10th Maths Quarterly Exam 2024 - Solutions
Tirunelveli District - Common Quarterly Examination - September 2024
PART - I (14 x 1 = 14)
1) If n(A×B) = 6 and A = {1, 3} then n(B) is
Solution:
Given, n(A×B) = 6 and A = {1, 3}.
We know, n(A) = 2.
Also, n(A×B) = n(A) × n(B).
So, 6 = 2 × n(B).
n(B) = 6 / 2 = 3.
Answer: c) 3
2) Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f: A→B given by f = {(1, 4), (2, 8), (3, 9), (4, 10)} is a
Solution:
Each element in the domain A has a distinct image in the co-domain B.
(1→4, 2→8, 3→9, 4→10). No two elements in A have the same image.
Therefore, it is a one-to-one function.
Answer: c) One to one function
3) Given F₁ = 1, F₂ = 3 and Fₙ = Fₙ₋₁ + Fₙ₋₂ then F₅ is
Solution:
Given F₁ = 1, F₂ = 3.
F₃ = F₂ + F₁ = 3 + 1 = 4.
F₄ = F₃ + F₂ = 4 + 3 = 7.
F₅ = F₄ + F₃ = 7 + 4 = 11.
Answer: d) 11
4) The next term of the sequence \( \frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, ... \) is
Solution:
This is a Geometric Progression (GP). Let's find the common ratio (r).
\( r = \frac{t_2}{t_1} = \frac{1/8}{3/16} = \frac{1}{8} \times \frac{16}{3} = \frac{2}{3} \).
The next term is \( t_5 = t_4 \times r = \frac{1}{18} \times \frac{2}{3} = \frac{2}{54} = \frac{1}{27} \).
Answer: b) 1/27
5) The square root of \( \frac{256x^8y^4z^{10}}{25x^6y^6z^6} \) is equal to
Solution:
\( \sqrt{\frac{256x^8y^4z^{10}}{25x^6y^6z^6}} = \sqrt{\frac{256}{25} x^{8-6} y^{4-6} z^{10-6}} \)
\( = \sqrt{\frac{256}{25} x^2 y^{-2} z^4} = \frac{16}{5} |x| |y^{-1}| z^2 = \frac{16|x|z^2}{5|y|} \).
Assuming variables are positive, the answer is \( \frac{16xz^2}{5y} \).
Answer: d) \( \frac{16}{5} \frac{xz^2}{y} \)
6) \( \sqrt{a^2x^2 + 2abx + b^2} \) square root
Solution:
Assuming the expression inside the square root is a perfect square. The expression seems to be a typo and should be \( (ax)^2 + 2(ax)(b) + b^2 \).
This is in the form of \( (p+q)^2 = p^2 + 2pq + q^2 \), where p = ax and q = b.
So, \( \sqrt{(ax+b)^2} = |ax+b| \).
Answer: b) |ax+b|
7) If ΔABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is
Solution:
Since ΔABC is an isosceles triangle with ∠C = 90°, the sides adjacent to the right angle are equal. So, AC = BC = 5 cm.
By Pythagoras 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.
Answer: d) 5√2 cm
8) The slope of the line which is perpendicular to a line joining the pts (0, 0) and (-8, 8) is
Solution:
Slope of the line joining (0,0) and (-8,8) is \( m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1 \).
The slope of the perpendicular line is \( m_2 = \frac{-1}{m_1} = \frac{-1}{-1} = 1 \).
Answer: b) 1
9) If \( (\sin \alpha + \text{cosec } \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = K + \tan^2 \alpha + \cot^2 \alpha \), then the value of K is equal to
Solution:
LHS = \( (\sin^2 \alpha + \text{cosec}^2 \alpha + 2\sin \alpha \text{cosec } \alpha) + (\cos^2 \alpha + \sec^2 \alpha + 2\cos \alpha \sec \alpha) \)
= \( (\sin^2 \alpha + \cos^2 \alpha) + \text{cosec}^2 \alpha + \sec^2 \alpha + 2(1) + 2(1) \)
= \( 1 + (1 + \cot^2 \alpha) + (1 + \tan^2 \alpha) + 4 \)
= \( 7 + \tan^2 \alpha + \cot^2 \alpha \)
Comparing with \( K + \tan^2 \alpha + \cot^2 \alpha \), we get K = 7.
= \( (\sin^2 \alpha + \cos^2 \alpha) + \text{cosec}^2 \alpha + \sec^2 \alpha + 2(1) + 2(1) \)
= \( 1 + (1 + \cot^2 \alpha) + (1 + \tan^2 \alpha) + 4 \)
= \( 7 + \tan^2 \alpha + \cot^2 \alpha \)
Comparing with \( K + \tan^2 \alpha + \cot^2 \alpha \), we get K = 7.
