k = 18x + 12 - k $

Given fog = gof:

$ 18x - 3k + 2 = 18x + 12 - k $

$ -3k + 2 = 12 - k $

$ 2 - 12 = -k + 3k $

$ -10 = 2k \implies k = -5 $

The value of k is -5.

18) Use Euclid's Division Algorithm to find HCF of 340 and 412

By Euclid's Division Algorithm, $ a = bq + r $, where $ 0 \le r < b $.

Step 1: $ 412 = 340 \times 1 + 72 $

Step 2: $ 340 = 72 \times 4 + 52 $

Step 3: $ 72 = 52 \times 1 + 20 $

Step 4: $ 52 = 20 \times 2 + 12 $

Step 5: $ 20 = 12 \times 1 + 8 $

Step 6: $ 12 = 8 \times 1 + 4 $

Step 7: $ 8 = 4 \times 2 + 0 $

The last non-zero remainder is the HCF.

HCF of 340 and 412 is 4.

19) Find x, y and z given that the numbers x, 10, y, 24, z are in A.P

Since the numbers are in an Arithmetic Progression (A.P), the common difference 'd' is constant.

Let the terms be $ a_1=x, a_2=10, a_3=y, a_4=24, a_5=z $.

We can find 'd' using known terms: $ d = a_4 - a_3 = a_3 - a_2 $. So, $ 24 - y = y - 10 $.

$ 2y = 34 \implies y = 17 $.

Now find d: $ d = y - 10 = 17 - 10 = 7 $.

Find x: $ x = a_1 = a_2 - d = 10 - 7 = 3 $.

Find z: $ z = a_5 = a_4 + d = 24 + 7 = 31 $.

x = 3, y = 17, z = 31.

20) Simplify $ \frac{5t^3}{4t-8} \times \frac{6t-12}{10t} $

Factor the expressions:

$ \frac{5t^3}{4(t-2)} \times \frac{6(t-2)}{10t} $

Cancel the common term $ (t-2) $:

$ \frac{5t^3}{4} \times \frac{6}{10t} $

Multiply the fractions: $ \frac{30t^3}{40t} $

Simplify the expression: $ \frac{3}{4} t^{3-1} = \frac{3}{4} t^2 $

Simplified form: $ \frac{3}{4} t^2 $

21) Find the sum and product of roots for $ 3 + \frac{1}{a} = \frac{10}{a^2} $

First, convert the equation to standard quadratic form $ Ax^2 + Bx + C = 0 $. Here the variable is 'a'.

Multiply the entire equation by $ a^2 $ to clear the denominators:

$ 3a^2 + a^2(\frac{1}{a}) = a^2(\frac{10}{a^2}) $

$ 3a^2 + a = 10 $

$ 3a^2 + a - 10 = 0 $

For a quadratic equation $ Ax^2+Bx+C=0 $:

Sum of roots = $ -\frac{B}{A} = -\frac{1}{3} $

Product of roots = $ \frac{C}{A} = \frac{-10}{3} $

Sum of roots = $ -\frac{1}{3} $, Product of roots = $ -\frac{10}{3} $

22) If the difference between the roots of the equation $ x^2 - 13x + k = 0 $ is 17, find K.

Let the roots be $ \alpha $ and $ \beta $.

From the equation, sum of roots $ \alpha + \beta = -(\frac{-13}{1}) = 13 $.

Product of roots $ \alpha \beta = \frac{k}{1} = k $.

Given, difference between roots $ |\alpha - \beta| = 17 $.

We have two equations:

$ \alpha + \beta = 13 $ --- (1)

$ \alpha - \beta = 17 $ --- (2)

Adding (1) and (2): $ 2\alpha = 30 \implies \alpha = 15 $.

Substitute $ \alpha=15 $ into (1): $ 15 + \beta = 13 \implies \beta = -2 $.

Now find k: $ k = \alpha \beta = 15 \times (-2) = -30 $.

The value of K is -30.

Part II Solutions (Questions 23-27)

23) In the figure AD is the bisector of $\angle A$, If BD=4cm DC=3cm and AB=6cm, find AC.

