Karnataka SSLC Maths Model Question Paper 2 with Answers

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Karnataka State Syllabus SSLC Maths Model Question Paper 2 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.
Euclids Division Lemma states that for any two positive integers a and b, there exists unique integers q and r such that a = bq + r, where r must satisfy.
a ) 0 < r < b
b) 0 < r<b
c) 1 < r < b
d) 0 < r < b
Answer:
a ) 0 < r < b

Question 2.
The value of Sin²29 + Sin²61 is ……..
a) 61
b) 29
c) 1
d) 0
Answer:
c) 1
Solution :
Sin²29 + Sin²61
sin²29 + sin²(90-29)
sin²29 + cos²29 = 1 [sin²θ + cos²θ = 1]

Question 3.
If 4 sides of a quadrilateral A BCD are tangents to a circle, then
a) AB + CD = AD + BC
b) AB + AD = BC + CD
c) AB + BC = AD + DC
d) AC + AD = BC + BD
Answer:
a) AB + CD = AD + BC

Question 4.
The volume of two shapes are in the ration 125 : 64. The ratio of their surface areas.
a) 1:5
b) 4:5
c) 5:4
d) 1:4
Answer:
c) 5:4
Solution:
\(\frac { 4 }{ 3 }\) πR³ \(\frac { 4 }{ 3 }\) πR³ = 125 : 64
R³: r³
5³.4³
\(\frac{R}{r}=\frac{5}{4}\)
4πR² : 4πR² = (5x)² : (4x)²
R²: r²
5³.4³
\(\frac{R^{2}}{r^{2}}=\frac{25}{16}
[latex]\frac{R}{r}=\frac{5}{4}\)

Question 5.
the degree of the polynomial in the graph given below is f
a) 0
b) 1
c) 2
d) 3
Answer:
d) 3
Solution :
3 since it is intersecting the x – axis at 3 points.

Question 6.
The 7th and 13th terms of an A.P. are 34 and 64 respectively. 1 Then its first term difference are :
a) 4,5
b) 5,4
c) 9,4
d) 4,9
Answer:
b) 5,4
Solution:

a + 12 d = 64
a + 12(5) = 64
a + 60 =  64
a = 64 – 60
a = 4

Question 7.
In the given fig AM : DN = 4 : 5 ∆ ABC ~ ∆ DEF and Area of ∆ DEF = 625 cm2 The area of ∆ ABC is
a) 16 cm²
b) 25 cm²
c) 81 cm²
d) 400 cm²
Answer:
d) 400 cm²
Solution:

Question 8.
The propability of picking a good apple in a lot of 400 apples is 0.035. The no of good apples in the lot is
a) 14
b) 15
c) 400
d) 35
Solution :
Let No of apples be = n(s)=400
Number of Good apples n (A) = ?
P(A) = \(\frac{n(A)}{n(S)}\)
0.035 = \(\frac{n(A)}{n(S)}\)
n(A) = 400 × 0.03 5 = 14

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

Question 9.
Find the remainder using remainder theorem, when 2x³ + 3x² + x + 1 is divided by \(x+\frac{1}{2}\)
Answer:
Let \(x+\frac{1}{2}\) = 0
x = – \(\frac{1}{2}\)
f(x) = 2x³ + 3x² + x + 1

Question 10.
If the sum of first n even natural number is 240. find the value of n.
Answer:
Sum of m even naturals = n(n+l)=240
n(n + 1) = 15 × 16
n = 15 or n + 1 = 16
n = 15 or n = 15

Question 11.
The sum of n natural numbers is 325. find n.
Answer:

n(n + 1 ) = 325 × 2
n(n + 1) = 650
n(n + l) = 25 × 26
n = 25, n + 1 = 26 or n = 25

Question 12.
Is every square similar to every Rectangle? Why?
Answer:
Every square is not similar to every Rectangle because even though are equiangular the corresponding sides are not propotional.

