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## Karnataka State Syllabus SSLC Maths Model Question Paper 5 with Answers

Time: 3 Hours

Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8)

Question 1.

In the following numbers, irrational number is

a) √16 – √19

b) 3/4

c) 0.3333……

d) 2 + √3

Answer:

d) 2 + √3

The value of 2+√3 is neither recurring nor terminating and cannot be expressed in the form p/q where P ∈ Z, Q∈N.

Question 2.

Is sin A = 1/√2 the magnitude of ∠A is

a) 90°

b)60

c) 30

d) 45°

Answer:

d) 45°

Question 3.

The maximum number of tangents that can be drawn to a circle from an external point is

a) 1

b)2

c) 3

d) 4

Answer:

b) 2

Question 4.

The formula used to find the curved surface are of a cone of radius (r), height (h) and slant height (l) is

a) CSA= πrl

b) CSA=2 π (r+1)

c) CSA=2 π r(r+h)

d) CSA= \(\frac{\pi r^{2} h}{3}\)

Answer:

a) CSA = πrl

Question 5.

If one of the zeros of the polynomial p(x)=x^{2} – x + k is 2 then the value of k is

a) 2

b) -2

c) -6

d) 6

Answer:

b) -2

p(x) = x^{2 }– x + k

0 = 22 – 2 + k

0 = 4 – 2 + k

k = -2

Question 6.

The 10th term of an A.P. 5,9,13,……. is

a) 36

b) 31

c) 41

d) 21

Answer:

c) 41

T_{10} = a + 9d

= 5 + 9 (4)

= 5 + 36 = 41

Question 7.

In the given figure ∆ABC, DE||BC. If DE = 5cm, BC = 8cm and AD = 3.5 cm, then the length of AB is

a) 5.6cm

b) 4.8cm

c) 5.2cm

d) 6.4cm

Answer:

a) 5.6cm

\(\frac{A B}{A D}=\frac{B C}{D E}\)

\(\frac{A B}{3.5}=\frac{8}{5}\)

AB = \(\frac{8}{5}\) × 3.5 = 5.6

Question 8.

The probability of an event ‘E’ is 0.05, then the probability of an event ‘NotE’ is

a) 0.05

b) 0.95

c) 1/0.05

d) 1/0.95

Answer:

b) 0.95

P(E) and P(Ē) are complementary

∴ P(E) + P(Ē) = 1

0.05 + P(Ē) = 1

P(Ē) = 1 – 0.05 = 0.95

II. Answer the following questions: ( 1 × 8 = 8 )

Question 9.

Write the number of zeros of the polynomial p(x)=x^{3} + 2x^{2} + x + 6.

Answer:

The degree of p(x) = 3

∴ There are 3 zeroes.

Question 10.

Write the ‘discrimant’ of the quadratic equation ax^{2 }+ bx + c=0

Answer:

Discriminant of ax^{2} + bx + c = 0 is

∆ = b^{2} – 4ac

Question 11.

If the first term and the common difference of an A.P. are 6 and 5 respectively, find its 3rd term.

Answer:

a=6 d=5

T_{3} = a + 2d

= 6 + 2(5)

= 6 + 10

= 16

Question 12.

State Basic proportionality theorem.

Answer:

Thales Theorem : “In a triangle, if a line is drawn parallel to one of its sides then the other two sides are divided proportionally”.

Question 13.

In Euclid’s division lemma, if a = 3q + r, then write all the possible values of r.

Answer:

a = 3q + r

where a = divident and b is divided where b = 3

∴ The possible remainders are 0,1,2

Question 14.

If Sin θ = \(\frac{3}{5}\) and Cos θ = \(\frac{4}{5}\), find the value of Sin^{2}θ + Cos^{2}θ .

Answer:

sinθ = \(\frac{3}{5}\)

Cos θ = \(\frac{4}{5}\)

Question 15.

Find the value of sin 30°+cos60°.

Answer:

sin 30° =\(\frac{1}{2}\)

cos 60° = \(\frac{1}{2}\)

sin30° + cos60°= \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

Question 16.

