COMMON QUARTERLY EXAMINATION-2024-25 - CLASS 10 MATHEMATICS - Solutions
Time Allowed: 3.00 Hours
Maximum Marks: 100
Here is the original question paper for the Common Quarterly Examination 2024-25 for Class 10 Mathematics.
SECTION - I (14 x 1 = 14)
Answer all of the following:
1. If the ordered pairs (a+2, 4) and (5, 2a+b) are equal then (a,b) is
a) (2,-2) b) (5,1) c) (2,3) d) (3,-2)
Solution:
Given that the ordered pairs are equal: $(a+2, 4) = (5, 2a+b)$.
This means the corresponding elements are equal.
$a+2 = 5 \implies a = 5 - 2 = 3$.
$4 = 2a+b$.
Substitute the value of $a=3$ into the second equation:
$4 = 2(3) + b$
$4 = 6 + b \implies b = 4 - 6 = -2$.
So, $(a,b) = (3, -2)$.
2. f(x) = (x+1)³ - (x-1)³ represents a function which is
a) Linear b) Cubic c) Reciprocal d) Quadratic
Solution:
We use the algebraic identities:
$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$f(x) = (x+1)^3 - (x-1)^3$
$f(x) = (x^3 + 3x^2(1) + 3x(1)^2 + 1^3) - (x^3 - 3x^2(1) + 3x(1)^2 - 1^3)$
$f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)$
$f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1$
$f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1+1)$
$f(x) = 6x^2 + 2$
Since the highest power of $x$ is 2, this is a quadratic function.
3. 7⁴ᵏ ≡ ______ (mod 100)
a) 1 b) 2 c) 3 d) 4
Solution:
Let's find the pattern of powers of 7 modulo 100.
$7^1 \equiv 7 \pmod{100}$
$7^2 = 49 \equiv 49 \pmod{100}$
$7^3 = 7^2 \times 7 = 49 \times 7 = 343 \equiv 43 \pmod{100}$
$7^4 = 7^3 \times 7 \equiv 43 \times 7 = 301 \equiv 1 \pmod{100}$
Now, we need to find $7^{4k} \pmod{100}$.
$7^{4k} = (7^4)^k$
Since $7^4 \equiv 1 \pmod{100}$,
$(7^4)^k \equiv 1^k \pmod{100}$
$7^{4k} \equiv 1 \pmod{100}$
4. The next term of the sequence 3/16, 1/8, 1/12, 1/18, ... is
a) 1/24 b) 1/27 c) 2/3 d) 1/81
Solution:
Let the terms be $t_1, t_2, t_3, t_4, ...$
$t_1 = 3/16, t_2 = 1/8, t_3 = 1/12, t_4 = 1/18$
Let's find the ratio between consecutive terms.
$r = t_2 / t_1 = (1/8) / (3/16) = (1/8) \times (16/3) = 16/24 = 2/3$.
$r = t_3 / t_2 = (1/12) / (1/8) = (1/12) \times 8 = 8/12 = 2/3$.
$r = t_4 / t_3 = (1/18) / (1/12) = (1/18) \times 12 = 12/18 = 2/3$.
Since the ratio is constant, the sequence is a Geometric Progression (G.P.) with common ratio $r = 2/3$.
The next term, $t_5$, is $t_4 \times r$.
$t_5 = (1/18) \times (2/3) = 2/54 = 1/27$.
5. Euclid division algorithm is a repeated application of division lemma until we get remainder as
a) 1 b) 3 c) 0 d) 2
Solution:
Euclid's division algorithm is a method to find the Highest Common Factor (HCF) of two integers. It repeatedly applies the division lemma, $a = bq + r$, where $0 \le r < b$. The process stops when the remainder, $r$, becomes 0. The HCF is the last non-zero remainder (or the divisor at that stage).
6. The solution of (2x - 1)² = 9 is equal to
a) -1 b) 2 c) -1, 2 d) None of these
Solution:
Given equation: $(2x - 1)^2 = 9$.
Take the square root of both sides:
$2x - 1 = \pm\sqrt{9}$
$2x - 1 = \pm 3$
We have two cases:
Case 1: $2x - 1 = 3 \implies 2x = 3 + 1 \implies 2x = 4 \implies x = 2$.
