Model Paper -7 (2016-17)SUMMATIVE ASSESSMENT - 2

CLASS XMATHEMATICSBLUE PRINT

S.No

Unit Chapter VSA SA-1 SA-2 LA Total

1 Mark Each

2 Marks Each

3 Marks Each

4 Marks Each

1 ALGEBRA QUADRATIC EQUATIONS

1(1) 2(4) 2(8) 8(23)

ARITHMETIC PROGRESSIONS

2(6) 1(4)

2 GEOMETRY CIRCLES 2(2) 1(2) 1(3) 1(4) 5(17)CONSTRUCTIONS

2(6)

3 TRIGONOMETRY

HEIGHTS AND DISTANCES

2(8) 3(8)

4 PROBABILITY

PROBABILITY

2(4) 1(4) 4(8)

5 COORDINATE GEOMETRY

COORDINATE GEOMETRY

1(1) 2(6) 1(4) 4(11)

6 MENSURATION

AREAS RELATED TO CIRCLES

2(6) 1(4) 7(23)

SURFACE AREAS AND VOLUMES

1(2) 1(3) 2(8)1*(4)

Total 4(4) 6(12)

10(30)

11(44)

31(90)

Model Paper -7 (2016-17)SUMMATIVE ASSESSMENT - 2

CLASS XMATHEMATICS

CLASS : XTime: 3hrs Max.Marks: 90General Instructions

1. All questions are Compulsory.2. 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, Section C comprises of 10 questions of 3 marks each and Section- D comprises of 11 questions of 4 marks each.3. There is no overall choice.4. Use of calculator is not permitted.

Section – A

Q1. Find the roots of the following quadratic equationx2-3x-10=0.

Q2. If the tangents PA and PB drawn from a point P to a circle with center O are

inclined to each other at an angle of 1100, find POA.

Q3.Find the mid- point of a line segment joining the point A (-2, 8) and B (-6, -4) ..

Q4.How many tangents can a circle have ?

SECTION – BQ5. Find the value of K such that the quadratic equation (K – 12) x2 – 2 (K-12) x + 2

= 0 has equal roots.

Q6.Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, andhence find the nature of its roots.Q7. Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.Q8. Two coins are tossed together. Find the probability of getting at least one tail..

Q9. Two friends were born in the year 2000. What is the probability that they have

the same birthday?

Q10 A chord AB of the larger of the two concentric circles is tangent to the smaller

circle at the point.P. show that Pis the midpoint of the chord AB.

Section – CQ11.Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Q12. Find the value of y for which the distance between the points P(2-3) and

Q(10,y) is 10 units.

Q13. Determine the sum of first 35 terms of AP if its second term is 2 &seventh

term is 22.

Q14.If (1,2),(4,b),(a,6),and (3,5) are the vertices of a parallelogram taken in order,

then find the values of a and b.

Q15. On a square handkerchief, nine circular designs each of radius 7cm are made.

Find the area of the remaining portions of the handkerchief.

Q16. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding side of ΔABC.

Q17. A cone of height 24 cm and radius of base 6 cm is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

Q18. How many terms of the series 54 + 51 + 48 + 45 + ----. must be taken in

orderto make the sum 513 ?

Q19. A circle touches the side BC of a ABC at P and the extended sides AB & AC

at Q and R, respectively. Prove that AQ =

12 [BC + CA + AB]

Q20. A paper is in the form of a rectangle ABCD in which AB = 16cm & BC =

12cm. A semi-circular portion with BC as diameter is cut off. Find the area of

remaining part.

Section – DQ21. A motor boat whose speed is 15 Km/h in still water, goes 30 Km downstream

& comes back in a total time of 4 h 30 minutes. Find the speed of the stream.

Q22. Find the sum of the odd numbers between 0 and 50.

Q23. ABCD is a square of side 4cm. At each corner of the square a quarter circle of

radius 1cm and the centre of a circle of radius 1cm are drawn as shown in the

figure. Find the area of the shaded region.

Q24. As observed from the top of light house, 100m high above sea level, the angle

of depression of a ship sailing directly towards it, changes from 300 to 600.

Determine the distance travelled by the ship during period of observation.

Q25. If the vertices of a triangle are (1, k), (4, -3), (-9, 7) and its area is 15 sq units, find the value of k..

Q26. Prove that the length of tangents drawn from an external point of a circle are equal.

Q27. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

28. A Cone of radius 8cm and height 12cm is divided into two parts by a plane

through the midpoint of its axis parallel to its base. Find the ratio of the volumes of

two parts.