A solid piece of iron is in the form of a cuboid of dimensions 10cmx5cmx2cm. find its volume.

Answer:

Volume of a cuboid = l × b × h

= 10 × 5 ×2

= 100 cm^{3}

III. Answer the following: ( 2 x 8 = 16 )

Question 17.

Prove that √2 + √3 is an irrational number.

Let √2 + √3 = a, where a is an Rational number.

√3 = a – √2

Squaring on both sides

(√3)^{2}=(a – √2)^{2}

3=(a)^{2}+(√2)^{2} – 2a(√2)

3 = a^{2} + 2 – 2a √2

2a =a^{2} + 2 – 3

√2 = \(\frac{a^{2}-1}{2 a}\)

It is contrary that \(\frac{a^{2}-1}{2 a}\) is rational and √2 is irrational. Hence our assumption is wrong.

∴ √2 +√3 is an Irrational number.

Question 18.

Solve : 10x + 3y = 75 and 6x – 5y = 11

(10x + 3y = 75) × 5

(6x – 5y = 11) × 3

50x + 15y = 375 .

18x – 15y = 33

68x = 408

x = \(\frac{408}{68}\) = 6

Consider

6x – 5y = 11

6(6) – 5y = 11

36 – 5y =11

36- 11 = 5y

25 = 5y

y = \(\frac{25}{5}\) = 5

y = 5

Question 19.

find the roots of the equation 6x^{2}+7x-10=0

Answer:

By factorisation method

6x^{2}+ 1x – 10 = 0

6x^{2}+ 12x – 5x – 10 = 0

6x(x+2) – 5 (x + 2) = 0 (x+2) (6x – 5) = 0

x + 2 = 0 OR 6x – 5=0

x = -2 or x = \(\frac{5}{6}\)

Question 20.

Find the distance between the points A(8,-3) and B(0,9) by using distance formula.

Answer:

Question 21.

The perimeters of two similar triangles are 25cm and 15cm. If one side of the first trinagle is 9cm, find the corresponding side of the second triangle.

Answer:

∆ ABC ~ ∆ DEF and AB = 9 cm.

Perimeter of ∆ ABC:Perimeter of ∆ DEF= AB: DE

25 : 15 = 9: DE

25 × DE = 15 × 9

DE = \(\frac{15 × 9}{25}=\frac{27}{5}\) = 5.4cm

OR

In the given figure ∆ ABC and ∆ DBC are on the same base BC. AD interesects BC at ‘O’. If AL⊥ BC and DM ⊥ BC,

prove that \(\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DBC}}=\frac{\mathrm{AO}}{\mathrm{DO}}\)

Answer:

In ∆ ALO and ∆ DMO

∴ ∆ ALO and ∆ DMO are equiangular

⇒ ∆ ALO ~ ∆ DMO

⇒ \(\frac{A L}{D A}=\frac{A O}{D O}\)

Hence proved.

Question 22.

Two cubical dice w;hose faces are numbe red 1 to 6 are rolled simultaneously once.

Find the probability that the sum of the two numbers occuring on their top faces is more than 7.

Answer:

If two dices of 6 faces are thrown simultaneously then S = {(1,1) (1,2) (1,3) (1,4) 0,5) (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

A = {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5} (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)}

n(A) = 15

Question 23.

Draw a circle of radius 3cm. Consruct a pair of tangents to it, from a point 8cm away from its center.

Answer:

Question 24.

If cos θ=0.6 show that 5 sin θ-3 tan θ=0

Answer:

Consider 5 sinθ – 3tanθ = θ

5cosθ – 3 = 0

(0.6) – 3 = 0

3 – 3 = 0

0 = 0

L.H.S = R.H.S.

