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## Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 1 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 – I

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

Question 1.

If {(a, 8),(6, b)} represents an identity function, then the value of a and b are respectively ……………. .

(1) (8,6)

(2) (8,8)

(3) (6,8)

(4) (6,6) .

Answer:

(1) (8,6)

Question 2.

If the H.C.F of 65 and 117 is expressible in the form of 65m -117, then the value of m is ……………. .

(1) 4

(2) 2

(3) 1

(4) 3

Answer:

(2) 2

Question 3.

The next term ol 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 4.

Which of the following should be added to make x^{4} + 64 a perfect square ……………… .

(1) 4x^{2}

(2) 16x^{2}

(3) 8x^{2}

(4) -8x^{2}

Answer:

(2) 16x^{2}

Question 5.

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 6.

Two poles of heights 6 m and 11 m stand vertically on a plane ground if the distance between their feet is 12m, what is the distance between their tops?

(1) 13 m

(2) 14 m

(3) 15 m

(4) 12.8 m

Answer:

(1) 13 m

Question 7.

If (5, 7), (3,p) and (6, 6) are collinear then the value of p is ……………… .

(1) 3

(2) 6

(3) 9

(4) 12

Answer:

(3) 9

Question 8.

The value of sin^{2}θ + \(\frac{1}{1+\tan ^{2} \theta}\) is equal to …………….. .

(1) tan^{2}θ

(2) 1

(3) cot^{2}θ

(4) 0

Answer:

(2) 1

Question 9.

If the radius of the base of a cone is tripled and the height is doubled then the volume is ……………… .

(1) made 6 times

(2) made 18 times

(3) made 12 times

(4) unchanged

Answer:

(2) made 18 times

Question 10.

Standard deviation of a collection of data is 2√2. If each value is multiplied by 3, then the standard deviation of the new data is

……………… .

(1) √l2

(2) 4√2

(3) 6√2

(4) 9√2

Answer:

(3) 6√2

Question 11.

A page is selected at random from a book. The probability that the digit at unit place of the page number chosen’ is less than 7 is …………. .

(1) \(\frac{3}{10}\)

(2) \(\frac{7}{10}\)

(3) \(\frac{3}{9}\)

(4) \(\frac{7}{9}\)

Answer:

(2) \(\frac{7}{10}\)

Question 12.

If f = {(6, 3), (8, 9), (5, 3), (-1, 6)} then the pre-images of 3 are ……………… .

(1) 5 and -1

(2) 6 and 8

(3) 8 and -1

(4) 6 and 5

Answer:

(4) 6 and 5

Question 13.

If a and p are the zeros 6f the polynomial P(x) = 4x^{2} + 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 14.

The probability that a-leap year will have 53 Fridays or 53 Saturdays is …………….. .

(1) \(\frac{2}{7}\)

(2) \(\frac{1}{7}\)

(3) \(\frac{4}{7}\)

(4) \(\frac{3}{7}\)

Answer:

(4) \(\frac{3}{7}\)

PART – II

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

Question 15.

A relation “f” is defined by f(x) = x^{2} – 2 where x ∈ {-2, -1, 0, 3}

(i) List the elements of f

(ii) If f a function?

Answer:

f(x) = x^{2} – 2 where x ∈ {-2,-1, 0, 3}

(i) f(-2) = (-2)^{2} – 2 = 2 ; f(-1) = (-1)^{2} – 2 = -1

f(0) = (0)^{2} – 2 = -2 ; f(3) = (3)^{2} – 2 = 7

Therefore, f = {(-2,2), (-1,-1), (0,-2), (3,7)}

(ii) We note that each element in the domain of f has a unique image.

Therefore f is a function.

Question 16.

Let f(x) = x^{2} – 1. Find fofof

Answer:

fofof = fof [(x)]

= fof (x^{2} – 1)

= f (x^{2} – 1)^{2} – 1

= f (x^{4} – 2x^{2} + 1 – 1)

= f (x^{4} – 2x^{2})

fofof = (x^{4} – 2x^{2})^{2} – 1

Question 17.

