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Karnataka State Syllabus SSLC Maths Model Question Paper 1 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.
x = 24 × 32, y=22 × 32 × 5, Z = 26 × 3, then H.C.F. of x, y, z is
a) 22 × 32 × 5
b) 26 × 32
c) 22 × 3
d) 22 × 32
Answer:
c) 22 × 3
Question 2.
\(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}\) is equal to ______
a) Tan θ + Sec θ
b) Tan θ – Sec θ
c) \(\frac{1}{\tan \theta}+\frac{1}{\sec \theta}\)
d) \(\frac{1}{\tan \theta}-\frac{1}{\sec \theta}\)
Answer:
a) Tan θ + Sec θ
Solution:
Tan θ + Sec θ
Question 3.
PA and PB are the two tangents drawn to a circle centered at o. from an external
point P. If ∠AOB = 150° then ∠APB is
a) 20°
b) 30°
c) 50°
d) 100°
Answer:
b) 30°
Solution:
AOBP is a cyclic quadrilateral
∠AOB + ∠APB = 180° (cycle quadrilateral)
150°+ ∠APB= 180°
∴ ∠APB= 180° – 150° = 30°
Question 4.
The formula to find the curved surface area of a sphere is
a) πr2
b) 2πr2
c) 4πr2
d) 3πr2
Answer:
c) 4πr2
Question 5.
The degree of the polynomial in the graph given below is _____
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3
Solution:
3 since it is intersecting the x – axis at 3 points.
Question 6.
(3x+2) (5x-3) and (4x+7) are the three consecutive terms of an A.P. then the value of x is
a) 1
b) 3
c) 5
d) 7
Answer:
c) 5
Question 7.
If ∆ ABC ~ ∆ DEF, BC = 3cm, EF = 4cm, and Area of ∆ ABC = 54cm2, then Area of ∆ DEF is
a) 96cm2
b) 86cm2
c) 76cm2
d) 66cm2
Answer:
a) 96cm2
Solution :
∆ ABC ~ ∆ DEF
Question 8.
Which among the following is not an example of a random experiment.
a) Tossing a coin
b) Throwing a die
c) Drawing a card from a well shufled pack of card
d) Determining the boiling point of water.
Answer:
d) Determining the boiling point of water.
II. Answer the following questions: ( 1 × 8 = 8 )
Question 9.
Write the quadiatic polynomial whose zeroes are 2 and -3
Answer:
Let the zeroes be α = 2 and β = -3
α + β = 2 – 3 = -1
α β = (2) (-3) = -6.
∴ Equation = x2 – (α + β )x + α β =
x2 – (1)x + (-6)
= x2 + x + (-6)
Question 10.
Mention the two roots of quadratic equation ax2 + bx + c = 0?
Answer:
The two roots of quadratic equation
ax2 + bx + c = 0
Question 11.
Find the 10th term from end of A.P. 4,9,14…. 254?
Answer:
254, 249, 244,………………. 14, 9,4
a = 254, d = -5, T10=?
Tn = a + (n – 1)d
T10 = 254 + (10 – 1) (-5)
= 254 – 45
= 209
Question 12.
State converse of Thales theorem.
Answer:
“If a straight line divides two sides of triangle proportionally, then the straight line is parallel to the third side.”
Question 13.
Find the H.C.F and L.C.M. of the integers 17, 23 and 29 by prime factorisation method.
Answer:
Factors of 17=1 × 17
Factors of 23 = 1 × 23
Factors of 29 = 1 × 29
∴ H.C.F of 17, 23 and 29 = 1
L.C.M of 17, 23 and 29 = 11339
Question 14.
Find the value of sin264 + sin226
Answer:
sin264 + sin226
sin264 + sin2 (90 – 64)
sin264 + cos264 = 1
Question 15.
Given Sin2θ + Cos2θ = 1, P.T. 1 + tan2θ = Sec2θ
Answer:
Sin2θ + Cos2θ = 1
Divide by Cos2θ
\(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+\frac{\cos ^{2} \theta}{\cos ^{2} \theta}=\frac{1}{\cos ^{2} \theta}\)
tan2θ + 1 = sec2θ
Question 16.