Answer: b) 7
10) If x = a tan θ and y = b sec θ then
Solution:
Given x = a tan θ ⇒ \( \tan \theta = \frac{x}{a} \).
Given y = b sec θ ⇒ \( \sec \theta = \frac{y}{b} \).
We know the trigonometric identity: \( \sec^2 \theta - \tan^2 \theta = 1 \).
Substituting the values: \( (\frac{y}{b})^2 - (\frac{x}{a})^2 = 1 \).
\( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).
Answer: a) \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
11) In ΔLMN, ∠L = 60°, ∠M = 50°. If ΔLMN ~ ΔPQR then the value of ∠R is
Solution:
In ΔLMN, sum of angles is 180°.
∠L + ∠M + ∠N = 180°.
60° + 50° + ∠N = 180°.
110° + ∠N = 180°.
∠N = 70°.
Since ΔLMN ~ ΔPQR, corresponding angles are equal.
∠R corresponds to ∠N. So, ∠R = ∠N = 70°.
Answer: b) 70°
12) If in triangles ABC and EDF, \( \frac{AB}{DE} = \frac{BC}{FD} \) then they will be similar when
Solution:
By SAS similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the triangles are similar.
The sides are AB, BC and DE, FD.
The included angle for sides AB and BC is ∠B.
The included angle for sides DE and FD is ∠D.
Therefore, for the triangles to be similar, ∠B must be equal to ∠D.
Answer: c) ∠B = ∠D
13) If \( 3\sqrt{x} = 9 \) find the value of 'x'
Solution:
\( 3\sqrt{x} = 9 \)
\( \sqrt{x} = \frac{9}{3} = 3 \)
Squaring both sides, \( (\sqrt{x})^2 = 3^2 \).
x = 9.
Answer: b) 9
14) If \( \sqrt{3} \sin \theta - \cos \theta = 0 \) then find the value of 'θ'
Solution:
\( \sqrt{3} \sin \theta - \cos \theta = 0 \)
\( \sqrt{3} \sin \theta = \cos \theta \)
\( \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \)
\( \tan \theta = \frac{1}{\sqrt{3}} \)
θ = 30°.
Answer: b) 30°
PART - II (10 x 2 = 20)
Answer any 10 questions. Question No. 28 is compulsory.
15) Define function.
Solution:
A function 'f' from a non-empty set A to a non-empty set B is a relation such that every element in set A has one and only one image in set B. It is denoted as f: A → B.
16) A = {0, 1}, B = {0, 1}, C = {0, 1} then find (A×B)×C
Solution:
First, find A×B:
A×B = {(0,0), (0,1), (1,0), (1,1)}
Now, find (A×B)×C: (A×B)×C = {((0,0),0), ((0,0),1), ((0,1),0), ((0,1),1), ((1,0),0), ((1,0),1), ((1,1),0), ((1,1),1)}
Now, find (A×B)×C: (A×B)×C = {((0,0),0), ((0,0),1), ((0,1),0), ((0,1),1), ((1,0),0), ((1,0),1), ((1,1),0), ((1,1),1)}
17) Find f.g and g.f when f(x) = 2x+1 and g(x) = x²-2
Solution:
Assuming 'f.g' means function composition f(g(x)):
f(g(x)) = f(x²-2) = 2(x²-2) + 1 = 2x² - 4 + 1 = 2x² - 3
g(f(x)) = g(2x+1) = (2x+1)² - 2 = (4x² + 4x + 1) - 2 = 4x² + 4x - 1
f(g(x)) = f(x²-2) = 2(x²-2) + 1 = 2x² - 4 + 1 = 2x² - 3
g(f(x)) = g(2x+1) = (2x+1)² - 2 = (4x² + 4x + 1) - 2 = 4x² + 4x - 1
18) Find the sum 3 + 1 + 1/3 + ... ∞
Solution:
The given series is an infinite Geometric Progression (GP).
First term, a = 3.
Common ratio, \( r = \frac{1}{3} \).
Since |r| < 1, the sum to infinity is given by \( S_\infty = \frac{a}{1-r} \).
\( S_\infty = \frac{3}{1 - 1/3} = \frac{3}{2/3} = 3 \times \frac{3}{2} = \frac{9}{2} \).
19) Solve: x+y = 1, x-y = 3.
Solution:
x + y = 1 ---(1)
x - y = 3 ---(2)
Adding (1) and (2):
(x+y) + (x-y) = 1 + 3
2x = 4 => x = 2
Substitute x=2 in (1):
2 + y = 1 => y = 1 - 2 = -1
Solution: x = 2, y = -1
x - y = 3 ---(2)
Adding (1) and (2):
(x+y) + (x-y) = 1 + 3
2x = 4 => x = 2
Substitute x=2 in (1):
2 + y = 1 => y = 1 - 2 = -1
Solution: x = 2, y = -1
20) The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.