What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?

Answer:

Find the L.C.M of 35, 56, and 91

35 = 5 × 7

56 = 2 × 2 × 2 × 7

91 = 7 × 13

L.C.M = 2^{3} × 5 × 7 × 13

= 3640

Since it leaves remainder 7

The required number = 3640 + 7 = 3647

∴ The smallest number is 3647

Question 18.

In a G.P. 729, 243, 81,… find t_{7}.

Answer:

The G.P. is 729,243,81, …………..

a = 729; r = \(\frac{t_{2}}{t_{1}}=\frac{243}{729}=\frac{81}{243}=\frac{27}{81}=\frac{3}{9}=\frac{1}{3}\)

t_{n} = a r^{n – 1}

t_{7} = 729 \(\left(\frac{1}{3}\right)^{7-1}\)

= \(729\left(\frac{1}{3}\right)^{6}=729 \times \frac{1}{729}=1\)

∴ The 7^{th} term is 1

Question 19.

Find the excluded values of the expression \(\frac{t}{t^{2}-5 t+6}\)

Answer:

The expression \(\frac{t}{t^{2}-5 t+6}\) is undefined

when t^{2} – 5t + 6 = 0

(t – 3) (t – 2) = 0

t – 3 = 0 or t – 2 = 0

t = 3 or t = 2

The excluded values are 2 and 3

Question 20.

Find the square root of 9x^{2} – 24xy + 30xz – 40yz + 25z^{2} + 16y^{2}

Answer:

\(\sqrt{9 x^{2}-24 x y+30 x z-40 y z+25 z^{2}+16 y^{2}}\)

\(\sqrt{(3 x)^{2}+(4 y)^{2}+(5 z)^{2}-2(3 x)(4 y)-2(4 y)(5 z)+2(3 x)(5 z)}\)

\(\sqrt{(3 x-4 y+5 z)^{2}}\) [using (a – b + c)^{2} = a^{2} + b^{2} + c^{2} -2ab -2bc + 2ac]

= |3x – 4y + 5z|

Question 21.

construct a 3 x 3 matrix whose elements are given by a_{ij} = \(\frac{(i+j)^{3}}{3}\)

Answer:

a_{ij} = \(\frac{(i+j)^{3}}{3}\)

Question 22.

If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm^{2} and the area of ∆DEF is 16 cm^{2} and BC = 2.1 cm. Find the length of EF.

Answer:

Given ∆ABC ~ ∆DEF

\(\frac{9}{16}=\frac{(2.1)^{2}}{\mathrm{EF}^{2}}\)

\(\left(\frac{3}{4}\right)^{2}=\left(\frac{2.1}{\mathrm{EF}}\right)^{2}\)

\(\frac{3}{4}=\frac{2.1}{\mathrm{EF}}\)

EF = \(\frac{4 \times 2.1}{3}\) = 2.8 cm

Length of EF = 2.8 cm

Question 23.

If the area of the triangle formed by the vertices (p,p), (5,6), (5, -2) is 32 sq.units. Find the value of ‘p’.

Answer:

Let the vertices be A (p,p), B (5, 6) and C (5, -2)

Area of a triangle = 32 sq. units

\(\frac{1}{2}\)[x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}] = 32

\(\frac{1}{2}\)[(6p – 10 + 5p) – (5p + 30 – 2p)] = 32

\(\frac{1}{2}\)[11p – 10 – 3p – 30] = 32

11p – 10 – 3p – 30 = 64

8p – 40 = 64

8p = 64 + 40 ⇒ 8p = 104

p = \(\frac{104}{8}\) = 13

The value of p = 13

Question 24.

Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 √3 m.

Answer:

Height of the tower (AC) = 10 √3 m

Distance between the base of the tower and point of observation (AB) = 30 m

Let the angle of elevation ∠ABC be θ

In the right ∆ ABC, tan θ = \(\frac{A C}{A B}\)

\(=\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°

∴ Angle of inclination is 30°

Question 25.

Find the diameter of a sphere whose surface area is 154m^{2}.

