[1]

Section – A

1. Find the value of x, if 3 cos sin .x x=2. Complete the missing entries in the following factor tree:

2

3

7

3. A number y is chosen at random from the numbers –2, –1, 0, 1, 2. What is the probability that y2 < 2?

4. The radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Find its slant height.

5. If the sum of first p terms of an A.P. is ap2 + bp, find the common difference.OR

Find the 9th term from the end of the A.P. : 5, 9, 13, .....185.

6. DABC ~ DPQR, if ar(DABC) = 2.25 m2 and ar(DPQR) = 6.25 m2, PQ = 0.5 m, find the length of AB.

OR In DABC, DE||BC, AD = x + 2, DB = 3, AE = x – 1, EC = 2. Find the value of x.

A

D

B

E

C

(2018-19)Date : ___________Duration : 3 Hrs. Max. Marks : 80

ClassX

Mathematics(Set - 1)

Instructions:4 All questions are compulsory.4 Section A contains 6 questions of one mark each.4 Section B contains 6 questions of two marks each.4 Section C contains 10 questions of three marks each.4 Section D contains 8 questions of four marks each.4 Allow 15 minutes extra time for reading question paper.

Sample Paper

[2]

Section – B

7. Find the values of k for which the following pair of linear equations have infinitely many solutions:

2x + 3y = 7 (k – 1)x + (k + 2)y = 3k8. Find the largest number which divides 318 and 739 leaving remainder 3 and 4 respectively.9. In the given figure, PA and PB are tangents to the circle from an external point P. CD is another

tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

C

D

P

B

A

O Q

10. If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal, then prove that 3x = 2y.OR

If (1, 5), (p, 1) and (4, 11) are collinear, find the value of p.

11. Find the values of k for which the equation 3x2 – k 3 x + 4 = 0 has real and equal roots.12. If a and b are the zeroes of the polynomial 3x2 + 5x + 2, then form a quadratic polynomial whose

zeroes are 2a and 2b.OR

If a and b are the zeroes of the polynomial x x2 4 3 3− + , then find the value of a + b - ab.

Section – C

13. Prove 3 is irrational and hence show that 2 3 1- is an irrational number.14. ABC is a right triangle, right angled at C and D is the mid-point of BC. Prove that: AB2 = 4AD2 – 3AC2

B C

A

D

OR ABC is a triangle right angled at A. BL and CM are medians to sides AC and AB. Prove that

4(BL2 + CM2) = 5 BC2.C

A

L

M B

15. If P(9a – 2, – b) divides the line segment joining A(3a + 1, –3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

16. From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of flagstaff and distance of the building from P.

[3]

The shadow of a tower standing on a level ground is found to be 50 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.

17. XY and X′Y′ are two parallel tangents to a circle with centre O. Another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

18. Evaluate:

sincos

sincos

costan tan tan

3258

2268

48 4228 25 30

°°+

°°−

° °° ° °

cosecttan tan62 65° °

OR

Geometrically prove sin60 32

° = .

19. Two different dice are thrown together. Find the probability that the numbers obtained have

(i) Even sum (ii) Even product (iii) Product is a perfect square 20. At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When

Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

OR Points A and B are 90 kms apart from each other on a highway. A car starts from A and another

from B at the same time. If they go in the same direction they meet in 9 hours and if they go in

opposite directions they meet in 97

hours. Find their speeds.

21. In a school, weights of teachers were recorded as shown in the following table:

Weight (in kg) 45 – 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75Number of Teachers 5 10 25 28 22 10

Find the modal weight.22. Solve for x and y: 43x + 67y = –24 67x + 43y = 24

Section – D

23. Obtain all other zeroes of the polynomial 4x4 + x3 – 72x2 – 18x, if two of its zeroes are 3 2 3 2 and - .

24. The following table gives the daily income of 50 workers of a factory. Draw both types of ogives and find the median from the graph.

Daily Income (in Rs.) 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200Number of Workers 12 14 8 6 10

OR

[4]

The median of the following data is 525. Find the values of x and y, if the sum of frequencies is 100.

Class Interval Frequency0 – 100

100 – 200200 – 300300 – 400400 – 500500 – 600600 – 700700 – 800800 – 900900 – 1000

25x

121720y974

25. State and prove Basic Proportionality Theorem.OR

In the given figure, the line segment XY||AC in DABC and it divides the triangle into two parts of equal areas. Prove that:

AXAB

=−2 12

A

CB Y

X

26. Divide 32 into 4 parts which are in A.P. such that the ratio of the product of the first and last term to the product of two middle terms is 7 : 15.

