Category: Class 10

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 5 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Paper 5 English Medium

Instructions

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III and IV are to be attempted separately.
• Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
• Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
• Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
• Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 100

PART – I

I. Choose the correct answer. Answer all the questions. [14 × 1 =14]

Question 1.
The range of the relation R = {(x, x²) x is a prime number less than 13} is ………… .
(1) {2, 3, 5, 7}
(2) (2, 3, 5, 7, 11}
(3) {4,9,25,49, 121}
(4) {1, 4, 9, 25, 49, 121}
Answer:
(3) {4,9,25,49, 121}

Question 2.
If g = {(1,1),(2, 3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are ……….. .
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)
Answer:
(2) (2,-1)

Question 3.
The sum of the exponents of the prime factors in the prime factorization of 1729 is ……….. .
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Question 4.
The next term of the sequence \(\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}\) is ……….. .
(1) \(\frac{1}{24}\)
(2) \(\frac{1}{27}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{1}{81}\)
Answer:
(2) \(\frac{1}{27}\)

Question 5.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of k is ……….. .
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5

Question 6.
The solution of (2x – 1)² = 9 is equal to ……….. .
(1) -1
(2) 2
(3) -1,2
(4) None of these
Answer:
(3) -1,2

Question 7.
In a given figure ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is ……….. .

(1) 25 : 4
(2) 25 : 7
(3) 25 :11
(4) 25 : 13
Answer:
(1) 25 : 4

Question 8.
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 }\). The value of ‘a’ is ……….. .
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2

Question 9.
If sin θ + cos θ = a and sec θ + cosec θ = b, then the value of b (a² – 1) is equal to ……….. .
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a

Question 10.
The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is ……….. .
(1) 60π cm²
(2) 68π cm²
(3) 120π cm²
(4) 136π cm²
Answer:
(4) 136π cm²

Question 11.
The sum of all deviations of the data from its mean is ……….. .
(1) always positive
(2) always negative
(3) zero
(4) non-zero integer
Answer:
(3) zero

Question 12.
If \(\left( \begin{matrix} x+y & x-y \\ 7 & 6 \end{matrix} \right) =\left( \begin{matrix} 10 & 2 \\ 7 & z \end{matrix} \right) \) then x, y, z are ……….. .
(1) 4, 6, 6
(2) 6, 6, 4
(3) 6, 4, 6
(4) 4, 4, 6
Answer:
(3) 6, 4, 6

Question 13.
If the n^th term of a sequence is 100n + 10 then the sequence is ……….. .
(1) an A.P.
(2) a G.P.
(3) a constant sequence
(4) neither A.P. nor G.P.
Answer:
(1) an A.P.

Question 14.
Probability of getting 3 heads or 3 tails in tossing a coin 3 times is ……….. .
(1) \(\frac { 1 }{ 8 }\)
(2) \(\frac { 1 }{ 4 }\)
(3) \(\frac { 3 }{ 8 }\)
(4) \(\frac { 1 }{ 2 }\)
Answer:
(2) \(\frac { 1 }{ 4 }\)

PART – II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
Let A = {x ∈ W / x < 2}, B = {x ∈ N / 1 < x ≤ 4} and C = {3, 5} find the value of (A × C) ∪ (B × C)
Answer:
A × C = {0,1} × {3,5}
= {(0,3 (0,5) (1,3) (1,5)}
B × C= (2,3,4) × (3,5)
= {(2,3) (2,5) (3,3) (3,5) (4,3) (4,5)}
(A × C) ∪ (B × C) = {(0,3) (0,5) (1,3) (1,5) (2,3) (2,5) (3,3) (3,5) (4,3) (4,5)}

Question 16.
Find k if fof(k) = 5 where f(k) = 2k – 1.
Answer:
fof(k) = f(f(k)) = 2(2k – 1)-1 = 4k – 3.
Thus, fof(k) = 4k – 3
But, it is given that fof(k) = 5
∴ 4k – 3 = 5 ⇒ k = 2.