Solution:
For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
\( \frac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔPQR}} = \frac{AB}{PQ} \)
\( \frac{36}{24} = \frac{AB}{10} \)
\( \frac{3}{2} = \frac{AB}{10} \)
\( AB = \frac{3 \times 10}{2} = 15 \) cm.
\( \frac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔPQR}} = \frac{AB}{PQ} \)
\( \frac{36}{24} = \frac{AB}{10} \)
\( \frac{3}{2} = \frac{AB}{10} \)
\( AB = \frac{3 \times 10}{2} = 15 \) cm.
21) Find the distance between from the points (3, 4) (5, 5)
Solution:
Let the points be A(x₁, y₁) = (3, 4) and B(x₂, y₂) = (5, 5).
The distance formula is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Substitute the values into the formula:
\( d = \sqrt{(5 - 3)^2 + (5 - 4)^2} \)
\( d = \sqrt{(2)^2 + (1)^2} \)
\( d = \sqrt{4 + 1} \)
\( d = \sqrt{5} \) units.
The distance between the points is \( \sqrt{5} \) units.
The distance formula is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Substitute the values into the formula:
\( d = \sqrt{(5 - 3)^2 + (5 - 4)^2} \)
\( d = \sqrt{(2)^2 + (1)^2} \)
\( d = \sqrt{4 + 1} \)
\( d = \sqrt{5} \) units.
The distance between the points is \( \sqrt{5} \) units.
22) Find the equation of a line which passes through (5, 7) and makes intercepts on the axes equal in magnitude but opposite in sign.
Solution:
The intercept form of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \), where 'a' is the x-intercept and 'b' is the y-intercept.
Given that the intercepts are equal in magnitude but opposite in sign, we have b = -a.
Substituting b = -a into the intercept form:
\( \frac{x}{a} + \frac{y}{-a} = 1 \)
\( \frac{x - y}{a} = 1 \)
\( x - y = a \) --- (1)
Since the line passes through the point (5, 7), substitute x = 5 and y = 7 into equation (1):
5 - 7 = a
a = -2
Substitute the value of a = -2 back into equation (1):
x - y = -2
The required equation of the line is x - y + 2 = 0.
Given that the intercepts are equal in magnitude but opposite in sign, we have b = -a.
Substituting b = -a into the intercept form:
\( \frac{x}{a} + \frac{y}{-a} = 1 \)
\( \frac{x - y}{a} = 1 \)
\( x - y = a \) --- (1)
Since the line passes through the point (5, 7), substitute x = 5 and y = 7 into equation (1):
5 - 7 = a
a = -2
Substitute the value of a = -2 back into equation (1):
x - y = -2
The required equation of the line is x - y + 2 = 0.
23) Prove that \( \frac{\sec \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \cot \theta \)
Solution:
LHS = \( \frac{\sec \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} \)
= \( \frac{1/\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} \)
= \( \frac{1}{\sin \theta \cos \theta} - \frac{\sin^2 \theta}{\sin \theta \cos \theta} \)
= \( \frac{1 - \sin^2 \theta}{\sin \theta \cos \theta} \)
Since \( \cos^2 \theta = 1 - \sin^2 \theta \),
= \( \frac{\cos^2 \theta}{\sin \theta \cos \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta \) = RHS.
Hence proved.
= \( \frac{1/\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} \)
= \( \frac{1}{\sin \theta \cos \theta} - \frac{\sin^2 \theta}{\sin \theta \cos \theta} \)
= \( \frac{1 - \sin^2 \theta}{\sin \theta \cos \theta} \)
Since \( \cos^2 \theta = 1 - \sin^2 \theta \),
= \( \frac{\cos^2 \theta}{\sin \theta \cos \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta \) = RHS.
Hence proved.
24) If the Highest Common Factor of 210 and 55 is expressible in the form 55x-325 find x.
Solution:
First, find the HCF of 210 and 55 using Euclid's algorithm.
210 = 3 × 55 + 45
55 = 1 × 45 + 10
45 = 4 × 10 + 5
10 = 2 × 5 + 0
The HCF is 5.
Given, HCF = 55x - 325.
5 = 55x - 325
55x = 325 + 5 = 330
x = 330 / 55 = 6.
210 = 3 × 55 + 45
55 = 1 × 45 + 10
45 = 4 × 10 + 5
10 = 2 × 5 + 0
The HCF is 5.
Given, HCF = 55x - 325.
5 = 55x - 325
55x = 325 + 5 = 330
x = 330 / 55 = 6.
25) If the first term of an infinite G.P. is 8 and its sum to infinity is \( \frac{32}{3} \) then find the common ratio.
Solution:
Given an infinite Geometric Progression (G.P.).
First term, a = 8.