Case 2: $2x - 1 = -3 \implies 2x = -3 + 1 \implies 2x = -2 \implies x = -1$.
The solutions are $x = -1$ and $x = 2$.
7. $\frac{3y-3}{y} \div \frac{7y-7}{3y^2}$ is
a) $\frac{9y}{7}$ b) $\frac{9y^3}{21y-21}$ c) $\frac{21y^2-42y+21}{3y^2}$ d) $\frac{7(y^2-2y+1)}{y^2}$
Solution:
To divide fractions, we multiply by the reciprocal of the second fraction.
$\frac{3y-3}{y} \div \frac{7y-7}{3y^2} = \frac{3y-3}{y} \times \frac{3y^2}{7y-7}$
Factor out the common terms from the numerators:
$= \frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)}$
Cancel out the common factor $(y-1)$:
$= \frac{3}{y} \times \frac{3y^2}{7}$
$= \frac{9y^2}{7y}$
Cancel out the common factor $y$:
$= \frac{9y}{7}$
8. If in triangle ABC and DEF, $\frac{AB}{DE} = \frac{BC}{FD}$, then they will be similar, when
a) ∠B = ∠E b) ∠A = ∠D c) ∠B = ∠D d) ∠A = ∠F
Solution:
According to the SAS (Side-Angle-Side) similarity criterion, two triangles are similar if two pairs of corresponding sides are in proportion and the included angles are equal.
In $\triangle ABC$, the angle included between sides AB and BC is $\angle B$.
In $\triangle DEF$, the sides given in the ratio are DE and FD. The angle included between sides DE and FD is $\angle D$.
For the triangles to be similar by SAS, the included angles must be equal.
Therefore, $\angle B = \angle D$.
9. If in $\triangle ABC$, DE||BC, AB = 3.6 cm, AC = 2.4 cm and AD = 2.1cm then the length of AE is
a) 1.4 cm b) 1.8 cm c) 1.2 cm d) 1.05 cm
Solution:
Given that in $\triangle ABC$, DE is parallel to BC.
By Thales' Theorem (or Basic Proportionality Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
$\frac{AD}{AB} = \frac{AE}{AC}$
We are given:
AD = 2.1 cm
AB = 3.6 cm
AC = 2.4 cm
Substituting the values:
$\frac{2.1}{3.6} = \frac{AE}{2.4}$
$AE = \frac{2.1 \times 2.4}{3.6}$
$AE = \frac{5.04}{3.6} = 1.4$ cm.
10. The slope of the line joining (12,3), (4,a) is 1/8. The value of 'a' is
a) 1 b) 4 c) -5 d) 2
Solution:
The formula for the slope (m) of a line joining two 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 $(4, a)$. The slope is $m = 1/8$.
$\frac{1}{8} = \frac{a - 3}{4 - 12}$
$\frac{1}{8} = \frac{a - 3}{-8}$
Cross-multiply:
$1 \times (-8) = 8 \times (a - 3)$
$-8 = 8a - 24$
$8a = 24 - 8$
$8a = 16 \implies a = \frac{16}{8} = 2$.
11. (2, 1) is the point of intersection of two lines
a) x-y-3= 0; 3x-y-7=0
b) x+y=3; 3x+y=7
c) 3x+y=3; x+y=7
d) x+3y-3=0; x-y-7=0
Solution:
To find the correct pair of lines, we substitute the point (2, 1) into the equations of each option.
a) $x-y-3=0 \implies 2-1-3 = -2 \ne 0$. Incorrect.
b) $x+y=3 \implies 2+1=3$. Correct.
$3x+y=7 \implies 3(2)+1 = 6+1=7$. Correct.
Since (2, 1) satisfies both equations, it is the point of intersection.
c) $3x+y=3 \implies 3(2)+1 = 7 \ne 3$. Incorrect.
d) $x+3y-3=0 \implies 2+3(1)-3 = 2 \ne 0$. Incorrect.
12. $tan\theta \ cosec^2\theta - tan\theta$ equal to
a) $sec\theta$ b) $cot^2\theta$ c) $sin\theta$ d) $cot\theta$
Solution:
Factor out $tan\theta$ from the expression:
$tan\theta \ cosec^2\theta - tan\theta = tan\theta(cosec^2\theta - 1)$
Using the Pythagorean identity $1 + cot^2\theta = cosec^2\theta$, we get $cosec^2\theta - 1 = cot^2\theta$.
