Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

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Tamilnadu 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
Max Marks: 100

PART – 1

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

Question 1.
If there are 10 relations from a set A = {1, 2, 3,4, 5} to a set B, then the number of elements in B is ………….. .
(1) 3
(2) 2
(3) 4
(4) 8
Answer:
(2) 2

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.
If 7^4k = ………….. (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1

Question 4.
The next term ot 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.
Find the matrix X if 2X + \(\left[ \begin{matrix} 1 & 3 \\ 5 & 7 \end{matrix} \right] =\left[ \begin{matrix} 5 & 7 \\ 9 & 5 \end{matrix} \right]\) ………….. .
(1) \(\left[ \begin{matrix} -2 & -2 \\ 2 & -1 \end{matrix} \right] \)
(2) \(\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right] \)
(3) \(\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix} \right] \)
(4) \(\left[ \begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix} \right] \)
Answer:
(2) \(\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right] \)

Question 7.
The two tangents from an external points P to a circle with centre at O are PA and PB. If ∠APB = 70° then the value of ∠AOB is ………….. .
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer:
(2) 110°

Question 8.
The equation of a line passing through the origin and perpendicular to the line 7x – 3y + 4 = 0 is ………….. .
(1)7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0

Question 9.
If x = a tan θ and y = b sec θ then ………….. .
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
(4) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)

Question 10.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is ………….. .
(1) 1:2:3
(2) 2:1:3
(3) 1:3:2
(4) 3:1:2
Answer:
(4) 3:1:2

Question 11.
The probability of getting a job for a person is \(\frac { x }{ 3 }\). If the probability of not getting the job is \(\frac { 2 }{ 3 }\) then the value of x is ………….. .
(1) 2
(2) 1
(3) 3
(4) 1.5
Answer:
(2) 1

Question 12.
Variance of the first 11 natural numbers is ………….. .
(1) √5
(2) √10
(3) 5√2
(4) 10
Answer:
(4) 10

Question 13.
If α and β are the roots of the equation ax² + bx + c = 0 then (α + β)² is ………….. .
(1) \(-\frac{b^{2}}{a^{2}}\)
(2) \(\frac{c^{2}}{a^{2}}\)
(3) \(\frac{b^{2}}{a^{2}}\)
(4) \(\frac{b c}{a}\)
Answer:
(3) \(\frac{b^{2}}{a^{2}}\)

Question 14.
If K(x) = 3x – 9 and L(x) = 7x – 10 then Lok is ………….. .
(1) 21x + 73
(2) -21x + 73
(3) 21x – 73
(4) 22x – 73
Answer:
(3) 21x – 73

PART – II

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

Question 15.
Let A = {1,2,3,4,…, 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.
Answer:
A = {1,2,3,4. . . .45} .
The relation is defined as “is square of’
R = {(1, 1) (2, 4) (3, 9) (4, 16) (5, 25) (6, 36)} .
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1,4,9,16,25,36}

Question 16.
If f(x) = 3x – 2, g(x) = 2x + k and if fog = gof, then find the value of k.
Answer:
f(x) = 3x – 2, g(x) = 2x + k
fog(x) = f(g(x)) = f(2x + k) = 3(2x + k) – 2 = 6x + 3k – 2
Thus, fog(x) = 6x + 3k – 2.
gof(x) = g(3x – 2) = 2(3x – 2) + k
Thus, gof(x) = 6x – 4 + k.
Given that fog = gof
Therefore, 6x + 3k – 2 = 6x – 4 + k
6x – 6x + 3k – k = – 4 + 2 ⇒ k = – 1

Question 17.
Find the rational form of the number \(0 . \overline{123}\).
Answer:
Letx = \(0 . \overline{123}\)
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P
Here a = 0.123, r = \(\frac{0.000123}{0.123}\) = 0.001
S_n = \(\frac{a}{1-r}=\frac{0.123}{1-0.001}=\frac{0.123}{0.999}=. \frac{41}{333}\)