Sum to infinity, \( S_\infty = \frac{32}{3} \).
The formula for the sum to infinity of a G.P. is \( S_\infty = \frac{a}{1-r} \).
Substituting the given values into the formula:
\( \frac{32}{3} = \frac{8}{1-r} \)
Now, we solve for 'r' (the common ratio).
Cross-multiplying gives:
32(1 - r) = 3 × 8
32 - 32r = 24
32 - 24 = 32r
8 = 32r
\( r = \frac{8}{32} \)
\( r = \frac{1}{4} \)
The common ratio is 1/4.
First term, a = 8.
Sum to infinity, \( S_\infty = \frac{32}{3} \).
The formula for the sum to infinity of a G.P. is \( S_\infty = \frac{a}{1-r} \).
Substituting the given values into the formula:
\( \frac{32}{3} = \frac{8}{1-r} \)
Now, we solve for 'r' (the common ratio).
Cross-multiplying gives:
32(1 - r) = 3 × 8
32 - 32r = 24
32 - 24 = 32r
8 = 32r
\( r = \frac{8}{32} \)
\( r = \frac{1}{4} \)
The common ratio is 1/4.
26) If ΔABC ~ ΔDEF such that area of ΔABC is 9 cm² and the area of ΔDEF is 16 cm² and BC = 2.1 cms. Find the length of EF.
Solution:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{\text{Area(ΔABC)}}{\text{Area(ΔDEF)}} = (\frac{BC}{EF})^2 \)
\( \frac{9}{16} = (\frac{2.1}{EF})^2 \)
Taking square root on both sides:
\( \sqrt{\frac{9}{16}} = \frac{2.1}{EF} \)
\( \frac{3}{4} = \frac{2.1}{EF} \)
\( 3 \times EF = 4 \times 2.1 = 8.4 \)
\( EF = \frac{8.4}{3} = 2.8 \) cm.
\( \frac{\text{Area(ΔABC)}}{\text{Area(ΔDEF)}} = (\frac{BC}{EF})^2 \)
\( \frac{9}{16} = (\frac{2.1}{EF})^2 \)
Taking square root on both sides:
\( \sqrt{\frac{9}{16}} = \frac{2.1}{EF} \)
\( \frac{3}{4} = \frac{2.1}{EF} \)
\( 3 \times EF = 4 \times 2.1 = 8.4 \)
\( EF = \frac{8.4}{3} = 2.8 \) cm.
27) Find the square root 361x⁴y²
Solution:
To find the square root of \( 361x^4y^2 \), we take the square root of each component of the term.
The expression is \( \sqrt{361x^4y^2} \).
\( \sqrt{361x^4y^2} = 19x^2y \)
The square root is 19x²y.
The expression is \( \sqrt{361x^4y^2} \).
- Square root of the numerical coefficient: \( \sqrt{361} = 19 \)
- Square root of the variable x: \( \sqrt{x^4} = x^{4/2} = x^2 \)
- Square root of the variable y: \( \sqrt{y^2} = y^{2/2} = y^1 = y \) (or more precisely \(|y|\))
\( \sqrt{361x^4y^2} = 19x^2y \)
The square root is 19x²y.
28) Area of a rectangle is \( \frac{(x-4)(x+3)}{3x-12} \) km², length is \( \frac{x-3}{3} \) km then find breadth.
Solution:
Area = \( \frac{(x-4)(x+3)}{3(x-4)} = \frac{x+3}{3} \) km² (for x ≠ 4).
Length = \( \frac{x-3}{3} \) km.
Breadth = Area / Length
Breadth = \( \frac{(x+3)/3}{(x-3)/3} = \frac{x+3}{3} \times \frac{3}{x-3} = \frac{x+3}{x-3} \) km.
Length = \( \frac{x-3}{3} \) km.
Breadth = Area / Length
Breadth = \( \frac{(x+3)/3}{(x-3)/3} = \frac{x+3}{3} \times \frac{3}{x-3} = \frac{x+3}{x-3} \) km.
PART - III (10 x 5 = 50)
Answer any 10 questions. Q.No. 42 is compulsory.
29) A function 'f' is defined by f(x) = 2x-3
(i) find \( \frac{f(0) + f(1)}{2} \)
(ii) find x such that f(x) = 0
(iii) find x such that f(x) = x
(iv) find x such that f(x) = f(1-x)
Solution:
Given f(x) = 2x - 3.
(i) f(0) = 2(0) - 3 = -3. f(1) = 2(1) - 3 = -1.
\( \frac{f(0) + f(1)}{2} = \frac{-3 + (-1)}{2} = \frac{-4}{2} = -2 \).
(ii) f(x) = 0 ⇒ 2x - 3 = 0 ⇒ 2x = 3 ⇒ x = 3/2.