Substitute this back into the expression:
$= tan\theta \times cot^2\theta$
We know that $cot\theta = \frac{1}{tan\theta}$.
$= tan\theta \times (\frac{1}{tan\theta})^2 = tan\theta \times \frac{1}{tan^2\theta} = \frac{1}{tan\theta}$
$= cot\theta$.
13. If $sin\theta = cos\theta$ then $2tan^2\theta + sin^2\theta - 1$ is equal to
a) -3/2 b) 3/2 c) 2/3 d) -2/3
Solution:
Given $sin\theta = cos\theta$. Dividing both sides by $cos\theta$ (assuming $cos\theta \ne 0$):
$\frac{sin\theta}{cos\theta} = 1 \implies tan\theta = 1$.
This implies $\theta = 45^\circ$.
Now evaluate the expression $2tan^2\theta + sin^2\theta - 1$.
Substitute $\theta = 45^\circ$:
$= 2tan^2(45^\circ) + sin^2(45^\circ) - 1$
We know $tan(45^\circ) = 1$ and $sin(45^\circ) = \frac{1}{\sqrt{2}}$.
$= 2(1)^2 + (\frac{1}{\sqrt{2}})^2 - 1$
$= 2(1) + \frac{1}{2} - 1$
$= 2 + \frac{1}{2} - 1 = 1 + \frac{1}{2} = \frac{3}{2}$.
14. f = {(2,a), (3,b), (4,b), (5,c)} is a
a) Identity function b) One-one function c) Many-one function d) Constant function
Solution:
A function is defined as a set of ordered pairs $(x, y)$ where each $x$ (pre-image) has exactly one $y$ (image).
The given function is f = {(2,a), (3,b), (4,b), (5,c)}.
Domain = {2, 3, 4, 5}. Range = {a, b, c}.
- One-one function: Different elements in the domain must have different images. Here, elements 3 and 4 both have the same image 'b'. So, it is not a one-one function.
- Many-one function: At least two different elements in the domain have the same image. Since f(3)=b and f(4)=b, it is a many-one function.
- Identity function: $f(x)=x$ for all $x$. Here f(2)=a, not 2. So it is not an identity function.
- Constant function: All elements in the domain map to the same single element in the range. Here, the images are a, b, and c. So it is not a constant function.
SECTION - II (10 x 2 = 20)
Answer any 10 questions. Question No. 28 is compulsory.
15. Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A x B and B x A.
Solution:
Given A = {1, 2, 3}.
B = {x | x is a prime number less than 10}. Prime numbers less than 10 are 2, 3, 5, 7. So, B = {2, 3, 5, 7}.
A x B (Cartesian Product):
A x B = {(1,2), (1,3), (1,5), (1,7), (2,2), (2,3), (2,5), (2,7), (3,2), (3,3), (3,5), (3,7)}
B x A (Cartesian Product):
B x A = {(2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (5,1), (5,2), (5,3), (7,1), (7,2), (7,3)}
16. If A = {-2, -1, 0, 1, 2} and f: A → B is an onto function defined by $f(x) = x^2+x+1$ then find B.
Solution:
Given A = {-2, -1, 0, 1, 2} and $f(x) = x^2+x+1$.
Since f is an onto function from A to B, the codomain B must be equal to the range of f.
Let's find the image for each element in A:
$f(-2) = (-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3$
$f(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1$
$f(0) = (0)^2 + 0 + 1 = 1$
$f(1) = (1)^2 + 1 + 1 = 1 + 1 + 1 = 3$
$f(2) = (2)^2 + 2 + 1 = 4 + 2 + 1 = 7$
The range of f is the set of all unique images, which is {1, 3, 7}.
Since f is an onto function, B = Range of f.
17. Find the first four terms of $a_n = n^3 - 2$.
Solution:
Given the general term $a_n = n^3 - 2$.
For $n=1$, $a_1 = 1^3 - 2 = 1 - 2 = -1$.
For $n=2$, $a_2 = 2^3 - 2 = 8 - 2 = 6$.
For $n=3$, $a_3 = 3^3 - 2 = 27 - 2 = 25$.
For $n=4$, $a_4 = 4^3 - 2 = 64 - 2 = 62$.