Q29 A die is thrown once. Find the probability of getting.

(i) a prime number

(ii) a number divisible by 2

(iii) a multiple of 3

Q30. Find the roots of the quadratic equation.

1x + 4

− 1x − 7

= 1130 , x -4, 7

Q31. Mahesh owns a juice shop, he sales the juice in glasses of 3 different shapes,

radius of each glass is 3cm and height is 12cm. (Use π = 3 . 14 )

(i)

Type A (cylindrical glass)

(ii)

Type B (hemispherical glass)

(iii)

Type C (conical glass)

Mahesh decided to serve juice to the customer in glass of type A.

\(a) Find the volume of the glass of type A.

(b) Which type of glass has minimum capacity?

(c) Which mathematical concept is generally used to solve the above problem?

(d) While serving, Mahesh chose glass of type A. What type of nature is depicted

by his choice of glass?

Model Paper -7 (2016-17)SUMMATIVE ASSESSMENT – 2

CLASS XMATHEMATICS

Marking Scheme

Sol.1 : Here x2-3x - 10 = 0

x2-5x+ 2x-10 = 0

x (x-5) +2 (x-5) = 0

(x-5)(x+2) = 0

x-5= 0 x+2 = 0

x=5 x = -2 (1)

Sol.2∵ OA ⊥ PA , ∠ APB = 90 0

∵ OP bi sec t ∠ APB

∴ ∠ OPA = 500

In Δ OAP

∠ POA =1800 −(∠ OAP +∠ OPA )

=1800 −( 900 + 550 )¿1800 − 1450 =350

(1)

Sol 3.Let P(x,y) be a midpointThen by mid point formula. x= -6-2/2 = -4

y= 8-4/2 = 2 (-4,2) (1)

SOL 4. Infinitely many tangents. (1)

SECTION -B

Sol 5. (K-12)x2 -2(K-12) x +2 =0

Here ,a=K-12 , b=-2(K-12) ,c=2

Since the quadratic equation has equal roots, therefore D=0

b2-4ac =0

[-2(K-12)]2 -4(K-12)(2) =0

(1)

4(K-12) -8(k-12) =0

4(K-12) [ K-12-2] =0

4(K-12)(K-14) =0

K=12 or K=14

(1/2)

When we put K=12 in a given quadratic equation coefficient of x2 becomes zero,

which is not possible , hence the value of K=14.

(1/2)

Sol.6 The given equation is of the form ax2 + bx + c = 0, where a = 2, b = – 4 and (1/2)c = 3. Therefore, the discriminantb2– 4ac = (– 4) 2– (4 × 2 × 3) = 16 – 24 = – 8 < 0 (1)So, the given equation has no real roots.

(1/2)

Sol.7 Given thatVolume of cubes = 64 cm3

(Edge) 3 = 64Edge = 4 cm (1/2)

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm. (1/2)

Sol.8 When two coins are tossed together, then the number of possible

outcomes= 4

i.e. ( HH, HT, TH, TT)

(1)

No. of favourable outcomes (at least one tail)

= 3 (HT, TH, TT)

(1)

Required Probability =

34

Sol 9.Total number of days in 2000 = 366

Total number of ways in which two friends have their birthdays

= 366x 366 (1)

Number of ways in which both have same birthday = 366

Probability that both have same birthdays

(1)

Sol 10. Given, two conentriccircles with centreO , a chord AB is the larger circle of a

tangent to the smaller circle at the point.P. (1/2)

To Prove : AP = PB

(1/2) Proof:- Since OP is the radius of the smaller circle and APB is

the tangent to the smaller circle OP

bisects AB

P is the mid pt. of the Chord AB (1)

AP =PB

Hence proved

Section – C

Sol 11.

=366366 × 366

= 1366

(3)

Sol 12.: Given P(2,-3)and Q(10,y)

P 10 units QPQ=10 (1)PQ2 =102 =100Therefore (10-2)2+(y-(-3))2=100 =>82+(y+3)2=100 =>64+y2+6y+9=100 =>y2+6y-27=0 (1) =>y2+9y-3y-27=0 =>y(y+9)-3(y+9)=0 =>(y+9)(y-3)=0 (1) =>Either y=-9 Or y=3Hence the required value of y can be -9 or 3

Sol .13 Let the AP be a, a + d, a + 2d……….. (1)

T2 = a + d = 2

T7 = a + 6d = 22

On subtracting T2 from T7, we get

5d = 20, d = 4

On Putting d = 4 in T2 , we get

a + 4 = 2 (1)

a = -2

S35 =

352

[−4 + 136 ]

Using

=

352

x 132 = 2310 (1)

Sol. 14.

