# Karnataka SSLC Maths Model Question Paper 4 with Answers

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## Karnataka State Syllabus SSLC Maths Model Question Paper 4 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 pair of linear equations x + y = 9 and x – y = 1,.the value of x and y are
a) 5 and 4
b) 4 and 5
c) 6 and 3
d) 3 and 6
a) 5 and 4
Solution :
x + y = 9….(1)
x – y =4……(2)
(1) + (2)
2x= 10
x = 10/2 = 5
x – y = 1
x – 1 = y
5 – 1 = y
∴ y = 4 Question 2.
The product of prime factors of 120 is
a) 23 × 32 × 51
b) 22 × 31 × 51
c) 23 × 31 × 52
d) 23 × 31 × 51
d) 23 x 31 x 51
Solution:
23 × 31 × 5
8 × 5 × 3
= 120 Question 3.
In the figure, the value of sin C is
a) 2/√3
b) √3/2
c) 1/2
d) 1
c) 1/2 Question 4.
I he distance between the point (4,3) and the Origin is
a) 7 units
b) 25 units
c) 5 units
d) 6 units
c) 5 units
Solution: Question 5.
In the figure BC is a
b). chord
c) Diameter
d) secant
b) chord Solution :
Aline segement which joins any two points on the circle.

Question 6.
If the nth term of an Arithmetic progression is 4n2-1, then the,8th term is,
a) 32
b) 31
c) 256
d) 255
d) 255
Solution:
Tn = 4n2-1
T = 4(8)2 – 1
= 4 × 64 – 1
= 256 – 1
T8 = 255 Question 7.
26 English alphabet cards (without repeating any alphabet) are put in a box and shuffled well. If a card is chosen at random then the probability that the card with an vowel is
a) 3/26
b) 5/26
c) 1/26
d) 21/26
b) 5/26
Solution :
Total Number of English alphabets = n(s) = 26 set of vowels in it form and event
V = {a,e,i,o,u} ∴ n(v)=5
P(v) = $$\frac{n(v)}{n(s)}=\frac{5}{26}$$

Question 8.
In the figure, the angle of elevation \$is
a) 30°
b) 45°
c) 90°
d) 60°
a) 30°
Solution :
Tanθ = 1/3
θ = 30° II. Answer the following questions: ( 1 x 8 = 8 )

Question 9.
The LCM of 24 and 36 is 48 and hence find their HCF.
A = 24, B = 36, L = 48, H = ?
L X H = AX B
H = $$\frac{\mathrm{AxB}}{L}=\frac{24 \times 36}{48}$$
∴ HCF = 18

Question 10.
Find the roots of the quadratic equation x2 + 7x + 12=0
Factorisation method
x2+ Tx + 12 = 0 ,
x2+ 4x + 3x + 12
x(x + 4) + 3(x + 4)=0
x + 4 = 0 OR x + 3 = 0
x = -4 or x = -3

Question 11.
Find the value of sin 90s + tan45°
Sin90°+ tan45°
= 1 + 1
= 2

Question 12.
Find the co-ordinates of the mid-point of the line sagments joint the points (6,2) and (4,4). the coordinate of mid P(x,y) Question 13.
If ‘A’ is an event of a random experiment, such that P(A): P(Ā)= 1:2, find the value of P(Ā)
P(A) and P(Ā)are complementory events.
P(A) and P(Ā) = 1
1x + 2x = 1
3x = 1
x = 1/3
∴ The value of P(Ā) = 2x = 2 × (1/3) = 2/3 Question 14.
If the perimeter and area of a circle are numerically equal, then find the radius of the circle.
Given
Perimeter of the Circle = Area of the circle , 2 7tr= nr2
2πr = πr2
2r = r2
r2 – 2r = 0
r(r – 2)=0
r = 0 OR r – 2 = 0
r = 2
The radius of the circle (r) = 2 units.

Question 15.
Write the formula to find the volume of the sphere.
volume of sphere = $$\frac{4}{3}$$ π r3

Question 16.
The perimeter of circle with center ‘O’ is 24cm, the angle formed by an are of the cirde at its centre is 30°. Find the length of the arc AB. III. Answer the following: ( 2 × 8 = 16 )

Question 17.
Prove that 7 + √5 is irrational.
Let us assume 7 + √5 = a, where ‘a’ is Rational number 7 = a – √5
Squaring on both sides
(7)2 = (a-√5)2
49=(a)2 +(√5)2 -2(a)(√5)
2a√5 =a2 + 5 – 49
√5 = $$\frac{a^{2}-44}{2 a}$$
a2 – 44 is Rational and √5 is Irrational. This leads to contradiction.
Our assumption that 7+√5 is rational is wrong.
∴ 7+√5 is Irrational.