OR

P.T. (sec^{4} θ – sec^{2} θ)=tan^{2} θ +tan^{4}θ

Consider tan^{2}θ + tan^{4}θ

tan^{2}θ(1 + tan^{2}θ)

tan^{2}θ (sec^{2}θ)

(sec^{2}θ – 1) (sec^{2}θ)

sec^{4}θ – sec^{2}θ = LHS

OR

Consider L.H.S = sec^{4} θ – sec^{2}θ = sec^{2}θ (sec^{2}θ – 1)

= (1 + tan^{2}θ)tan^{2}θ

= tan^{2}θ+tan^{4}θ = R.H.S.

IV. Answer the following : ( 3 × 9 = 27 )

Question 25.

The sum of the numerator and the demonimator of a given fraction is 12. ; If 3 is added to its denominator, then the fraction becomes 1/2. Find the given fraction.

Answer:

Let the fraction be x/y

Given, the sum of numerator and denominator = 12

x + y = 12 → (1)

Given : If 3 is added to Denominator of \(\frac{x}{y}=\frac{1}{2}\)

\(\frac{x}{y+3}=\frac{1}{2}\)

-2x = y + 3

2x – y = 3 → (2)

From (1) and (2) by Elimination method

Considcr x + y= 12

5 + y = 12

y = 12 – 5

y = 7

OR

Seven times a two digit number is equal to four times the number obtained by reversing the places of its digits. If the difference between the digits is 3, find the number.

Answer:

Let the numbers be x and y where x > y

The two digit number = 10x + y

The Number obtained by reversing the digits = 10y + x

Given : 1 times a two digit number = 4 times

the number obtained by reversing the digits.

7(10x + y) = 4 (10y + x)

70x +7y = 40y + 4x

70x – 4x + 7y – 40y = 0

66x – 33y =0

Dividing by 33.

2x – y = 0

2x = y

Given : The difference between the digits = 3.

x – y = 3

substitute y=2x in this

x – 2x = 3

-x = 3

x = -3

y = 2x

y = 2(-3)=-6

∴ Two digit number = 10x + y

= 10(-3) + (-6)

=-30 – 6 = -36.

Question 26.

If 3 and -3 are two zeros of the polynomial p(x)=x^{4} + x^{3} – 11x^{2} – 9x + 18, then find the remaining two zeros of the polynomial.

Given Polynomial

P(x) = x^{4} + x^{3} – 11x^{2} – 9x + 18,

It is of degree 4

This has 4 zeroes.

Let the two zeroes given by x = -3 and x = 3

(x – 3)=0, (x + 3) = 0

(x – 3) (x + 3) = 0

x^{2} = 9 = 0

Divide x^{4} + x^{3} – 11x^{2} – 9x + 18 by x^{2} – 9

Factorize the quotient x^{2} + x – 2 to get two more zeroes.

x^{2} + x – 2 = 0

x2 + 2x – x – 2 = 0

x(x + 2) – 1(x + 2)=0

(x + 2) (x – 1) = 0

x+2 = 0 or x – 1 = 0

x = -2 x = 1

∴ The two more roots are – 2 and 1.

Question 27.

The sum of the areas of two squares is 640 m^{2}. If the difference between their perimeters is 64m, then find sides of the square.

Answer:

Let the areas of the two squares of sides be x and y be x^{2} and y^{2}.

Given ;Sum of the areas of two squares = 640m^{2}

∴ x^{2}+y^{2}=640

Perimeters of the two squares with the sides x and y will be 4x and 4y.

The difference of the perimeter = 64 m.

4x – 4y = 64

4(x – y) = 64

x – y = \(\frac{64}{4}\) = 16

x – y= 16

x = 16 + y

Consider, x^{2} + y^{2} = 640

(16 + y)^{2} + y^{2}=640

16^{2} + y2 + 2(16)(y) + y^{2} = 640

2y^{2} + 32y + 256 – 640 = 0

2y^{2} + 32y – 384 = 0

y^{2} + 16y – 192 = 0

y^{2} + 24y – 8y – 192 = 0

y(y + 24) – 8(y + 24)=0

(y+24) (y-8) = 0

y + 24 = 0 OR y – 8 = 0

y = -24 y = 8

But x = 16 + y

x = 16 + 8

x = 24

OR

x = 16 + y

x = 16 – 24

x = -8

∴ The dimensions of the squares are x and y

x = 24 m, y = 8m.