27. Construct a triangle ABC with sides BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a

triangle whose sides are 34

times of the corresponding sides of DABC.

28. In the figure, ABC is a triangle right angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

A

B C

6 cm8 cm

29. If tan ,θ =ab

prove that:

2 12

2 2seccos

θθ++

=+a bb

OR

[5]

If x = tan A + sin A y = tan A – sin A

Prove that : x yx y

x y+−

−

+

=

2 2

21.

30. Water in a canal, 8.4 m wide, 1.5 m deep, is flowing with a speed of 5 km/hr. How much area will it irrigate in 20 minutes, if 12 cm of standing water is needed?

OR

(2018-19)Date : ___________Duration : 3 Hrs. Max. Marks : 80

ClassX

Mathematics(Set - 1)

1. 3 cos sinx x=

⇒ 3 = tanx ⇒ tan 60° = tan x

⇒ x = 60°.

2. 42

2

21

3

7

3. P y2 2 35

<( ) =4. Frustum :

h = 40 cm

r1 = 20 cm

r2 = 11 cm

l h r r= + −( )2

1 22

= ( ) + −( ) = + =40 20 11 1600 81 16812 2

= 41 cm.

5. S1 = a(1)2 + b(1) = a + b

S2 = a(2)2 + b(2) = 4a + 2b

a1 = S1 = a + b

a2 = S2 – S1 = 4a + 2b – a – b = 3a + b

d = a2 – a1

d = 2a.

OR a = 185 d = – 4 a9 = a + 8d = 185 + 8(–4) = 185 – 32 = 153.

6. DABC ~ DPQR

arar

∆∆ABCPQR

ABPQ

( )( )

=2

2

2 256 25 0 5

2

2.. .

=( )AB

AB0 5

1 52 5...

=

AB =×

=1 5 0 5

2 50 3. .

..

AB = 0.3 cm

Hints/Solutions to Sample Paper

[7]

OR In DABC, Q DE||BC,

ADDB

AEEC

(By BPT)=

x x+=

−23

12

2x + 4 = 3x – 3 x = 7.

7. For infinitely many solutions:

aa

bb

cc

12

12

12

= =

21

32k k−

=+

2k + 4 = 3k – 3

– k = – 7

\ k = 7. 8. Required positive integer is the HCF of 318 – 3 = 315 and 739 – 4 = 735.

3 315

3 105

5 35

7 7

1

3 735

5 245

7 49

7 7

1

315 = 32 × 51 × 71

735 = 3 × 5 × 72

\ HCF (315, 735) = 3 × 5 × 7 = 105 Hence, the required number = 105.

9.

A

B

C

D

PQO

Given : PA and PB are the tangents to the circle

PA = 12 cm, QC = QD = 3 cm

To find: PC + PD

Sol. PA = PB = 12 cm (The length of tangents

QC = AC = 3 cm drawn from an external point

QD = BD = 3 cm to a circle are equal)

Now, PC = PA – AC = 12 – 3 = 9 cm

PD = PB – BD = 12 – 3 = 9 cm

\ PC + PD = 18 cm.

10. Using distance formula

PA = −( ) + −( )x x y y2 1

22 1

2

PA = PB

x y x y−( ) + −( ) = +( ) + −( )5 1 1 52 2 2 2

Squaring both sides,

x y x y−( ) + −( ) = +( ) + −( )5 1 1 52 2 2 2

x x y y x x y y2 2 2 210 25 2 1 2 1 10 25− + + − + = + + + − +

–10x – 2x = – 10y + 2y

–12x = – 8y

⇒ 3x = 2y.

[8]

OR Let the points be A(1, 5), B(p, 1), C(4, 11). Q Points are collinear \ ar(DABC) = 0.

12

01 2 3 2 3 1 3 1 2x y y x y y x y y−( ) + −( ) + −( ){ } =

– 10 + 6p + 16 = 0 6p + 6 = 0 p = – 1.