Question 18.
How many consecutive odd integers beginning with 5 will sum to 480?
Answer:
First term (a) = 5
Common difference (d) = 2 (consecutive odd integer)
S_n = 480

\(\frac { n }{ 2 }\) [2a + (n – 1)d] = 480
\(\frac { n }{ 2 }\) [10 + (n – 1)2] = 480
\(\frac { n }{ 2 }\) [10 + 2n – 2] = 480
\(\frac { n }{ 2 }\) (8 + 2 n) = 480
n( 4 + n) = 480
4n + n² – 480 = 0
n² + 4n – 480 = 0
(n + 24) (n – 20) = 0
n + 24 = 0 or n – 20 = 0
n = -24 or n = 20 [number of terms cannot be negative]
∴ Number of consecutive odd integers is 20

Question 19.
Simplify \(\frac{5 t^{3}}{4 t-8} \times \frac{6 t-12}{10 t}\)
Answer:

Question 20.
Solve the following quadratic equations by factorization method √2x² + 7x + 5 √2 = 0
Answer:
√2x² + 7x + 5√2 = 0
√2x² + 2x + 5x + 5√2 = 0
√2x(x + √2) + 5(x + √2) = 0
(x + √2) + 5(x + √2) = 0

(x + √2) or 5(x + √2) = 0 (equate the product of factors to zero)
x = -√2 or √2x = -5 x ⇒ x = \(\frac{-5}{\sqrt{2}}\)
The roots are -√2, \(\frac{-5}{\sqrt{2}}\)

Question 21.
Find the value of a, b, c, d from the equation \(\begin{pmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{pmatrix}=\begin{pmatrix} 1 & 5 \\ 0 & 2 \end{pmatrix}\)
Answer:
The given matrices are equal. Thus all corresponding elements are equal.
Therefore, a – b = 1 …(1)
2a + c = 5 …(2)
2a – b = 0 ….(3)
3c + d = 2 …(4)
(3) gives 2a – b = 0 …(4)
2 a = b …(5)
Put 2a = b in equation (1), a – 2a = 1 gives a = -1
Put a = -1 in equation (5), 2(-1) = b gives b = -2 .
Put a = -1 in equation (2), 2(-1) + c = 5 gives c = 7
Put c = 7 in equation (4), 3(7) + d = 2 gives = -19
Therefore, a = -1, b = -2, c = 7, d = -19

Question 22.
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC If AD = 8x -7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.
Answer:

Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{8 x-7}{5 x-3}=\frac{4 x-3}{3 x-1}\)
(8x – 7) (3x -1) = (4x -3) (5x -3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 20x² – 29x + 27x + 7 – 9 = 0
4x² – 2x – 2 = 0 .
2x² – x – 1 = 0 (Divided by 2)
2x² – 2x + x – 1 = 0

2x(x – 1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = -1 ⇒ x = \(\frac { 1 }{ 2 }\) (Negative value will be omitted)
The value of x = 1

Question 23.
The hill in the form of a right triangle has its foot at (19, 3) . The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Answer:

Slope of AB (m) = tan 45° = 1
Equation of the hill joining the foot and the top is 45°
y – y₁ = m(x – x₁
y – 3 = 1(x – 19)
y – 3 = x – 19
– x + y – 3 + 19 = 0
– x + y + 16 = 0
x – y – 16=0
The required equation is x – y – 16 = 0

Question 24.
Prove that \(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}\) = sec θ + tan θ
Answer:

Question 25.
If the total surface area of a cone of radius 7cm is 704 cm², then find its slant height.
Answer:
Given that, radius r = 7 cm
Now, total surface area of the cone = πr(l + r)sq. units
T.S.A = 704 cm²
704 = \(\frac { 22 }{ 7 }\) × 7(l + 7)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