Question 18.
In the adjoining figure, XY || BC . AX = p-3; BX = 2p-2 and $$\frac{\mathbf{A} \mathbf{Y}}{\mathbf{C} \mathbf{Y}}=\frac{1}{4}$$. Find the value of p.
In ∆ ABC, XY ]| BC By B.P.T.
$$\frac{\mathrm{AX}}{\mathrm{BX}}=\frac{\mathrm{A.Y}}{\mathrm{CY}}$$
$$\frac{(P-3)}{(2 p-2)}=\frac{1}{4}$$
4(P-3)=2P-2
4P-12 = 2P-2
4P-2P=-2+12 2p = 10
P = 10/2
P=5

OR

In the figure, PC || QK and BC || HK. If AQ = 6cm, QH = 4cm, HP = 5cm and KC = 18cm, then find the lengths of AK and AB.
In ∆ APC, QK||PC
By B,P,T AK = 12cm
In ABC, BC || HK
By corollary of BPT
$$\frac{A B}{A H}=\frac{A C}{A K}$$  Question 19.
Write the general form of the following.
a) Linear polynomial –
f(x)= ax+b, where a,b are, constants, a ≠ 0 and x is a variable with degree 1.
b) Cubic polynomial – f(x) = ax3 + bx2 + cx + d, where a,b,c,d are= constants a ≠ 0. x is a variable with degree 3. Question 20.
Draw a circle of radius 4cm, and construct a pair of tangents to the circle from a point 8cm away from its center. Question 21.
If α and β are the zeros of the polynomial p(x) = 3x2 – 12 + 15, find the value of α2 + β2
P(X) = 3x2 – 12x + 15
a = 3, b = -12, c = 15 Sum of the zeroes
(α + β)2 = α2 + β2 + 2αβ
(+4)22 + β2+ 2(5)
16 = α2 + β2 +10
α2 + β2 = 16-10
α2 + β2 = 6

Question 22.
The angle of elevation of the top of a vertical tower on a level ground from point, at a distance of 9√3 from its foot on the same ground is 60°. find the height of the tower.
In ∆ ABC,∠B = 90°, ∠A – 60° .
Tan 60° = $$\frac{B C}{A B}$$
√3 = $$\frac{\mathrm{BC}}{9 \sqrt{3}}$$
BC = √3 × 6√3 = 9 × 3
Height of the tower = BC = = 27cm. OR
Find the diameter of the circular base of right circular cone whose slant height is- 8cm and semi vertex angle is 60° ∴ Diameter of Circular bax of right circular cone = 8√3 m.

Question 23
Find the area of the shaded region in the given circle of radius 7cm and sector
θ = 30, r = 7cm
Area of the shaded region $$\frac{\theta}{360} x \pi r^{2}$$  Question 24.
Curved surface area of right circular cylinder is 440cm2 and the radius of its circular base is 7cm. Find the volume of the cylinder.
C.S.A. of cylinder = 2 π rh = 440cm2
r = 7cm, V = ? Volume of the cylinder = π r2 h IV. Answer the following: ( 3 × 9 = 27 )

Question 25.
A fraction becomes $$\frac{8}{11}$$ ,if 3 is added to both the numerator , and the denominator, also if 3 is suhstracted from the numerator and the denominator it becomes $$\frac{2}{5}$$. Find the fraction.
Let the fraction be $$\frac{x}{y}$$
If “3” is added to both the numerator and denominator, the fraction obtained is $$\frac{8}{11}$$.
$$\frac{x+3}{y+3}=\frac{8}{11}$$
11 (x+3)=8(y+3)
11x+33=8y+24
11x-8y=24-33
11x – 8y = -9 ……(1)
If “3” is subtracted to both the numerator and denominator, the fraction obtained is $$\frac{2}{5}$$
$$\frac{x-3}{y-3}=\frac{2}{5}$$
5(x-3)=2(y-3)
5x – 15 = 2y – 6
5x – 2y = -6 + 15
5x – 2y = 9 ….. (2)
From (1) and (2) consider 5x – 2y = 9
5(5) – 9 = 2y
16 = 2y
y = 8
∴ The fraction is $$\frac{x}{y}=\frac{5}{8}$$

OR

10 years hence, the age of x will be 2 times that of age of y, 10 years ago, the age of x was six times that of age of y. what are their present ages?
Let the present age of x be “p” years and the age of y be “q” years.
Given : 10 years hence
Age of x = 2× age of y
p + 10 = 2(q+10)
p+ 10 = 2q + 20
p – 2q = 20 – 10
P – 2q = 10 …. (1)
10 years ago,
Age of x = 6 × Age of y
p-10 = 6 (q-10)
p- 10 = 6q+ 60
p – 6q = -60+50
P-6q=-50 … (2)
Consider Eqn (1) and (2)
P-2q = 10
P – 6q = -50
4q = 60
q = 60/4
q = 15
Consider,
P – 2 q = 10
P-2 (15)= 10
P-30= 10
P = 10 + 30
P = 40
Present Age of x = P = 40 years
Present Age of y = q = 15 years.