Answer:

Let r be the radius of the sphere.

Given that, surface area of sphere = 154 m^{2}

4πr^{2} = 154

4 × \(\frac{22}{7}\) × r^{2} = 154

gives r^{2} = \(154 \times \frac{1}{4} \times \frac{7}{22}\)

hence, r^{2} = \(\frac{49}{4}\) we get r = \(\frac{7}{2}\)

Therefore, diameter is 7m

Question 26.

Find the standard deviation and the variance of first 23 natural numbers.

Answer:

Standard deviation of first “n” natural numbers = \(\sqrt{\frac{n^{2}-1}{12}}\)

Standard deviation of first 23 natural numbers = \(\sqrt{\frac{23^{2}-1}{12}}\)

= \(\sqrt{\frac{529 – 1}{12}}\)

\(=\sqrt{\frac{528}{12}}\)

= √44 (variance)

Variance = 44 and standard deviation = 6.63

Question 27.

Form the quadratic equation whose roots are 3 + √7 ; 3 – √7

Answer:

Sum of the roots = 3 + √7 + 3 – √7

= 6

Product of the roots = (3 + √7)(3 – √7)

= 9 – 7 = 2 .

The required equation is

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

x^{2} – (6) x + 2 = 0

x^{2} – 6x + 2 = 0

Question 28.

Two coins are tossed together. What is the probability of getting at most one head?

Answer:

Sample space = {HH, HT, TH, TT}

n(S) = 4

Let A be the event of getting atmost one head A = {HT, TH, TT}

n(A) = 3

P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)

The probability is \(\frac{3}{4}\)

PART – III

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

Question 29.

Answer:

f(x) = 6x + 1 ; x = {-5, -4, -3, -2, -1, 0, 1}

f(x) = 5x^{2} – 1 ; x = {2,3,4,5}

f(x) = 3x – 4; x = {6,7,8,9}

(i) f(-3) + f(2)

f(x) = 6x+1

f(3) = 6(-3) + 1 = -18 + 1 = -17

f(x) = 5x^{2} – 1

f(2) = 5(2)^{2} – 1 = 20 – 1 = 19

f(-3) + f(2) = -17 + 19 = 2

(ii) f(7) – f(1)

f(x) = 3x – 4

f(7) = 3(7) – 4= 21 – 4 = 17

f(x) = 6x + 1

f(1) = 6(1)+ 1 = 6 + 1 = 7

f(7) – f(1) = 17 – 7

= 10

(iii) 2f(4) + f(8)

f(x) = 5x^{2} – 1

f(4) = 5(4)^{2} – 1 = 15(16) – 1

= 80 – 1 = 79

f(x) = 3x – 4

f(8) = 3(8) – 4 = 24 – 4 = 20

2f(4) + f(8) = 2(79) + 20

= 158 + 20 .

= 178

(iv) \(\frac{2 f(-2)-f(6)}{f(4)+f(-2)}\)

f(x) = 6x + 1

f(-2) = 6(-2) + 1 = -12 + 1 = -11

f(x) = 3x – 4

f(6) = 3(6) – 4 = 18 – 4 = 14

f(x) = 5x^{2} – 1

f(4) = 5(4)^{2} – 1 = 5(16) – 1

= 80 – 1 = 79

f(x) = 6x+1

f(-2) = 6(-2) + 1 = -12 + 1 = -11

\(\frac{2 f(-2)-f(6)}{f(4)+f(-2)}\) = \(\frac{2(-11)-14}{79-11}\)

= \(\frac{-22-14}{68}=\frac{-36}{68}=\frac{-9}{17}\)

Question 30.

If f(x) = 2x + 3, g(x) = 1 – 2x and h(x) = 3x Prove that fo(goh) = (fog)oh

Answer:

f(x) = 2x + 3, g(x) = 1 – 2x, h(x) = 3x

Now, (fog)(x) = f(g(x)) = f(1 – 2x) = 2(1 – 2x) + 3 = 5 – 4x

Then, (fog)oh(x) = (fog) (h(x)) = (fog)(3x) = 5 – 4(3x) = 5 – 12x ….(1)

(goh)(x) = g(h(x)) = g(3x) = 1 – 2(3x) = 1 – 6x

So, fo (goh)(x) = f(1 – 6x) = 2(1 – 6x) + 3 = 5 – 12x …….. (2)

From (1) and (2), we get (fog)oh = fo(goh)

Question 31.