By the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side in the same ratio as the other two sides.

So, we have the proportion:

$\frac{AB}{AC} = \frac{BD}{DC}$

Substitute the given values into the equation:

$\frac{6}{AC} = \frac{4}{3}$

Now, solve for AC by cross-multiplying:

$4 \times AC = 6 \times 3$

$4 \times AC = 18$

$AC = \frac{18}{4} = 4.5$ cm

The length of AC is 4.5 cm.

24) Find the area of the triangle, formed by the (1,−1), (-4, 6) and (-3, -5)

Let the vertices be A(1, -1), B(-4, 6), and C(-3, -5).

The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:

Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_1) - (y_1x_2 + y_2x_3 + y_3x_1)|$

Substitute the coordinates:

Area = $\frac{1}{2} |((1)(6) + (-4)(-5) + (-3)(-1)) - ((-1)(-4) + (6)(-3) + (-5)(1))|$

Area = $\frac{1}{2} |(6 + 20 + 3) - (4 - 18 - 5)|$

Area = $\frac{1}{2} |(29) - (-19)|$

Area = $\frac{1}{2} |29 + 19| = \frac{1}{2} |48|$

Area = 24

The area of the triangle is 24 square units.

25) Find the intercepts made by the line $4x-9y+36=0$ on the coordinate axes

The given equation of the line is $4x - 9y + 36 = 0$.

To find the x-intercept, set $y = 0$:

$4x - 9(0) + 36 = 0$

$4x = -36 \implies x = -9$

So, the x-intercept is -9. The point is (-9, 0).

To find the y-intercept, set $x = 0$:

$4(0) - 9y + 36 = 0$

$-9y = -36 \implies y = 4$

So, the y-intercept is 4. The point is (0, 4).

The x-intercept is -9 and the y-intercept is 4.

26) Show that the straight lines $x-2y+3=0$ and $6x+3y+8=0$ are perpendicular

Two lines are perpendicular if the product of their slopes is -1 (i.e., $m_1 \times m_2 = -1$).

First line: $x - 2y + 3 = 0$. Let's find its slope ($m_1$).

Rearrange to the form $y=mx+c$: $2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2}$.

So, the slope $m_1 = \frac{1}{2}$.

Second line: $6x + 3y + 8 = 0$. Let's find its slope ($m_2$).

Rearrange to the form $y=mx+c$: $3y = -6x - 8 \implies y = -2x - \frac{8}{3}$.

So, the slope $m_2 = -2$.

Now, multiply the slopes:

$m_1 \times m_2 = (\frac{1}{2}) \times (-2) = -1$.

Since the product of the slopes is -1, the lines are perpendicular.

Hence, the given straight lines are perpendicular.

27) Prove that $\frac{\cos \theta}{1 + \sin \theta} = \sec\theta - \tan\theta$

We start with the Left Hand Side (LHS): $\frac{\cos \theta}{1 + \sin \theta}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 - \sin \theta)$:

LHS = $\frac{\cos \theta}{1 + \sin \theta} \times \frac{1 - \sin \theta}{1 - \sin \theta}$

LHS = $\frac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{\cos \theta (1 - \sin \theta)}{1 - \sin^2 \theta}$

Using the identity $\sin^2\theta + \cos^2\theta = 1$, we know that $1 - \sin^2 \theta = \cos^2 \theta$.

LHS = $\frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta}$

Cancel out a $\cos \theta$ term:

LHS = $\frac{1 - \sin \theta}{\cos \theta}$

Split the fraction into two parts:

LHS = $\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}$

Using the identities $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$:

LHS = $\sec\theta - \tan\theta$

Thus, LHS = RHS.

Hence Proved.

28) Find the sum $ 3+1+\frac{1}{3}+.....\infty $

This is an infinite geometric progression (G.P.).

First term $ a = 3 $.

Common ratio $ r = \frac{1}{3} $.

Since $ |r| = |\frac{1}{3}| < 1 $, the sum to infinity exists.

The formula for the sum to infinity is $ S_\infty = \frac{a}{1-r} $.