18. Find the sum 3 + 1 + 1/3 + ...
Solution:
The given series is 3 + 1 + 1/3 + ...
First term, $a = 3$.
Common ratio, $r = \frac{1}{3} = \frac{1/3}{1} = 1/3$.
Since the common ratio $|r| = |1/3| < 1$, the series is a convergent geometric series, and we can find the sum to infinity.
The formula for the sum to infinity of a G.P. is $S_\infty = \frac{a}{1-r}$.
$S_\infty = \frac{3}{1 - 1/3} = \frac{3}{2/3} = 3 \times \frac{3}{2} = \frac{9}{2}$.
19. Find the 12th term from the last term of the A.P -2, -4, -6, ..., -100.
Solution:
Method 1: Find the total number of terms.
The A.P is -2, -4, -6, ... , -100.
First term $a = -2$, common difference $d = -4 - (-2) = -2$, last term $l = -100$.
Let $n$ be the number of terms. $l = a + (n-1)d$.
$-100 = -2 + (n-1)(-2) \implies -98 = (n-1)(-2) \implies 49 = n-1 \implies n = 50$.
The 12th term from the last is the $(n - 12 + 1)^{th}$ term from the beginning.
$(50 - 12 + 1) = 39^{th}$ term.
$a_{39} = a + (39-1)d = -2 + 38(-2) = -2 - 76 = -78$.
Method 2: Reverse the A.P.
The reversed A.P is -100, -98, ..., -4, -2.
Here, the new first term is $a' = -100$ and the common difference is $d' = -98 - (-100) = 2$.
We need to find the 12th term of this new A.P.
$a'_{12} = a' + (12-1)d' = -100 + 11(2) = -100 + 22 = -78$.
20. Find the LCM of $8x^4y^2, 48x^2y^4$.
Solution:
First, find the LCM of the numerical coefficients 8 and 48.
$8 = 2^3$
$48 = 16 \times 3 = 2^4 \times 3$
LCM(8, 48) = $2^4 \times 3 = 16 \times 3 = 48$.
Next, find the LCM of the variable parts by taking the highest power of each variable.
LCM of $x^4$ and $x^2$ is $x^4$.
LCM of $y^2$ and $y^4$ is $y^4$.
Combining these results, the LCM is $48x^4y^4$.
21. Determine the quadratic equation whose sum and product of roots are -9 and 20.
Solution:
Given: Sum of roots $(\alpha + \beta) = -9$.
Product of roots $(\alpha \beta) = 20$.
The general form of a quadratic equation is:
$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - (-9)x + 20 = 0$.
$x^2 + 9x + 20 = 0$.
22. Check whether AD is bisector of $\angle A$ of $\triangle ABC$ in the following. AB = 5cm, AC = 10cm, BD = 1.5cm and CD = 3.5cm.
Solution:
According to the Angle Bisector Theorem, if AD is the angle bisector of $\angle A$, then it divides the opposite side BC in the ratio of the other two sides.
That is, $\frac{AB}{AC} = \frac{BD}{CD}$.
Let's check if this condition is satisfied.
LHS = $\frac{AB}{AC} = \frac{5}{10} = \frac{1}{2}$.
RHS = $\frac{BD}{CD} = \frac{1.5}{3.5} = \frac{15}{35} = \frac{3}{7}$.
Since LHS $\ne$ RHS (i.e., $\frac{1}{2} \ne \frac{3}{7}$), the condition for the Angle Bisector Theorem is not satisfied.
23. What is the inclination of a line whose slope is 0?
Solution:
The relationship between the slope ($m$) of a line and its inclination ($\theta$) is given by $m = tan\theta$.
Given slope $m=0$.
$tan\theta = 0$.
For $0^\circ \le \theta < 180^\circ$, the value of $\theta$ for which $tan\theta = 0$ is $\theta = 0^\circ$.
An inclination of $0^\circ$ means the line is horizontal (parallel to the x-axis).
24. Show that the straight lines 2x+3y-8=0 and 4x+6y+18=0 are parallel.
Solution:
Two lines are parallel if their slopes are equal.
The slope of a line in the form $Ax+By+C=0$ is given by $m = -A/B$.
For the first line, $2x+3y-8=0$:
$A=2, B=3$. Slope $m_1 = -2/3$.
For the second line, $4x+6y+18=0$:
$A=4, B=6$. Slope $m_2 = -4/6 = -2/3$.