Question 17.
Find the sum of the following 6 + 13 + 20 +…. + 97
Answer:
Here a = 6, d = 13 – 6 = 7, l = 97
n = \(\frac{l-a}{d}+1\)
= \(\frac{97-6}{7}+1\) = \(\frac{91}{7}+1\) = 13 + 1 = 14
S_n = \(\frac{n}{2}\)(a + l)
S_n = \(\frac{14}{2}\)(6 + 97) = 7 × 103 = 721

Question 18.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …,.24 cm. How much area can be decorated with these colour papers?
Answer:
Area of 15 square colour papers
= 10² + 11² + 12² + …. + 24²
= (1² + 2² + 3² + …. + 24²) – (1² + 2² + 9²)
= \(\frac{24 \times 25 \times 49}{6}-\frac{9 \times 10 \times 19}{6}\)
=4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm²

Question 19.
Simplify \(\frac{x+2}{4 y} \div \frac{x^{2}-x-6}{12 y^{2}}\)
Answer:

x² – x – 6 = (x – 3) (x + 2)
= \(\frac{x+2}{4 y} \div \frac{x^{2}-x-6}{12 y^{2}}=\frac{x+2}{4 y}+\frac{(x-3)(x+2)}{12 y^{2}}\)
= \(\frac{(x+2)}{4 y} \times \frac{12 y^{2}}{(x-3)(x+2)}\)
= \(\frac{3 y}{x-3}\)

Question 20.
Solve \(\frac{x}{x-1}+\frac{x-1}{x}=2 \frac{1}{2}\)
Answer:
Let y = \(\frac{x}{x-1}\) then \(\frac{1}{y}=\frac{x-1}{x}\)
Therefore, \(\frac{x}{x-1}+\frac{x-1}{x}=2 \frac{1}{2}\) becomes \(y+\frac{1}{y}=\frac{5}{2}\)
2y² – 5y + 2 = 0 then, y = \(\frac{1}{2}, 2\)
\(\frac{x}{x-1}=\frac{1}{2}\) we get, 2x = x – 1 implies x = – 1
\(\frac{x}{x-1}=2\) we get, x = 2x – 2 implies x = 2
∴ The roots are x = -1, 2

Question 21.
Construct a 3 × 3 matrix whose elements are given by a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is
A = \(\left( \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right)\)
a₁₁ = |1 – 2(1)| = |1 – 2| = |-1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = |-3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = |-5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = |-2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = |-4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = |1| = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = |-1| = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = |-3| = 3
The required matrix A = \(\left[ \begin{matrix} 1 & 3 & 5 \\ 0 & 2 & 4 \\ 1 & 1 & 3 \end{matrix} \right] \)

Question 22.
The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?
Answer:

Let the radius AB be r. In the right ∆ ABO,
OB² = OA² + AB²
25² = 24² + r²
25² – 24² = r²
(25 + 4) (25 – 24) = r²
r = √49 = 7
Radius of the circle = 7 cm

Question 23.
If the straight lines 12y = – (p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’.
Answer:
Slope of the first line 12y = -(p + 3)x + 12
y = \(-\frac{(p+3) x}{12}+1\) (comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac{-(p+3)}{12}\)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac{-a}{b}=\frac{-12}{-7}=\frac{12}{7}\)
Since the two lines are perpemdicular.
m₁ × m₂ = -1
\(\frac{-(p+3)}{12} \times \frac{12}{7}=-1 \Rightarrow \frac{-(p+3)}{7}=-1\)
-(p + 3) = -7
– p – 3 = – 7 ⇒ -p = -7 + 3
– p = -4 ⇒ p = 4
The value of p = 4

Question 24.
Prove that 1 + \(\frac{\cot ^{2} \theta}{1+\csc \theta}\) = cosec θ
Answer:

Question 25.
Find the standard deviation of first 21 natural numbers.
Answer:
Here n = 21
Standard deviation of the first ‘n’ natural numbers,
= \(\sqrt{\frac{n^{2}-1}{12}}\)
1,2,3,4, ……. , 21 = \(\sqrt{\frac{21^{2}-1}{12}}=\sqrt{\frac{441-1}{12}}=\sqrt{\frac{440}{12}}\)
= \(\sqrt{36.666}\) = \(\sqrt{36.67}\)
= 6.055 = 6.06
Standard deviation of first 21 natural numbers = 6.06

Question 26.
How many litres of water will a hemispherical tank hold whose diameter is 4.2m?
Answer:
Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m
Volume of the hemisphere = \(\frac{2}{3}\) πr³ cu. unit
= \(\frac{2}{3} \times \frac{22}{7}\) × 2.1 × 2.1 × 2.1 m³
= 19.404 m³
= 19.4o4 × 1000 litre
= 19,4o4 litres

Question 27.
A two digit number is formed with the digits 2, 5, 9 (repetition is allowed). Find the probability that the number is divisible by 2.
Answer:
Sample space (S) = {22, 25, 29, 55, 59, 52, 99, 92, 95}
n(S) = 9
Let A be the event of getting number divisible by 2
A = {22,52,92}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{9}=\frac{1}{3}\)

Question 28.
Solve √x + 5 = 2x + 3 using formula method.
Answer:
\(\sqrt{x+5}\) = 2x + 3
\((\sqrt{x+5})^{2}\) = (2x + 3)²
x + 5 = 4x² + 9 + 12x
4x² + 11x + 4 = 0
Here a = 4, b = 11, c = 5
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-11 \pm \sqrt{121-64}}{8}\)
= \(\frac{-11 \pm \sqrt{57}}{8}\)
∴ x = \(\frac{-11+\sqrt{57}}{8}\) ; x = \(\frac{-11-\sqrt{57}}{8}\)

PART – III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the comers and turning up the sides as shown. Express the volume V of the box as a function of x.