Question 26.
The first term of an A.P is 6 and the common difference is 5. Find the A.P and its general term.
Answer:
Given a = 6, d = 5
General term t_n = a + (n – 1) d
= 6 + (n – 1)5
= 6 + 5n – 5
= 5n + 1
The general form of the A.P is a, a + d, a + 2d ………
The A.P. is 6, 11, 16, 21 …. 5n + 1

Question 27.
If θ is an acute angle and tan θ + cot θ = 2 find the value of tan⁷θ + cot⁷θ
Answer:
Given tan θ + cot θ = 2
tan θ + \(\frac{1}{\tan \theta}\) = 2
\(\frac{\tan ^{2} \theta+1}{\tan \theta}\) = 2
tan²θ + 1 = 2 tan θ
tan²θ – 2 tan θ + 1 = 0
(tan θ – 1)² = 0
∴ tanθ – 1 = 0
tanθ = 1
tanθ = tan45 ⇒ θ = 45°
tan⁷θ + cot⁷θ = tan⁷45° + cot ⁷45°
= (1)⁷ + (1)^{7
= 2}

Question 28.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly one card is drawn from this box. Find the probability that the number on the card is a number which is a perfect square.
Answer:
Sample space = {2, 3, 4,… 101}
n(s) = 100
Let A be the event of getting perfect square numbers
A= {4, 9, 16, 25, 36, 49, 64, 81, 100}
n(A) = 9
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{100}\)
Probability of getting a card marked with a number which is a perfect square is \(\frac{9}{100}\)

PART – III

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

Question 29.
Consider the functions f(x) = x², g(x) = 2x and h(x) = x + 4 Show that (fog)oh = fo(goh)
Answer:
f(x) = x² ; g(x) = 2x and h(x) = x + 4
(fog) x = f[g(x)]
= f( 2X)
= (2x)²
= 4x²
(fog) oh (x) = fog [h(x)]
= fog(x + 4)
= 4(x + 4)²
= 4[x² + 8x + 16]
= 4x² + 32x + 64 ….(1)
goh (x) = g[h(x)]
= g(x + 4)
= 2(x + 4)
= 2x + 8
fo(goh)x = fo[goh(x)}
= f[2x + 8}
= (2x + 8)²
= 4x² + 32x + 64 ……. (2)
From (1) and (2) we get (fog) oh = fo(goh)

Question 30.

(i) f(4)
(ii) f(-2)
(iii) f(4) + 2f(1)
(iv) \(\frac{f(1)-3 f(4)}{f(-3)}\)
Answer:
The function f is defined by three values in intervals I, II, III as shown by the side

For a given value of x = a, find out the interval at which the point a is located, there after find f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x – 2 ; f(4) = 3(4) – 2 = 10

(ii) x = -2 lies in the second interval.
Therefore, f(x) = x² – 2 ; f(-2) = (-2)² – 2 = 2

(iii) From (i), f(4) = 10.
To find f(1), first we see that x = 1 lies in the second interval.
Therefore,f(x) = x² – 2 => f(1) = 1² – 2 = -1
So, f(4) + 2f(1) = 10 + 2 (-1) = 8

(iv) We know that f(1) = -1 and f(4) = 10.
For finding f(-3), we see that x = -3 , lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, \(\frac{f(1)-3 f(4)}{f(-3)}=\frac{-1-3(10)}{1}=-31\)

Question 31.
If (m + 1)^th term of an A.P. is twice the (n + 1)^th> term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Answer:
t_n = a + (n – 1)d
Given t_m+1 = 2 t_n+1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md = 2a + 2nd
md – 2nd = a
d(m – 2n) = a ……. (1)
To prove that t_{3m + 1} = 2(t_{3m + n + 1})
L.H.S. = t_{3m + 1}
= a + (3m + 1 – 1 )d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)

