Download question paper of maths for class x

Model Paper -6 (2016-17)
SUMMATIVE ASSESSMENT - 2
CLASS X
MATHEMATICS
BLUE PRINT
Sr.
No.
Name of unit
1 mark
2 marks
3 marks
4 marks
Total marks
1.
Algebra
1
2
2
3
23
2
Geometry
1
0
4
1
17
3
Co-ordinate
Geometry
0
0
1
2
11
4
Trigonometry
2
8
5
Probability
1
2
1
0
8
6
Mensuration
1
2
2
3
23
total
4(1)
6(2)
10(3)
11(4)
90
MODEL PAPER-6 (2016-17)
SUMMATIVE ASSESSMENT - 2
CLASS X
MATHEMATICS
Time: 3hrs
General Instruction:-
Max. Marks: 90
1. All questions are Compulsory.
The question paper consists of 31 questions divided into 4 sections, A,B,C and D. Section –
A comprises of 4 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks
each and Section -C consists 10 questions of 3 marks each and Section- D comprises of 11
questions of 4 marks each.
Question numbers 1 to 4 in Section –A are very short answer questions There is no overall
choice. Use of calculator is not permitted.
SECTION-A
1. If P (E) = 0.05, what is the probability of ‘not E’?.
2. Find 30 th term of A.P 10,7,4...........
3. How many tangents can a circle have?
4.Find the area of the quadrant of a circle of radius 14 cm.
SECTION-B
5. Find the values of k for each of the following quadratic equations, so that they have two
equal roots.
2x2 + kx + 3 = 0
6. Write the common difference of an A.P whose nth term is 3n+5
7. A die is thrown once. Find the probability of getting
(i) a prime number
(ii)a number beween 2 and 6
8. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
9. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O.
If angle AOB = 30°, find the area of the shaded region.
10. Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting
cuboid.
SECTION-C
11. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides
are 2/3 of the corresponding sides of the first triangle.
12. Find the roots of the following equation:
13. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
14. Find the point on the x-axis which is equidistant from (2, − 5) and (− 2, 9).
15. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a
distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q
16. (i)A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is
the probability that this bulb is defective?..
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at
random from the rest. What is the probability that this bulb is not defective?
17. Prove that the parallelogram circumscribing a circle is a rhombus.
18. In the given figure, ABCD is a square of side 14 cm. With centers A, B, C and D, four circles
are drawn such that each circle touches externally two of the remaining three circles. Find the
area of the shaded region.
19.Metallic spheres of radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid
sphere. Find the radius of the resulting sphere.
20 In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another
tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
SECTION D
21. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have
taken 1 hour less for the same journey. Find the speed of the train.
22. Find the sum of first 15 multiples of 8.
23. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
24. Find the roots of the following quadratic equations, if they exist, by the method of completing
square.
2x2 – 7x + 3 = 0
25. Prove that the lengths of tangent segment drawn from an external point to a circle are equal
26. Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose
vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
27. A chord of a circle of radius 12 cm subtends an angle of 1200 at the Centre. Find the area of the
corresponding segment of the circle.
28.A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the
hemisphere can have? Find the surface area of the solid.
29.The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular
ends are 18 cm and 6 cm. find the curved surface area of the frustum.
30. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of
elevation of the top of the statue is 60° and from the same point the angle of elevation of the top
of the pedestal is 45°. Find the height of the pedestal.
31(i).A tree breaks due to storm and the broken part bends so that the top of the tree touches the
ground making an angle 30 ° with it. The distance between the foot of the tree to the point
where the top touches the ground is 8 m. Find the height of the tree.
(ii) What is the importance of plants.
Sample Paper -14 (2014-15)
SUMMATIVE ASSESSMENT - 2
CLASS X
MATHEMATICS
Marking Scheme
SECTION-A
Ans 1
1 mark
P(E) = 0.05
Ans.2.
1 mark
30th term of the A.P: 10, 7, 4, …, is
Here
Therefore
Ans.3.Infinitely many
1 mark
Ans .4 Area of quadrant=1/4 πr2
½ mark
On solving we get,area of quadrant =154 cm2
½ mark
SECTION-B
Ans 5.
