Set - A

MT EDUCARE LTD.SUMMATIVE ASSESSMENT - II

(2015-16) CBSE - X Set - A

• Please check that this question paper contains 5 printed pages.• Code number given on the right hand side of the question paper should be

written on the title page of the answer-book by the candidate.• Please check that this question paper contains 31 questions.• Please write down the serial number of the question before attempting it.

MATHEMATICS

General Instructions:i) All questions are compulsory.

ii) The question paper consists of 31 questions divided in four 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.

iii) There is no overall choice.

iv) Use of calculator is not permitted.

Time allowed : 3 hours Maximum Marks : 90

Series RLH

Roll No. Code No. 31/1

Set - A

SECTION - AQuestion number 1 to 4 carry 1 marks each.

1. If one root of the quadratic equation kx2 – 7x + 12 = 0 is 3, then find thevalue of k.

2. A point P is 13cm from the centre of the circle. The length of the tangentdrawn from P to the circle is 12 cm. Find the radius of the circle.

3. Find the ratio in which the y - axis divides the line segment joining thepoints (5, – 6) and (–1, – 4).

4. A rectangular solid metallic cuboid 9 cm × 8 cm × 2 cm is melted andrecast into solid cube. Find volume of the cube.

SECTION - BQuestion number 4 to 10 carry 2 marks each.

5. In figure, two tangents RQ and RP are drawn froman external point R to the circle with centre O.If PRQ = 120º, then prove that OR = PR + RQ.

6. The string of a kite is 100 metres long and it makes an angle of 60º withthe horizontal. Find the height of the kite, assuming that there is no slackin the string.

7. A bag contains lemon flavoured candies only, Malini takes out onecandy without looking into the bag. What is the probability that shetakes out.(i) an orange flavoured candy(ii) a lemon flavoured candy ?

8. If the mid point of the segment joining the points P(6, b – 2) and Q (– 2, 4)is (2, –3), find the value of b.

9. Two tangents PA and PB are from an external point P to a circle withcentre O, such that APB = x and AOB = y. Prove that opposite anglesare supplementray.

... 2 ...

RO

Q

P

Set - A... 3 ...

10. A right circular cone made of iron is of 8 cm height and has base radius2cm. It is melted and recast into a sphere. Determine the radius of thesphere.

SECTION - CQuestion numbers 11 to 20 carry 3 marks each.

11. Find the roots of the following quadratic equation by factrisation :

2 x2 + 7x + 5 2 = 0

12. If the seven times the seventh term of an A. P. is equal to eleven timesits eleventh term, show that its eighteenth term is zero.

13. Draw a pair of tangents to a circle of radius 5 cm which are inclined toeach other at an angle of 60º.

14. Two circles with centres O and O’ of radii 3cm and 4 cm, respectively intersect at twopoints P and Q such that OP and O’P aretangents to the circles. Find the length of thecommon chord PQ.

15. A vertical tower stands on a horizontal plane and is surmounted by avertical flag - staff of height h. At a point on the plane, the angles ofelevation of the bottom and the top of the flag - staff are and

respectively. Prove that the height of the tower ish tan

tan tan

.

16. The angle of elevation of the top of a tower from a point A on the ground is30º. On moving a distance of 30 metre towards the foot of the tower to apoint B the angle of elevation increases to 60º.Find the height of the tower and the distance of the tower from the point A.

17. Three coins are tossed simultaneously. Find the probaility of getting :(a) Three tails (b) Exactly 2 tails(c) At least 2 tails.

18. (i) A lot of 20bulbs contain 4 defective ones. One bulb is drawn atrandom from the lot. What is the probability that this bulb is defective ?

T5 cm4 cm

3cm

Q

P

O O’

Set - A... 4 ...

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Nowone bulb is drawn at random from the rest. What is the probability thatthis bulb is not defective ?

19. A cylindrical pipe inner diameter of 7cm and water flows through it at192.5 litres per minute. Find the rate of flow in kilometers per hour.

20. In the given figure, find the area of the shadedregion, where AC = 13 cm, AB = 12 cm, O iscentre of the circle and ABC = DAB = 90º.(Use = 3.14)

SECTION - DQuestion numbers 21 to 31 carry 4 marks each.