11. Q The equation has real and equal roots,

\ D = 0

b2 – 4ac = 0

−( ) − ( ) ( ) =k 3 4 3 4 0

2

3k2 – 48 = 0

k2 = 16

k = ± 4.12. P(x) = 3x2 + 5x + 2

α β+ = − = −

ba

53

αβ = =

ca

23

Required polynomial = k [x2 – (sum of zeroes) x + product of zeroes]

= k[x2 – (2a + 2b)x + 2a × 2b]

= k[x2 – 2(a + b)x + 4ab]

= + +

k x x2 103

83

If k = 3

Required polynomial = 3x2+ 10x + 8

OR

p x x x( ) = − +2 4 3 3

α β+ = − =

ba

4 3

αβ = =

ca

3

\ α β αβ+( ) − = −4 3 3.

13. Let 2 3 1- be a rational number.

\ 2 3 1 0− = ≠ abb, , a and b are co-prime integers

2 3 1= +ab

2 3 =+a bb

3

2=

+a bb

LHS is irrational but

RHS is rational.

\ Our assumption is wrong

\ 2 3 1- is an irrational number.

Also, to prove 3 to be irrational, refer NCERT, example 9, Page 13

[9]

14. Given: DABC is a right triangle. D is the mid-point of BC.

B D C

A

TP: AB2 = 4AD2 – 3AC2

Pf: In right DABC, AB2 = AC2 + BC2

AB2 = AC2 + (2DC)2

AB2 = AC2 + 4DC2 ...(i)

In right DADC, AD2 = AC2 + DC2

DC2 = AD2 – AC2 ...(ii)

Substitute (ii) in (i),

AB2 = AC2 + 4(AD2 – AC2)

AB2 = AC2 + 4AD2 – 4AC2

AB2 = 4AD2 – 3AC2.OR

Given: BL and CM are medians of the ΔABC in which ∠A = 90°. TP: 4(BL2 + CM2) = 5BC2 From DABC, BC2 = AB2 + AC2 ...(Pythagoras Theorem) ...(1) From DABL, BL2 = AL2 + AB2

or, BL2 = (AC/2)2 + AB2 ...(L is the mid-point of AC) or, BL2 = AC2/4 + AB2

or, 4BL2 = AC2 + 4AB2 ...(2) From DCMA, CM2 = AC2 + AM2

or, CM2 = AC2 + (AB/2)2 ...(M is the mid-point of AB) or, CM2 = AC2 + AB2/4 or 4CM2 = 4AC2 + AB2 ...(3) Adding (2) and (3), we have 4(BL2 + CM2) = 5(AC2 + AB2) i.e., 4(BL2 + CM2) = 5BC2 ...[From (1)]

15. P divides AB in the ratio 3 : 1 P(9a – 2, – b)

B(8a, 5)A(3a + 1, –3) By section formula,

P x y m x m x

m mm y m ym m

, ,( ) = ++

++

1 2 2 1

1 21 2 2 1

1 2

9 2

3 8 1 3 13 1

aa a

− =( ) + +( )

+ 4(9a – 22) = 24a + 3a + 1 36a – 8 = 27a + 1 9a = 9 a = 1.

− =

( ) + −( )+

b3 5 1 3

3 1

− =

−b 15 34

− =b 12

4 b = – 3.16. Refer to NCERT, Page 200, example 4.

OR In figure, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation

of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is 30°.

[10]

Now, let AB be h m and BC be x m. According to the question, DB is 50 m longer than BC. So, DB = (50 + x) m

50 m

Now, we have two right triangles ABC and ABD. In ΔABC, tan 60° = AB/BC or, √3 = h/x ...(1) [½] In ΔABD, tan 30° = AB/BD i.e., 1/√3 = h/x + 50 ...(2) [½] From (1), we have h = x√3 Putting this value in (2), we get (x√3)√3 = x + 50, i.e., 3x = x + 50 i.e., x = 25

So, h = 25√3 ...[From (1)] Therefore, the height of the tower is 25√3 m.