Question 13.
Find the HCF of 105 and 1515 by prime factorization method.
Factors of 105 = {3, 5, 7}
Factors of 1515 = 3, 5, 101.
∴ H.C.F. of 105 and 1515 = {3,5}

Question 14.
If Sinθ = \(\frac { 4 }{ 5 }\) and Cosθ = \(\frac { 3 }{ 5 }\) find the
value of Sin² θ + Cos²θ

Question 15.
Find the value of 4sin²60 + 3 tan² 30 – 8sin45. cos45
Answer:
4sin²60 + 3 tan²30 – 8sin45. cos45

= 3 + 1 = 4
= 4 – 4 = 0

Question 16.
Find the volume of the hemisphere of radius 21 cm.
Answer:
2 Volume of hemisphere =\(\frac{1}{2}\)πr³

= 441 × 44 = 19404 cm³

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

Question 17.
Prove that n² – n is divisible by 2 for every positive integer n.
Answer:
When n = 2q, an even number.
n² – n = (2q)² – (2q)
= 4q² – 2q = 2q (2q – 1)
⇒ n² – n = 2r, where r = q (2q – 1)
⇒ n²– n is divisible by 2.
when n = 2q + 1, an odd number.
n²– n = (2q + 1)² – (2q + 1)
⇒ n²– n = 2r where r = q(2q + 1)
n²– n is divisible by 2
Hence, n² – n is divisible by 2 for every positive integer.

Question 18.
Solve for x and y
41x + 53y = 135
53x + 41y = 147
Answer:
Consider 41x + 53y = 135 ………(1)
53x + 41y = 147 ……. (2)
Add (1) and (2)
94x + 94y = 282
Divide by 94 x + y= 3 …….. (3)
Subtract (1) and (2) -12x + 12y = -12
Divide by 12 -x + y = -1 ……(4)

y = \(\frac{2}{2}\) = 1
y = 1
Substitute the value of y in (3)
x + y = 3
x + 1 = 3
x = 3 – 1 = 2
x = 2

Question 19.
Find the roots of the quadratic equation 9×2 – 3x-20 = 0 by formula method.
Answer:

Question 20.
If A(-2,-1) B(a,0) C(4,b) and D(1,2) are the vertices of a parallelogram. Find a and b.
Answer:
ABCD is a parallelogram Diagonals AC and BD intersect each other at right angles.
Mid point of AC= Mid point of BD.

Question 21.
In given figure, \(\frac{A O}{O C}=\frac{B O}{O D}=\frac{1}{2}\) and AB= 5cm. Find the value of DC.
Answer:

OR

Equilateral triangles are drawn on the sides of a right triangle show that the area of triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

Answer:
Given : ABC is a right angled at A at B Equilateral triangle PAB, QBC and RAC are on the sides AB, BC & CA ‘
T.P.T. : Area of A PAB + Area of A QBC
= Area of A RAC.
Proof: Since traingles PAB, QBC and RAC are equilateral.
They are quiangular.

⇒ ARea of ∆ PAB + Area of ∆ QBC
= Area of ∆ RAC

Question 22.
A child has a dice whose six faces show the letters are given below :
A B C D E A
The die is thrown once, what is the probability of getting (i) A (ii) D.
Answer:
A die has 6 faces, marked A, B, C,D, E, A n,(s) = 6
∴ The event of getting A is (A, A) n(A) = 2
n(A) = 2
P(A) = \(\frac{n(A)}{n(s)}=\frac{2}{6}=\frac{1}{3}\)
(ii) An event of getting a letter D is only one n(B) = 1
∴ P (B) = \(\frac{n(B)}{n(s)}=\frac{1}{6}\)

Question 23.
Draw a circle of radius 4cm. Draw a tangent to this circle making an angle of 30° with a line passing through the centre.
Answer:

Question 24.
Prove that \(\frac{\sin \theta}{1-\cos \theta}\) = cosec θ + cot θ
Answer:

OR

Prove that \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ
Answer:

IV. Answer the following; ( 3 x 9 = 27 )

Question 25.
The sum of digits of a two digit number is 7. If the digits are reversed and the resulting number is decreased by 2, twice the original number is obtained, find the original no.
Answer:
Let the original number be
\(\overline{y x}\) ⇒ 10y + x
Sum of digits of a two digit number = 7
x + y = 7 _(1)
Reversing the digits = \(\overline{x y}\)
⇒ 10x + y
Resulting number is decreased by 2 = ⇒ 10x + y – 2
= Twice the original number 2(10y + x)
∴ 10x + y – 2 = 2 (10y + x)
10x + y – 2 = 20y + 2x
10x – 2x = 20y – y + 2
8x – 19y = 2 …. (2)
Solve (1) and (2)
x + y = 7 … (3)
8x – 19y = 2 … (4)
Multiply (3) by 8 and (4) by 1

Substitute the value of y in eq (3)
x + y = 7 .
x + 2 = 7
x= 7 – 2 = 5
x = 5
∴ The original number = yx = 25

OR

A boat takes 5 hours to go 300 km down stream. It takes the same time to go 150 km upstream. Calculate,
(i) the speed of the boat in still water.
(ii) the speed of the stream.
Answer:
Let the .speed of the boat in still water be x km/hr. Let the speed of the stream be y km/hr. Downstream, the relative speed of the boat is (x+y) km/hr and upstream, its relative speed is (x-y) km/hr.
Distance = speed x time
300 = (x + y)5 and 150 = (x – y)5
\(\frac{300}{5}\) = x + y
\(\frac{150}{5}\) x – y
x + y =60 …….(1)
x – y =30 …….(2)
Solve (1) and (2)

Substitute x in (1) x + y = 60
45 + y = 60
y = 60 – 45
y = 15
∴ Speed of the boat in still water is 45km/hr. Speed of the stream is 15km/hr.

Question 26.
The zeroes of the cubic polynomial x3- 6×2+3x+10 are in A.P. for some real numbers a and d. Find the zeroes of the given polynomial.
Answer:
Sum of the zeroes = \(\frac{-\cos \theta \cdot \theta x^{2}}{\cos \theta \cdot \theta f x^{3}}\)
Let the zeroes of the polynomial be a, a+d, a+2d
a, a+d, a+2d = -(-6)/1 = 6
3a + 3d = 6
a +d = 6/3 = 2
a +d = 2 ……… (1)
a + d is one of the zeroes = 2
x = 2
x – 2 = 0
Divide x³ – 6x² + 3x + 10 by x – 2

Factors of x² – 4x – 5 = 0
x² – 5x + 1x – 5 = 0
x(x – 5) + 1(x – 5) = 0
(x – 5) (x + 1) = 0
x – 5 = 0, x + 1 = 0
x = 5 or x = -1
∴ zeroes of the given polynomial are 5,-1,2

Question 27.
In a class test, the sum of Sharma age marks in Mathematics and English is 30. Had he got 2 marks more in maths and 3 marks less in English, the product of their marks would have been 210. Find his marks in two subjects.
Answer:
Let the marks in Mathematics be x and that of science be y.
Sum of Sharma marks in Mathematics and science = 30.
⇒ x + y= 30
x = 30 – y ……(1)
If he gets 2 marks more in Maths, it will be x + 2 and 3 marks less in English = (y-3) Product of their marks = 210
(x+2) (y-3) = 210
(30-y+2) (y-3)=210
(32-y) (y-3) = 210
32y – y² – 96 + 3y = 210
-y² + 35y – 96-210 = 0
-y² – 35y – 306 = 0
y²-35y +306 = 0
y² – 18y – 17y +306 = 0
y(y-18)-17(y-18)=0
(y – 18)(y – 17) = 0
y – 18 = 0 y – 17 = 0
y = 18 and y = 17
⇒ when y = 18, x = 30 – 18 = 12
y = 17, x – 30- 17= 13.