21. Solve for x : 1

1 2x x + 1

2 3x x + 1

3 4x x =16

22. If the sum of the first of n terms an AP is 4n – n2, what is the first term (Thatis S1) ? What is the sum of first two terms ? What is the second term?Similarly, find the 3rd, the 10th and the nth terms.

23. Three consecutive odd natural numbers are such that the product of thefirst and third is greater than four times the middle by 1. Find the numbers.

24. A contract on construction job specifies a penalty for delay of completionbeyond a certain date as follows : Rs 200 for the first day 250 for thesecond day, Rs 300 for the third day, etc., the penalty for each succeedingday being Rs 50 more than for the preceding day. How much money thecontractor has to pay as penalty, if he has dealyed the work by 30 days ?

25. The radii of two concentric circles are 16cmand 10 cm. AB is a diameter of the bigger circle.BD is tangent to the smaller circle touching itat D. Find the length of AD.

O

D C

BA

B

A

ED

O

Set - A... 5 ...

26. Two circles touch each other externally. The sum of their areas is149 cm2 and the distance between their centres is 17 cm, Find the radiiof the circles.

27. If A (5, –1), B (–3, –2) and C (–1, 8) are the vertices of triangle ABC, find thelength of median through A and the coordinates of the centroid.

28. Show that the points A (1, 2), B (5, 4), C(3, 8) and D (–1, 6) are the verticesof a square.

29. In figure ABC is a right triangle right angled at A. Find the area of shadedregion if AB = 6cm, BC = 10cm and O is the centre of the incircle of ABC(Take = 3.14)

30. A container open at the top, is in the form of a frustum of a cone of height24 cm with radii of its lower and upper circular ends as 8 cm and 20 cmrespectively. Find the cost of milk which can completely fill the container

at the rate of Rs 21 per litre22Use7

31. A solid metalic right circular cone 20cm high with vertical angle 60º iscut into two parts at the middle point of its height by a plane parallel tothe base. If the frustrum, so obtained be drawn into a wire of diameter

116 cm, find the length of the wire.

All the Best

O

A

BC

30º 30º10cm

10cm

A’O’

B’

OBA

r2

r1

V

Set - A

Any method mathematically correct should be given full credit of marks.SECTION - A

Question number 1 to 4 carry 1 mark each.1. kx2 – 7x + 12 = 0

x = 3 is root of given quadratic equation.So it satisfies the given equationSubstituting x = 3 in the given equation.

k(3)2 – 7(3) + 12 = 0 ½ k(9) – 21 + 12 = 0 9k – 9 = 0 9k = 9

k =99

k = 1 ½

2. Since tangent to a circle is perpendicular to the radius through the pointof contact. OTP = 90º OP2 = OT2 + PT2

132 = OT2 + 122 ½ OT2 = 132 – 122

= (13 – 12) (13 + 12) = 25 ½ OT = 5Hence radius of the circle is 5 cm.

3.

Let y - axis divides the line segment joining the points P (5,– 6) andQ(–1, –4) in the ratio m1 : m2 at R and the coordinates of the required pointbe (0, y)

MT EDUCARE LTD.CBSE X

Date :

SUBJECT : MATHEMATICS

SUMMATIVE ASSESSMENT - IIMODEL ANSWER PAPER

Set - A

Marks : 90

Time : 3 hrs.

O

T

P

12 cm

13 cm

(5, –6) (0, y) (–1, –4)P R Q

Set - A... 2 ...

x =1 2 2 1

1 2

m x +m xm +m ½

0 = 1 2

1 2

m -1 +m 5m +m

0 = –m1 + 5m2

m1 = 5m2

1

2

mm =

51 ½

m1: m2 = 5 : 1

4. Volume of the rectangular solid metallic cuboid = 9 × 8 × 2= 144 cm3

Volume of cube = volume of cuboid. Volume of cube = 144 cm3 1

SECTION - BQuestion number 5 to 10 carry 2 marks each.