17. Given : XY, X'Y', AB are tangents X Y

X' Y'B

A

E

D

O C

XY||X'Y' TP: ∠AOB = 90°

Cons: Draw radii OD, OE, OC

Pf: In DAOD and DAOC,

OD = OC (radii of the same circle)

OA = OA (common)

∠ODA = ∠OCA (Radius is perpendicular to tangent at point of contact)

\ DAOD ≅ DAOC (RHS)

\ ∠AOD = DAOC (CPCT)

Similarly, ∠BOC = ∠BOE

Q XY||X'Y' \ Points D, O, E are collinear

∠DOC + ∠COE = 180° (straight angle)

2∠AOC + 2∠BOC = 180°

∠AOC + ∠BOC = 90°

⇒ ∠AOB = 90°

18.

sincos

sincos

costan tan tan

3258

2268

48 4228 25 30

°°+

°°−

° °° ° °

cosecttan tan62 65° °

coscos

coscos

sin90 3258

90 2268

90 48 42° − °( )°

+° − °( )

°−

° − °( ) °cosec

ccot tan cot tan90 28 62 90 25 65 13

° − °( ) ° ° − °( ) ° ×

=

°°+

°°−

° °° °

coscos

coscos

sincot tan cot

5858

6868

42 42 362 62

cosec 665 65° °tan

= + −1 1 3 (Qcos q = sin(90 – q), tanq = cot(90 – q), tanq X cot q =1)

= −2 3

OR Consider an equilateral DABC

D CB

A

60°

a

2a

Draw AD ⊥ BC

Let AB = BC = AC = '2a'

DABD ≅ DACD (RHS)

BD = DC = 'a'

AD2 = AB2 – BD2 (Pythagoras Theorem)

= (2a)2 – (a)2

= 3a2

AD = 3a

In right DABD,

sin 60° = ADAB

[11]

=3

2a

a

sin60 32

° =

19. Sample space when two dice are tossed: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

(i) Total number of outcomes = 36

Number of favourable outcomes = 18

P even sum Number of favorable outcomesTotal no. of possi

( ) =bble outcomes

= =1836

12

(ii) Total number of outcomes = 36

Number of favourable outcomes = 27

P even product( ) = =2736

34

(iii) Total number of outcomes = 36

Number of favourable outcomes = 8

P(product is a perfect square) = =836

29

20. Let Nisha's age be x

Asha's age = x2 + 2

Difference of their ages = x2 + 2 – x ATQ, x2 + 2 + x2 + 2 – x = 10x – 1

2x2 – x + 4 = 10x – 1

2x2 – 11x + 5 = 0

2x2 – 10x – x + 5 = 0

2x (x – 5) – 1(x – 5) = 0

(2x – 1)(x – 5) = 0

x =12

5,

x ¹12

as it is not possible

\ x = 5.

\ Nisha's age = 5 years

Asha's age = 27 years.

OR Let the speed of car A be 'x' km/hr and speed of car B be 'y' km/hr.

Case: (i) When they go in same direction :

Distance travelled by A = 9x

Distance travelled by B = 9y

\ 9x – 9y = 90

x – y = 10 ...(i)

Case(ii) When they go in opposite directions:

Distance travelled by A =97

x

Distance travelled by B =97

y

97

97

90x y+ =

x + y = 70 ...(ii)

[12]

Solving (i) and (ii)

x = 40, y = 30

\ Speed of car A is 40 km/hr

Speed of car B is 30 km/hr

21.

Weight (in kg) Number of teachers

45 – 50 5

50 – 55 10

55 – 60 25 → f0

60 – 65 28 → f1

65 – 70 22 → f2

70 – 75 10

Modal class is 60 – 65

Mode = l f ff f f

h+−

− −

×1 0

1 0 22

= +

−× − −

×60 28 25

2 28 25 225

= +

−

×60 3

56 475

= + = + =60 15

960 1 66 61 66. .

22. 43x + 67y = – 24 ...(i)

67x + 43y = 24 ...(ii)

Adding (i) and (ii),

100x + 100y = 0

x + y = 0 ...(iii)

Subtracting (i) and (ii),

–24x + 24y = – 48

x – y = 2 ...(iv)

Solving (iii) and (iv),

2x = 2

x = 1, y = – 1.23. P(x) = 4x4 + x3 – 72x2 – 18x

Since zeroes are 3 2 3 2 and -

\ x x−( ) +( )3 2 3 2 is a factor

x2 – 18 is a factor

x2

– 184 + 72 18x x – x – x

4 3 2

4 +x x2

4 – 72x x4 2

(–) (+)

x x3

– 18

x x3

– 18

(–) (+)

0

P(x) = (x2 – 18)(4x2 + x)

4x2 + x = 0

x(4x + 1) = 0

[13]

x =

−0 14

,

\ The other two zeroes are 0 14

, -

24. Less than ogive table :

x�

y�

y

x

100 120 140 160 180 200

60

50

40

30

20

10

Daily Income

Nu

mber

of

Work

ers

(120, 12)