$ S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2} $.

The sum is $ \frac{9}{2} $ or 4.5.

Part III Solutions (Questions 29-31)

29) Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that $(A \cap B) \times C = (A \times C) \cap (B \times C)$

First, let's write the sets in roster form:

A = {1, 2, 3, 4, 5, 6, 7}

B = {2, 3, 5, 7}

C = {2}

LHS: $(A \cap B) \times C$

First, find $A \cap B$:

$A \cap B = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 3, 5, 7\} = \{2, 3, 5, 7\}$

Now, find the Cartesian product with C:

$(A \cap B) \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\}$ --- (1)

RHS: $(A \times C) \cap (B \times C)$

First, find $A \times C$:

$A \times C = \{1, 2, 3, 4, 5, 6, 7\} \times \{2\} = \{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)\}$

Next, find $B \times C$:

$B \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\}$

Now, find the intersection of these two sets:

$(A \times C) \cap (B \times C) = \{(2, 2), (3, 2), (5, 2), (7, 2)\}$ --- (2)

From (1) and (2), we can see that LHS = RHS.

Hence, $(A \cap B) \times C = (A \times C) \cap (B \times C)$ is verified.

30) Let f: A → B be a function defined by $f(x) = \frac{x}{2} - 1$ where A = {2, 4, 6, 10, 12} B = {0, 1, 2, 4, 5, 9}. Represent f by (i) Set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph

Given function: $f(x) = \frac{x}{2} - 1$. Let's find the image for each element in A.

$f(2) = \frac{2}{2} - 1 = 1 - 1 = 0$

$f(4) = \frac{4}{2} - 1 = 2 - 1 = 1$

$f(6) = \frac{6}{2} - 1 = 3 - 1 = 2$

$f(10) = \frac{10}{2} - 1 = 5 - 1 = 4$

$f(12) = \frac{12}{2} - 1 = 6 - 1 = 5$

(i) Set of ordered pairs:

$f = \{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\}$

(ii) A table:

x	f(x)
2	0
4	1
6	2
10	4
12	5

(iii) An arrow diagram:

Draw two ovals. Label the first 'A' and the second 'B'.

Inside A, list the elements: 2, 4, 6, 10, 12.

Inside B, list the elements: 0, 1, 2, 4, 5, 9.

Draw arrows from each element in A to its corresponding image in B:

2 → 0, 4 → 1, 6 → 2, 10 → 4, 12 → 5

(iv) A graph:

Plot the ordered pairs on a coordinate plane with the x-axis and y-axis. The points to be plotted are:

(2, 0), (4, 1), (6, 2), (10, 4), and (12, 5).

31) If $f(x)=x^2$, $g(x) = 2x$ and $h(x) = x + 4$ show that $(fog)oh = fo(goh)$

LHS: $(fog)oh$

First, find $fog(x)$:

$fog(x) = f(g(x)) = f(2x) = (2x)^2 = 4x^2$

Now, find $(fog)oh(x)$:

$(fog)oh(x) = (fog)(h(x)) = (fog)(x+4)$

Substitute $(x+4)$ into the expression for $fog(x)$:

$(fog)oh(x) = 4(x+4)^2$ --- (1)

RHS: $fo(goh)$

First, find $goh(x)$:

$goh(x) = g(h(x)) = g(x+4) = 2(x+4)$

Now, find $fo(goh)(x)$:

$fo(goh)(x) = f(goh(x)) = f(2(x+4))$

Substitute $2(x+4)$ into the function $f(x)$:

$fo(goh)(x) = [2(x+4)]^2 = 4(x+4)^2$ --- (2)

From (1) and (2), we see that the expressions are identical.

LHS = RHS. Thus, $(fog)oh = fo(goh)$ is shown.

Part III: Answer any 10 questions. 42th question is compulsory (10 x 5 = 50)

32) Find the sum of all natural numbers between 300 and 600 which are divisible by 7.

The sequence of numbers between 300 and 600 divisible by 7 forms an A.P.