5. PRO =120º

2= 60º,

POR = 90º – 60º 1= 30º

PROR = sin 30º

PROR =

12

OR = 2 PR 1PR = QR

OR = PR + QR

6. Let OA be the horizontal ground, and let K be theposition of the kite at a height h above the ground.Then, AK = h.It is given that OK = 100 metres and AOK = 60º.Thus , in OAK, we have hypotenuse OK = 100mertes and AOK = 60º and we wish to find theperpendicular AK. So, we use the trigonometricratio involving perpendicular and hypotenuse.Clearly, sine is such a ratio. So, we take the sine ofAOK . In AOK, we have,

P

R

Q

O

O A60º

100m h

K

Set - A... 3 ...

sin 60º =AKOK 1

sin 60º = 100h

h = 100 sin 60º

h = 10032

= 50 3 1

= 86.60 metres.

7. A box contains only lemon flavoured candies. Let it be xTotal candies = x(i) Let A be the event of drawing an orange flavoured candy No. of outcomes favourable to A = 0

P(A) =0x

= 0 1

(ii) Let B be the event of drawing a lemon flavoured candy No. of outcomes favourable to B = x

P(B) =xx

= 1 1

8. According to question PR = RQ or R is mid point of PQ.Using mid-pint formula

x = 1 2

2x x

y = 1 2

2y y

1

2 42

b = – 3

b + 2 = – 6 1 b = – 6 – 2 b = – 8

9. In APBOA + B + x + y = 360º 1

90º + 90 + x + y = 360º 180 + x + y = 360º

x + y = 180º 1

A

B

OP x yr

r

(6, b – 2) (2, –3) (–2, 4)P R Q

Set - A... 4 ...

10. Volume of cone = Volume of sphere13 r

2h =43 R

3 1

[Where, R is the radius of sphere, r is the radius of cone and h is theheight of cone]

(2)2 (8) = 4R3

8 = R3 1 R = 2 Radius of the sphere is 2 cm.

SECTION - CQuestion numbers 10 to 20 carry 3 marks each.

11. 2 x² + 7x + 5 2 = 0 2 x² + 2x + 5x + 5 2 = 0 1 2 x (x + 2 ) + 5 (x + 2 ) = 0 (x + 2 ) ( 2 x + 5) = 0 1 x + 2 = 0 or 2 x + 5 = 0

x = – 2 or x =52

The roots of the given quadratic equation are52

and – 2 . 1

12. Let the first term be a and common difference be d. Then,7 × t7 = 11 × t11 1 7 × (a + 6d) = 11 × (a + 10d) 7a + 42d = 11a + 110d 7a – 11a = 110d – 42d – 4a = 68 d 1 a = – 17d ..... (1)Now, t18 = a + 17d = – 17d + 17d = 0 [Using (1)]Hence, eighteenth term is zero. 1

13. In order to draw the pair of tangents,we follow the following steps.Steps of construction 2

60º 120º O

A

B

P

Set - A... 5 ...

STEP I Take a point on the plane of the paper and draw a circle withcentre O and radius OA = 5cm.

STEP II At O construct radii OA and OB such thatAOB equal to 120º i.e. supplement of theangle between the tangents.

STEP III Draw perpendiculars to OA and OB at A and B respectivelysuppose these perpendiculars intersect at P. Then, PA and PBare required tangents. 1

Justification : In quadrilateral OAPB, we have OAP = OBP = 90º and AOB = 120º OAP + OBP + AOB + APB = 360º 90º + 90º + 120º + APB = 360º APB = 360º – (90º + 90º + 120º) APB = 60º

14. OP is the tangent of the circle having center O’So, OPO’ = 90º [ Radius and tangent are to each other at the

point of contact]In right - angled OPO’

OP = 4 cm [Given]O’P = 3cm [Given] 1

OO’2 = OP2 + O’P2

= 42 + 32

= 16 + 9 = 25 OO’ = 5 cm.

If two circles intersect each other, then line joining the two centres isalways bisect the common chord.

OO’ PQ and PT = TQ

Area of OO’P =12 × base × altitude.