Since $m_1 = m_2$, the two lines are parallel.
25. Prove that $sec\theta - cos\theta = tan\theta \ sin\theta$.
Solution:
Starting with the Left Hand Side (LHS):
LHS = $sec\theta - cos\theta$
We know that $sec\theta = \frac{1}{cos\theta}$.
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 $sin^2\theta = 1 - cos^2\theta$.
LHS = $\frac{sin^2\theta}{cos\theta}$
We can write this as:
LHS = $\frac{sin\theta}{cos\theta} \times sin\theta$
Since $\frac{sin\theta}{cos\theta} = tan\theta$,
LHS = $tan\theta \ sin\theta$ = RHS.
Hence proved.
26. Prove that $\sqrt{\frac{1+sin\theta}{1-sin\theta}} = sec\theta + tan\theta$.
Solution:
Starting with the Left Hand Side (LHS):
LHS = $\sqrt{\frac{1+sin\theta}{1-sin\theta}}$
Multiply the numerator and denominator inside the square root by the conjugate of the denominator, which is $(1+sin\theta)$:
LHS = $\sqrt{\frac{1+sin\theta}{1-sin\theta} \times \frac{1+sin\theta}{1+sin\theta}}$
LHS = $\sqrt{\frac{(1+sin\theta)^2}{1^2 - sin^2\theta}}$
LHS = $\sqrt{\frac{(1+sin\theta)^2}{1 - sin^2\theta}}$
Using the identity $cos^2\theta = 1 - sin^2\theta$:
LHS = $\sqrt{\frac{(1+sin\theta)^2}{cos^2\theta}}$
Take the square root:
LHS = $\frac{1+sin\theta}{cos\theta}$
Split the fraction:
LHS = $\frac{1}{cos\theta} + \frac{sin\theta}{cos\theta}$
LHS = $sec\theta + tan\theta$ = RHS.
Hence proved.
27. Show that the points P(-1.5, 3), Q(6, -2), R(-3, 4) are collinear.
Solution:
Three points are collinear if the slope of the line segment joining any two pairs of points is the same.
Let's find the slope of PQ:
$m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 3}{6 - (-1.5)} = \frac{-5}{6 + 1.5} = \frac{-5}{7.5} = \frac{-50}{75} = -\frac{2}{3}$.
Now, let's find the slope of QR:
$m_{QR} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{4 - (-2)}{-3 - 6} = \frac{4 + 2}{-9} = \frac{6}{-9} = -\frac{2}{3}$.
Since the slope of PQ is equal to the slope of QR, and Q is a common point, the points P, Q, and R lie on the same straight line. Therefore, they are collinear.
28. If $P = \frac{x}{x+y}$, $Q = \frac{y}{x+y}$ then find $\frac{1}{P-Q} - \frac{2Q}{P^2-Q^2}$.
Solution:
The expression is $\frac{1}{P-Q} - \frac{2Q}{P^2-Q^2}$.
We know that $P^2-Q^2 = (P-Q)(P+Q)$.
So, the expression becomes $\frac{1}{P-Q} - \frac{2Q}{(P-Q)(P+Q)}$.
Take a common denominator, which is $(P-Q)(P+Q)$:
$= \frac{1(P+Q) - 2Q}{(P-Q)(P+Q)} = \frac{P+Q-2Q}{(P-Q)(P+Q)} = \frac{P-Q}{(P-Q)(P+Q)}$
Cancel the common term $(P-Q)$:
$= \frac{1}{P+Q}$
Now, let's find the value of $P+Q$ using the given values:
$P+Q = \frac{x}{x+y} + \frac{y}{x+y} = \frac{x+y}{x+y} = 1$.
Substitute this value back into our simplified expression:
$\frac{1}{P+Q} = \frac{1}{1} = 1$.
SECTION - III (10 x 5 = 50)
Answer any 10 questions. Question No. 42 is compulsory.
29. If the function f is defined by $f(x) = \begin{cases} x+2 & \text{if } x > 1 \\ 2 & \text{if } -1 \le x \le 1 \\ x-1 & \text{if } -3 < x < -1 \end{cases}$ Find the values of i) f(3), ii) f(0), iii) f(-1.5), iv) f(2) + f(-2).