We have; a = 2, b = k and c = 3
For equal roots; D should be zero.
Hence;
1mark
1 mark
Ans.6:
Given that, an =3n+5
Put n=1, 2, 3, 4......
a1= 8
a2=11
a3=14 and so on
d= a2-a1
1 mark
or a3-a2 =3
Therefore, the common difference is 3.
1 mark
Ans.7:
Total number of events = 6
Number of prime numbers = 3
(i)
1 mark
(ii) Number of numbers between 2 and 6 = 3
1 mark
Ans.8: Total number of events = 52
(i)Number of red king = 2
1 mark
(ii) Number of face cards in a pack = 12
1 mark
Ans.9:
Area of shaded region
= Area of sector of bigger circle – Area of sector of smaller circle
Area of sector of bigger circle
Area of sector of smaller circle
1 mark
Area of shaded region
1 mark
Ans. 10:
Given that
Volume of cubes = 64 cm3
(Edge) 3 = 64
Edge = 4 cm
1 mark
If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.
1 mark
SECTION-C
Ans.11:
2 marks for correct construction and 1 mark for steps
Ans.12:
We have; a = 1, b = - 3 and c = - 1
Root can be calculated as follows:
1 mark
2 marks
Ans.13
Given, a11 = 38 and a16 = 73
We know that an = a + (n – 1)d
Hence, a11 = a + 10d = 38
And, a16 = a + 15d = 73
Subtracting 11th term from 16th term, we get following:
a + 15d – a – 10d = 73 – 38
Or, 5d = 35
Or, d = 7
1 mark
Substituting the value of d in 11th term we get;
a + 10 x 7 = 38
Or, a + 70 = 38
Or, a = 38 – 70 = - 32
Now 31st term can be calculated as follows:
a31 = a + 30d
= - 32 + 30 x 7
1 mark
= - 32 + 210 = 178
1mark
Ans. 14:
We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
`1 mark
By the given condition, these distances are equal in measure.
Therefore, the point is (− 7, 0).
2 marks
Ans. 15:
Draw a circle with radius 3 cm.
2 marks for correct construction and 1 mark for steps
Ans.16
(i) Total Number of events = 20
Number of favourable events = 4
1.5 marks
(ii) Total number of events = 19
Number of favourable events = 15
1.5 marks
Ans. 17
Since ABCD is a parallelogram,
AB = CD …(1)
BC = AD …(2)
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus
Ans. 18:
1 mark
1 mark
1 mark
Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius.
Area of each sector
1 mark
Area of square ABCD = (Side)2 = (14)2 = 196 cm2
Area of shaded portion = Area of square ABCD − 4 × Area of each sector
1 mark
1 mark
Ans. 19:
Radius (r1) of 1st sphere = 6 cm
Radius (r2) of 2nd sphere = 8 cm
Radius (r3) of 3rd sphere = 10 cm
Let the radius of the resulting sphere be r.
The object formed by recasting these spheres will be same in volume as the sum of the volumes of
these spheres.
Volume of 3 spheres = Volume of resulting sphere
1 mark
2 marks
Ans. 20:
In ΔAPO and ΔACO
AP = AC (tangents from a point)
OP = OC (radii)
OA = OA (common side)
Hence; ΔAPO ≈ ΔACO
So, ∠ PAO = ∠ CAO
1.5 marks
Hence; AO is the bisector of ∠ PAC.
Similarly, it can be proved that
BO is the bisector of ∠ QBC
Now, XY || X’Y’ (given)
So, ∠ AOB = Right angle
1.5 marks
SECTION-D
Ans. 21:
Let us assume, speed of train = x
We know; time = distance/speed
In case of normal speed;
1 mark
In case of increased speed;
1 mark
From above equations;
2 marks
x = 40 and x = - 45
Discarding the negative value; we have speed of train = 40 km/h
Ans. 22:
The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d=8
S15 =?
marks
1.5
2.5 marks
= 960
Ans. 23:
Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a
parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.
Therefore, O is the mid-point of AC and BD.