(iii) f(x) = x ⇒ 2x - 3 = x ⇒ 2x - x = 3 ⇒ x = 3.
(iv) f(x) = f(1-x) ⇒ 2x - 3 = 2(1-x) - 3 ⇒ 2x - 3 = 2 - 2x - 3 ⇒ 2x - 3 = -2x - 1 ⇒ 4x = 2 ⇒ x = 1/2.
(i) f(0) = 2(0) - 3 = -3. f(1) = 2(1) - 3 = -1.
\( \frac{f(0) + f(1)}{2} = \frac{-3 + (-1)}{2} = \frac{-4}{2} = -2 \).
(ii) f(x) = 0 ⇒ 2x - 3 = 0 ⇒ 2x = 3 ⇒ x = 3/2.
(iii) f(x) = x ⇒ 2x - 3 = x ⇒ 2x - x = 3 ⇒ x = 3.
(iv) f(x) = f(1-x) ⇒ 2x - 3 = 2(1-x) - 3 ⇒ 2x - 3 = 2 - 2x - 3 ⇒ 2x - 3 = -2x - 1 ⇒ 4x = 2 ⇒ x = 1/2.
30) f(x) = x², g(x) = 2x and h(x) = x+4 prove that fo(goh) = (fog)oh
Solution:
LHS = f o (g o h)
First, (g o h)(x) = g(h(x)) = g(x+4) = 2(x+4) = 2x+8.
Now, f o (g o h)(x) = f((g o h)(x)) = f(2x+8) = (2x+8)² = 4x² + 32x + 64.
RHS = (f o g) o h
First, (f o g)(x) = f(g(x)) = f(2x) = (2x)² = 4x².
Now, (f o g) o h(x) = (f o g)(h(x)) = (f o g)(x+4) = 4(x+4)² = 4(x² + 8x + 16) = 4x² + 32x + 64.
Since LHS = RHS, it is proved.
First, (g o h)(x) = g(h(x)) = g(x+4) = 2(x+4) = 2x+8.
Now, f o (g o h)(x) = f((g o h)(x)) = f(2x+8) = (2x+8)² = 4x² + 32x + 64.
RHS = (f o g) o h
First, (f o g)(x) = f(g(x)) = f(2x) = (2x)² = 4x².
Now, (f o g) o h(x) = (f o g)(h(x)) = (f o g)(x+4) = 4(x+4)² = 4(x² + 8x + 16) = 4x² + 32x + 64.
Since LHS = RHS, it is proved.
31) If the function f:R→R is defined by \( f(x) = \begin{cases} 2x + 7 & x < -2 \\ x^2 - 2 & -2 \leq x < 3 \\ 3x - 2 & x \geq 3 \end{cases} \) then find the value of (i) f(4) (ii) f(-2) (iii) f(4)+2f(1) (iv) \( \frac{f(1)-3f(4)}{f(-3)} \)
Solution:
(i) f(4): Since 4 ≥ 3, use f(x) = 3x - 2.
f(4) = 3(4) - 2 = 12 - 2 = 10.
(ii) f(-2): Since -2 is in -2 ≤ x < 3, use f(x) = x² - 2. f(-2) = (-2)² - 2 = 4 - 2 = 2.
(iii) f(4) + 2f(1): We need f(1). Since 1 is in -2 ≤ x < 3, use f(x) = x² - 2. f(1) = 1² - 2 = -1. f(4) + 2f(1) = 10 + 2(-1) = 10 - 2 = 8.
(iv) \( \frac{f(1)-3f(4)}{f(-3)} \): We need f(-3). Since -3 < -2, use f(x) = 2x + 7. f(-3) = 2(-3) + 7 = -6 + 7 = 1. The expression is \( \frac{-1 - 3(10)}{1} = \frac{-1 - 30}{1} = -31 \).
(ii) f(-2): Since -2 is in -2 ≤ x < 3, use f(x) = x² - 2. f(-2) = (-2)² - 2 = 4 - 2 = 2.
(iii) f(4) + 2f(1): We need f(1). Since 1 is in -2 ≤ x < 3, use f(x) = x² - 2. f(1) = 1² - 2 = -1. f(4) + 2f(1) = 10 + 2(-1) = 10 - 2 = 8.
(iv) \( \frac{f(1)-3f(4)}{f(-3)} \): We need f(-3). Since -3 < -2, use f(x) = 2x + 7. f(-3) = 2(-3) + 7 = -6 + 7 = 1. The expression is \( \frac{-1 - 3(10)}{1} = \frac{-1 - 30}{1} = -31 \).
32) The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Solution:
Let the three consecutive terms in A.P. be a-d, a, a+d.
Sum of terms: (a-d) + a + (a+d) = 27
3a = 27 ⇒ a = 9.