To find the first term (a), divide 300 by 7: $ 300 \div 7 = 42 $ with remainder 6. So, the first number is $ 300 + (7-6) = 301 $. So, $ a=301 $.

To find the last term (l), divide 600 by 7: $ 600 \div 7 = 85 $ with remainder 5. So, the last number is $ 600 - 5 = 595 $. So, $ l=595 $.

The common difference $ d = 7 $.

Number of terms $ n = \frac{l-a}{d} + 1 = \frac{595-301}{7} + 1 = \frac{294}{7} + 1 = 42 + 1 = 43 $.

Sum of the A.P. is $ S_n = \frac{n}{2}(a+l) $.

$ S_{43} = \frac{43}{2}(301+595) = \frac{43}{2}(896) = 43 \times 448 = 19264 $.

The sum is 19264.

Part III Solution (Question 33)

33) If a, b, c are three consecutive terms of an A.P and x,y,z are three consecutive terms of a G.P. Then prove that $x^{b-c} y^{c-a} z^{a-b} = 1$

Step 1: Analyze the given conditions.

Since a, b, c are in an Arithmetic Progression (A.P.), the common difference is constant. Let the common difference be 'd'.

$b - a = d$ and $c - b = d$.

From these, we can express the exponents in terms of 'd':

$b - c = -(c - b) = -d$
$c - a = (c - b) + (b - a) = d + d = 2d$
$a - b = -(b - a) = -d$

Since x, y, z are in a Geometric Progression (G.P.), the common ratio is constant. Let the first term be x and the common ratio be 'r'.

$y = xr$

$z = xr^2$

Step 2: Substitute these relationships into the Left Hand Side (LHS) of the equation.

LHS = $x^{b-c} y^{c-a} z^{a-b}$

First, substitute the expressions for the exponents derived from the A.P. condition:

LHS = $x^{-d} y^{2d} z^{-d}$

Next, substitute the expressions for y and z derived from the G.P. condition:

LHS = $x^{-d} (xr)^{2d} (xr^2)^{-d}$

Step 3: Simplify the expression using the laws of exponents.

Using the rule $(ab)^n = a^n b^n$:

LHS = $x^{-d} \cdot (x^{2d} r^{2d}) \cdot (x^{-d} (r^2)^{-d})$

Using the rule $(a^m)^n = a^{mn}$:

LHS = $x^{-d} \cdot x^{2d} \cdot r^{2d} \cdot x^{-d} \cdot r^{-2d}$

Group the terms with the same base (x and r) and use the rule $a^m a^n = a^{m+n}$:

LHS = $(x^{-d + 2d - d}) \cdot (r^{2d - 2d})$

Simplify the exponents:

LHS = $x^{0} \cdot r^{0}$

Since any non-zero number raised to the power of 0 is 1:

LHS = $1 \cdot 1 = 1$

Step 4: Conclude the proof.

We have shown that LHS = 1, which is equal to the Right Hand Side (RHS).

Hence, $x^{b-c} y^{c-a} z^{a-b} = 1$ is proved.

34) Find the sum of $ 9^3 + 10^3 + ... + 21^3 $

We can write the sum as $ (1^3 + 2^3 + ... + 21^3) - (1^3 + 2^3 + ... + 8^3) $.

Using the formula for the sum of cubes of first n natural numbers: $ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 $.

For n=21: $ \left( \frac{21(21+1)}{2} \right)^2 = \left( \frac{21 \times 22}{2} \right)^2 = (21 \times 11)^2 = 231^2 = 53361 $.

For n=8: $ \left( \frac{8(8+1)}{2} \right)^2 = \left( \frac{8 \times 9}{2} \right)^2 = (4 \times 9)^2 = 36^2 = 1296 $.

Required sum = $ 53361 - 1296 = 52065 $.

The sum is 52065.

Part III Solutions (Questions 35-41)

35) Find the LCM of each pair of the following Polynomials $a^2+4a-12$, $a^2-5a+6$ whose GCD is $a-2$

Let the two polynomials be $p(x) = a^2+4a-12$ and $q(x) = a^2-5a+6$.