OR

If the roots of the equation

(a^{2} + b^{2}) x^{2} + 2(bc – ad)x + c^{2} + d^{2} = 0 are equal, show that ac+bd=0

Answer:

It is given that the roots of the quadratic equation are equal

∴ ∆ =0

b^{2}-4ac=0

b^{2}=4ac

Given : a = (a^{2} + b^{2})

b = 2(bc-ad)

c = (c^{2} + d^{2})

[2(bc – ad)]^{2} = 4(a^{2} + b^{2}) (c^{2} + d^{2})

4(bc – ad)^{2}=4(a^{2} + b^{2})(c^{2} + d^{2})

b^{2}c^{2 }+ a^{2}d^{2} – 2abcd = a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}

a^{2}c ^{2}+ b^{2}d^{2}+ 2abcd = 0

(ac – bd)^{2}=0

ac – bd=√0

ac – bd=0

Question 28.

Find the ratio in which the point P(2,x) divides the line joining the points A(-2,2) and B(3,7) internally. Also find the value of x.

Answer:

Consider,

2 = \(\frac{3 m-2 n}{m+n}\)

3m – 2n = 2(m + n)

3m – 2n = 2m + 2n

3m – 2m = 2n + 2n

m = 4n

\(\frac{m}{n}=\frac{4}{1}\)

m:n = 4: 1

Consider,

or

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2,2) B(4,4) and C(2,6)

Answer:

P, Q, R are the mid points of AB, AC, BC respectively. Using the mid point formula.

Now, the triangle obtained is

Let P(3,3) = (x_{1}y_{1})

Q(2,4) = (x_{2}y_{2}) R(3,5) = (x_{3}y_{3})

Area of PQR = \(\frac{1}{2}\)Σx_{1}(y_{2} – y_{3})

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

= \(\frac{1}{2}\) [3(4 – 5) + 2(5 – 3) + 3(3 – 4)]

= \(\frac{1}{2}\) [3(-1) + 2(2) + 3(-1)]

= \(\frac{1}{2}\) [-3 + 4 – 3]

= \(\frac{1}{2}\) [-2]

The area of a triangle in, co-ordinate geometry is an absolute value ∴ Area of ∆=1 sq.units.

Question 29.

Prove that the, “tangents drawn to a circle from an external point are equal”.

Data: “P” is an external point to a circle centre at “O”

PA and PB are tangents.

T.P.T. :PA=PB

Construction : Join OA, OB and OP

Answer:

In ∆ OAP and ∆ OBP

∠OAP=∠OBP = 90°

(Radius Tangent drawn at the point of contact)

OP = OP, OA = OB (∵ Radii of same circle)

According to R.H.S. Postulate ∆OAP ≅ ∆OBP

⇒ PA = PB (∵ CPCT)

Question 30.

In the figure ABCD is a square, whose vertices lie on the circle. Find the area of the shaded region, if the perimeter of the circle is 88 cm.

Answer:

Given : Perimeter of the circle = 88cm

2 π r = 88

In ∆ BDC, B=90 and BC – CD

BD^{2} = BC^{2} + CD^{2}

(2r)^{2} = a^{2} + a^{2}

4r^{2} = 2a^{2}

a^{2} =\(\frac{4 r^{2}}{2}\) = 2r^{2}

a = √2r^{2}

a = √2 × r

a = √2 × \(\frac{14}{2}\) = 7√2

Area of square = a^{2}

= (7√2)^{2}

49 ×2 = 98cm^{2}

Area of shaded region = Area of circle – Area of square

π^{2} r^{2} – a^{2}

= \(\frac{22}{7}\) × 14^{2} – 98

= \(\frac{22}{7}\) × 14 × 14 – 98

= 616 – 98

= 518cm^{2}

OR

∆ ABC is right angled at A. The sides AB, BC and AC are the tangents to the circle with centre ‘O’ as shown in the figure. If AB = 6cm, BC = 8cm, find the area of the shaded region.