We know that the diagonals of a parallelogram bisect each other therefore, mid

points coincide each other.

Let P (1, 2) Q (4,b) R (a, 6) S(3,5) are the vertices of a parallelogram PQRS

Now using midpoint formula For

diagonal PR

⇒M [ a + 1

2, 6 + 2

2 ] (1)

M [ a + 12

, 4 ]For diagonal QS

M [4 + 32

, b + 52 ]

(1)

Sol. 15Radius of Circle = 7cm

Length of the side of square = 3 (diameter)

Length of the side of square = 3 (diameter) (1)

=3(14) = 42cm

Area of Square = (Side)2 = 422 = 1746cm2

Area of 1 circle = r2

22 (7 )7

2

= 154 cm2

Area of a circles = 9 x 154 = 1386 cm2

(1)

Remaining Portion = Area of handkerchief - Area of Design

= 1764 - 1386 = 378cm2

Hence Area of remaining portion = 378cm2

(1)

Sol.16. 1)

The required triangle can be drawn as follows.

Step 1Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.Step 2Draw a ray BX making an acute angle with BC on the opposite side of vertex A.Step 3Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.Step 4Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C'.Step 5Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.

Sol17.

(1)

(1)

Sol 18. a = 54, d = 51 -54 = - 3

Let the sum of n terms of given series be 513

Then Sn =

n2 [2a + (n − 1)d ]

513 =

n2 [2 × 54 + ( n − 1 )−3 ]

(1)

⇒1026 = 111n – 3n2

⇒3n2 – 111n + 1026 = 0

⇒n2 – 37n + 342 = 0

⇒n – 18n -19n + 342 = 0

⇒ (n – 18) (n -19) = 0 (1)

⇒ n = 18, 19

18th term = 54 + (18 – 1) (-3) = 3

19th term = 54 + (19 – 1) (-3) = 0

The sum of 18th& 19th will be 513.

(1)

Sol.19. Here BQ = BP [length of tangents drawn from an external point to a circle

are equal] (1)

Similarly

CP = CR and AQ = AR

(1)

⇒2AQ = AQ + AR

2AQ = AB + BP + AC + CP

⇒2AQ = (BP + CP) + (AC + AB)

⇒2AQ = BC + CA + AB

⇒AQ = ½ [BC + CA + AB]

(1)

Sol.20. Length of rectangle ABCD = 16 cm

Breadth of rectangle ABCD = 12 cm

Area of rectangle = (16 x 12)cm2 = 192cm2

(1)

Radius of semi circle = 6cm

Area of remaining part

Area of rectangle ABCD - Area of Semi-circle with radius 6cm (1)

⇒192cm2 -

12π r2cm2

⇒192cm2– 1/2 x(22/7)x7x7

⇒192cm2 – 11 x 7cm2

⇒192cm2 - 77cm2

¿115cm2

(1)

Section – D

Sol.21. . Let the speed of stream be x km/h

(1)

Speed downstream = (15 + x) Km/h

Speed upstream = (15 – x) Km/h

Distance downstream = 30 Km

Distance upstream = 30 Km

Total time taken = 4h 30m

(1)

= (4 +

12)h =9

2h

Time taken by boat in downstream and upstream be

P= 30

15+x , Q= 30

15−x respectively.

Acc to Question

(1)

3015 +x

+ 3015 − x

= 92❑❑

1015 +x

+ 1015 − x

= 32

10 (15 − x + 15 + x )(15 + x ) (15 − x )

= 32

200 = 225 – x2

X2 = 25

x = +5

As speed = -5 is neglected

Speed of stream = 5km/h (1)

Sol22.The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49 (1)Therefore, it can be observed that these odd numbers are in an A.P.a = 1d = 2

l = 49 (1)l = a + (n − 1) d49 = 1 + (n − 1)248 = 2(n − 1)n − 1 = 24 (1)n = 25

= 625 (1)

Sol 23. Area of a square with side 4 cm = 4×4 = 16cm2 (1)

Area of smaller circle of radius

22

= 1cm ]

π ×1 ×1 = 3 .14 cm2

(1)

Area of sector APQ =

π × AP2 × θ3600

=3.14 x 1 x 1 x

900

3600 = 14

×3.14 cm 2

(1)