We know the relationship between LCM, GCD, and the polynomials:

$LCM \times GCD = p(x) \times q(x)$

$LCM = \frac{p(x) \times q(x)}{GCD}$

First, let's factorize the polynomials:

$p(x) = a^2+4a-12 = (a+6)(a-2)$

$q(x) = a^2-5a+6 = (a-3)(a-2)$

Given, $GCD = a-2$.

Now, substitute the factored polynomials and GCD into the formula:

$LCM = \frac{(a+6)(a-2) \times (a-3)(a-2)}{a-2}$

Cancel one of the $(a-2)$ terms from the numerator and denominator:

$LCM = (a+6)(a-3)(a-2)$

The LCM is $(a-2)(a-3)(a+6)$.

36) If $9x^4+12x^3+28x^2+ax + b$ is a perfect square Find the values of 'a' and 'b'

We can find the values of 'a' and 'b' using the long division method for finding the square root.

                    3x² + 2x  + 4
                  _____________________
            3x²   | 9x⁴ + 12x³ + 28x² + ax + b
                  |-(9x⁴)
                  |____________________
            6x²+2x|       12x³ + 28x²
                  |     -(12x³ +  4x²)
                  |     _________________
            6x²+4x+4|             24x² + ax + b
                    |           -(24x² + 16x + 16)
                    |           __________________
                    |                    0
            

Since the polynomial is a perfect square, the remainder must be 0.

This means $(24x^2 + ax + b) - (24x^2 + 16x + 16) = 0$.

$(a-16)x + (b-16) = 0$.

Equating the coefficients of x and the constant term to zero:

$a - 16 = 0 \implies a = 16$

$b - 16 = 0 \implies b = 16$

The values are a = 16 and b = 16.

37) If $\alpha, \beta$ are the roots of $7x^2+ax+2=0$ and if $\beta - \alpha = -\frac{13}{7}$. Find the values of 'a'

For the quadratic equation $7x^2+ax+2=0$, the sum and product of roots are:

Sum of roots: $\alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{a}{7}$

Product of roots: $\alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{2}{7}$

Given: $\beta - \alpha = -\frac{13}{7}$

We use the identity: $(\alpha+\beta)^2 = (\beta-\alpha)^2 + 4\alpha\beta$

Substitute the known values into the identity:

$(-\frac{a}{7})^2 = (-\frac{13}{7})^2 + 4(\frac{2}{7})$

$\frac{a^2}{49} = \frac{169}{49} + \frac{8}{7}$

To add the fractions on the right, find a common denominator (49):

$\frac{a^2}{49} = \frac{169}{49} + \frac{8 \times 7}{7 \times 7} = \frac{169}{49} + \frac{56}{49}$

$\frac{a^2}{49} = \frac{169+56}{49} = \frac{225}{49}$

$a^2 = 225$

$a = \pm\sqrt{225} = \pm 15$

The values of 'a' are 15 and -15.

38) State and prove Angle Bisector Theorem

Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.

Given: In $\triangle ABC$, AD is the internal bisector of $\angle A$ which meets the side BC at D.

To Prove: $\frac{AB}{AC} = \frac{BD}{DC}$

Construction: Draw a line CE parallel to AB. Extend the line segment AD to meet the line CE at E.

Proof:

Statement	Reason
In $\triangle DCE$ and $\triangle DBA$, $\angle DCE \cong \angle DBA$ and $\angle DEC \cong \angle DAB$. Thus $\triangle DCE \sim \triangle DBA$.	AA Similarity (Since CE \|\| AB, corresponding angles are equal). This approach is complex. Let's use angles directly.

Proof (Alternative and Standard Method):

Statement	Reason
$\angle BAE = \angle AEC$ (i.e. $\angle 1 = \angle 3$)	Since AB \|\| CE and AE is the transversal, alternate interior angles are equal.
$\angle CAD = \angle ACE$ (i.e. $\angle 2 = \angle 4$)	Since AB \|\| CE and AC is transversal, this is incorrect. The correct reason is for corresponding angles. Let's restart the angles.

