Karnataka SSLC Maths Model Question Paper 5 with Answers

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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
Karnataka SSLC Maths Model Question Paper 5 - 01

Karnataka SSLC Maths Model Question Paper 5 with Answers

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)=x2 – x + k is 2 then the value of k is
a) 2
b) -2
c) -6
d) 6
Answer:
b) -2
p(x) = x2 – 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
T10 = a + 9d
= 5 + 9 (4)
= 5 + 36 = 41

Karnataka SSLC Maths Model Question Paper 5 with Answers

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
Karnataka SSLC Maths Model Question Paper 5 - 1
\(\frac{A B}{A D}=\frac{B C}{D E}\)
\(\frac{A B}{3.5}=\frac{8}{5}\)
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)=x3 + 2x2 + x + 6.
Answer:
The degree of p(x) = 3
∴ There are 3 zeroes.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 10.
Write the ‘discrimant’ of the quadratic equation ax2 + bx + c=0
Answer:
Discriminant of ax2 + bx + c = 0 is
∆ = b2 – 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
T3 = 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 Sin2θ + Cos2θ .
Answer:
sinθ = \(\frac{3}{5}\)
Cos θ = \(\frac{4}{5}\)
Karnataka SSLC Maths Model Question Paper 5 - 2

Karnataka SSLC Maths Model Question Paper 5 with Answers

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 cm3

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 = a2 + 2 – 2a √2
2a =a2 + 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.

Karnataka SSLC Maths Model Question Paper 5 with Answers

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 6x2+7x-10=0
Answer:
By factorisation method
6x2+ 1x – 10 = 0
6x2+ 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 = \(\frac{5}{6}\)

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 20.
Find the distance between the points A(8,-3) and B(0,9) by using distance formula.
Answer:
Karnataka SSLC Maths Model Question Paper 5 - 3
Karnataka SSLC Maths Model Question Paper 5 - 4

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:
Karnataka SSLC Maths Model Question Paper 5 - 5
∆ 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 \(\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DBC}}=\frac{\mathrm{AO}}{\mathrm{DO}}\)
Karnataka SSLC Maths Model Question Paper 5 - 6
Answer:
In ∆ ALO and ∆ DMO
Karnataka SSLC Maths Model Question Paper 5 - 7
∴ ∆ ALO and ∆ DMO are equiangular
⇒ ∆ ALO ~ ∆ DMO
⇒ \(\frac{A L}{D A}=\frac{A O}{D O}\)
Karnataka SSLC Maths Model Question Paper 5 - 8
Hence proved.

Karnataka SSLC Maths Model Question Paper 5 with Answers

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
Karnataka SSLC Maths Model Question Paper 5 - 9

Question 23.
Draw a circle of radius 3cm. Consruct a pair of tangents to it, from a point 8cm away from its center.
Answer:
Karnataka SSLC Maths Model Question Paper 5 - 10

Question 24.
If cos θ=0.6 show that 5 sin θ-3 tan θ=0
Answer:
Consider 5 sinθ – 3tanθ = θ
Karnataka SSLC Maths Model Question Paper 5 - 11
Karnataka SSLC Maths Model Question Paper 5 - 12
5cosθ – 3 = 0
(0.6) – 3 = 0
3 – 3 = 0
0 = 0
L.H.S = R.H.S.

OR

P.T. (sec4 θ – sec2 θ)=tan2 θ +tan4θ
Consider tan2θ + tan4θ
tan2θ(1 + tan2θ)
tan2θ (sec2θ)
(sec2θ – 1) (sec2θ)
sec4θ – sec2θ = LHS
OR
Consider L.H.S = sec4 θ – sec2θ = sec2θ (sec2θ – 1)
= (1 + tan2θ)tan2θ
= tan2θ+tan4θ = R.H.S.

Karnataka SSLC Maths Model Question Paper 5 with Answers

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
Karnataka SSLC Maths Model Question Paper 5 - 13
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)=x4 + x3 – 11x2 – 9x + 18, then find the remaining two zeros of the polynomial.
Given Polynomial
P(x) = x4 + x3 – 11x2 – 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
x2 = 9 = 0
Divide x4 + x3 – 11x2 – 9x + 18 by x2 – 9
Karnataka SSLC Maths Model Question Paper 5 - 14
Factorize the quotient x2 + x – 2 to get two more zeroes.
x2 + 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.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 27.
The sum of the areas of two squares is 640 m2. 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 x2 and y2.
Given ;Sum of the areas of two squares = 640m2
∴ x2+y2=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, x2 + y2 = 640
(16 + y)2 + y2=640
162 + y2 + 2(16)(y) + y2 = 640
2y2 + 32y + 256 – 640 = 0
2y2 + 32y – 384 = 0
y2 + 16y – 192 = 0
y2 + 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
(a2 + b2) x2 + 2(bc – ad)x + c2 + d2 = 0 are equal, show that ac+bd=0
Answer:
It is given that the roots of the quadratic equation are equal
∴ ∆ =0
b2-4ac=0
b2=4ac
Given : a = (a2 + b2)
b = 2(bc-ad)
c = (c2 + d2)
[2(bc – ad)]2 = 4(a2 + b2) (c2 + d2)
4(bc – ad)2=4(a2 + b2)(c2 + d2)