If S_{1} , S_{2} , S_{3}, ……. S_{m}are the sums of n terms of m A.P.,s whose first terms are 1,2, 3, …….. m and whose common differences are 1,3,5,….(2m – 1) respectively, then show that (S_{1} + S_{2} + S_{3} + …….. + Sm) = \(\frac { 1 }{ 2 }\) mn(mn + 1)

Answer:

First terms of an A.P and 2, 3 ……….. m

The common difference are 1, 3, 5 , ………….. (2m – 1)

S_{n} = \(\frac { n }{ 2 }\)[2a+(n – 1)d]

S_{1} = \(\frac { n }{ 2 }\)[2+(n – 1)(1)]

= \(\frac { n }{ 2 }\)[2 + n – 1]

S_{1} = \(\frac { n }{ 2 }\)[n + 1] …….. (1)

S_{2} = \(\frac { n }{ 2 }\)[2(2) + (n – 1)3]

= \(\frac { n }{ 2 }\)[4 + 3n – 3]

S_{2} = \(\frac { n }{ 2 }\)(3n – 2) ……. (2)

S_{3} = \(\frac { n }{ 2 }\)[2(3) + (n – 1)5]

= \(\frac { n }{ 2 }\)[6 + 5n – 5}

= \(\frac { n }{ 2 }\)[5n + 1] …….. (3)

S_{m} = \(\frac { n }{ 2 }\)[2m + (n – 1)(2m – 1)]

= \(\frac { n }{ 2 }\)[2m + 2mn – n – m – 2m + 1]

= \(\frac { n }{ 2 }\)[n(2m – 1) + 1]

By adding (1) (2) (3) we get

S_{1} + S_{2} + S_{3} + ……….. + S_{m} = \(\frac { n }{ 2 }\)(n + 1) + \(\frac { n }{ 2 }\)(3n + 1) + \(\frac { n }{ 2 }\)(5n + 1)+ …………+ \(\frac { n }{ 2 }\)[n(2m – 1) + 1]

= \(\frac { n }{ 2 }\)[n + 1 + 3n + 1 + 5n + 1 …………… + n(2m – 1) + m]

= \(\frac { n }{ 2 }\)[n + 3n + 5n + ………….. + n(2m – 1) + m]

= \(\frac { n }{ 2 }\)[n\((\frac { m }{ 2 })\)(2m) + m]

= \(\frac { n }{ 2 }\)[nm^{2} + m]

S_{1} + S_{2} + S_{3} + ……… + S_{m} = \(\frac { mn }{ 2 }\)[nm + 1]

Hint:

1 + 3 + 5 + …….. + 2m – 1

S_{n} = \(\frac { n }{ 2 }\)(a + 1)

= \(\frac { m }{ 2 }\) (1 + 2m – 1)

= \(\frac { m }{ 2 }\) (2m)

Question 32.

Find the sum to n terms of the series 0.4 + 0.44 + 0.444 + ………. to n terms

Answer:

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 2^{nd} 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 3rd 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.

Find the values of a and b if the following polynomials are perfect squares. ax^{4} + bx^{3} + 361x^{2} + 220.x + 100

Answer:

Re-arrange the order we get .

100 + 220x + 361x^{2} + by^{3} + ax^{4}

Since it is a perfect square

b – 264 = 0

b = 264

a – 144 = 0

a = 144

∴ The value of a = 144 and b = 264

Question 35.

State and prove Pythagoras Theorem

Answer:

Statement

In a right angle triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Proof

Given : In ∆ABC, ∠A = 90°.