Correct Proof:

In $\triangle BCE$, since AD || CE, by Thales Theorem (BPT), we have $\frac{AB}{AE} = \frac{BD}{DC}$ --- (1)

Since AB || CE and AC is the transversal, $\angle BAE = \angle AEC$ (Alternate angles).

Also, with transversal BE, $\angle DAB = \angle AEC$ (Corresponding angles). Wait, construction is different. Let's use the standard construction: Draw CE || DA to meet BA extended at E.

Correct Standard Proof:

Construction: Through C, draw a line CE parallel to AD, such that it meets the line BA extended at E.

Proof:

Since AD || CE, by Basic Proportionality Theorem on $\triangle BCE$, we have $\frac{BA}{AE} = \frac{BD}{DC}$ --- (i)

Now, since AD || CE and AC is the transversal, $\angle DAC = \angle ACE$ (Alternate interior angles).

Since AD || CE and BE is the transversal, $\angle BAD = \angle AEC$ (Corresponding angles).

But we are given that AD is the angle bisector, so $\angle BAD = \angle DAC$.

Therefore, $\angle ACE = \angle AEC$.

In $\triangle ACE$, since the angles opposite to sides AE and AC are equal, the sides themselves are equal. Hence, $AE = AC$.

Substitute $AC$ for $AE$ in equation (i):

$\frac{AB}{AC} = \frac{BD}{DC}$. Hence the theorem is proved.

39) If the points A(-3, 9) B(a, b) and C(4, -5) are collinear and if a+b=1, then find 'a' and 'b'

Since the points A, B, and C are collinear, the slope of AB must be equal to the slope of BC.

Slope of AB = $\frac{b - 9}{a - (-3)} = \frac{b-9}{a+3}$

Slope of BC = $\frac{-5 - b}{4 - a}$

Equating the slopes: $\frac{b-9}{a+3} = \frac{-5-b}{4-a}$

Cross-multiply: $(b-9)(4-a) = (a+3)(-5-b)$

$4b - ab - 36 + 9a = -5a - ab - 15 - 3b$

The term '-ab' cancels from both sides.

$4b - 36 + 9a = -5a - 15 - 3b$

Rearrange the terms to form a linear equation:

$9a + 5a + 4b + 3b = 36 - 15$

$14a + 7b = 21$

Divide the entire equation by 7: $2a + b = 3$ --- (1)

We are also given: $a + b = 1$ --- (2)

Subtract equation (2) from equation (1):

$(2a + b) - (a + b) = 3 - 1$

$a = 2$

Substitute $a=2$ into equation (2):

$2 + b = 1 \implies b = 1 - 2 = -1$

The values are a = 2 and b = -1.

40) If the points A(2, 2) B(-2, −3) C(1, −3) and D(x, y) form a parallelogram then find the value of 'x' and 'y'

A property of a parallelogram is that its diagonals bisect each other. This means the midpoint of diagonal AC is the same as the midpoint of diagonal BD.

Midpoint Formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Midpoint of AC = $(\frac{2+1}{2}, \frac{2+(-3)}{2}) = (\frac{3}{2}, -\frac{1}{2})$

Midpoint of BD = $(\frac{-2+x}{2}, \frac{-3+y}{2})$

Equating the x-coordinates and y-coordinates of the midpoints:

For x-coordinate: $\frac{-2+x}{2} = \frac{3}{2} \implies -2+x = 3 \implies x = 5$

For y-coordinate: $\frac{-3+y}{2} = -\frac{1}{2} \implies -3+y = -1 \implies y = 2$

The value of x is 5 and the value of y is 2. The coordinates of D are (5, 2).