Question 26.
Find the two consecutive positive integers, whose sum of their squares is 365.
let the no. be  x and x+1
then we have x²+(x+1)²=365
2x²+2x – 364=0
x² + x – 182=0
x² + 14x -13x – 182=0
(x + 14) (x – 13)=0
That gives x=13 Avoiding negative value
one number is 13 and other one 14.

Question 27.
Find the perimeter of the triangle whose vertices are (-2,1), (4,6) and (6,3)  The perimeter of ABC
= AB + BC = CA
= (√61 + √31 + √68) units

OR

Three consecutive vertices the parallelogram are A(1,2), B(2,3) and C(8,5). Find the fourth vertex.
The diagonals AC and BD of a parallelogram bisect each other at O.
‘O’ is the mid-point of AC and BD.  The co-ordinates of midpoint of AC The co-ordinates of midpoint of BD 4.5 = $$\frac{2+x}{2}$$
2 + x = 9
x = 7
3.5 = $$\frac{3+y}{2}$$
3 + y = 7
y = 7 – 3
y = 4
∴ The coordinates of D are (7,4) Question 28.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Data : ‘O’ is the centre of the circle, AB is the tangent drawn to the circle at C. OC is the radius. To prove that: and $$\overrightarrow{\mathrm{OC}} \perp \overrightarrow{\mathrm{AB}}$$
Construction : Mark a point ‘D’ on AB and let it cuts the circle at E.
Proof: ‘D’ is a point on AB outside the circle ∴ OE < OD OE is the radius of the circle ∴ OE = OC (∵ Radii of the same circle) => OC < OD
Of all the points marked on AB, OC is the shortest distance from external point O to the line AB
$$\overrightarrow{\mathrm{OC}} \perp \overrightarrow{\mathrm{AB}}$$

Question 29.
A flower vase is in the form of a frustum of a cone. The perimeters of its bases are 44cm and 8.4 n cm. if the depth is 14cm, then find how much soil it can hold.
Perimeter of the top circle.
= 2πR=44
2x $$\frac{22}{7}$$ R = 44
R = $$\frac{44 x 7}{44}$$ = 7cm
Perim eter of the bottom .circle = 2πR= 8.4π
r = 8.4 / 2
r = 4.2 cm. The quantity of soil that the vase can hold = Volume of the frustum of the cone. OR

A toy is in the form of a cone mounted on a hemisphere both are of same radius. The diameter of the conical portion is 6cm and its height is 4cm. Determine the surface area of the solid, (taken π = 3.14)
d = 6cm
r=6/2=3cm, h = 4cm, l = ?
P = r2 + h2
P = 32 + 42
l = $$\sqrt{9+16}$$ = √25 l = 5cm
diameter = 6cm
r = $$\frac{6}{2}$$ h =4 cm Surface area of the solid
= C.S.A. of cone + C.S.A of Hemisphere =πrl + 2πr2 Question 30.
Solve graphically : 5x + y = 7 and 2x – 2y = 2
5x +y = 7
y = 7-5x 2x – 2y = 2
2y = 2x – 2
∴y = $$\frac{2(x-1)}{2}$$
∴y = x – 1 2x – 2y = 2
– 2y = 2 – 2x
∴y = $$\frac{(2-2x)}{-2}$$  Question 31.
The following distribution gives the daily income of 50 workers of a factory. Convert the distribution above to a less than type cumulative frequency distribution, and draw it’s ogive.   Question 32.
The sum of first ‘n’ terms of an arith¬metic progression is 210 and sum of its first (n-1) terms is 171. If the first term 3, then write the arithmetic progression.
Given Sn =210, Sn-1 =171 a=3
Tn =Sn – Sn-1 .
Tn = 39 = l n = 10
Now, Tn=a+(n-l)d
T10
=3+(10-1)d
39=3+9d
9d= 39-3
d = $$\frac{36}{9}$$ = 4
∴ d = 4
∴ The Arithmatic progression is 3, 7, 10,13, 39

