10th Maths - Quarterly Exam 2024 - English Medium Original Question Paper | Vellore District
Part - I: Multiple Choice Questions (14 x 1 = 14)
Choose the correct answer:
1. A = {a,b,p}, B = {2,3}, C = {p,q,r,s} then n[(A∪C) x B] is
Answer: c) 12
Solution:
Given, $A = \{a,b,p\}$, $B = \{2,3\}$, $C = \{p,q,r,s\}$
First, find $A \cup C$:
$A \cup C = \{a,b,p\} \cup \{p,q,r,s\} = \{a,b,p,q,r,s\}$
Next, find the number of elements in $A \cup C$ and $B$:
$n(A \cup C) = 6$
$n(B) = 2$
Now, calculate $n[(A \cup C) \times B]$:
$n[(A \cup C) \times B] = n(A \cup C) \times n(B)$
$= 6 \times 2 = 12$
2. If $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$, then fog is
Answer: c) $\frac{2}{9x^2}$
Solution:
$fog(x) = f(g(x))$
Substitute $g(x) = \frac{1}{3x}$ into $f(x)$:
$f(g(x)) = f(\frac{1}{3x})$
Now apply the function $f(x) = 2x^2$ to the input $\frac{1}{3x}$:
$= 2 \left(\frac{1}{3x}\right)^2 = 2 \left(\frac{1}{9x^2}\right) = \frac{2}{9x^2}$
3. A function f: R→R defined by $f(x) = ax^2 + bx + c, (a \neq 0)$ is called a
Answer: d) quadratic function
Solution:
The standard form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a, b,$ and $c$ are constants and $a \neq 0$. This is the definition of a quadratic function.
4. $7^{4K} \equiv \underline{\hspace{1cm}} (\text{mod } 100)$
Answer: a) 1
Solution:
Let's find the pattern for powers of 7 modulo 100:
- $7^1 \equiv 7 \pmod{100}$
- $7^2 = 49 \equiv 49 \pmod{100}$
- $7^3 = 49 \times 7 = 343 \equiv 43 \pmod{100}$
- $7^4 = 43 \times 7 = 301 \equiv 1 \pmod{100}$
So, $7^4 \equiv 1 \pmod{100}$.
Therefore, $7^{4K} = (7^4)^K \equiv 1^K \equiv 1 \pmod{100}$.
5. The sum of first n natural numbers are also called
Answer: c) Triangular numbers
Solution:
The sum of the first $n$ natural numbers, given by the formula $S_n = \frac{n(n+1)}{2}$, represents the number of dots in an equilateral triangle. These numbers (1, 3, 6, 10, ...) are known as triangular numbers.
6. The value of $(1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15)$ is
Answer: c) 14280
Solution:
We use the formulas for the sum of the first $n$ natural numbers and the sum of the cubes of the first $n$ natural numbers.
Sum of first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
Sum of cubes of first $n$ natural numbers: $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$
Here, $n=15$.
Sum of cubes: $(1^3 + \dots + 15^3) = \left(\frac{15(15+1)}{2}\right)^2 = \left(\frac{15 \times 16}{2}\right)^2 = (15 \times 8)^2 = 120^2 = 14400$.
Sum of numbers: $(1 + \dots + 15) = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120$.
The required value is $14400 - 120 = 14280$.
7. $\frac{3y-3}{y} \div \frac{7y-7}{3y^2}$ is
Answer: a) $\frac{9y}{7}$
Solution:
Division of rational expressions is multiplication by the reciprocal.
$\frac{3y-3}{y} \div \frac{7y-7}{3y^2} = \frac{3y-3}{y} \times \frac{3y^2}{7y-7}$
Factor out the common terms:
$= \frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)}$
Cancel the common factor $(y-1)$ and simplify:
$= \frac{3}{y} \times \frac{3y^2}{7} = \frac{9y^2}{7y} = \frac{9y}{7}$
8. Graph of a linear equation is a ______.
Answer: a) straight line
Solution:
A linear equation in two variables (e.g., $ax + by = c$) always represents a straight line when plotted on a Cartesian coordinate system.