Area of such 4 sectors at each corner of square = 4 × 1

4× 3 .14 = 3014cm2

Area of shaded portion = area of square – (area of smaller circle + area of 4

such sectors) = 16 – (3.14 + 3.14)

= 16 – 6.28

= 9.72cm2

(1)

Sol. 24. Let DC = 100m be the height of light house (1)

Let A and B are the positions of two ships such that CBD = 600

In right angle BCD (1)

BCDC

= Cot 600 = BC100

= 1√3

BC =

100m√3

In rt. Angled ACD

(1)

ACDC

= Cot 300 = AC100

= √3 = AC = 100√3m

Distance b/w two ships = AB

AC – BC

(1)

= 100√3 − 100

√3= 200 × √3

3= 115 . 47m

Sol25.

Let A(1, k) ,B(4, -3)and C(-9, 7) be the vértices of triangle

Area of ABC =

12

[x1 (y2-y3)+x2(y3-y1) + x1(y1-y2)]

=

12 [1(-3-7)+4(7-k)+(-9)(k+3)] = 15

-10 + 28 – 4k – 9k – 27 = 30- 9 – 13k = 30

k = -3

Sol.26.

Given: a circle with centre O, a point P lying outside the circle (1/2) and two tangents PQ, PR on the circle from P (see Fig.). To prove: PQ = PR. (1/2)

Const: join OP, OQ and OR.

(1/2)Proof: < OQP and <∠ ORP are right angles ,because these are angles between the radii and tangents,

Now in right triangles OQP and ORP, (1/2)OQ = OR (Radii of the same circle)OP = OP (Common)Therefore, OQP ≅ ORP (RHS)This gives PQ = PR (CPCT) (2)

Sol27.

(1)

Let AB be a building and CD be a cable tower.In ΔABD,

(1)

In ΔACE,AC = BD = 7 m

(1)

Therefore, the height of the cable tower is7(3+1)m

Sol28.

Let ORN be the cone, then given radius of the base of cone r1 = 8cm

The height of cone h = OM = 120

(1)

Let P be the midpoint of OM,

OP = PM =

122

= 6cm

Now OPD &OMN (1)

OPOM

= PDMN

612

= PD8

= 12

= PD8

PD = 4cm

The plane CD divides the came into two parts namely

(i) A smaller cone of radius 4cm & height 6cm

(ii) Frustum of a cone for which radius of the top of the frustum r1 = 8cm

Radius of bottom r2 = 4cm

(1)

Height of frustum = 6cm

Volume of smaller cone

13

× π × 6 [82 + 42 + 8 × 4 ]cm3

2 π (64 + 16 + 32 ) + 224 π cm2

Required ratio = 32 : 224 =1:7

(1)

Sol29. There are 6 outcomes (1, 2, 3, 4, 5, 6) in a single throw of a die.

(1)

(i) 2, 3, 5 and prime number

(1)

P (a prime number)

= 36

= 12

(ii) The number divisible by 2 are 2, 4, 6

P (a number divisible by 2)= 3

6= 1

2 (1)

(iii) Multiple of 3 are 3 and 6

P (a number multiple of 3) = 26 =

13 (1)

Sol30

Sol.31 Given radius of glass, r = 3cm

height of glass, h = 12cm

Now ,

(i) Volume of glass of type A = r2h (1/2)

3.14 x 3 x 3 x 12

339.12cm3

(ii) Volume of the hemisphere =

23πr3

(½)

23

× 3 . 14×3×3×3

2 × 3 .14 × 956 . 52cm3

Volume of glass of type B = 339.12 – 56.52

= 282.6cm3

(iii) Volume of cone =

13π r2h

(1)

13

× 3 . 14 × 3 × 3 × 0 .5

= 3. 14 × 3 × 0 .5= 4 . 71cm2

Volume of glass of type C = 339.12 – 4.71

= 334.41cm3

(a) Hence, the volume of glass to type A is 339.12cm3

(1/2)

(b) From the above solutions, we get volume of glass to type B has minimum

volume i.e., 282.6cm3

(1/2)

(c)As Mahesh chose glass to type A to serve juice to his customer, it shows his

honesty. (1)

Group-8

1. Mrs Anita Jolly K V SUNJUWAN (GL)

2. MrSuman Kumar K V SARAIKHAS

3. Mrs Anita Bhardwaj K V 2 JALANDHAR

4. MrsHarminderKaur K V 3 JALANDHAR

5. MrsBalwinderKaur K V 4 JALANDHAR

6. MrsSumanKumari K V SURANUSSI