Question 33.
A man drives his car with uniform speed from palce A to the place B which is 150 km away. Again he returns to the place A by increasing the speed of the car lOkm/hour and there by reaches 30 min¬utes earlier than the time taken in his forward journey, find the total time taken by him in forward and return journey.
Let the speed of the car be “x” km/hr d= 150 km l = ?
S = d/t
x = 150/t
t1 = ( 150/x ) hour
If the speed car is increased by 10 kmph, then the total speed (x+10) kmph to cover a differ-ence of 150 km return journey.
S = d/t
t2 = $$\frac{d}{s}=\left(\frac{150}{x+10}\right)$$ hour
t1 – t2 = The difference in time taken = 30min. x2 +10x = 3000
x2 + 10 x – 3000 = 0
x2 + 60x – 50x – 3000 = 0
x(x + 60)-50(x + 60)=0
(x + 60) (x – 50)=0
x + 60 = 0 OR
x = -60
x – 50 = 0
x = 50
t1 + t2 = 3 +2.5 = 5.5 hour

OR

A, B and P are the three non – collinear points on a plane. The distance between the point A and P is 2m more than the distance between the points B and P. if the distance between points A and B is 10m and AB is the longest side of the triangle ABC. Is ABC a right angled triangle or not, justify your an¬swer using the discriminant of quadratic equation and also find the measure of APand BP. A,B and P are three non – col¬B linear points which form AABP. AB is 10 units Let BP = x units then AP = BP + AP = (x +2 )
In ∆ABP
If AB > AP > BP
AB2 = AP2 + BP2
102=(x + 2)2+x2
100=x2+ 4x + 4 + x2
100=2x2 + 4 x+ 4
2x2 + 4x + 4 – 100=0 .
2x2 + 4x – 96 = 0
Dividing by 2
x2 + 2x – 48=0
by factorisation
x2 + 8x – 6x-48 = 0
x (x + 8) – 6 (x + 8)=0
(x + 8) (x – 6)=0
x + 8 = 0 OR x – 6 = 0
x = -8
x = 6
∴ BP = x = 6
AP = x + 2
= 6 + 2 = 8
AB= 10
by equation (1)
AB2 = AP2 + BP2
102 = 82 + 62
100 = 64 + 36
100 =100
∴ AP & BP form the right angle
=> A ABP form Right angled A le with
ZAPB=90°
Justification
Consider
x2 + 2x – 48 = 0
a = 1, b = 2, c = -48
∆ = b2-4ac
∆ = (2)2-4(l)(-48)
∆ = 4+192
∆ = 196
∴ The side of A APB has a real value. V. Answer the following. ( 4 × 4 = 16 )

Question 34.
If the sum of first 8 terms of arithmetic progressions is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.
Given S8 = 136, S15=465, S25=?  From (2) and (1) Consider a + 7d = 31
= 3 + 7 (d) = 31
7d=31 -3 = 28
d = 28/7 = 4 OR

The sum of the 5th and 9th terms of an arithmetic progression is 40 and the sum of the 8th and 14th term is 64. find the sum of first 20 terms.
Given T5 + T9 = 40
a + 4d + a + 8d=40
2a + 12d=40
a + 6d=20 …. (1)
T8 + T14 = 64
a + 7d + a + 13d = 64
2a + 20d = 64
a+10d=32….(2)
(2) From (1) and (2)
a + 10d = 32
a + 6d = 20
4d = 12
d = 3
Consider
a+ 10d = 32
a + 10(3)  =32
a + 30 = 32
a = 32 – 30 = 2 = 10 (4 + 57)
10 × 61
=610

Question 35.
The mode of the following distribution
table is 15. Find the mean of this data, and then find the median value by using empirical formula relating mean, median and mode.  = 9 + 2 = 11
True Mode = 3 Median -2 Mean
=3(11) – 2(9.8)
=33 – 19.6
= – 13.4
Mode $$L+\left[\frac{F_{1}-F_{0}}{2 F_{1}-F_{0}-F_{2}}\right] x h$$
C.I =13 – 17 has max Fre =.8
L= 13, h = 4, f1 = 8, F0 = 2 F2 = 1  Question 36.
$$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \csc \theta$$
Consider L.H.S  Question 37.
Draw a right traingle in which the sides (other than hypotenuse) are of lengths 8cm and 6cm, then construct another traingle whose sides are $$\frac{5}{3}$$ times the cor-responding sides of the given traingle.  VI. Answer the following ( 1 × 5 = 5 )

Question 38.
State and prove : The converse of the Pythagoras theorem. If the square on the longest side of a tri¬angle is equal to the sum of the squares on the other two sides, then those two sides contain a right angle”
Data : In ∆ ABC, AC2 = AB2 + BC2
T.P.T. : ∠ABC -90°
Construction : Draw 4 perpendicular on AB at B, Select a point D. such that, DB = BC. Join A and D.
Proof: ∴ AD2 = AB2 + BD2(Pythagoras)
But in ∆ ABC,
AC2 = AB2 + BC2 (Data)  