9. The square root of $\frac{256x^8y^4z^{10}}{25x^6y^6z^6}$ is equal to
Answer: d) $\frac{16}{5}\frac{xz^2}{y}$
Solution:
First, simplify the expression inside the square root:
$\frac{256x^8y^4z^{10}}{25x^6y^6z^6} = \frac{256}{25} x^{8-6} y^{4-6} z^{10-6} = \frac{256}{25} x^2 y^{-2} z^4 = \frac{256x^2z^4}{25y^2}$
Now, take the square root:
$\sqrt{\frac{256x^2z^4}{25y^2}} = \frac{\sqrt{256}\sqrt{x^2}\sqrt{z^4}}{\sqrt{25}\sqrt{y^2}}$
Assuming the variables represent positive quantities:
$= \frac{16xz^2}{5y}$
10. If $\triangle ABC$ is an isosceles triangle with $\angle C = 90^\circ$ and $AC = 5$cm, then AB is
Answer: d) $5\sqrt{2}$ cm
Solution:
Given an isosceles triangle $\triangle ABC$ with $\angle C = 90^\circ$. In an isosceles right-angled triangle, the sides adjacent to the right angle are equal.
So, $AC = BC = 5$ cm.
By the Pythagorean theorem, $AB^2 = AC^2 + BC^2$.
$AB^2 = 5^2 + 5^2 = 25 + 25 = 50$
$AB = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ cm.
11. In a $\triangle ABC$, AD is the bisector of $\angle BAC$. If AB = 8 cm, BD = 6 cm and DC = 3 cm. The length of the side AC is
Answer: b) 4 cm
Solution:
According to the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides.
So, $\frac{AB}{AC} = \frac{BD}{DC}$
Substitute the given values:
$\frac{8}{AC} = \frac{6}{3}$
$\frac{8}{AC} = 2$
$AC = \frac{8}{2} = 4$ cm.
12. The area of triangle formed by the points (-5,0), (0,-5) and (5,0) is
Answer: b) 25 sq.units
Solution:
Let the vertices be $A(-5,0)$, $B(0,-5)$, and $C(5,0)$.
The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area = $\frac{1}{2} |-5(-5 - 0) + 0(0 - 0) + 5(0 - (-5))|$
Area = $\frac{1}{2} |-5(-5) + 0 + 5(5)|$
Area = $\frac{1}{2} |25 + 0 + 25| = \frac{1}{2} |50| = 25$ sq.units.
13. The slope of the line joining (12,3), (4,a) is $\frac{1}{8}$. The value of 'a' is
Answer: d) 2
Solution:
The formula for the slope (m) of a line joining points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Given points are (12,3) and (4,a), and slope $m = \frac{1}{8}$.
$\frac{1}{8} = \frac{a - 3}{4 - 12}$
$\frac{1}{8} = \frac{a - 3}{-8}$
Multiply both sides by -8:
$-1 = a - 3$
$a = 3 - 1 = 2$
14. $\tan\theta \csc^2\theta - \tan\theta$ is equal to
Answer: d) $\cot\theta$
Solution:
Start by factoring out $\tan\theta$ from the expression:
$\tan\theta \csc^2\theta - \tan\theta = \tan\theta(\csc^2\theta - 1)$
Using the Pythagorean identity $1 + \cot^2\theta = \csc^2\theta$, we get $\csc^2\theta - 1 = \cot^2\theta$.
Substitute this into the expression:
$= \tan\theta(\cot^2\theta)$
Since $\cot\theta = \frac{1}{\tan\theta}$, we have:
$= \tan\theta \left(\frac{1}{\tan^2\theta}\right) = \frac{1}{\tan\theta} = \cot\theta$
Part - II: Short Answer Questions (10 x 2 = 20)
Answer any 10 questions. (Q.No.28 is compulsory)
15. A relation R is given by the set $\{(x,y) / y = x + 3, x \in \{0,1,2,3,4,5\}\}$. Determine its domain and range.
Solution:
The domain is the set of all possible input values for $x$.
Given $x \in \{0,1,2,3,4,5\}$.
Domain = $\{0,1,2,3,4,5\}$
The range is the set of all possible output values for $y$. We find $y$ for each $x$ using $y = x+3$.
- If $x=0$, $y=0+3=3$
- If $x=1$, $y=1+3=4$
- If $x=2$, $y=2+3=5$
- If $x=3$, $y=3+3=6$
- If $x=4$, $y=4+3=7$
- If $x=5$, $y=5+3=8$
Range = $\{3,4,5,6,7,8\}$
16. Given the function $f: x \to x^2 - 5x + 6$, evaluate i) $f(-1)$ ii) $f(2a)$
Solution:
The function is $f(x) = x^2 - 5x + 6$.
i) $f(-1)$
Substitute $x = -1$ into the function:
$f(-1) = (-1)^2 - 5(-1) + 6$
$= 1 + 5 + 6 = 12$
$f(-1) = 12$
ii) $f(2a)$
Substitute $x = 2a$ into the function:
$f(2a) = (2a)^2 - 5(2a) + 6$
$= 4a^2 - 10a + 6$
$f(2a) = 4a^2 - 10a + 6$
17. Find k if $fof(k) = 5$ where $f(k) = 2k - 1$
Solution:
Given $f(k) = 2k-1$.