Product of terms: (a-d) * a * (a+d) = 288
Substitute a=9: (9-d) * 9 * (9+d) = 288
(9-d)(9+d) = 288 / 9 = 32
81 - d² = 32
d² = 81 - 32 = 49
d = ±7
Case 1: d = 7 The terms are 9-7, 9, 9+7, which are 2, 9, 16.
Case 2: d = -7 The terms are 9-(-7), 9, 9-7, which are 16, 9, 2.
Therefore, the three terms are 2, 9, and 16.
Sum of terms: (a-d) + a + (a+d) = 27
3a = 27 ⇒ a = 9.
Product of terms: (a-d) * a * (a+d) = 288
Substitute a=9: (9-d) * 9 * (9+d) = 288
(9-d)(9+d) = 288 / 9 = 32
81 - d² = 32
d² = 81 - 32 = 49
d = ±7
Case 1: d = 7 The terms are 9-7, 9, 9+7, which are 2, 9, 16.
Case 2: d = -7 The terms are 9-(-7), 9, 9-7, which are 16, 9, 2.
Therefore, the three terms are 2, 9, and 16.
PART - III (Continued)
33) The product of three consecutive terms of a Geometric Progression is 343 and their sum is \( \frac{91}{3} \). Find the three terms.
Solution:
Let the three consecutive terms in a Geometric Progression (G.P.) be \( \frac{a}{r}, a, ar \).
Step 1: Use the product of the terms.
Given, product = 343
\( (\frac{a}{r}) \times a \times (ar) = 343 \)
\( a^3 = 343 \)
\( a^3 = 7^3 \)
\( a = 7 \)
Step 2: Use the sum of the terms.
Given, sum = \( \frac{91}{3} \)
\( \frac{a}{r} + a + ar = \frac{91}{3} \)
Substitute a = 7:
\( \frac{7}{r} + 7 + 7r = \frac{91}{3} \)
Divide the entire equation by 7:
\( \frac{1}{r} + 1 + r = \frac{13}{3} \)
Step 3: Solve for the common ratio 'r'.
\( \frac{1 + r + r^2}{r} = \frac{13}{3} \)
Cross-multiply:
\( 3(1 + r + r^2) = 13r \)
\( 3 + 3r + 3r^2 = 13r \)
\( 3r^2 - 10r + 3 = 0 \)
Factorize the quadratic equation:
\( 3r^2 - 9r - r + 3 = 0 \)
\( 3r(r - 3) - 1(r - 3) = 0 \)
\( (3r - 1)(r - 3) = 0 \)
So, \( r = 3 \) or \( r = \frac{1}{3} \).
Step 4: Find the terms.
If a = 7 and r = 3, the terms are \( \frac{7}{3}, 7, 7(3) \), which are \( \frac{7}{3}, 7, 21 \).
If a = 7 and r = 1/3, the terms are \( \frac{7}{1/3}, 7, 7(\frac{1}{3}) \), which are \( 21, 7, \frac{7}{3} \).
In both cases, the three terms are \( \frac{7}{3}, 7, 21 \).
Step 1: Use the product of the terms.
Given, product = 343
\( (\frac{a}{r}) \times a \times (ar) = 343 \)
\( a^3 = 343 \)
\( a^3 = 7^3 \)
\( a = 7 \)
Step 2: Use the sum of the terms.
Given, sum = \( \frac{91}{3} \)
\( \frac{a}{r} + a + ar = \frac{91}{3} \)
Substitute a = 7:
\( \frac{7}{r} + 7 + 7r = \frac{91}{3} \)
Divide the entire equation by 7:
\( \frac{1}{r} + 1 + r = \frac{13}{3} \)
Step 3: Solve for the common ratio 'r'.
\( \frac{1 + r + r^2}{r} = \frac{13}{3} \)
Cross-multiply:
\( 3(1 + r + r^2) = 13r \)
\( 3 + 3r + 3r^2 = 13r \)
\( 3r^2 - 10r + 3 = 0 \)
Factorize the quadratic equation:
\( 3r^2 - 9r - r + 3 = 0 \)
\( 3r(r - 3) - 1(r - 3) = 0 \)
\( (3r - 1)(r - 3) = 0 \)
So, \( r = 3 \) or \( r = \frac{1}{3} \).
Step 4: Find the terms.
If a = 7 and r = 3, the terms are \( \frac{7}{3}, 7, 7(3) \), which are \( \frac{7}{3}, 7, 21 \).
If a = 7 and r = 1/3, the terms are \( \frac{7}{1/3}, 7, 7(\frac{1}{3}) \), which are \( 21, 7, \frac{7}{3} \).
In both cases, the three terms are \( \frac{7}{3}, 7, 21 \).
34) Find the sum of n terms of the series 5+55+555+...
Solution:
Let \( S_n = 5 + 55 + 555 + ... \) to n terms.