OR

The Sum of the reciprocals of Adithya ages 3 years ago and 5 years from now is \(\frac{1}{3}\) Find his present age.
Answer:
Let the present age be = x years.
3 years ago, Adithya’s age was = (x-3) years.
5 years from now, Adithya’s age will be (x+5)years.
Sum of their reciprocals = \(\frac{1}{3}\)

3(2x+2) = (x-3) (x+5)
6x + 6 = x2 – 3x + 5x – 15
x² + 2x – 15 – 6x – 6 = 0
x² – 4x – 21 =0
x² – 1x + 3x – 21 = 0
x(x – 7) + 3(x – 7) = 0
(x – 7) (x + 3) = 0
x – 7 = 0,
x + 3 = 0
x = 7,
x = -3
∴ The present age of Adithya = 7 years.

Question 28.
What type of triangle is A ABC, with the co – ordinates of the vertices are A(-1,0)
B(1,0) and C(0, √3 ). Calculate its Area.
Answer:

Since the length of AB, BC and CA are the same, the triangle is an equilateral triangle.
Area of an equilateral triarigle = \(\frac{a^{2} \sqrt{3}}{4}\)
= \(\frac{(2)^{2} \times \sqrt{3}}{4}=\frac{4 \sqrt{3}}{4}\)
∴ Area of an triangle = 3cm²

OR

Find the values of k so that the area of the triangle with vertices (1, -1) (-4, 2k) and (-k, -5) is 24 sq. units.
Answer:

Area of the triangle
= \(\frac { 1 }{ 2 }\) [x₁(y₂ – y₃) +x₂ (y₃ – y₁) + x₃ (y₁– y₂) = 0
24 = \(\frac { 1 }{ 2 }\) [(2k + 5)+ (-4)(-5 + 1) + (-k)( -1 – 2t)]
24 = \(\frac { 1 }{ 2 }\) [2k + 5 – 4 (-4) – k(-1 – 2k)]
24 × 2 = [2k + 5 +16 + k + 2k²]
48 = 2k² + 3k + 21
2k² + 3k+ 21 – 48 = 0
2k² + 3k – 27 = 0 3
2k² + 9k – 6k – 27= 0
2k² – 6k + 9k – 27= 0
2k(k – 3) + 9(k – 3) = 0
(k – 3) (2k + 9) = 0
k – 3 = 0 and 2k + 9 = 0
K= 3 k = -9/2

Question 29.
Prove that the tangents drawn to a circle from an external point are equal.

Data : ‘O’ is the centre of the Circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.

( ∵ angle between the radius and tangent at the point of contact is 90°)
OP = OP (∵ common side)
OA = OB (∵ Radius of the same circle)
According to RHS postulate)

—»opposite angles of OAPB quadilatiral are supplementary
∴ OAPB is a cyclic quadilateral

∴ Hence proved.

Question 30.
Find the area of the shaded region in fig if radii of the two concentric circles with centre O are 7cm and 14cm respectively and ∠AOC = 40°
Answer:

Area of the shaded region ABDC = Area of sector AOC – Area of sector BOD.
= \(\frac{616}{9}-\frac{154}{9}=\frac{462}{9}\) = 51.33sq.cms

OR

Find the area of the shaded region in fig. if ABCD is a square of side 14cm and APD and BPC are semi circles.
Answer:
ABCD is a square of side 14cm.
Area of sq. ABCD (1 4)² = 1 96sq.cms.
∴ APD and BPC are semicircles of diameter 14cms.

Area of shaded region = Area of square ABCD – Area of semicircle region. =196-154=42 sq.cms.

Question 31.
During the medical check-up of 35 students of a class, their weights were recorded as follows.