Here, base = 4cm,altitude = 3cm. 1

Area =12 × 4 × 3

= 6 cm2

But if base OO’ = 5cm and

T5 cm4 cm

3cm

Q

P

O O’

Set - A... 6 ...

altitude = PT

Then, Area POO’ =12 × 5 × altitude

Comparing (i) and (ii), we get

6 =12 × 5 × PT

2 × 6

5 = PT 1

125 = PT

PQ = 2PT

=2 × 12

5 =245 cm

So, Length of common chord =245 cm = 4.8 cm.

15. Let PQ = H be the height of the tower. PR = h is the height of the flag - staff.PAQ =

and RAQ =

Suppose, AQ = x

In PAQ,Hx = tan 1

H = x tan ...... (1)

In RAQ,H + h

x = tan

H + h = x tan b ......(2)Dividing (2) by (1), we get

H + hH =

tantan

1

H tan + h tan = H tan h tan = H tan – H tan

H (tan – tan ) = h tan 1

H = h tan

tan tan

Ax Q

H

h

P

R

Set - A... 7 ...

16. Let height of tower = h mand distance BC = x m

In BDC,hx = tan 60º 1

h = 3x ...... (i)

In ADC

30h

x = tan 30º 1

30h

x =13

3h = x + 30 ...... (ii)

Substituting the value of h from eq. (i) in eq. (ii), we get

3x = x + 30 1 x = 15 m

AC = 45 m.

Height of tower = 3 15 = 15 3 m

17. When three coins are tossed simultaneously, then the number of

possible outcomes = 8,

(i.e., HHH, HTH, THH, TTH, HHT,HTT, THT, TTT)

(a) Number of favourable outcomes(three tails) = 1 (i.e. TTT)

Required probability =18 1

(b) Number of favourable outcomes(exactly 2 tails) = 3 (i.e. HTT, TTH, THT)

required probability =38 1

(c) Number of favourable outcomes(at least two tails) = 4 (i.e. HTT, TTH, THT, TTT)

Required probability =48 =

12 1

30m x m30º 60º

A B C

D

C

h m

Set - A... 8 ...

18. (i) Total bulbs = 20No. of all possible outcomes = 20Defective bulbs = 4

Non-defective bulbs = 20 – 4 = 16 1(i) Let A be the event that the bulb taken out is defective No. of outcomes favourable to A = 4

P (A) =4

20 =15 1

(ii) Since the bulb drawn in (i) is not defective and is not replaced Total defective bulbs = 4

Total non defective bulbs = 16 – 1 = 15Total no. of possible outcomes = 4 + 15 = 19Let B be the event that this bulb is not defective

No. of outcomes favourable to B = 15

P(B) =1519

1

19. We have,Volume of water that = (192.50 × 60) litres ....(i)

flows per hour= (192.50 × 60 × 1000) cm3 1

Inner diameter of the pipe = 7 cm

Inner radius of the pipe =72 cm = 3.5cm

Let h cm be the length of the column of water that flows in one hour.Clearly, water column forms a cylinder of radius 3.5 cm and length h cm.

Volume of water that flows in one hour.= Volume of the cylinder of radius 3.5 cm and length h cm

=222 (3.5) h

7

cm3 ....(ii) 1

From (i) and (ii), we have227 × 3.5 × 3.5 × h = 192.50 × 60 × 1000

h =192.50 60 1000 7

22 3.5 3.5 cm 1

= 300000 cm= 3 km

Hence, the rate of flow of water is 3 km per hour.

Set - A... 9 ...

20. In right-angled ABC, AC2 = AB2 + BC2 1 (13)2 = (12)2 + (BC)2

BC2 = 25 BC = 5 cm

Now, radius of circle =132

= 6.5 cmArea of circle = r2

= 3.14 × 6.5 × 6.5= 132.665 cm2 1

Area of the rectangle = 12 × 5= 60 cm2

Area of the shaded part = (132.665 – 60) cm2

= 72.665 cm2 1

SECTION - DQuestion numbers 21 to 31 carry 4 marks each.