Question 30.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?

Question 31.
The present value of a machine is ₹40,000 and its value depreciates each year by 10%. Find the estimated value of the machine in the 6th year.

Question 32.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number?

Question 33.
Find the square root of the expression \(\frac{4 x^{2}}{y^{2}}+\frac{20 x}{y}+13-\frac{30 y}{x}+\frac{9 y^{2}}{x^{2}}\)

Question 34.
If α, β are the roots of 7x² + ax + 2 = 0 and if β – α = \(\frac { -13 }{ 7 }\) . Find the values of a.

Question 35.
State and prove Angle Bisector Theorem.

Question 36.
A quadrilateral has vertices at A(-4, -2) , B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.

Question 37.
Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively.
If the distance between the ships is 200 \(\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)\) meters, find the height of the light house.

Question 38.
A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.

Question 39.
Two dice are rolled together. Find the probability of getting a doublet or sum of faces as 4.

Question 40.
Given f(x) = x – 2 ; g(x) = 3x + 5; h(x) = 2x – 3 verify that (goh) of = go (hof)

Question 41.
A 20 m deep well with inner diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22m by 14m. Find the height of the platform.

Question 42.
The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 7 }{ 3 }\) of the corresponding sides of the triangle PQR (scale factor \(\frac { 7 }{ 3 }\)).

[OR]

(b) Draw a circle of diameter 6 cm from a point P, which is 8 cm away from its centre. Draw the two tangents PA and PB to the circle and measure their lengths.

Question 44.
(a) Draw the graph of y = x² – 5x – 6 and hence solve x² – 5x – 14 = 0.

[OR]

(b) Solve graphically 2x² + x – 6 = 0.

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamil Nadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

Instructions

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III and IV are to be attempted separately.
• Question numbers 1 to 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
• Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
• Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
• Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 100

PART – I

I. Choose the correct answer. Answer all the questions. [14 × 1 = 14]

Question 1.
If A = {1,2}, B = {1, 2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true ………….. .
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Answer:
(1) (A × C) ⊂ (B × D)

Question 2.
If f(x) = 2x² and g(x) = \(\frac { 1 }{ 3x }\), then jog is ………….. .
(1) \(\frac{3}{2 x^{2}}\)
(2) \(\frac{2}{3 x^{2}}\)
(3) \(\frac{2}{9 x^{2}}\)
(4) \(\frac{1}{6 x^{2}}\)
Answer:
(3) \(\frac{2}{9 x^{2}}\)

Question 3.
Given F₁ = 1,F₂ = 3 and F_n = F_n-1 + F_n-2  then F₅ is ………….. .
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11

Question 4.
The value of (1³ + 2³ + 3³ +. . . .+ 15³) – (1 + 2 + 3 +….+ 15) is ………….. .
(1) 14400
(2) 14200
(3) 14280
(4) 14520
Answer:
(3) 14280

Question 5.
The solution of the system x + y – 3z = -6 , – 7y + 7z = 7, 3z = 9 is ………….. .
(1) x = 1, y = 2, z = 3
(2) x = -1, y = 2, z = 3
(3) x = -1, y = -2, z = 3
(4) x = 1, y = 2, z = 3
Answer:
(1) x = 1, y = 2, z = 3

Question 6.
If number of columns and rows are not equal in a matrix then it is said to be a ………….. .
(1) diagonal matrix
(2) rectangular matrix
(3) square matrix
(4) identity matrix
Answer:
(2) rectangular matrix

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (-8, 8) is ………….. .
(1) -1
(2) 1
(3) \(\frac { 1 }{ 3 }\)
(4) -8
Answer:
(2) 1

Question 8.
If sin θ + cos θ = a and sec θ + cosec θ = 6, then the value of b (a² – 1} is equal to ………….. .
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a

Question 9.
If two solid hemispheres of same base radius r units are joined together along their bases, then curved surface area of this new solid is ………….. .
(1) 4π r² sq. units
(2) 6π r² sq. units
(3) 3π r² sq. units
(4) 8π r² sq. units
Answer:
(1) 4π r² sq. units