Solution:
i) f(3):
Since $3 > 1$, we use the definition $f(x) = x+2$.
$f(3) = 3 + 2 = 5$.
ii) f(0):
Since $-1 \le 0 \le 1$, we use the definition $f(x) = 2$.
$f(0) = 2$.
iii) f(-1.5):
Since $-3 < -1.5 < -1$, we use the definition $f(x) = x-1$.
$f(-1.5) = -1.5 - 1 = -2.5$.
iv) f(2) + f(-2):
First, find f(2). Since $2 > 1$, $f(2) = 2+2 = 4$.
Next, find f(-2). Since $-3 < -2 < -1$, $f(-2) = -2 - 1 = -3$.
$f(2) + f(-2) = 4 + (-3) = 1$.
30. Let $f(x) = x^2-1$. Find (i) fof (ii) fofof
Solution:
(i) fof(x)
$fof(x) = f(f(x))$
Substitute $f(x)$ into itself:
$f(f(x)) = f(x^2-1)$
Now, replace $x$ with $(x^2-1)$ in the definition of $f(x)$:
$= (x^2-1)^2 - 1$
$= (x^4 - 2x^2 + 1) - 1$
$= x^4 - 2x^2$.
(ii) fofof(x)
$fofof(x) = f(fof(x))$
From part (i), we know $fof(x) = x^4 - 2x^2$.
$f(fof(x)) = f(x^4 - 2x^2)$
Replace $x$ with $(x^4 - 2x^2)$ in the definition of $f(x)$:
$= (x^4 - 2x^2)^2 - 1$
$= (x^4)^2 - 2(x^4)(2x^2) + (2x^2)^2 - 1$
$= x^8 - 4x^6 + 4x^4 - 1$.
(ii) $fofof(x) = x^8 - 4x^6 + 4x^4 - 1$
31. Find the HCF of 396, 504, 636.
Solution:
We will use Euclid's division algorithm to find the HCF.
Step 1: Find HCF of 504 and 396.
$504 = 1 \times 396 + 108$
$396 = 3 \times 108 + 72$
$108 = 1 \times 72 + 36$
$72 = 2 \times 36 + 0$
The HCF of 504 and 396 is 36.
Step 2: Find HCF of the result (36) and the third number (636).
$636 = 17 \times 36 + 24$
$36 = 1 \times 24 + 12$
$24 = 2 \times 12 + 0$
The HCF of 36 and 636 is 12.
Therefore, the HCF of 396, 504, and 636 is 12.
32. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24cm. How much area can be decorated with these colour papers?
Solution:
The sides of the square papers are 10 cm, 11 cm, ..., 24 cm.
The area of a square is side². So, the areas are $10^2, 11^2, ..., 24^2$.
Total area = $10^2 + 11^2 + 12^2 + ... + 24^2$.
This can be calculated as $(\sum_{n=1}^{24} n^2) - (\sum_{n=1}^{9} n^2)$.
The formula for the sum of the squares of the first n natural numbers is $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$.
For $n=24$:
$\sum_{n=1}^{24} n^2 = \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24(25)(49)}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900$.
For $n=9$:
$\sum_{n=1}^{9} n^2 = \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9(10)(19)}{6} = 3 \times 5 \times 19 = 285$.
Total area = $4900 - 285 = 4615$ cm².
33. Solve x + 2y - z = 5, x - y + z = -2, -5x - 4y + z = -11.
Solution:
Let the equations be:
(1) $x + 2y - z = 5$
(2) $x - y + z = -2$
(3) $-5x - 4y + z = -11$
Step 1: Eliminate z from (1) and (2).
Add equation (1) and (2):
$(x + 2y - z) + (x - y + z) = 5 + (-2)$
$2x + y = 3$ ---(4)
Step 2: Eliminate z from (2) and (3).
Subtract equation (3) from (2):
$(x - y + z) - (-5x - 4y + z) = -2 - (-11)$
$x - y + z + 5x + 4y - z = -2 + 11$
$6x + 3y = 9$
Divide by 3: $2x + y = 3$ ---(5)
Since equations (4) and (5) are identical, the system has infinitely many solutions. We can express the solution in terms of a parameter.
Let $y = k$. From equation (4), $2x + k = 3 \implies 2x = 3-k \implies x = \frac{3-k}{2}$.
