Mathematics CLASS : X - K.V. No. 3 Amritsar · Mathematics CLASS : X ... However, internal choice...

SAMPLE PAPER –1 (SA II) MRS.NALNI SARAF KV BANTALAB

Mathematics

CLASS : X

Time: 3hrs Max. Marks: 90

General Instruction:-

1. All questions are Compulsory.

1. The question paper consists of 34 questions divided into 4 sections, A,B,C and D. Section – A comprises of 8 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each and Section- D comprises of 10 questions of 4 marks each.

2. Question numbers 1 to 8 in Section –A multiple choice questions where you are to select one correct option out of the given four.

3. There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.

4. Use of calculator is not permitted.

SECTION –A

Question numbers 1 to 8 carry 1 mark each. For each of the question numbers 1 to 8, four alternative choices have been provided, of which only one is correct. Select the correct choice.

Q1. If the numbers n - 2, 4n - 1 and 5n + 2 are in A.P., then the value of n is : (A) 1 (B) 2 (C) 3 (D) 0 Q2.To divide a line segment PQ in the ratio 3 : 4, first a ray PX is drawn so that QPX an acute angle and then at equal distances points are marked on the ray PX such that the minimum number of these points is :

(A) 7 (B) 4 (C) 5 (0) 3

Q3. In the given figure, if O is the centre of a circle, PQ is a chord and

the tangent PR at P makes an angle of 50° with PQ, then POQ is

equal to:

(A) 100o (B) 80°

(C) 90° (D) 75°

Q4.The tops of two poles with heights 25 m and 35 m are

connected by a wire, which makes an angle of elevation of 30o at

the top of 25 m pole. Then the length of the wire is:

(A)26m (B)35m (C)15m (D)20m

O

P

Q

R 50o

40o

100o 40o

Q5. One coin is tossed thrice. The probability of getting neither 3 heads nor 3 tails, is:

(A) 1/3 (B) ¾ (C) ½ (D) 2/3

Q6.Value of p so that the point (3, p) lies on the line 2x – 3y = 5 is :

(A) 12 (B)3 (C) 1/3 (D) ½

Q.7 For a race of 1540 m, number of rounds one have to take on a circular track of radius

35m:

(A) 5 (B) 6 7 (C) 7 (D) 10

Q8. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of

base diameter 2 cm and height 16 cm. The diameter of each sphere is:

(A) 4 cm (B) 3 cm (C) 2 cm (D) 6 cm

SECTION-B

Question numbers 9 to 14 carry 2 marks each.

Q9. Solve the following quadratic equation by factorisation:

2 - 5 + 2 = 0; (x 0) x2 x Or

Find the roots of the following quadratic equation : (x +3)(x- 1) = 3 x - 1 3 Q10. Find the value of m for which the point with coordinates (3, 5), (m, 6) and

( 1 , 15) are collinear. 2 2

Q11. A horse is tethered to one corner of rectangular grass field 40m by 24 m, by a rope

14m long. Over how much area of the field can it graze?

Q12. In two concentric circles, prove that a chord of a larger

circle which is tangent to smaller circle is bisected at the

point of contact.

Q13. Find the common difference of an A.P. whose first term is ½ and the 8th term is 17 . Also write its 4th term. 6

Q14. Prove that the lengths of the tangents drawn from an external point to a circle are

equal.

SECTION-C

Q15. Onkar gets pocket money, from his father every week and saves 15 in

and on each successive week he increases his saving by 5 .

(1) Find the amount saved by Omkar in One month

(2) Find the amount saved in one year.

(3) Which quality of Onkar is referred in the given question?

Q16. Find the root of the equation 4 - 3 = 5 , x 0, -3

X (2x +3) 2

Or

Find two consecutive positive even integers, the sum of whose squares, the sum of

whose squares is 340.

Q17. Draw a circle of radius 3 cm. From a point 5 cm away

from its center. Construct the pair of tangents of the circle and

measure their length.

Q18. A card is drawn at random from a well shuffled deck of 52 cards. Find the probability of getting:

(i) a king (ii) a king of red suit

Or

Two dice are thrown at the same time. Determine the probability that the

Difference of the number on the two dice is 2.

Q19. On seeing a child in first floor of a burning house, a

man climbed from the ground along the rope stretched

from the top of a vertical tower and tied at the ground.