21. We have, 1 1

1 2 2 3x x x x

+ 1

3 4x x =16

3 1

1 2 3x x

x x x

+ 1

3 4x x =16

2 4

1 2 3x

x x x

+ 1

3 4x x =16

2

1 3x x + 1

3 4x x =16 1

2 8 11 3 4x x

x x x

=16

3 3

1 3 4x

x x x

=16

3

1 4x x =16 1

(x – 1) (x – 4) = 18 x2 – 5x + 4 = 18

O

D C

BA

Set - A... 10 ...

x2 – 5x – 14 = 0Using middle term splitting we get. x2 – 7x + 2x – 14 = 0 1 x (x – 7) + 2(x – 7) = 0 (x – 7) (x + 2) = 0If x – 7 = 0, then x = 7 and if x + 2 = 0, then x = – 2Hence, x = 7, – 2. 1

22. Now, Sn = 4n – n² S1 = 4 (1) – (1)2

= 4 – 1

S1 = 3 a = 3 1 S2 = 4 (2) – (2)2

= 8 – 4

S2 = 4

Now a2 = S2 – S1

= 4 – 3

a2 = 1 1Now d = a2– a1

= 1 – 3= – 2

a3 = a2 + d= 1 + (– 2) ½

a3 = – 1

Now a10 = 3 + (10 – 1) (–2)= 3 + 9 × (–2)= 3 – 18

a10 = – 15 ½ an = a + (n – 1) d

= 3 + (n – 1) (–2)= 3 + (–2n) + 2

an = 5 – 2n 1

23. Let the third consecutive odd natural number be x, x + 2 and x + 4As per the given condition,

x (x + 4) = 4 (x + 2) + 1 1 x2 + 4x = 4x + 8 + 1 x2 = 9 1

Set - A... 11 ...

Taking square root on both the sides we get,x = + 3

x is a natural number x – 3 1Hence, x = 3x + 2 = 3 + 2 = 5 and x + 4 = 3 + 4 = 7

The 3 consecutive odd natural numbers are 3, 5and 7 respectively. 1

24. The AP is in the form of 200, 250, 300, ........ a30.Now a = 200, d = 250 – 200, n = 30, S30 = ?

= 50 1Now, an = a + (n – 1) d a30 = 200 + (30 – 1) 50

= 200 + 29 × 50= 200 + 1450= 1650 1

The penalty for the delay of work on the 30th day is Rs. 1650

Sn = 2n

[a + an)

S30 =302 [200 + 1650] 1

= 15 × 1850= 27750

The contractor has to pay Rs. 27,750 if he delays the work by 30 days. 1

25. Produce BD so that it meets the bigger circleat E as shown in figure Join OD and AE. Wehave, BEA = 90º (Angle in the semicirclediameter AB)BDO = 90º (radius OD to tangent BD at

point D)In BAE and BOD

BEA = BDO (each = 90º)ABE = OBD (each = B)

BAE BOD (By AA similarity criterion)

BEBD =

AEOD =

BABO

BEBD =

AE10cm =

32cm16cm 1

BD E

A

O

Set - A... 12 ...

BEBD =

AE10cm = 2

BEBD = 2, i.e.

BE = 2 × BD ....(i)

andAE10 = 2 i.e. AE = 20cm ....(ii)

BD = DE [From (i)] 1Now, from right angled OBD

OB2 = BD2 + OD2

(16)2 = BD2 + (10)2

BD2 = 256 – 100= 156 1

BD = 156 cm

DE = 156 cm (BD = DE) ....(iv)

From right ADE, AD2 = AE2 + DE2 (By pythagoras therorem)

= (20)2 + 2156 [From (ii) and (iv)]

= 400 + 156= 556 1

AD = 556 cm

AD = 2 139 cm

26. Let the radii of the two circles be r1 cm

and r2 cm where r1 > r2.

Distance between the centers if the circles.

r1 + r2 = 17 ....(1)

Sum of the areas of the two circles

= 149 cm2 (Given) 1 r2

1 + r22 = 149

r21 + r2

2 = 149 ....(2)

We have an identity

(r1 + r2)2 + (r1 – r2)

2 = 2 (r21 + r2

2) (By [1] and [2])

(17)2 + (r1 – r2)2 = 2 × 149

C1 C2

Pr1 r2

Set - A... 13 ...