To prove : AB^{2} + AC^{2} = BC^{2}

Construction : Draw AD ⊥ BC

Adding (1) and (2) we get

AB^{2} + AC^{2} = BC × BD + BC × DC

= BC (BD + DC) = BC × BC

AB^{2} + AC^{2} = BC^{2}

Hence the theorem is proved.

Question 36.

Find the equation of the median and altitude of AABC through A where the vertices are A(6, 2), B(-5, -1) and C(1,9).

Answer:

(i) To find median

Mid point of BC(D) = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

\(=\left(\frac{-5+1}{2}, \frac{-1+9}{2}\right)\)

\(=\left(\frac{-4}{2}, \frac{8}{2}\right)\) = (-2, 4)

Equation of the median AD is

\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)

\(\frac{y-2}{4-2}=\frac{x-6}{-2-6}\)

\(\frac{y-2}{2}=\frac{x-6}{-8}\)

= 2(x – 6) = 8(y – 2)

2x – 12 = -8y + 16

2x + 8y – 28 = 0

(÷by 2) x + 4y – 14 = 0

(ii) To find the equation of the altitude

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(=\frac{9+1}{1+5}\)

\(=\frac{10}{6}=\frac{5}{3}\)

Slope of the altitude = \(-\frac{3}{5}\)

Equation of the altitude AD is

y – y_{1} = m(x – x_{1})

y – 2 = \(-\frac{3}{5}\)(x – 6)

-3 (x – 6) = 5 (y – 2)

-3x + 18 = 5y – 10

-3x – 5y + 18 + 10 = 0

-3x – 5y + 28 =0

3x + 5y – 28 = 0

∴ Equation of the median is x + 4y – 14 = 0

Equation of the altitude is 3x + 57 – 28 = 0

Question 37.

Two ships are sailing in the sea on either sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the ships are 30° and 45° respectively. If the lighthouse is 200 m high, find the distance between the two ships. (√3 = 1.732 )

Answer:

Let AB be the lighthouse. Let C and D be the positions of the two ships.

Then, AB = 200 m

∠ACB = 30°, ∠ADB = 45°

In right triangle BAC, tan 30° = \(\frac{A B}{A C}\)

\(\frac{1}{\sqrt{3}}=\frac{200}{\mathrm{AC}}\) gives AC = 200√3

In right triangle BAD, tan 45° = \(\frac{A B}{A D}[latex]

1 = [latex]\frac{200}{\mathrm{AD}}\) gives AD = 200

Now, CD = AC + AD = 200√3 + 200 [by (1) and (2)]

CD = 200 (√3 + 1) = 200 × 2.732 = 546.4

Distance between two ships is 546.4 m

Question 38.

Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Answer:

Radius of the cylinder = \(\frac{4.5}{2} \mathrm{cm}\)

Height of the cylinder = 10 cm

Volume of the cylinder = π r^{2} h cu. units

Radius of the coin (r) .= \(\frac{1.5}{2} \mathrm{cm}\)

Thickness of the coin (h) = 2 mm = \(\frac{2}{10} \mathrm{cm}\)

Volume of one coin = π r^{2} h cu. units

Number of coins = 450

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.

Answer:

Sample space = {3, 5, 7,9 ,… ,35, 37} \(\left[\frac{37-1}{2}=\frac{36}{2}=18\right]\)

n (S) = 18

Let A be the event of getting a multiple of 7

A = {7,21,35}

n(A) = 3

p(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{18}\)

‘ Let B be the event of getting a prime number

B = {3,5, 7, 11, 13, 17, 19, 23.29,31.37}

n (B) = 11

p(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{11}{18}\)

A∩B = {7}

n(A ∩ B) = 1

P(A ∩ B) = \(\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{1}{18}\)

p(A∪B) = P(A) + P(B) – P(A ∩ B)

\(=\frac{3}{18}+\frac{11}{18}-\frac{1}{18}=\frac{3+11-1}{18}=\frac{13}{18}\)

Probability of getting a muliple of 7 or a priime number = \(\frac{13}{18}\)

Question 40.