Calculate the volume of a cylinder of base radius 7cm and height 7cm.
Solution:
Volume of a cylinder -π r2h
= \(\frac{22}{7}\) × 7 × 7 × 7
= 22 × 49
= 1078 cm3
III. Answer the following : ( 2 x 8 = 16 )
Question 17.
Prove that √5 is irrational
Answer:
Let us assume that √5 is a rational number. There exists integers p and q such that
√5 = p/q
Where p and q are co-prime to each other
∴ √5 x q = p
Squaring on both sides, 5q2 = p2
⇒ 5 divides p2
∴ 5 divides p.
Let p = 5k where k is an integer divides a
5q2 = (5p)2 = 25p2
q2 = \(\frac{25}{5} \mathrm{P}^{2}\) or q2 is divisible by 5
∴ From (1) and (2) we can conclude that 5 is a common factor of both a and b. but this contradicts our supposition that a and b are co-primes.
∴ Hence √5 is irrational.
Question 18.
Solve for x and y
37x + 41y = 70
41x + 37y = 86
Answer:
Consider 37 x + 41y = 70 ….. (1)
41x + 37y = 86 ……(2)
Add (1) and (2)
78x + 78y =156
Divide by 78
x + y = 2 …….(3)
Subtract (1) and (2)
-4x + 4y = -16
divide by 4
-x + y = -4 ……..(4)
Solve (3) and (4)
substitute y in (3)
x + y = 2 .
x – 1 = 2
x + 2 + 1 = 3
x = 3
Question 19.
Find the roots of the quadratic equation
9y2 – 3y = 2
Answer:
9y2 – 3y – 2 = 0
9y2 – 6y+ 3y – 2 = 0
3y(3y – 2) + 1 (3y – 2) = 0
(3y – 2) (3y + 1) = 0
3y – 2 = 0, 3y + 1 = 0
y = 2/3 OR y = -1/3
Question 20.
Find the distance between the points A(3,4) and B(6,-3)
Answer:
Distance = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
A(3, 4) = A(x1; y1)
x1 = 3 y1 = 4
B(6, -3) = B (x2y2)
x2 = 6, y2 = -3
Question 21.
E is a point on the side AD produced of a parallelogram ABCD and BE inter sects CD at F, show that
Answer:
∆ ABE ~ ∆ CFB
In ∆ABE and ∆CFB
∠A = ∠C
∠AEB = ∠CBE (AE || BC, Alternate ∠s)
∠ABE = ∠CFB (Remaining angles)
∴ ∆ ABE and CFB are equiangular.
∴ ∆ ABE ~ ∆ CFB (AAA similarity criteria)
OR
The permiters of two similar triangles ABC and PQR are respectively 36cm and 24cm. If PQ = 10cm, find AB.
Answer:
Since the ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.
Question 22.
Find the probability that a number selected at random from the numbers
1,2,3,4,5,…. 34,35 is a (i) Prime number
(ii) Multiples of 7.
Answer:
Total number of possible outcomes = 35
i) Out of the given numbers, prime numbers
are 2,3,5,7, 11, 13, 17, 19,23,29,31
Let A be the event of getting prime no.
∴ n(A) = 11
⇒ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{11}{35}\)
ii) Out of the given numbers, the multiples of 7 are 7, 14, 21, 28, 35
Let B the event of getting a multiple of 7
Then, no of favourable outcomes = 5
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{5}(\mathrm{35})}=\frac{1}{7}\)
Question 23.
Draw a circle of radius 3.5cm and construct a tangent to this circle making an angle of 30° with a line passing through the centre of the circle.
Answer:
Question 24.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of SinAcosC+cosAsinC
Answer:
In ABC
AC2 = AB2 + BC2
= (√3)2 + (1)2
= 3 + 1
= 4
∴ AC = √4 = 2
OR
Prove the identity tan2θ – sin2θ = tan2θ – sin2θ
Answer:
IV. Answer the following : ( 3 x 9 = 27 )
Question 25.