1 mark
If O is the mid-point of AC, then the coordinates of O are
1mark
If O is the mid-point of BD, then the coordinates of O are
1 mark
Since both the coordinates are of the same point O,
1 mark
Ans.24: 2x2 – 7x + 3 = 0
Checking the existence of roots:
We know;
Since D > 0; hence two different roots are possible for this equation.
Now; 2x2 – 7x + 3 can be written as follows:
1 mark
1 mark
Assuming x = a and 7/4 = b; the LHS of equation is made in the form of (a – b)2
1 mark
Case 1:
Case 2:
Hence; x = 3 and x = ½
Ans. 25:
Given ; A circle with centre O
1 mark
To Prove: PQ = PR
Construction: Draw a circle with centre O. From a point P outside the circle, draw two tangents P
and R
1.5 marks
Proof: In Δ POQ and Δ POR
OQ = OR (radii)
PO = PO (common side)
∠ PQO = ∠ PRO (Right angle)
Hence; Δ POQ ≈ Δ POR
PQ = PR
proved
Ans. 26
A = (0, -1), B = (2, 1), C = (0, 3)
Coordinates of midpoint of AB can be calculated as follows:
Coordinates of midpoint of BC:
2.5 marks
Coordinates of midpoint of AC:
1 mark
We have; D = (1, 0), E = (1, 2), F = (0, 1
Area of triangle DEF:
= ½ [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= ½ [1(2 – 1) + 1 (1 – 0) + 0(0 – 2)]
= ½ (1 + 1) = 1 sq unit
1 mark
Area of triangle ABC:
= ½ [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= ½ [0(1 – 3) + 2(3 + 1) + 0( - 1 – 1)]
= ½ (0 + 8 + 0 ) = 4 sq unit
1 mark
Ratio of areas = 1 : 4
.
1 mark
Answer:27
O
A
D
B
Here OA = OB = r = 12cm , <AOB =1200
In triangle AOB ,AO= BO = 12cm
Draw OD Perpendicular to AB
Since triangle AOB is an isosceles triangle and OD is perpendicular to AB
Therefore OD is the angle bisector as well as median .
Now in right triangle ADO < D = 900
< AOD = 600
OD/OA = Cos 600 = ½
OD = AO X ½ = 12 X ½ = 6cm
And AD/AO = Sin600
AD = 12 x√3/2 =6√3
AB =2AD =12√3cm
2 marks
Area of minor segment=Area of sector AOB – Area of triangle AOB
= 𝜃/3600 X 𝜋r2 – ½ X AB X OD
= 1200/3600 X 3.14 X 12 X12- ½ X 12√3 X 6
= 150.72 – 62.28 = 88.44 cm2
marks
2
Ans.28
From the figure, it can be observed that the greatest diameter possible for such hemisphere is
equal to the cube’s edge, i.e., 7cm.
Radius (r) of hemispherical part =
= 3.5cm
1 mark
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
− Area of base of hemispherical part
= 6 (Edge)2
−
= 6 (Edge)2 +
1.5
marks
1.5 marks
Ans. 29:
Perimeter of upper circular end of frustum = 18
2πr1 =18
Perimeter of lower end of frustum = 6 cm
2πr2 = 6
2 marks
Slant height (l) of frustum = 4
CSA of frustum = π (r1 + r2) l
2 marks
Ans. 30
In this figure; AD is the height of the statue which is 1.6 m, DB is the height of the pedestal and C
is the point where observer is present.
In Δ ABC;
2 marks
In Δ DBC;
From equation (1) and (2);
½ mark
1.5 marks
Ans.31
Let AC was the original tree. Due to storm, it was broken into two parts. The broken part
making 30° with the ground.
In
,
(2)
Height of tree =
+ BC
Hence, the height of the tree is
(ii) One mark for correct answer.
.
is
Group No. 7
1.Mr. Ram Lal KV Samba. (GL)
2.Mr. Som Raj KV Tibri Cantt.
3.Mr Sunil Aggarwal KV 1 Jalandhar.
4.Mrs Radha Devi. KV 4 Pathankot.
5 Mr. Sohan Lal KV No 3 Pathankot