Find X and Y if X – Y = \(\left[ \begin{matrix} 2 & 1 \\ 4 & 3 \\ 0 & 6 \end{matrix} \right]\) and X + Y = \(\left[ \begin{matrix} 12 & 13 \\ 6 & 5 \\ 4 & 8 \end{matrix} \right] \)

Answer:

Question 41.

A lead pencil is in the shape of a right circular cylinder. The pencil is 28 cm long and its radius is 3 in. If the led is of radius one mm (1 mm) then find the volume of the wood used in the pencil.

Answer:

Length of a pencil (h) = 28 cm

Radius of a pencil (R) = 3mm \(\left(\frac{3}{10} \mathrm{cm}\right)\)

Radius of a led (r) = 1 mm \(\left(\frac{1}{10} \mathrm{cm}\right)\)

Volume of the wood used = Volume of the pencil – Volume of the led

= πh(R^{2} – r^{2}cu. unit

Volume of the wood used in the pencil = 7.04 cm^{3}

Question 42.

The following table shows the marks obtained by 48 students in a quiz competition in Mathematics. Calculate the variance and standard deviation.

Answer:

Let us form the following table using the given data.

PART – IV

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

Question 43.

(a) Draw a circle of radius 4.5 cm. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem. Rough Diagram

Answer:

Radius of the circle = 4.5 cm

Steps of construction:

- With O as centre, draw a circle of radius 4.5 cm.
- Take a point L on the circle. Through L draw any chord LM.
- Take a point M distinct from L and N on the circle, so that L, M, N are in anti-clockwise direction. Join LN and NM.
- Through “L” draw tangent TT’ such that ∠TLM = ∠MNL.
- TT’ is the required tangent.

[OR]

(b)

Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.

Answer:

Steps of construction

- Draw a line segment RQ = 5 cm.
- At R draw RE such that ∠QRE = 40°.
- At R. draw RF such that ∠ERF = 90°.
- Draw the perpendicular bisector to RQ, which intersects RF at O and RQ at G.
- With O as centre and OP as radius draw a circle.
- From G mark arcs of radius 4.4 cm on the circle. Mark them as P and S.
- Join PR and PQ. Then ∆PQR is the required triangle.
- From P draw’ a line PN which is perpendicular to RQ it meets at N.
- Measure the altitude PN.

PN = 2.2 cm.

Question 44.

(a) Draw the graph of y = x^{2} – 4x + 3 and use it to solve x^{2} – 6x + 9 = 0.

Answer:

Step 1: Draw the graph of y = x^{2} – 4x + 3 by preparing the table of values as below.

Step 2: To solve x^{2} – 6x + 9 = 0, subtract x^{2} – 6x + 9 = 0 from y = – x^{2} – 4x + 3

The equation y = 2x – 6 represent a straight line. Draw the graph of y = 2x – 6 form in the table of values as below.

The line y = 2x – 6 intersect y = x^{2} – 4x + 3 only at one point.

Step 3: Mark the point of intersection of the curve y = x^{2} – 4.r 3 and y = 2x – 6 that is (3,0).

Therefore, the x coordinate 3 is the only solution for the equation x^{2} – 6x + 9 = 0.

[OR]

(b) Draw the Graph of y = x^{2} – 4x + 4

Answer:

Let y = x^{2} – 4x + 4

(i) Prepare the table of values for y = x^{2} – 4x + 4

(ii) Plot the points (-3, 25) (-2, 16) (-1, 9) (0, 4) (1, 1) (2, 0), (3, 1) and (4, 4)

(iii) Join the points by a free hand smooth curve.

(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.

(v) Since there is only one point of intersection with X-axis (2, 0).

The solution set is 2.

∴ The Quadratic equation x^{2} – 4x + 4 = 0 has real and equal roots. 1 English Medium – 33″ width=”573″ height=”72″ />

The line y = 2x – 6 intersect y = x^{2} – 4x + 3 only at one point.

Step 3: Mark the point of intersection of the curve y = x^{2} – 4.r 3 and y = 2x – 6 that is (3,0).

Therefore, the x coordinate 3 is the only solution for the equation x^{2} – 6x + 9 = 0.