Anirudh can row downstream 20km in 2 hours, and upstream 4km in 2 hours. Find his speed of rowing in still water and speed of the current.
Answer:
Let the speed of Anirudh in still water be x km/ hr. and the speed of current by y km / hour.
The speed down stream = (x + y)km/hr
The speed up stream = (x – y) km/hr.
\(\frac{20}{x+y}\) = 2 and \(\frac{4}{x-y}[latex] = 2
2(x + y)=20 and (x – y)2 = 4
x + y =10 ……(1)
x – y = 2 ………(2)
solve (1) and (2)
x = 12/2
x = 6
Substitute x in (1)
x + y = 10
6 + y = 10
y = 10 – 6
y = 4
∴ The speed of Anirudh in still water = 6km / hr
speed of current = 4 km/hr.
OR
Adithya travels 300km to her house partly by train and partly by bus. She takes 4 hours if she travels 60km by train and the remaining by bus. If she travels 100km by train and the remaining by bus, she takes lOmin longer. Snd the speed of the train and the bus separately.
Answer:
Let the speed of the train be x kmph and the speed of the bus is y km/hr.
c d W.K.T. S = d/t
t = d/s Total distance travelled = 300 km.
He takes 4 hours for 60km and remaining 240 km by bus.
Here, t = 4 hours
60p + 240q = 4 …….. (1)
If she travels 100km by train and the remaining 200 km by bus she takes 1 Omin means 4 hours 10min.
100 p + 200 q= 25/9
600 p + 1200q = 25 ….. (2)
Solve (1) and (2)
(60p + 240q = 4) ÷ 4
(600p+ 12000q=25) ÷ 25
∴ x = 60kmph
y = 80kmph
∴ speed of train = x = 60km/hr
speed of bus = y = 80km/hr
Question 26.
Find the zeroes of the polynomial
f(y)= y3 – 5y2 – 2y + 24, if it is given that the product of its two zeroes is 12.
Answer:
Let α, β , y be the zeroes of the polynomial F(x) such that α, β = 12.
Substituting αβ = 12 in αβ r = -24,
12 r = -24
r = -24/12 = -2
α + β + r = 5
α + β – 2 = 5
α + β = 5 + 2 = 7
(α – β)2 – (α + β)2 – 4αβ
= (7)2 – 4(12)
= 49 – 48
= ±1
∴ α – β = +1
α + β= 7 and α – β= 1 …… (1)
α + β = 7 and α – β = 1 ….. (2)
solve (1)
α = 4
α + β = 7
4 + β = 7
β = 7 – 4
β = 3
solve (2)
α = 3
α + β = 7
3 + β = 7
β = 7 – 3
β = 4
∴ The zeroes of the polynomial are 3, 4, -2
Question 27.
Two pipes running to can fill a cylindrical tank in 3[latex]\frac{1}{13}\)min. If one pipe takes 3 minutes more than the other to fill it, find the time, in which each pipe would fill the cylindrical tank.
Answer:
Let the bigger pipe takes x minutes to fill the cylindrical tank, then the smaller pipe would take (x+3) min to fill the tank.
Since, the time taken to fill both the pipe 3\(\frac{1}{13}\)
40(2x + 3)= 13(x) (x + 3)
80x + 120 = 13 (x2 + 3x)
80x + 120 = 13x2 + 39x
13×2+ 39x – 80x – 120 = 0
13×2 – 41x – 120 = 0
solve by formula method
∴ Bigger pipe takes 5min to fill the tank smaller pipe takes (x+3) = 5+3=8 min to fill the tank.
OR
The difference of squares of two numbers is 180, The square of the smaller number is 8 times the larger number, find the two numbers.
Answer:
Let the bigger number be y and smaller number be x
y2 – x2 = 180
x2 = 8y
Consider, y2 – x2 = 180
y2 – 8y – 180=0
y2 – 18y+ 10y – 180=0
y(y – 18) + 10(y – 18) = 0
(y – 18) (y + 10) =0
y – 18 = 0 or y + 10=0
y = 18 or y = -10
We consider the positive value y = 18. Consider
x2 = 8y
x2 = 8(18)
x2 =144
x = √44
x = 12
∴ The two numbers are x = 12, y = 18
Question 28.