41) Prove that $\frac{\cos^3 A – \sin^3 A}{\cos A - \sin A} - \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = 2 \sin A \cos A$

We will simplify the Left Hand Side (LHS) using the sum and difference of cubes formulas:

$a^3 - b^3 = (a-b)(a^2+ab+b^2)$

$a^3 + b^3 = (a+b)(a^2-ab+b^2)$

First term of LHS: $\frac{\cos^3 A – \sin^3 A}{\cos A - \sin A}$

$= \frac{(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A)}{\cos A - \sin A}$

Cancel the $(\cos A - \sin A)$ term. Since $\cos^2 A + \sin^2 A = 1$, the expression becomes:

$= 1 + \cos A \sin A$

Second term of LHS: $\frac{\cos^3 A + \sin^3 A}{\cos A + \sin A}$

$= \frac{(\cos A + \sin A)(\cos^2 A - \cos A \sin A + \sin^2 A)}{\cos A + \sin A}$

Cancel the $(\cos A + \sin A)$ term. Since $\cos^2 A + \sin^2 A = 1$, the expression becomes:

$= 1 - \cos A \sin A$

Now, substitute these simplified parts back into the LHS expression:

LHS = $(1 + \cos A \sin A) - (1 - \cos A \sin A)$

LHS = $1 + \cos A \sin A - 1 + \cos A \sin A$

LHS = $2 \cos A \sin A$

LHS = RHS

Hence, the identity is proved.

42) If A(-3, 0) B(10, −2) and C(12, 3) are the vertices of ∆ABC. Find the equation of the altitude through 'A'.

The altitude through A is a line segment from vertex A that is perpendicular to the opposite side BC.

First, find the slope of the side BC.

Slope of BC $ (m_1) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2} $.

The altitude from A is perpendicular to BC, so its slope $ (m_2) $ is the negative reciprocal of the slope of BC.

$ m_2 = -\frac{1}{m_1} = -\frac{1}{5/2} = -\frac{2}{5} $.

Now, we have the slope of the altitude $ m_2 = -2/5 $ and a point it passes through, A(-3, 0).

Using the point-slope form of a line, $ y - y_1 = m(x - x_1) $:

$ y - 0 = -\frac{2}{5}(x - (-3)) $

$ y = -\frac{2}{5}(x + 3) $

$ 5y = -2(x + 3) \implies 5y = -2x - 6 $

$ 2x + 5y + 6 = 0 $

The equation of the altitude through A is $ 2x + 5y + 6 = 0 $.

Part IV: Answer all the questions (2 x 8 = 16)

43) Construct a triangle similar to a given triangle PQR with its sides equal to 7/4 of the corresponding sides of the triangle PQR (scale factor 7/4 > 1)

Steps of Construction:

Draw a triangle PQR with any suitable measurements.
Draw a ray PX from P on the side opposite to vertex R, making an acute angle with PQ.
Since the scale factor is 7/4, locate 7 (the greater of 7 and 4) points P₁, P₂, P₃, P₄, P₅, P₆, P₇ on the ray PX such that PP₁ = P₁P₂ = ... = P₆P₇.
Join the 4th point (P₄, corresponding to the denominator) to Q.
Extend the line segment PQ. Draw a line from P₇ parallel to P₄Q, which intersects the extended line PQ at Q'.
Extend the line segment PR. Draw a line from Q' parallel to QR, which intersects the extended line PR at R'.
The triangle PQ'R' is the required similar triangle whose sides are 7/4 of the corresponding sides of $\triangle PQR$.

Construction of similar triangle with scale factor 7/4

Justification: By construction, Q'R' is parallel to QR. Therefore, $\triangle PQ'R' \sim \triangle PQR$. Also, $\frac{PQ'}{PQ} = \frac{PP_7}{PP_4} = \frac{7}{4}$. Thus, the corresponding sides are in the ratio 7/4.

(OR) Construct a $\triangle PQR$ which the base PQ=4.5 cm $\angle R.= 35°$ and the median RG from R to PQ is 6 cm

Steps of Construction:

Draw a line segment PQ = 4.5 cm.
At point P, draw a line PE such that $\angle QPE = 35°$.
At point P, draw a line PF perpendicular to PE (i.e., $\angle EPF = 90°$).
Draw the perpendicular bisector of the line segment PQ. Let it intersect PQ at G and the line PF at O.
With O as the center and OP (or OQ) as the radius, draw a circle. All points on the major arc of this circle will subtend an angle of 35° at the segment PQ.
G is the midpoint of PQ. From G, with a radius of 6 cm (the length of the median), draw an arc.
This arc intersects the circle at two points. Label one of these intersection points as R.
Join PR and QR.
The triangle PQR is the required triangle.