R.H.S. = 2 (t_{m + n + 1})
= 2[a + (m + n + 1 – 1) d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2md – nd] = 2d (2m – n)
R.H.S = L.H.S
∴ t_{(3m + 1)}= 2t_{(m + n + 1)}

Question 32.
Find the sum to n terms of the series 5 + 55 + 555 + ….
Answer:
The series is neither Arithmetic nor Geometric series. So it can be split into two series and then find the sum.
5 + 55 + 555 + …. + n terms = 5[1 + 11 + 111 + …. + n terms]
= \(\frac { 5 }{ 9 }\)[9 + 99 + 999 + …. + « terms]
= \(\frac { 5 }{ 9 }\)[(10 – 1) + (100 – 1) + (1000 – 1) + …. + n terms)]
= \(\frac { 5 }{ 9 }\)[(10 + 100 + 1000 + …. +n terms) -n]
= \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{(10-1)}-n\right]=\frac{50\left(10^{n}-1\right)}{81}-\frac{5 n}{9}\)

Question 33.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Answer:
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
\(\frac{x+y+z}{3}=53\)
x + y + z = 159….(1)
By the given 2nd condition.
\(\frac{1}{2} z+\frac{1}{3} y+\frac{1}{4} x=65\)
Multiply by 12
6z + 4y +3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3^rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
-4x + z = -16 + 4
4x – z = 12 ….(3)

Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-Z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51
Vani age = 24 years
Vani’s father age = 51 years
Vani – grand father age = 84 years

Question 34.
If A = \(\left( \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right)\) and B = \(\left( \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right)\) show that (AB)^T = B^TA^T
Answer:

Question 35.
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.
Answer:

In the right ∆ OTP,
PT² = OT² – OP²
= 13² – 5²
= 169 – 25 = 144
PT = √144 = 12 cm
Since lengths of tangent drawn from a point to circle are equal.
∴ AP = AE = x .
AT = PT – AP
= (12 – x) cm
Since AB is the tangent to the circle E.
∴ OE ⊥ AB .
∠OEA= 90°
∠AET = 90°
In ∆AET, AT² = AE² + ET²
In the right triangle AET,
AT² = AE² + ET²
(12 – x)² = x² + (13 – 5)²
144 – 24x + x² = x² + 64
24x = 80 ⇒ x = \(\frac{80}{24}=\frac{20}{6}=\frac{10}{3}\)
BE = \(\frac{10}{3}\) cm
AB = AE + BE
= \(\frac{10}{3}+\frac{10}{3}=\frac{20}{3}\)
∴ Length of AB = \(\frac{20}{3}\) cm

Question 36.
Find the area of the quadrilateral whose vertices are at (-9, 0), (-8, 6), (-1, -2) and (-6,-3)
Answer:

Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB
= \(\frac { 1 }{ 2 }\) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄ )]

= \(\frac { 1 }{ 2 }\) [27 + 12- 6 + 0 -(0 + 3 + 16 – 54)]
= \(\frac { 1 }{ 2 }\) [33 -(-35)]
= \(\frac { 1 }{ 2 }\) [33 + 35] = \(\frac { 1 }{ 2 }\) × 68 = 34 sq. units.
Area of the Quadrilateral = 34 sq. units

Question 37.
A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression to the point ‘A’ from the top of the tower is 45°. Find the height of the tower. (√3 = 1.732)
Answer:
Let BC be the height of the tower and CD be the height of the pole.
Let‘A’be the point of observation.
Let BC = x and AB = y.
From the diagram,

∠BAD = 60° and ∠XCA = 45° = ∠BAC ,
In right triangle ABC, tan 45° = \(\frac{B C}{A B}\)
gives 1 = \(\frac { x }{ y }\) so, x = y …… (1)
In right triangle ABD, tan60° = \(\frac{B D}{A B}=\frac{B C+C D}{A B}\)
gives √3 = \(\) so, √3y = x + 5
we get √3x = x + 5 [From (1)]

Hence, height of the tower is 6.83 m.