Step 1: Take 5 as a common factor.
\( S_n = 5(1 + 11 + 111 + ...) \)
Step 2: Multiply and divide by 9.
\( S_n = \frac{5}{9}(9 + 99 + 999 + ...) \)
Step 3: Express the terms inside the bracket in terms of powers of 10.
\( S_n = \frac{5}{9}((10-1) + (100-1) + (1000-1) + ...) \)
Step 4: Separate the terms into two groups.
\( S_n = \frac{5}{9}((10 + 10^2 + 10^3 + ...) - (1 + 1 + 1 + ...)) \)
The first group is a G.P. with first term a = 10, common ratio r = 10. The second group is the sum of 'n' ones.
Step 5: Apply the sum formula for a G.P., \( S_n = \frac{a(r^n - 1)}{r-1} \).
\( S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{10-1} - n \right] \)
\( S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{9} - n \right] \)
Final Answer: \( S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9} \)
Step 1: Take 5 as a common factor.
\( S_n = 5(1 + 11 + 111 + ...) \)
Step 2: Multiply and divide by 9.
\( S_n = \frac{5}{9}(9 + 99 + 999 + ...) \)
Step 3: Express the terms inside the bracket in terms of powers of 10.
\( S_n = \frac{5}{9}((10-1) + (100-1) + (1000-1) + ...) \)
Step 4: Separate the terms into two groups.
\( S_n = \frac{5}{9}((10 + 10^2 + 10^3 + ...) - (1 + 1 + 1 + ...)) \)
The first group is a G.P. with first term a = 10, common ratio r = 10. The second group is the sum of 'n' ones.
Step 5: Apply the sum formula for a G.P., \( S_n = \frac{a(r^n - 1)}{r-1} \).
\( S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{10-1} - n \right] \)
\( S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{9} - n \right] \)
Final Answer: \( S_n = \frac{50}{81}(10^n - 1) - \frac{5n}{9} \)
35) Solve: 3x - 2y + z = 2, 2x + 3y - z = 5, x + y + z = 6.
Solution:
Let the equations be:
3x - 2y + z = 2 ---(1)
2x + 3y - z = 5 ---(2)
x + y + z = 6 ---(3)
Step 1: Eliminate 'z' using equations (1) and (2).
Add (1) and (2):
(3x - 2y + z) + (2x + 3y - z) = 2 + 5
5x + y = 7 ---(4)
Step 2: Eliminate 'z' using equations (2) and (3).
Add (2) and (3):
(2x + 3y - z) + (x + y + z) = 5 + 6
3x + 4y = 11 ---(5)
Step 3: Solve the new system of equations (4) and (5).
Multiply equation (4) by 4:
20x + 4y = 28 ---(6)
Subtract equation (5) from (6):
(20x + 4y) - (3x + 4y) = 28 - 11
17x = 17
x = 1
Step 4: Substitute x = 1 into equation (4) to find y.
5(1) + y = 7
y = 7 - 5
y = 2
Step 5: Substitute x = 1 and y = 2 into equation (3) to find z.
1 + 2 + z = 6
3 + z = 6
z = 3
The solution is x = 1, y = 2, z = 3.
3x - 2y + z = 2 ---(1)
2x + 3y - z = 5 ---(2)
x + y + z = 6 ---(3)
Step 1: Eliminate 'z' using equations (1) and (2).
Add (1) and (2):
(3x - 2y + z) + (2x + 3y - z) = 2 + 5
5x + y = 7 ---(4)
Step 2: Eliminate 'z' using equations (2) and (3).
Add (2) and (3):
(2x + 3y - z) + (x + y + z) = 5 + 6
3x + 4y = 11 ---(5)
Step 3: Solve the new system of equations (4) and (5).
Multiply equation (4) by 4:
20x + 4y = 28 ---(6)
Subtract equation (5) from (6):
(20x + 4y) - (3x + 4y) = 28 - 11
17x = 17
x = 1
Step 4: Substitute x = 1 into equation (4) to find y.
5(1) + y = 7
y = 7 - 5
y = 2
Step 5: Substitute x = 1 and y = 2 into equation (3) to find z.
1 + 2 + z = 6
3 + z = 6
z = 3
The solution is x = 1, y = 2, z = 3.
36) Find the square root of 64x⁴ − 16x³ + 17x² − 2x + 1.
Solution:
We use the long division method to find the square root.
8x² -x +1
____________________
8x² | 64x⁴ -16x³ +17x² -2x +1
| 64x⁴
|____________________
16x²-x| -16x³ +17x²
| -16x³ + x²
|____________________
16x²-2x+1| 16x² -2x +1
| 16x² -2x +1
|____________________
| 0
The square root of the given polynomial is |8x² - x + 1|.