The vertices of A PQR are P(6, -2) Q(0,-6) and R(4,8) find the co-ordinates of mid points of PQ & PR and QR.
Answer:
Let the mid points of PQ, QR, PR be (x1, y1) (x2, y2)(x3, y3)
∴ The co-ordinates of D
∴ The co-ordinates of the mid-points of AB, BC and AC are D(3, -4) E(2, 1) F (5, 3)
OR
If point C(-1, 2) divides internally the line segments joinng the points A(2,5) and B(x,y) in the ratic 3:4. find the value
∴ (x, y) = (-5, -2)
x2 + y2 = (-5)2 + (-2)2
= 25 + 4
=29
Question 29.
Prove that the tangents drawn to a circle from an external point are equal and are equally inclined from the line joining the centre and the external point.
Data: PA and PB are the line tangets drawn from an external point P to the circle centred O.
OA and OB are radius. OP joins the centre and the external point.
To prove that (1) PA = PB
(2 ) ∠AOP – ∠BOP
Proof: In A AOP and A BOP
∠OAP = ∠OBP = 90° (∵ OA ⊥ AP, OB ⊥ BP)
OP = OP (∵ common side)
OA = OB (∵ Radii of the same circle)
By RHS postulate
∆AOP ≅ ∆BOP
1) AP = BP
2) ∠AOP =∠BOP
3) ∠OPA =∠OPB
Question 30.
Find the area of the shaded region. In the given figure common between the two quadrants of circles of radius 8cm each. r, 8cm each.
Answer:
OR
In fig. ABC is a quadrant of a circle of radius 14cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. B
Answer:
In ABC, ∠A = 90°
BC2 = AB2 + AC2
BC2 = 142 + 142
BC2= 196 + 196
BC2 =2 x 196
BC = \(\sqrt{2 \times 196}=14 \sqrt{2}\) cm
Area of semi circle taking BC as diameter
Area of quadrant ABDC with radius 14cm.
Area of shaded region = Area of semicircle – (Area of sector ABDC) – (Area of ∆ ABC)
= 154 – (154 – 98)
= 154 – 154 + 98
= 98 sq.cms.
Question 31.
The following table gives production yield per hectare of wheat of 100 gharms of cillage
| Production yield (in kg/ha) | Number of farms |
| 50-55 | 2 |
| 55-60 | 8 |
| 60 – 65 | 12 |
| 65 – 70 | 24 |
| 70 – 75 | 38 |
| 75 – 80 | 16 |
Change the distribution to a more than type distribution, and draw its ogive.
Scale
X axis = 5 cm Y axis = 10 cm
Question 32.
Calculate the mean for the following data
Answer:
Question 33.
Construct a triangle ABC in which AB = 8cm, BC = 10cm and AC = 6cm. Then construct another triangle whose side are 4/5 of the corresponding sides of ∆ ABC.
Answer:
V. Answer the following ( 4 x 4 = 16 )
Question 34.
Solve graphically : 2x – y = 2 and 4x – y = 4
Answer:
Solve the pair of linear equations graphically
2x – y=2
x = \(\frac{2+y}{2}\)
4x – y = 4 .
x = \(\frac{4+y}{4}\)
x = 2/2 = 1
2x – y = 2
2(1) – y = 2
2 – y = 2
-y2 = 2
-y = 0
y = 0
Question 35.
If the sum of first 7 term of an A.P is 49 and that of 17 terms is 289. Find the sum of first “n” terms.
Answer:
S = 49 S17 =289 Sn =?
