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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 x^{2} – 2x – 24 and x^{2} – 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 ax^{2} + 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 + n^{2} – 480 = 0

n^{2} + 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 √2x^{2} + 7x + 5 √2 = 0

Answer:

√2x^{2} + 7x + 5√2 = 0

√2x^{2} + 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)

24x^{2} – 8x – 21x + 7 = 20x^{2} – 12x – 15x + 9

24x^{2} – 20x^{2} – 29x + 27x + 7 – 9 = 0

4x^{2} – 2x – 2 = 0 .

2x^{2} – x – 1 = 0 (Divided by 2)

2x^{2} – 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 – y_{1} = m(x – x_{1}

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 cm^{2}, 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 cm^{2}

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 tan^{7}θ + cot^{7}θ

Answer:

Given tan θ + cot θ = 2

tan θ + \(\frac{1}{\tan \theta}\) = 2

\(\frac{\tan ^{2} \theta+1}{\tan \theta}\) = 2

tan^{2}θ + 1 = 2 tan θ

tan^{2}θ – 2 tan θ + 1 = 0

(tan θ – 1)^{2} = 0

∴ tanθ – 1 = 0

tanθ = 1

tanθ = tan45 ⇒ θ = 45°

tan^{7}θ + cot^{7}θ = tan^{7}45° + cot ^{7}45°

= (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) = x^{2}, g(x) = 2x and h(x) = x + 4 Show that (fog)oh = fo(goh)

Answer:

f(x) = x^{2} ; g(x) = 2x and h(x) = x + 4

(fog) x = f[g(x)]

= f( 2X)

= (2x)^{2}

= 4x^{2}

(fog) oh (x) = fog [h(x)]

= fog(x + 4)

= 4(x + 4)^{2}

= 4[x^{2} + 8x + 16]

= 4x^{2} + 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}

= 4x^{2} + 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) = x^{2} – 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) = x^{2} – 2 => f(1) = 1^{2} – 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^{T}A^{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,

PT^{2} = OT^{2} – OP^{2}

= 13^{2} – 5^{2}

= 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, AT^{2} = AE^{2} + ET^{2}

In the right triangle AET,

AT^{2} = AE^{2} + ET^{2}

(12 – x)^{2} = x^{2} + (13 – 5)^{2}

144 – 24x + x^{2} = x^{2} + 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

= \(\frac { 1 }{ 2 }\) [(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4} )]

= \(\frac { 1 }{ 2 }\) [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 πr^{2}

= π [6.18 (2.5 + 1) + 2 × 1^{2}] cm^{2}

= \(\frac { 22 }{ 7 }\)[6.18 × 3.5 + 2]

= \(\frac { 22 }{ 7 }\) × (21.63 + 2)

= \(\frac { 22 }{ 7 }\) × 23.63 cm^{2} = 74.26 cm^{2}

External surface area = 74.26 cm^{2}

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 = a^{2} + b^{2}

a^{2} + b^{2 }= 100 ⇒ (a + b)^{2} – 2ab = 100 [a^{2} + b^{2} = (a+b)^{2} – 2 ab]

14^{2} – 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 ⇒ a^{2} + 48 = 14a

a^{2} – 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

= πr^{2}h – \(\frac { 1 }{ 3 }\)πR^{2}H

= π(r^{2}h – \(\frac { 1 }{ 3 }\)R^{2}H)

= \(\frac { 22 }{ 7 }\) [6 × 6 × 10 – \(\frac { 1 }{ 3 }\) × 6 × 6 × 10] cm^{2}

= \(\frac { 22 }{ 7 }\) [360 – 120] cm^{3}

= \(\frac { 22 }{ 7 }\) × 240 cm^{3}

= \(\frac { 5280 }{ 7 }\) cm^{3} = 754.29 cm^{3}

(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 + πr^{2}

= π[2 × 6 × 10 + 6 × 11.66 + 6 × 6]

= \(\frac { 22 }{ 7 }\)[120 + 69.96 + 36] cm^{2}

= \(\frac { 22 }{ 7 }\) × 225.96 cm^{2}

= \(\frac { 4971.12 }{ 7 }\) cm^{2}

= 710.16 cm^{2}

(i) Volume of the remaining solid = 754.29 cm^{3}

(ii) whole surface area = 710.16 cm^{2}

Question 41.

If a and p are the roots of the equation 3x^{2} – 6x + 1=0 form the equation whose roots are 2α + β and 2β + α

Answer:

α and β are the roots of 3x^{2} – 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 + 2α^{2} + 2β^{2} + αβ

= 5αβ + 2(α^{2} + β^{2})

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 x^{2} – (sum of the roots)x – product of the roots = 0

x^{2} – 6x + \(\frac { 25 }{ 3 }\) = 0

3x^{2} – 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:

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

Verification: In the right ∆ OAP

PA^{2} = OP^{2} – OA^{2}

= 10^{2} – 5^{2} = \(\sqrt{100-25}\) = √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

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

Question 44.

(a) Draw the graph of y = x^{2} and hence solve x^{2} – 4x – 5 = 0.

Answer:

Given equations are y = x^{2} and x^{2} – 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 – x^{2} -5x – 6 and hence solve x^{2} – 5x – 14 = 0

Answer:

Let y = x^{2} – 5x – 6

(i) Draw the graph of y = x^{2} – 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 x^{2} – 5x – 14 = 0, subtract x^{2} – 5x – 14 = 0 from y = x^{2} – 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