Draw a less than type ogive for given data. Hence, obtain the median weight from the graph and verify the result by using the formula
Answer:
Scale
X axis 1cm = 2 units
Y axis 1cm = 5 units

Question 32.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows :

Determine the median number of letters in the surnames. Find the mean number of letters in the surnammes? Also, find the modal size of the surnames.
Answer:

To calculate mode Max frequency = 40
∴ C.L =7 – 10
L = 7, h = 3 , F₁ = 40 , F₀ = 30 F₂ = 16

Question 33.
Draw a right triangle in which (other than the hypoten use) are of lengths 4cm and 3cm. Then construct another triangle whose sides are 5/3 times the cor-responding sides of the given triangle.
Answer:

Answer:
Verification BC = 3cm

Question 34.
Solve the pair of linear equations graphically: 2x – y = 2 and 2x – 3y = -6 using graphical method.
Answer:
2x -y = 2 ,
2x – 3y = -6
Taking Eq 1 2x – y = 2
y = 2x – 2
If x = 0
y = 2x – 2
y = 2(0) – 2 =-2

x = 1
y = 2x – 2
= 2(1 )-2
= 2 – 2 = 0
x = 2
= 2(2  – 2
= 4 – 2
= 2
x = 3
y = 2(3 ) – 2
= 6 – 2 = 4 .

x = 4
y = 2(4) – 2
= 8 – 2 =6

2x + 6 = 3y
y = \(\frac{2 x+6}{3}\)
Karnataka SSLC Maths Model Question Paper 2 with Answers – 44

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

Question 35.
Find the 60th term of an A.P 8, 10, 12, ….. if it has a total of 60 terms and hence
find the sum of its last 10 terms.
Answer:
a_n = a + (n-1)d
t₆₀ = 8 + (60-1)2
= 8 + 59×2
= 8+118
= 126
sum of last 10 terms = (t₅₁ +t₅₂ + ….. t₆₀)
t₅₁ = 8 + 50 x 2 = 108
S_n = \(\frac{n}{2}\) [ a + l]
S₁₀ = \(\frac{10}{2}\) [ 108 + 126]
= 5[243]
S₁₀ = 1170

OR

Find the sum of all natural numbers between 200 and 300 which are exactly divisible by 6.
Answer:
The arithmetic series is, 204 + 210 + 216 ……. + ……. +294
∴ a = 204, d = 210 – 204=6, Tn=294, S_n=?
To find S_n, we should first find the value of ‘n’
Tn = 294
a + (n-l)d=294
204 + (n-1)6=294
204 + 6n – 6 = 294
198 + 6n = 294
6n = 294- 198 = 96
n = 16
S_n = \(\frac{n}{2}\) [ a + T_n]
S₁₆ = \(\frac{16}{2}\) [ 204 + 294] = \(\frac{16}{2}\) × 492 = 3984

Question 36.
The angle of elevation of the top of a cliff as seen from the top and bottom of a building are 450 and 600 respectively. If the height of the buildings is 24m, find the height of the cliff.
Answer:
Let the height of the cliff be bx from the ground
:.OD = x – 24
In triangle OCD, the triange OCD,’
tan 45 = \(\frac{x-24}{y}\)
1 = \(\frac{x-24}{y}\)
y = x – 24 ……. (1)
In triangle OAB,
tan 60 = x/y
√3y = x ….. (2)
y = x/√3
substituting (2) in (1)

Question 37.
State and prove Basic proportionality theorem
Answer:
Statement: If a line is drawn parallel to one of the sides then the other two sides are divided proportionally.
Data: ∆ ABC, DE || BC

∆ BDE and ∆ CDE are on the same base DE and between same parallel DE||BC
Area of ∆ DBE = Area of ∆ CDE
From (1) and (2)
\(\frac{A D}{B D}=\frac{A E}{C E}\)
Hence proved.

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

Question 38.
A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the conical portion is 6cm and its height is 4cm. Determine the surface area and volume of the solid.
Answer:

Answer:
Conical portion
r = 3 cm, h = 4cm
P = r² + h².
= 3² + 4²
= 9 + 16
l² = 25
L = √25 = 5cm.
Surface Area of the toy = C.S.A of hemisphere + C.S.A. of cone
= 2πr² + πrl

= 103.71 cm²
Surface Area of the toy = 103.71 cm²
Volume of the toy = volume of hemisphere + volume of cone

∴ Volume of the toy = 94.28cm3