Construction of triangle with given base, vertex angle, and median

44) Draw the graph of xy=24, x,y>0. Using the graph find (i) x when y = 6 (ii) y when x = 3

The equation is $xy=24$, which represents an indirect variation. We can write it as $y = \frac{24}{x}$. Since $x, y > 0$, the graph will be in the first quadrant.

Table of Values:

x	1	2	3	4	6	8	12
y	24	12	8	6	4	3	2

Plotting the Graph:

Plot the points (1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2) on a graph sheet. Join the points with a smooth curve. This curve is a rectangular hyperbola.

Scale: X-axis: 1 cm = 2 units, Y-axis: 1 cm = 2 units.

From the graph:

(i) To find x when y = 6: On the Y-axis, locate the point for y=6. Draw a horizontal line from this point to meet the curve. From the point of intersection on the curve, draw a vertical line down to the X-axis. The line meets the X-axis at x = 4.

(ii) To find y when x = 3: On the X-axis, locate the point for x=3. Draw a vertical line from this point to meet the curve. From the point of intersection on the curve, draw a horizontal line to the Y-axis. The line meets the Y-axis at y = 8.

(i) When y = 6, x = 4.
(ii) When x = 3, y = 8.

(OR) A bus is travelling at a uniform speed of 50 km/hr Draw the distance time graph and hence find
i) the constant of variation
ii) how far will it travel in 90 minutes?
iii) the time required to cover a distance of 300 km from the graph.

The relationship between distance (y) and time (x) at a uniform speed is a direct variation, given by the equation: Distance = Speed × Time.

So, $y = 50x$, where y is the distance in km and x is the time in hours.

Table of Values:

Time (x) in hours	1	2	3	4	5
Distance (y) in km	50	100	150	200	250

Plotting the Graph:

Plot the points (1, 50), (2, 100), (3, 150), etc., on a graph sheet. Join the points to get a straight line passing through the origin.

Scale: X-axis: 1 cm = 1 hour, Y-axis: 1 cm = 50 km.

Answering the questions:

i) The equation is $y=50x$. Comparing this with the direct variation equation $y=kx$, the constant of variation (k) is 50.

ii) To find the distance travelled in 90 minutes: 90 minutes = 1.5 hours. From the graph, draw a vertical line from x=1.5 on the X-axis to the graph line. Then draw a horizontal line to the Y-axis. It meets the Y-axis at y=75. So, the distance is 75 km.

iii) To find the time to cover 300 km: From y=300 on the Y-axis, draw a horizontal line to the graph line. Then draw a vertical line down to the X-axis. It meets the X-axis at x=6. So, the time required is 6 hours.

i) The constant of variation is 50 km/hr.
ii) The bus will travel 75 km in 90 minutes.
iii) The time required to cover 300 km is 6 hours.

youth to old age	up or down the hill	to hurry nor move away
high adventure	joyful	mourn	looking ahead

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10th Maths Quarterly Exam 2024 Question Paper with Solutions | Tenkasi District Samacheer Kalvi

10th Maths Quarterly Exam 2024 Question Paper & Solutions - Tenkasi District

Exam Details

Part I: Answer all the following questions (14 x 1 = 14)

Part II: Answer any 10 questions. 28th question is compulsory (10 x 2 = 20)

Part II Solutions (Questions 23-27)

Part III Solutions (Questions 29-31)

Part III: Answer any 10 questions. 42th question is compulsory (10 x 5 = 50)

Part III Solution (Question 33)

Part III Solutions (Questions 35-41)

Part IV: Answer all the questions (2 x 8 = 16)

chapter 1.2 - Life [Latest edition] Samacheer Kalvi solutions for English Class 10th SSLC Tamil Nadu State Board

Chapter 1: Life

Samacheer Kalvi solutions for English Class 10th SSLC Tamil Nadu State Board Chapter 1 Life Exercise [Pages 18 - 19]

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