Use the formula Sn = \(\frac{n}{2}\) [2a+(n-1)d]
S7 = \(\frac{7}{2}\) [2a + (7 – 1)d]
S7 =\(\frac{7}{2}\) [2a + 6d]
S7= \(\frac{7}{2}\) x 2 [a+3d]
S7 = 7(a+3d)
49/7 = a + 3d
a + 3d = 7 …… (1)
S17 = 289
\(\frac{17}{2}\) [2a+(17 – 1)d)}=289
\(\frac{17}{2}\) [2a+16d]=289
\(\frac{17}{2}\) x 2(a + 8d)=289
solve (1) and (2)
d = 10/15
d = 2
Consider
a + 3d = 7
a + 3 (2) = 7
a + b = 7
a = 7 – 6
a = 1
Sum of n terms
Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
Sn = \(\frac{n}{2}\) [2(1) + (n – 1)2]
Sn = \(\frac{n}{2}\) [2 + 2n – 2]
Sn = \(\frac{n}{2}\) 2n
Sn = n2
OR
The sum of the third and seventh terms of an AP is 6 and their product is 8. find the sum of first sixteen terms of the A.P.
Answer:
Let the seven terms of an A.P. be (a-4d), (a- 3d), (a-2d), (a-d), a(a+d), (a+2d)
Given :
T3 + T7 = 6
a – 2b + a + 2d = 6
2a = 6
a = 6/2
a = 3
Given : T3 X T7 = 8
(a-2d) (a+2d) = 8
a2 – (2d)2 = 8
(3)2 – 4d2 = 8
9 – 8 = 4d2
4d2 = 1
d2 = 1/4
d = ± \(\sqrt{\frac{1}{4}}\)
T16 + (16 – 1 )d
T16 = 1 + 15(1/2)
d = ± \(\frac{1}{2}\)
If d = \(\frac{1}{2}\) and T1 = (a-4d)
∴ Ti = a1 = (a – 4d)
= ( 3-4 x \(\frac{1}{2}\))
= (3 – 2)
T1 = a1 = 1
Question 36.
A person, walking 20mts from a point towards a flagpost along a horizontal passing through its base, observes that its angle of elevation changes from 30° to 45° Find the height of the flag – post.
Answer:
Let AB be the flag post and OA be the horizontal passing through A. Let the man start from m and go towards A.
Let MN = 20 mtrs
from AAMB;
tan 30° = AB/AM AB = AM. tan 30°
From ANB; tan 45° = AB/AN
AB = AN tan 45°
∴ AM tan 30° = AN, tan 45°
But AM = MN + AN
(MN + AN) tan 30° = AN tan 45°
Question 37.
In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other sides. Prove the statement.
Answer:
Refer CPC Model Question Paper – 6 (Q-No-38)
In a right angled triangle, square on the hypotenuse is equal to sum of A the squares on the other sides.
Data: ∆ ABC,∠B = 90°
T.P.T : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof: In ∆ ABC and ∆ ABD
∠ABC = ∠ADB = 90° [∵ Data & construction]
∠A = ∠A (Common angle)
∠ACB = ∠ABD[∵ Re maining angle]
∆ ABC and ∆ ABD are equiangular
∆ ABC ∼ ∆ ABD
AB2 = AC x AD ………(1)
In ∆ABC and ∆BDC
∠ABC = ∠BDC = 90°[∵ Data&construcnon]
∠C=∠C=9O° (∵ Common angle)
∠BAC = ∠DBC (∵ Rcmaining angle)
∆ ABC and ∆ BDC are equiangular
∆ ABC ∆ BDC
AB2 + BC – AC (AD + DC)
AB2 + BC2 = AC( AC)
AB2 + BC2 = AC2
AC2 = AB2 + BC2
VI. Answer the followings : ( 5 x 1 = 5 )
Question 38.
A frustum of a cone has its radii of 15cm and 7cm with a height of 6cm is completely filled with water, find the volume of water and curved surface area of frusturm (Take π 3.14)
Answer:
r = 15cm r = 7cm, h = 6cm, v =?
V = \(\frac{1}{3}\) π h(R2 + r2 + Rr)
= \(\frac{1}{3}\) x 3.14 x 62(152 + 72 +15 x 7)
=6.28(225 + 49 + 105)
=6.28(379)
=2380.12cm3 =2.380.12L
Slant height of frustum (l)
Curved surface area of frustum = πl(R + r)
= 3.14 x 10 x (15+7) = 31.4 x 22 = 690.8cm2