289 + (r1 – r2)2 = 298 1

(r1 – r2)2 = 9

= (3)2

r1 – r2 = 3 ....(3)

Adding (1) and (3), we get

2r1 = 17 + 3

= 20 1 r1 = 10

From (1), 10 + r2 = 17

r2 = 7.

Hence, the radii of the two circles are 10 cm and 7 cm. 1

27. Let AD be the median through thevertex A of ABC. Then, D is themid-point of BC. So, the coordinates

of D are3 1 2 8,2 2

i.e. (–2, 3). 1

AD = 2 25 2 1 3

= 49 16

= 65 units 1

Let G be the centroid of ABC. Then, G lies on median AD and divides itin the ratio 2:1. So, coordinates of G are

2 2 1 5 2 3 1 1,2 1 2 1

=4 5 6 1,3 3

=1 5,3 3

2

28. A(1, 2), B(5, 4), C(3, 8) and D (–1, 6)

AB = 2 24 2 = 16 4 = 20 1

BC = 2 22 4 = 4 16 = 20

CD = 2 24 2 = 16 4 = 20 1

B(–3, –2) C (–1, 8)

A (5, –1)

G

2

1

D (–2, 3)

Set - A... 14 ...

DA = 2 22 4 = 4 16 = 20Here, AB = BC = CA = DA

AC = 2 22 6 = 40 1

BD = 2 26 2 = 40

All sides of quadrilateral are equal and diagonals are equal. ABCD is square. 1

29. Let radius be ‘r’

In ABC, A = 90º

AC2 + AB2 = BC2

AC2 + 62 = 102

AC2 = 100 – 36 = 64 1 AC = 8cm

AP = AQ = x {Tangent from external point

CP = CR = y are equal}

BQ = BR = z

Perimeter of ABC = AB + BC + AC

= 6 + 10 + 8

2x + 2y + 2z = 24

x + y + z = 12 1y + z = 10

x + 10 = 12

x = 2

OPAQ is rectangle (by defination) 1

x = rArea of shaded portion = A (Triangle ABC) – A (circle)

=12 × AC × AB – r2

=12 × 8 × 6 – 3.14 (2)2

= 24 – 12.56= 11.44 cm2 1

Q

RC

O

P

B

A

y

x

10 cm

y zz

x

6 cm

y

PQ

Set - A... 15...

30. r1 = 8cmr2 = 20cmh = 24 cm

Volume of container =13 (

21r + 2

2r + r1r2) h 1

=13 ×

227 × [82 + 202 + 8 (20)] 24

=227 × (624) × 8

=109824

7 cm3 1

=109824

7000 l [ 1l = 1000 cm3]

Rate of milk = Rs 21 per litre 1

Total cost = 21 ×109824

7000

=3294721000 1

= Rs 329.47

31. Let VAB the solid metallic rightcircular cone of height 20cm. Supposethis cone is cut by a plane parallel toits base at a point O’ such thatVO’ = O’O i.e. O’ is the mid - point ofVO. Let r1 and r2 be the radii iscircular ends of the frustrum ABB’ A’.In triangles VOA and VO’ A’, we have

tan 30º =OAVO and tan 30º =

O'A'VO'

13 = 1r

20 and13 = 2r

101

30º 30º10cm

10cm

A’O’

B’

OBA

r2

r1

V

Set - A

r1 =203 cm and r2 =

103 cm

Volume of the frustum =13 2 2

1 2 1 2r +r +r r h

Volume of the frustum = 3 400 100 200

3 3 3

× 10cm2

=7000

9 cm2 1

Let the length of the wire of1

16 cm diameter be l cm. Then,

Volume of the metal used in wire = ×21

32

× l cm2

1radius = cm32

Volume of the metal used in wire = 1024l

cm2 1

Since the frustum is recast into a wire of length l cm and diameter1

16 cm

Volume of the metal used in wire = Volume of the frustum

1024l

=7000

9

l =7000

9

×1024

cm 1

=7000

9 × 1024 cm

l = 7694.4 m Length of wire is 7694.4 m.