Question 10.
The standard deviation of a data is 3. If each value is multiplied by 5 then the new variance is ………….. .
(1) 3
(2) 15
(3) 5
(4) 225
Answer:
(4) 225

Question 11.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac { 1 }{ 9 }\), then the number of tickets bought by Kamalam is ………….. .
(1) 5
(2) 10
(3) 15
(4) 20
Answer:
(3) 15

Question 12.
If α and β are the zeros of the polynomial p(x) = 4x² + 3x + 7 then \(\frac{1}{\alpha}+\frac{1}{\beta}\) is equal to …… .
(1) \(\frac{7}{3}\)
(2) \(-\frac{7}{3}\)
(3) \(\frac{3}{7}\)
(4) \(-\frac{3}{7}\)
Answer:
(4) \(-\frac{3}{7}\)

Question 13.
The common ratio of the G.P. a^m-n, a^m, a^{m + n} is ………….. .
(1) a^m
(2) a^-m
(3) aⁿ
(4) aⁿ
Answer:
(3) aⁿ

Question 14.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ………….. .
(1) 1200 π cm²
(2) 600 π cm²
(3) 300 π cm²
(4) 600 π m²
Answer:
(2) 600 π cm²

PART – II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
Let A = {1,2, 3, 4} and B = N . Let f: A → B be defined by f(x) = x³ then,
(i) find the range of f
(ii) identify the type of function
Answer:
A = {1,2,3,4}
B = (1,2,3,4,5, ………}
f(x) = x³
f(1) = 1³ = 1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
(i) Range = {1, 8,27, 64)
(ii) one -one and into function.

Question 16.
If f(x) = 2x -1, g (x) = \(\frac{x+1}{2}\), show that fog = gof = x
Answer:
f(x) = 2x – 1 : g(x) = \(\frac{x+1}{2}\)
fog = f[g(x)]
= \(f\left[\frac{x+1}{2}\right]\)
= \(2\left(\frac{x+1}{2}\right)-1\)
= x + 1 – 1
= x

gof = g[f(x)]
= g (2x – 1)
= \(\frac{2 x-1+1}{2}\)
= \(\frac{2 x}{2}\)
= x
∴ fog = gof = x
Hence it is proved.

Question 17.
Determine the general term of an A.P. whose 7^th term is -1 and 16^th term is 17.
Answer:
Let the AP. be t₁, t₂, t₃, t₄, ………
It is given that t₇ = -1 and t₁₆ = 17
a + (7 – 1)d = -1 and a + (16 – 1) d = 17
a + 6d = – 1 ……. (1)
a + 15d = 17 ……… (2)
Subtracting equation (1) from equation (2), we get 9d = 18 ⇒ d = 2
Putting d = 2 in equation (1), we get a + 12 = -1 So, a = -13
Hence, general term t_n = a + (n – 1) d = -13 + (n – 1) × 2 = 2n – 15

Question 18.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.
Answer:
\(\frac{t_{2}}{t_{1}}=\frac{x+12}{x+6}\), \(\frac{t_{3}}{t_{2}}=\frac{x+15}{x+12}\)
Since it is a G.P.
\(\frac{x+12}{x+6}=\frac{x+15}{x+12}\)
(x + 12)² = (x + 6) (x + 15)
x² + 24x + 144 = x² + 21x + 90
3x = -54 ⇒ x = \(\frac{-54}{3}\) = – 18

Question 19.
Find the excluded values of the expression \(\frac{x^{2}+6 x+8}{x^{2}+x-2}\)
Answer:
x² + 6x + 8 = (x + 4)(x + 2)
x² + x – 2 = (x + 2)(x – 1)


The expression \(\frac{x+4}{x-1}\) is undefined
when x – 1 = 0 ⇒ x = 1
The excluded value is 1

Question 20.
Solve 3p² + 2√5p – 5 = 0 by formula method.
Answer:
Compute 3p² + 2√5p – 5 = 0 with the standard form ax² + bx + c = O
a = 3, b = 2√5, c = -5
p = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Substituting the values of a, b and e in the formula we get,

Therefore, x = \(\frac{\sqrt{5}}{3},-\sqrt{5}\)

Question 21.
Write the expression \(\frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha}\) in terms of α + β and αβ.
Answer:

Question 22.
Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
Answer:
Let the side of the rhombus be “x”. Since PQRB is a Rhombus PQ || BC
By basic proportionality theorem
\(\frac{A P}{A B}=\frac{P Q}{B C} \Rightarrow \frac{12-x}{12}=\frac{x}{6}\)
12x = 6(12 – x)
12x = 72 – 6x
12x + 6x = 72
18x = 72 ⇒ x = \(\frac { 72 }{ 18 }\) = 4
Side of a rhombus = 4cm
PQ = RB = 4cm

Question 23.
Find the equation of a straight line which is parallel to the line 3x – 7y = 12 and passing through the point (6,4)
Answer:
Equation of the straight line, parallel to 3x – 7y – 12 = 0 is 3x – 7y + k= O
Since it passes through the point (6,4)
3(6) – 7(4) + k = 0
k = 28 – 18= 10
Therefore, equation of the required straight line is 3x – 7y + 10 = 0.