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.
Answer:

Radius of the lower end of the frustum (r) = 1 cm
Radius of the upper end of the frustum (R) = 2.5 cm
Height of the frustum (h) = 6 cm
Let “l” be the slant height of the frustum
l = \(\sqrt{h^{2}+(\mathrm{R}-r)^{2}}\)
= \(\sqrt{6^{2}+(2.5-1)^{2}}\)
= \(\sqrt{36+2.25}\) = \(\sqrt{38.25}\)
= 6.18 cm
External surface area of shuttle cock = C.S.Aof the frustum + C.S.Aof a hemisphere
= πl(R + r) + 2 πr²
= π [6.18 (2.5 + 1) + 2 × 1²] cm²
= \(\frac { 22 }{ 7 }\)[6.18 × 3.5 + 2]
= \(\frac { 22 }{ 7 }\) × (21.63 + 2)
= \(\frac { 22 }{ 7 }\) × 23.63 cm² = 74.26 cm²
External surface area = 74.26 cm²

Question 39.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4,10,12 and 14, then find the remaining two observations.
Answer:
Let the missing two observation be ‘a’ and ‘b’
Arithmetic mean = 8

\(\frac{2+4+10+12+14+a+b}{7}\) = 8 ⇒ \(\frac{42+a+b}{7}\) = 8
a + b + 42 = 56
a + b = 56 – 42
a + b = 14 ……… (1)
Variance = 16
Variance = \(\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}\)

560 – 460 = a² + b²
a² + b² = 100 ⇒ (a + b)² – 2ab = 100 [a² + b² = (a+b)² – 2 ab]
14² – 2 ab = 100 ⇒ 196 – 2 ab = 100 [a + b = 14(from (1)]
196 – 100 = 2ab
96 = 2ab ⇒ ab = \(\frac{96}{2}\) = 48
∴ b = \(\frac{48}{a}\) …… (2)
Substitute thr value of b = \(\frac{48}{a}\) in (1)
a + \(\frac{48}{a}\) = 14 ⇒ a² + 48 = 14a
a² – 14a + 48 = 0 ⇒ (a – 6) (a – 8) = 0
a = 6 or 8
when a = 6
b = \(\frac{48}{a}=\frac{48}{6}=8\) = 8

when a = 8
b = \(\frac{48}{a}=\frac{48}{6}=8\) = 6

∴ Missing observation is 8 and 6 (or) 6 and 8

Question 40.
From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find (i) the volume of the remaining solid also, find (ii) the whole surface area.
Answer:
Circular cylinder
Radius of the base (r) = 6 cm
Height of the cylinder (h) = 10 cm

Circular Cone
Radius of the base (R) = 6 cm
Height of the cone (H) = 10 cm

(i) Volume of the remaining solid = Volume of the cylinder – Volume of the cone

= πr²h – \(\frac { 1 }{ 3 }\)πR²H
= π(r²h – \(\frac { 1 }{ 3 }\)R²H)
= \(\frac { 22 }{ 7 }\) [6 × 6 × 10 – \(\frac { 1 }{ 3 }\) × 6 × 6 × 10] cm²
= \(\frac { 22 }{ 7 }\) [360 – 120] cm³
= \(\frac { 22 }{ 7 }\) × 240 cm³
= \(\frac { 5280 }{ 7 }\) cm³ = 754.29 cm³

(ii) Slant height of a cone = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{10^{2}+6^{2}}\)
= \(\sqrt{100+36}\)
= \(\sqrt{136}\) = 11.66 cm
Whole surface area of the solid = curved surface area of the cylinder + curved surface area of the cone + base area
= 2 πrh + πRI + πr²
= π[2 × 6 × 10 + 6 × 11.66 + 6 × 6]
= \(\frac { 22 }{ 7 }\)[120 + 69.96 + 36] cm²
= \(\frac { 22 }{ 7 }\) × 225.96 cm²
= \(\frac { 4971.12 }{ 7 }\) cm²
= 710.16 cm²
(i) Volume of the remaining solid = 754.29 cm³
(ii) whole surface area = 710.16 cm²