Substitute $x$ and $y$ into equation (1):
$\frac{3-k}{2} + 2k - z = 5$
$z = \frac{3-k}{2} + 2k - 5 = \frac{3-k+4k-10}{2} = \frac{3k-7}{2}$.
35. If $4x^4 - 12x^3 + 37x^2 + bx + a$ is a perfect square, find the values of 'a' and 'b'.
Solution:
We use the long division method to find the square root.
2x² - 3x + 7
_____________________
2x² | 4x⁴ - 12x³ + 37x² + bx + a
| 4x⁴
|--------------------
4x²-3x | -12x³ + 37x²
| -12x³ + 9x²
|--------------------
4x²-6x+7 | 28x² + bx + a
| 28x² - 42x + 49
|--------------------
| (b+42)x + (a-49)
For the given polynomial to be a perfect square, the remainder must be 0.
Remainder = $(b+42)x + (a-49) = 0$.
This implies that the coefficients must be zero.
$b + 42 = 0 \implies b = -42$.
$a - 49 = 0 \implies a = 49$.
36. State and prove Thales Theorem.
Solution:
Statement (Basic Proportionality Theorem):
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In $\triangle ABC$, a line DE is drawn parallel to BC, intersecting AB at D and AC at E.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
Proof:
In ∆ABC, D is a point on AB and E is a point on AC.
To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)
Construction: Draw a line DE || BC
| No. | Statement | Reason |
|---|---|---|
| 1. | ∠ABC = ∠ADE = ∠1 | Corresponding angles are equal because DE || BC |
| 2. | ∠ACB = ∠AED = ∠2 | Corresponding angles are equal because DE || BC |
| 3. | ∠DAE = ∠BAC = ∠3 | Both triangles have a common angle |
| ∆ABC ~ ∆ADE | By AAA similarity | |
| \(\frac{AB}{AD} = \frac{AC}{AE}\) | Corresponding sides are proportional | |
| \(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\) | Split AB and AC using the points D and E. | |
| \(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\) | On simplification | |
| \(\frac{DB}{AD} = \frac{EC}{AE}\) | Cancelling 1 on both sides | |
| \(\frac{AD}{DB} = \frac{AE}{EC}\) | Taking reciprocals. Hence proved. |
41. Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A∩C) x (B∩D) = (AxB) ∩ (CxD) is true?
Solution:
LHS: (A∩C) x (B∩D)
First, find the intersections:
A ∩ C = {1, 2, 3} ∩ {3, 4} = {3}
B ∩ D = {2, 3, 5} ∩ {1, 3, 5} = {3, 5}
Now, find the Cartesian product:
(A ∩ C) x (B ∩ D) = {3} x {3, 5} = {(3, 3), (3, 5)}.
RHS: (A x B) ∩ (C x D)
First, find the Cartesian products:
A x B = {(1,2), (1,3), (1,5), (2,2), (2,3), (2,5), (3,2), (3,3), (3,5)}
C x D = {(3,1), (3,3), (3,5), (4,1), (4,3), (4,5)}
Now, find the intersection of these two sets:
(A x B) ∩ (C x D) = {(3, 3), (3, 5)}.
Since LHS = RHS = {(3, 3), (3, 5)}, the given statement is true.
42. Find the sum of all natural numbers between 100 and 10,000 which are divisible by 11.
Solution:
We need to find the sum of an Arithmetic Progression (A.P.).
The first number greater than 100 divisible by 11 is $11 \times 10 = 110$. So, $a = 110$.
The last number less than 10,000 divisible by 11 is found by dividing 10000 by 11.
$10000 \div 11 = 909$ with a remainder of 1. So, $10000 - 1 = 9999$.
$9999 = 11 \times 909$. So, the last term $l = 9999$.
The common difference $d = 11$.
First, find the number of terms, $n$.
$l = a + (n-1)d$
$9999 = 110 + (n-1)11$
$9999 - 110 = (n-1)11$
$9889 = (n-1)11$
$n-1 = \frac{9889}{11} = 899$
$n = 899 + 1 = 900$.
Now, find the sum using the formula $S_n = \frac{n}{2}(a+l)$.
$S_{900} = \frac{900}{2}(110 + 9999)$
$S_{900} = 450(10109)$
$S_{900} = 4549050$.
SECTION - IV (2 x 8 = 16)
Answer the following.