(140, 26)

(160, 16)

(180, 10)

(140, 26)

(160, 34)

(180, 40)

(200, 50)(100, 50)

Scale

-axis – 1cm = 20 unitsx

-axis – 1 cm = 10 unitsy

(120, 38)

Daily Income (in `) Number of Workers (cf )

Less than 120 12

Less than 140 26

Less than 160 34

Less than 180 40

Less than 200 50

More than ogive table :

Daily Income (in `) Number of Workers (cf )

More than 100 50

More than 120 38

More than 140 24

More than 160 16

More than 180 10

Point of intersection of less than ogive and more than ogive gives the median of data

Median is 140.

OR Refer to NCERT example 8, Page 284

25. Refer or NCERT Page 124, Theorem 6.1

OR Refer to NCERT Page 143, Example 9.

26. Let the numbers be a – 3d, a – d, a + d, a + 3d

a – 3d + a – d + a + d + a + 3d = 32

4a = 32

a = 8

( )( )a d a da d a d− +−( ) +( )

=3 3 7

15

64 964

715

2

2−

−=

dd

⇒ 960 – 135 d2 = 448 – 7d2

–128d2 = – 512

d2 = 4

d = ± 2. Case 1: When a = 8, d = 2

a – 3d = 8 – 3 × 2 = 2

a – d = 8 – 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3 × 2 = 14

Case 2: When a = 8, d = – 2

a – 3d = 8 – 3(– 2) = 14

a – d = 8 – (– 2) = 10

a + d = 8 – 2 = 6

a + 3d = 8 + 3(– 2) = 2

\ The four numbers are 2, 6, 10, 14.

[14]

27. In DABC,

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 30°.

45°

105°

A�A

C� C

30°

B

B1

B2

B3

B4

X

\ A'BC' is the required D.

28. Area of shaded region = Area of DABC + Area of semicircle AXB + Area of semicircle AYC- Area of semicircle BAC

In right DBAC, BC2 = AB2 + AC2

A

B C

6cm

6cm

YX

8cm

BC = 10 cm.

Area of semicircle AXB r= =

( )=

π π π12 2

2

232

92

cm

Area of semicircle AYCr

= =( )

=π π

π22 2

2

242

8 cm

Area of semicircle BACr

= =( )

=π π π3

2 22

252

252

cm

ar ABC∆( ) = × × =12

8 6 243

cm2

\ Required area = 24 92

8 252

+ + −π

ππ

= 24 + 12.5 p – 12.5 p

= 24 cm2.

29. If tan tanθ θ= ⇒ =ab

a b

TP: 2 12

2 2seccos

θθ++

=+a bb

RHS : a bb

2 2+

LHS: 2 1

2sec

cosθθ++

=

+b bb

2 2 2tan θ

=× +

+

2 1 1

2cos

cosθ

θ =

b

b

2 2 1tan θ +( )

=+( )

+( )2

2cos

cos cosθ

θ θ =

bb

2 2sec θ

=1

cos θ =

bb

sec θ

= sec q = sec q

Q LHS = RHS, Hence Proved.

OR x = tan A + sin A

y = tan A – sin A

LHS: x yx y

x y+−

−

+

2 2

2

⇒ tan sin tan sintan sin tan sin

tan sin tanA A A AA A A A

A A+ + −+ − +

−

+ + AA A−

sin2

2

[15]

⇒ 22

22

2 2tansin

tanAA

A

−

⇒ sincos sin

tanAA A

A×

− ( )1 2

2

⇒ sec2 A – tan2A

⇒ 1

Q LHS = RHS

Hence proved.

30. Canal is in the shape of a cuboid where

b = 8.4 m

h = 1.5 m

Speed = 5km/hr

Length of canal in 1 hr = 5 km

Length of canal in 60 mins = 5000 m

Length of canal in 1 min. =500060

m

Length of canal in 20 mins = ×500060

20 m

Volume of water flowing in canal = l × b × h

= × ×

50003

8 4 1 5 3. . m

Volume of water flowing = Volume of area irrigated

50003

8 4 1 5× × =. . Area irrigated × height

5000 8 4 1 5 12

100× × =. . Area irrigated ×

Area irrigated = 5000 84 512× ×

= 175000 m2 or

= 17.5 hectares.

\ Area of land irrigated = 17.5 hectares.