Answer:

Let a = AB = 6cm, b=AC=8cm, c=BC = ?

In ∆ ABC, ∠A =90°

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

= 6^{2}+8^{2}

=36 + 64

BC^{2} =100

BC = √100 = 10cm c = 10cm

In – radius of a right angled ∆

∴ r = 2 cm

Area of shaded region

=Area of ∆ ABC – Area of circle

Question 31.

The following table gives the production yield per hectare of wheat of 100 farms of a village. Draw more than type Ogive

Answer:

Question 32.

Find the mean of the following data :

Answer:

mean \((\bar{\pi})=\frac{\sum f x}{\sum f}\)

= 625/25

= 25

Question 33.

Draw a triangle ABC with side base BC=8cm and altitude 4cm, and then con struct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the isoceles triangle ABC.

Answer:

V. Answer the following ( 4 × 4 = 16 )

Question 34.

Solve the pair of linear equations graphically : x – 2y= 0 and 3x + 4y = 20

Solution:

x – -2y = 0

-2y = 0 – x = -x

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

3x +4y = 20

4y = 20 – 3x

y = \(\frac{20-3 x}{4}\)

Question 35.

The sum of the first three terms of an A.P. is 33. If the product of the first term and third term exceeds the 2nd term by 29, then find the A.P.

Answer:

Let the first three terms of A.P. be a-d, a, a+d

given : Sum of these three terms = 33

a-d + a + a + d = 33

3a = 33

a = 33/3

a= 11

Given : Product of first and third term term + 29

(a-d) (a+d) = a + 29

a^{2} – d^{2} = a +29

a^{2}– a – 29 =d^{2}

d^{2} = 11^{2} – 11 – 29

d^{2}= 121 -40

d^{2} = 81

d= √8l = ±9

If a = 11, d = 9

T_{1} = a – d

T_{1} = 11 – 9

T_{2} = a = 11

T_{3} = a + d

= 11+9 = 20

or

∴ The three terms of A.P are 2,. 11, 20 OR 20,11,2

Or

The pth, qth and rth term of an A.P. are a, b and c respectively. Prove that a(q-r)+b(r-p)+c(p-q)=0

Answer:

Given :T_{p} = a,T_{q} = b, T_{r} = C

Use the formula

(a-b) (q-r) = (b-c) (p-q)

aq – bq – ar + br = bp – bq – cp + cq aq – ar – bq + bq + br – bp + cp – cq = 0

(aq – ar) + (br – bp) + cp – cq = 0

a(q-r) + b(r-p) + c(p-q) = 0

Question 36.

The angle of elevation of the top of an unfinished verticle building on a ground, at a point which is 100m from the base of the building is 45°. How much height the building must be raised, so that its angle of elevation from the same point

be 60°. (Take √3 = 1.73)

In OBC, \(\mathrm{lOCB}=90^{\circ} \mathrm{IBOC}=45^{\circ}]\)

Tan 45° = BC/OC

1 = BC/OC

∴ BC = 100m

In A OAC, ∠OCA = 90°and ∠AOC = 60°

Tan 60 = AC/OC

√3 = \(\frac{A B+B C}{100}\)

AB + BC = 100√3

AB + 100 = 100 √3

AB= 100√3 – 100

AB =100 (√3 -1)

AB = 100 (1.73-1)

AB= 100×0.73

AB = 73.00

AB = 73 m.

To reach maximum height the building should be raised to 73m.

Question 37.

Prove that “the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides”

Answer:

VI. Answer the following : ( 5 × 1 = 5 )

Question 38.

A cone of radius 10cm is cut into two parts by a plane through the mid-point of its vertical axis parallel to the base. Find the ratio of the volumes of the smaller cone and frustum of the cone.

∆AOD~∆ABC

Answer:

Given AO = \(\frac{\mathrm{AB}}{2}\)

Ab = 2AO

AB = h_{1} AO = h_{2}

BC=R= 10cm

OD = r