Question 41.
If a and p are the roots of the equation 3x² – 6x + 1=0 form the equation whose roots are 2α + β and 2β + α
Answer:
α and β are the roots of 3x² – 6x + 1 = 0
α + β = \(\frac { 6 }{ 3 }\) = 2
αβ = \(\frac { 1 }{ 3 }\)
Given the roots are 2α + β; 2β + α
Sum of the roots = 2α + β + 2β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6
Product of roots = (2α + β) (2β + α)
= 4αβ + 2α² + 2β² + αβ
= 5αβ + 2(α² + β²)
5αβ + 2[(α + β)² – 2αβ]
= 5 × \(\frac { 1 }{ 3 }\) + 2 (4 – 2 × \(\frac { 1 }{ 3 }\))
= \(\frac { 5 }{ 3 }\) + \(2\left(\frac{12-2}{3}\right)\)
= \(\frac { 5 }{ 3 }\) + 2 × \(\frac { 10 }{ 3 }\)
= \(\frac{5}{3}+\frac{20}{3}\)
= \(\frac { 25 }{ 3 }\)
The quadratic polynomial is x² – (sum of the roots)x – product of the roots = 0
x² – 6x + \(\frac { 25 }{ 3 }\) = 0
3x² – 18x + 25 = 0

Question 42.
Tw o dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.
Answer:
Sample space = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3.1) , (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5.1) , (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S) = 36
Let A be the event of getting the sum is divisible by 3 ‘
A = { (1,2) (2,1) (1,5) (5,1) (2,4) (4,2) (3,3) (3,6) (6,3) (4,5) (5,4) (6,6)}
n (A) = 12
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{12}{36}\)
Let B be the event of getting a sum is divisible by 4.
B = {(1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6,6)} n (B) = 9
n(B) = 9
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{9}{36}\)
A ∩ B = {(6,6)}
n(A ∩ B) = 1
P(A ∩ B) = \(\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{1}{36}\)
P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
\(=\frac{12}{36}+\frac{9}{36}-\frac{1}{36}\)
\(=\frac{12+9-1}{36}=\frac{20}{36}\)
Neither divisible by 3 nor by 4
P(A’ ∩ B’) = P(A ∪ B)’
= 1 – p(A ∪ B) = 1 – \(\frac{20}{36}=\frac{36-20}{36}\)
= \(\frac{16}{36}=\frac{4}{9}\)

PART-IV

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

Question 43.
(a) Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer:


Steps of construction:

1. With. O as centre, draw a circle of radius 5 cm.
2. Draw a line OP = 10 cm.
3. Draw a perpendicular bisector of OP, which cuts OP at M.
4. With M as centre and MO as radius draw a circle which cuts previous circle at A and B.
5. Join AP and BP. AP and BP are the required tangents.

Verification: In the right ∆ OAP
PA² = OP² – OA²
= 10² – 5² = \(\sqrt{100-25}\) = √75 = 8.7 cm.
Length of the tangent is 8.7 cm

[OR]

(b) Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer:


Steps of construction

1. Draw a line segment PQ = 4.5 cm
2. At P, draw PE such that ∠QPE = 60°
3. At P, draw PF such that ∠EPF = 90°
4. Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
5. With O as centre and OP as radius draw a circle.
6. From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
7. Join PR and RQ. PQR is the required triangle.
8. 

Question 44.
(a) Draw the graph of y = x² and hence solve x² – 4x – 5 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (-4, 16), (- 3, 9), (-2,4), (-1, 1), (0, 0), (1, 1), (2,4), (3, 9), (4, 16), (5, 25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations


(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.

[OR]

(b) Draw the graph of y – x² -5x – 6 and hence solve x² – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5, -6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7