First, find the composition $fof(k)$:
$fof(k) = f(f(k)) = f(2k-1)$
Now, substitute $(2k-1)$ into the function $f$:
$f(2k-1) = 2(2k-1) - 1 = 4k - 2 - 1 = 4k - 3$
We are given $fof(k) = 5$.
So, $4k - 3 = 5$
$4k = 8$
$k = \frac{8}{4} = 2$
$k = 2$
18. Find the HCF of 252525 and 363636
Solution:
We can use Euclid's Division Algorithm to find the HCF.
Step 1: Divide 363636 by 252525.
$363636 = 1 \times 252525 + 111111$
Step 2: Divide 252525 by 111111.
$252525 = 2 \times 111111 + 30303$
Step 3: Divide 111111 by 30303.
$111111 = 3 \times 30303 + 20202$
Step 4: Divide 30303 by 20202.
$30303 = 1 \times 20202 + 10101$
Step 5: Divide 20202 by 10101.
$20202 = 2 \times 10101 + 0$
The remainder is now 0. The last non-zero remainder is the HCF.
HCF = 10101
Alternatively, by observation:
$252525 = 25 \times 10101$
$363636 = 36 \times 10101$
The Highest Common Factor is 10101.
19. What is the time 15 hours before 11 p.m?
Solution:
11 p.m. in 24-hour format is 23:00.
We need to find the time 15 hours before 23:00.
23:00 - 15:00 = 8:00
The time is 8:00 in the morning.
The time is 8 a.m.
20. Find the sum $3 + 1 + \frac{1}{3} + \dots \infty$
Solution:
This is a 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 of a G.P. 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.
21. Subtract $\frac{1}{x^2+2}$ from $\frac{2x^3+x^2+3}{(x^2+2)^2}$
Solution:
We need to calculate: $\frac{2x^3+x^2+3}{(x^2+2)^2} - \frac{1}{x^2+2}$
To subtract, we need a common denominator, which is $(x^2+2)^2$.
$\frac{1}{x^2+2} = \frac{1 \times (x^2+2)}{(x^2+2) \times (x^2+2)} = \frac{x^2+2}{(x^2+2)^2}$
Now subtract the fractions:
$\frac{2x^3+x^2+3}{(x^2+2)^2} - \frac{x^2+2}{(x^2+2)^2} = \frac{(2x^3+x^2+3) - (x^2+2)}{(x^2+2)^2}$
$= \frac{2x^3+x^2+3 - x^2 - 2}{(x^2+2)^2}$
$= \frac{2x^3+1}{(x^2+2)^2}$
Result: $\frac{2x^3+1}{(x^2+2)^2}$
22. Solve $x^2 + 2x - 2 = 0$ by formula method.
Solution:
The given quadratic equation is $x^2 + 2x - 2 = 0$.
Comparing with $ax^2+bx+c=0$, we have $a=1, b=2, c=-2$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2(1)}$
$x = \frac{-2 \pm \sqrt{4 + 8}}{2}$
$x = \frac{-2 \pm \sqrt{12}}{2}$
$x = \frac{-2 \pm \sqrt{4 \times 3}}{2} = \frac{-2 \pm 2\sqrt{3}}{2}$
Factor out 2 from the numerator:
$x = \frac{2(-1 \pm \sqrt{3})}{2} = -1 \pm \sqrt{3}$
The solutions are $x = -1 + \sqrt{3}$ and $x = -1 - \sqrt{3}$.
23. If $\triangle ABC$ is similar to $\triangle DEF$ such that $BC = 3$ cm, $EF = 4$ cm and area of $\triangle ABC = 54$ cm². Find the area of $\triangle DEF$.
Solution:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2$
Substitute the given values:
$\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$
Rearrange to find Area($\triangle DEF$):
Area($\triangle DEF$) = $\frac{54 \times 16}{9}$
Area($\triangle DEF$) = $6 \times 16 = 96$ cm².