b2c2 + a2d2 – 2abcd = a2c2 + a2d2 + b2c2 + b2d2
a2c 2+ b2d2+ 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:
Karnataka SSLC Maths Model Question Paper 5 - 15
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,
Karnataka SSLC Maths Model Question Paper 5 - 16
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.
Karnataka SSLC Maths Model Question Paper 5 - 17
Karnataka SSLC Maths Model Question Paper 5 - 18
Now, the triangle obtained is
Karnataka SSLC Maths Model Question Paper 5 - 18(i)
Let P(3,3) = (x1y1)
Q(2,4) = (x2y2) R(3,5) = (x3y3)
Area of PQR = \(\frac{1}{2}\)Σx1(y2 – y3)
= \(\frac{1}{2}\) [x1 (y2 – y3) + x2(y3 -y1)+x3(y1 -y2)]
= \(\frac{1}{2}\) [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.

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 29.
Prove that the, “tangents drawn to a circle from an external point are equal”.
Karnataka SSLC Maths Model Question Paper 5 - 19
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.
Karnataka SSLC Maths Model Question Paper 5 - 20
Answer:
Given : Perimeter of the circle = 88cm
2 π r = 88
Karnataka SSLC Maths Model Question Paper 5 - 21
In ∆ BDC, B=90 and BC – CD
BD2 = BC2 + CD2
(2r)2 = a2 + a2
4r2 = 2a2
a2 =\(\frac{4 r^{2}}{2}\) = 2r2
a = √2r2
a = √2 × r
a = √2 × \(\frac{14}{2}\) = 7√2
Area of square = a2
= (7√2)2
49 ×2 = 98cm2
Area of shaded region = Area of circle – Area of square
π2 r2 – a2
= \(\frac{22}{7}\) × 142 – 98
= \(\frac{22}{7}\) × 14 × 14 – 98
= 616 – 98
= 518cm2

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:
Karnataka SSLC Maths Model Question Paper 5 - 22
Let a = AB = 6cm, b=AC=8cm, c=BC = ?
In ∆ ABC, ∠A =90°
BC2 = AB2 + AC2
= 62+82
=36 + 64
BC2 =100
BC = √100 = 10cm c = 10cm
In – radius of a right angled ∆
Karnataka SSLC Maths Model Question Paper 5 - 23
∴ r = 2 cm
Area of shaded region
=Area of ∆ ABC – Area of circle
Karnataka SSLC Maths Model Question Paper 5 - 24

Karnataka SSLC Maths Model Question Paper 5 with Answers

Question 31.
The following table gives the production yield per hectare of wheat of 100 farms of a village. Draw more than type Ogive
Karnataka SSLC Maths Model Question Paper 5 - 25
Answer:
Karnataka SSLC Maths Model Question Paper 5 - 26
Karnataka SSLC Maths Model Question Paper 5 - 27

Question 32.
Find the mean of the following data :
Karnataka SSLC Maths Model Question Paper 5 - 28
Answer:
Karnataka SSLC Maths Model Question Paper 5 - 29
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:
Karnataka SSLC Maths Model Question Paper 5 - 30

Karnataka SSLC Maths Model Question Paper 5 with Answers

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 = \(\frac{-x}{-2}=\frac{x}{2}\)
Karnataka SSLC Maths Model Question Paper 5 - 31
3x +4y = 20
4y = 20 – 3x
y = \(\frac{20-3 x}{4}\)
Karnataka SSLC Maths Model Question Paper 5 - 32
Karnataka SSLC Maths Model Question Paper 5 - 33

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
a2 – d2 = a +29
a2– a – 29 =d2
d2 = 112 – 11 – 29
d2= 121 -40
d2 = 81
d= √8l = ±9

If a = 11, d = 9
T1 = a – d
T1 = 11 – 9
T2 = a = 11
T3 = 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 :Tp = a,Tq = b, Tr = C
Use the formula
Karnataka SSLC Maths Model Question Paper 5 - 34
(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

Karnataka SSLC Maths Model Question Paper 5 with Answers

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.
Karnataka SSLC Maths Model Question Paper 5 - 35

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”
Karnataka SSLC Maths Model Question Paper 5 - 36
Answer:
Karnataka SSLC Maths Model Question Paper 5 - 37
Karnataka SSLC Maths Model Question Paper 5 - 37(i)

Karnataka SSLC Maths Model Question Paper 5 with Answers

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 = \(\frac{\mathrm{AB}}{2}\)
Karnataka SSLC Maths Model Question Paper 5 - 38
Ab = 2AO
AB = h1 AO = h2
BC=R= 10cm
OD = r
Karnataka SSLC Maths Model Question Paper 5 - 39
Karnataka SSLC Maths Model Question Paper 5 - 40

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