Question 24.
Prove that \(\frac{\tan ^{2} \theta-1}{\tan ^{2} \theta+1}\) = 1 – 2 cos² θ
Answer:

= sin² θ – cos² θ = 1 – cos² θ – cos² θ
= 1 – 2 cos²θ
L.H.S = R.H.S
Hence it is proved.

Question 25.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer:
Standard deviation (σ) = 1.2
Coefficient of variation = 25.6
\(\frac{\sigma}{x} \times 100\) = 25.6
\(\frac{1.2}{\bar{x}} \times 100\) = 25.6 ⇒ 25.6 × x̄ = 1.2 × 100
x̄ = \(\frac{120}{25.6}=\frac{120 \times 10}{256}\) = 4.687 = 4.69
Value of mean = 4.69

Question 26.
Three dice are thrown simultaneously. Find the probability of getting the same number.
Answer:
Sample space = {(1, 1, 1) (1, 1,2) (1, 1,3). .. . (6,6, 6)}
n(S) = 216
Let A be the event of getting the same number on all the three dice
A = {(1,1,1),(2,2,2),(3,3,3),(4,4,4)(5,5,5)(6,6,6fl
n(A) = 6
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{216}=\frac{1}{36}\)

Question 27.
If the curved surface area of a solid hemisphere is 2772 sq. cm, then find its total surface area.
Answer:
Curved surface area ofa hemisphere = 2772 sq.cm
2πr² = 2772
πr² = \(\frac{2772}{2}\) = 1386
T.S.A of the hemisphere = 3πr² sq. units
= 3 × 1386 cm²
= 4158 cm²

Question 28.
Which term of the geometric sequence 5,2, \(\frac{4}{5}, \frac{8}{25} \cdots \text { is } \frac{128}{15625}\) ?
Answer:
The given G.P is 5,2, \(\frac{4}{5}, \frac{8}{25} \cdots \text { is } \frac{128}{15625}\)
Here a = 5, r = \(\frac { 2 }{ 5 }\)
t_n = \(\frac{128}{15625}\)
a.r_n-a = \(\frac{128}{15625}\)

PART – III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
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
Answer:
(i) f = {(2,0)(4, 1)(6,2)(10,4)(12,5)

(ii)

(iii)

(iv)

Question 30.
Given f(x) = x – 1, g(x) = 3x + 1 and h(x) = x² show that (fog)oh = fo(goh)

Question 31.
The product of three consecutive terms of a Geometric Progression is 343 and their sum is \(\frac{91}{3}\) Find the three terms.

Question 32.
Find the sum of all natural numbers between 602 and 902 which are not divisible by 4?

Question 33.
There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. But when first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.

Question 34.
A passenger train takes 1 hr more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of passenger train is less than that of an express train by 20 km per hour. Find the average speed of both the trains.

Question 35.
ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.

Question 36.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).

Question 37.
From the top of a tower 50 m high, the angles of depression of the top and bottom of a tree and observed to be 30° and 45° respectively. Find the height of the tree. ( √3 = 1.732)

Question 38.
A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical 1 vessel.

Question 39.
A box contains cards numbered 3, 5, 7,9,… 35,37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.

Question 40.
The function f: [-7, 6] → R is defined as follows.

Question 41.
The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator determine the fraction.

Question 42.
If for distribution Σx – 7 = 3 ; (Σx – 7)² = 57 and total number of the item is 20; find the mean and standard deviation.

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Draw a circle of diameter 6 cm from a point P, which is 8 cm away from its centre. Draw the two tangents PA and PB to the circle and measure their lengths.

[OR]

(b) Construct a triangle ∆PQR such that QR = 5 cm, ∠P = 30° and the altitude from P to QR is of length 4.2 cm.

Question 44.
(a) Draw the graph of y = x² and hence solve x² – 4x – 5 = 0.

[OR]

(b) Draw the graph of y = x² + 3x + 2 and use it to solve the equation x² + 2x + 4 = 0.