38) Two poles of height 'a' metres and 'b' metres are 'p' metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by \( \frac{ab}{a+b} \) metres.
Solution using similar triangles:
Step 1: Consider similar triangles ΔABD and ΔEFD.
\( \triangle EFD \sim \triangle ABD \)
\( \frac{EF}{AB} = \frac{FD}{BD} \)
\( \frac{h}{a} = \frac{p-x}{p} \) --- (1)
Step 2: Consider similar triangles ΔCDB and ΔEFB.
\( \triangle EFB \sim \triangle CDB \)
\( \frac{EF}{CD} = \frac{FB}{DB} \)
\( \frac{h}{b} = \frac{x}{p} \) --- (2)
Step 3: Solve the equations for h.
From (1), \( \frac{h}{a} = 1 - \frac{x}{p} \).
From (2), \( \frac{x}{p} = \frac{h}{b} \).
Substitute \( \frac{x}{p} \) in the modified equation (1):
\( \frac{h}{a} = 1 - \frac{h}{b} \)
\( \frac{h}{a} + \frac{h}{b} = 1 \)
Take 'h' common:
\( h \left( \frac{1}{a} + \frac{1}{b} \right) = 1 \)
\( h \left( \frac{b+a}{ab} \right) = 1 \)
\( h = \frac{ab}{a+b} \)
Hence, the height of the intersection point is \( \frac{ab}{a+b} \) metres.
Let the poles be AB and CD with heights 'a' and 'b' respectively. The distance between them is BD = 'p'. Let the lines AC and BD intersect at point E. Let the height of E from the ground be EF = h. Let BF = x, so FD = p-x.
Step 1: Consider similar triangles ΔABD and ΔEFD.
\( \triangle EFD \sim \triangle ABD \)
\( \frac{EF}{AB} = \frac{FD}{BD} \)
\( \frac{h}{a} = \frac{p-x}{p} \) --- (1)
Step 2: Consider similar triangles ΔCDB and ΔEFB.
\( \triangle EFB \sim \triangle CDB \)
\( \frac{EF}{CD} = \frac{FB}{DB} \)
\( \frac{h}{b} = \frac{x}{p} \) --- (2)
Step 3: Solve the equations for h.
From (1), \( \frac{h}{a} = 1 - \frac{x}{p} \).
From (2), \( \frac{x}{p} = \frac{h}{b} \).
Substitute \( \frac{x}{p} \) in the modified equation (1):
\( \frac{h}{a} = 1 - \frac{h}{b} \)
\( \frac{h}{a} + \frac{h}{b} = 1 \)
Take 'h' common:
\( h \left( \frac{1}{a} + \frac{1}{b} \right) = 1 \)
\( h \left( \frac{b+a}{ab} \right) = 1 \)
\( h = \frac{ab}{a+b} \)
Hence, the height of the intersection point is \( \frac{ab}{a+b} \) metres.
PART - IV (2 x 8 = 16)
43) (OR) Construct a \( \triangle PQR \) such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.
Steps of Construction:
- Draw a line segment QR = 6.5 cm.
- At Q, draw a line QX such that ∠RQX = 60°.
- Draw a line QY perpendicular to QX at Q.
- Draw the perpendicular bisector of the line segment QR. Let it intersect QR at M and the line QY at O.
- With O as the center and OQ (or OR) as the radius, draw a circle. This circle will pass through Q and R. The major arc of this circle makes an angle of 60° at any point on it.
- Draw a line segment EF parallel to QR at a distance of 4.5 cm from QR. This line represents the locus of points at an altitude of 4.5 cm from the base QR.
- This parallel line EF intersects the circle at two points, P and P'.
- Join PQ and PR (or P'Q and P'R).
- \( \triangle PQR \) is the required triangle.
44) Draw the graph of xy = 24. x, y > 0 using the graph find (i) y when x = 3 and (ii) x when y = 6.
Solution:
Step 2: Plot the points and draw the graph.
Step 3: Find the required values from the graph.
The equation xy = 24 represents a rectangular hyperbola. Since x, y > 0, we will plot the graph only in the first quadrant.
Step 1: Create a table of values.| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y = 24/x | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
Step 2: Plot the points and draw the graph.
Plot the points (1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2) on a graph paper and join them with a smooth curve.
Step 3: Find the required values from the graph.
(i) Find y when x = 3:
On the graph, locate x = 3 on the x-axis. Draw a vertical line up to the curve. From that point on the curve, draw a horizontal line to the y-axis. The line meets the y-axis at y = 8.
Therefore, when x = 3, y = 8.
(ii) Find x when y = 6:
On the graph, locate y = 6 on the y-axis. Draw a horizontal line to the curve. From that point on the curve, draw a vertical line down to the x-axis. The line meets the x-axis at x = 4.
Therefore, when y = 6, x = 4.