The area of $\triangle DEF$ is 96 cm².
24. In $\triangle ABC$, D and E are points on the sides AB and AC respectively such that $DE||BC$. If $\frac{AD}{DB} = \frac{3}{4}$ and $AC = 15$ cm, find AE.
Solution:
Since $DE || BC$, by Thales' theorem (Basic Proportionality Theorem), the line DE divides the sides AB and AC in the same ratio.
$\frac{AD}{AB} = \frac{AE}{AC}$
We are given $\frac{AD}{DB} = \frac{3}{4}$. Let $AD=3k$ and $DB=4k$.
Then $AB = AD + DB = 3k + 4k = 7k$.
The ratio $\frac{AD}{AB} = \frac{3k}{7k} = \frac{3}{7}$.
Now substitute this into the theorem's equation:
$\frac{3}{7} = \frac{AE}{15}$
$AE = \frac{3 \times 15}{7} = \frac{45}{7}$ cm.
$AE = \frac{45}{7}$ cm.
25. Show that the points (-2,5), (6,-1) and (2,2) are collinear.
Solution:
Points are collinear if the slope between any two pairs of points is the same.
Let $A=(-2,5)$, $B=(6,-1)$, and $C=(2,2)$.
Slope of AB:
$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{6 - (-2)} = \frac{-6}{8} = -\frac{3}{4}$
Slope of BC:
$m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{2 - (-1)}{2 - 6} = \frac{3}{-4} = -\frac{3}{4}$
Since the slope of AB is equal to the slope of BC, the points A, B, and C lie on the same straight line.
Hence, the points are collinear.
26. Find the slope and y-intercept of $\sqrt{3}x + (1-\sqrt{3})y = 3$
Solution:
The equation of a line in slope-intercept form is $y = mx+c$, where $m$ is the slope and $c$ is the y-intercept.
Rearrange the given equation:
$(1-\sqrt{3})y = -\sqrt{3}x + 3$
$y = \frac{-\sqrt{3}}{1-\sqrt{3}}x + \frac{3}{1-\sqrt{3}}$
Slope (m):
$m = \frac{-\sqrt{3}}{1-\sqrt{3}}$. Rationalize the denominator:
$m = \frac{-\sqrt{3}(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})} = \frac{-\sqrt{3}-3}{1-3} = \frac{-(\sqrt{3}+3)}{-2} = \frac{\sqrt{3}+3}{2}$
y-intercept (c):
$c = \frac{3}{1-\sqrt{3}}$. Rationalize the denominator:
$c = \frac{3(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})} = \frac{3(1+\sqrt{3})}{1-3} = \frac{3(1+\sqrt{3})}{-2}$
Slope = $\frac{3+\sqrt{3}}{2}$
y-intercept = $-\frac{3(1+\sqrt{3})}{2}$
27. Prove that $\sec\theta - \cos\theta = \tan\theta \sin\theta$
Solution:
Left Hand Side (LHS): $\sec\theta - \cos\theta$
Rewrite $\sec\theta$ as $\frac{1}{\cos\theta}$:
LHS = $\frac{1}{\cos\theta} - \cos\theta$
Take a common denominator:
LHS = $\frac{1 - \cos^2\theta}{\cos\theta}$
Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, we have $1 - \cos^2\theta = \sin^2\theta$.
LHS = $\frac{\sin^2\theta}{\cos\theta}$
Split the term:
LHS = $\frac{\sin\theta}{\cos\theta} \cdot \sin\theta$
Since $\frac{\sin\theta}{\cos\theta} = \tan\theta$:
LHS = $\tan\theta \sin\theta$
LHS = Right Hand Side (RHS). Hence proved.
28. (Compulsory) Find the excluded values of the following expression: $\frac{7P+2}{8P^2+13P+5}$
Solution:
The excluded values are the values of P for which the denominator is equal to zero.
Set the denominator to zero:
$8P^2 + 13P + 5 = 0$
Factor the quadratic expression. We need two numbers that multiply to $8 \times 5 = 40$ and add to 13. The numbers are 8 and 5.
$8P^2 + 8P + 5P + 5 = 0$
$8P(P+1) + 5(P+1) = 0$
$(8P+5)(P+1) = 0$
This gives two possible solutions:
$8P+5 = 0 \implies 8P = -5 \implies P = -\frac{5}{8}$
$P+1 = 0 \implies P = -1$
The excluded values are -1 and $-\frac{5}{8}$.