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to 3/5 of the corresponding sides of the ΔPQR (scale factor 3/5 < 1)
Solution:
Steps of Construction:
- Construct the given triangle PQR with any arbitrary measurements.
- Draw a ray QX starting from Q, making an acute angle with QR on the side opposite to vertex P.
- On the ray QX, mark 5 (the larger number in the scale factor 3/5) equally spaced points, say $Q_1, Q_2, Q_3, Q_4, Q_5$.
- Join the last point $Q_5$ to R.
- From the point $Q_3$ (the smaller number in the scale factor), draw a line parallel to $Q_5R$. This line will intersect QR at a point, let's call it R'. (To draw a parallel line, make an arc at $Q_5$ and a same-sized arc at $Q_3$, then measure the angle at $Q_5$ and cut the arc at $Q_3$).
- From R', draw a line parallel to PR. This line will intersect PQ at a point, let's call it P'.
- The triangle $\triangle P'QR'$ is the required triangle, which is similar to $\triangle PQR$ and its sides are 3/5 of the corresponding sides of $\triangle PQR$.
(OR)
b) Construct a ΔABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4.5 cm.
Solution:
Steps of Construction:
- Draw a line segment AB of length 5.5 cm.
- At point A, draw a line AX such that $\angle BAX = 25^\circ$.
- Draw a line AY perpendicular to AX.
- Draw the perpendicular bisector of the line segment AB. Let it intersect AB at M and the line AY at O.
- With O as the center and OA as the radius, draw a circle. This circle will pass through points A and B. Any angle subtended by the arc AB in the major segment will be $25^\circ$.
- Now, draw a line GH parallel to AB at a distance of 4.5 cm from AB. This line represents all possible positions for vertex C, as the altitude from C is 4.5 cm.
- The line GH will intersect the circle at two points. Label these points as C and C'.
- Join AC and BC (or AC' and BC').
- $\triangle ABC$ (or $\triangle ABC'$) is the required triangle.
44. a) Graph the following linear function y=1/2x. Identify the constant of variation and verify it with the graph. Also (i) Find y when x=9, (ii) Find x when y = 7.5.
Solution:
The given linear function is $y = \frac{1}{2}x$. This is in the form $y=kx$, which represents a direct variation.
Constant of Variation:
Comparing $y = \frac{1}{2}x$ with $y=kx$, the constant of variation is $k = \frac{1}{2}$.
Table of Values:
| x | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| y = (1/2)x | 0 | 1 | 2 | 3 | 4 | 5 |
Graphing:
Plot the points (0, 0), (2, 1), (4, 2), (6, 3), (8, 4), (10, 5) on a graph sheet and draw a straight line passing through them. The line will pass through the origin.
Verification from Graph:
From the graph, when x = 10; y = 5
Now, k = y/x = 5/10 = 1/2
Hence verified.
(i) From the graph, when x = 9, y = 4.5
(i) From the graph, when y = 7.5, x = 15
(OR)
b) A bus is travelling at a uniform speed of 50 km/hr. Draw the distance-time graph and hence find i) the constant of variations ii) how far will it travel in 90 minutes? iii) the time required to cover a distance of 300 km from the graph.
Solution:
The relationship between distance (d), speed (s), and time (t) is $d = s \times t$.
Let x be the time taken in minutes and y be the distance travelled in km.
| Time taken x (in minutes) | 60 | 120 | 180 | 240 |
|---|---|---|---|---|
| Distance y (in km) | 50 | 100 | 150 | 200 |
(i) Observe that as time increases, the distance travelled also increases. Therefore, the variation is a direct variation. It is of the form y = kx.
Constant of variation:
\(k = \frac{y}{x} = \frac{50}{60} = \frac{100}{120} = \frac{150}{180} = \frac{200}{240} = \frac{5}{6}\)
Hence, the relation may be given as \(y = \frac{5}{6}x\).
(ii) From the graph, if x = 90 minutes (\(1\frac{1}{2}\) hours), then \(y = \frac{5}{6} \times 90 = 75\) km. The distance travelled is 75 km.
(iii) From the graph, if y = 300 km, then \(300 = \frac{5}{6}x \implies x = \frac{300 \times 6}{5} = 360\) minutes (or) 6 hours. The time taken to cover 300 km is 6 hours.