The height of the tower is 24 m and the angle made by

the rope to the ground is 30o. Calculate the distance covered by the man

to reach the top of the pole. What do you consider the act done by the

man to save the child?

Q20. Two cones with same base radius 8 cm and height 15 cm are joined

together along their bases. Find the surface area of the shape

Or

An ice-cream cone having radius 5 cm and height 10 cm as shown in figure. Calculate the volume of ice-cream, provided that its 1 part is left unfiled with ice-cream. 6

Q21. A square OABC is inscribed in the quadrant OPBQ. If OA = 10

cm, then find the area of the shaded region.

Q22. In the given fig., ABCD is a trapezium with AB||DC,

AB=18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A,B,C and D have been drawn, then find the area of the shaded region of the figure.

Q23. A spherical glass vessel has a cylindrical neck 8cm long, 2cm in

diameter, the diameter of the spherical part is 8.5cm. By measuring

the amount of water it holds, a child finds its volume to be 345 cm3.

Check whether she is correct, taking the above as the inside

measurements, and = 3.14.

Q24. A storage oil tanker consists of a cylindrical portion 7 m in

diameter with two hemispherical ends of the same

diameter. The oil tanker lying horizontally. If the total length

of the tanker is 20m, then find the capacity of the container.

10

52

B

D C

A 18cm

18cm

32cm

7cm 7cm

SECTION-D


Q25. The lower portion of a haystack is an inverted cone frustum and upper part is a cone. Find the total volume of the haystack, If AB = 3 m and CD = 2 m.

Or

A bucket of height 8 cm is made up of copper sheet is in the form of frustum of cone with radii of its lower and upper ends as 3 cm and 9 cm respectively. Calculate:

(i) The height of the cone of which the bucket is a part.

(ii) The volume of the water which can be filled in the bucket.

(iii) The area of the copper sheet required to make the

bucket.

Q26. Find four terms in an A.P., whose sum is 20 and the sum of whose squares is 120.

Q27. A man on the top of a vertical tower observes

a car moving at a uniform speed coming

directly towards him. If it takes 12 minutes for

the angle of depression to change from 30o to

45°, how soon after this, will the car reach the

tower? Give your answer to the nearest

second.

Q28. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

h

h

h (3-1) h 3 cm

A

B

C

45o

30o

45o 30o D

Q29. In the figure, OP is equal to the diameter of the circle. Prove

that ABP is an equilateral triangle.

Or

In the figure, PO QO. The tangents to the circle with centre O at P and Q intersect at a point T. Prove That PQ and OT are right bisectors of each other.

Q30. Swati can row her boat at a speed of 5 km/h in still water. If it

takes her 1 hour more to row the boat 5.25 km upstream, then to return

downstream, find the speed of the stream.

Q31. A child's game has 8 triangles of which 5 are blue and rest are red and ,10 squares

of which 6 are blue and rest are red. One piece is low at random. Find the

probability that it is a:

(i) triangle (ii) square

(iii) square of blue colour (iv) triangle of red colour.

Q32. Find the area of the triangle ABC With A(1, -4) and the mid-

points of sides through A being (2, -1) and (0, -1)

Q33. The two opposite vertices of a square are (-1, 2) and

(3, 2). Find the coordinates of the other two vertices.

Q34. Amar and Gugu together have 45 marbles, Both of them lost 5 marbles each and

the product of the number of marbles they now have is 124. We would like to find

out how many marbles they had to start with. While playing in the ground you

found a purse containing money and some documents what will you do ?

P

SOLUTION SAMPLE PAPER 1

SECTION –A

Question numbers 1 to 8 carry 1 mark each. For each of the question numbers 1 to 8, four alternative choices have been provided, of which only one is correct. Select the correct choice. Sol.1 (A) 1 [Since Here, n - 2, 4n - 1 and 5n + 2 are in A.P.

=> 4n - 1 - (n - 2) = 5n + 2 - (4n - 1) => 4n - 1 - n + 2 = 5n + 2 - 4n + 1

=> 3n-n = 3-1

=> 2n = 2 => n = 1] Sol.2 (A) 7

Sol.3 (A) 100°

[ Here, OPR 90o and RPQ = 50°

OPQ = 90o – 50o = 40o

OPQ = OQP 40o

Now, POQ = 180o - 40 o - 40 o = 100 o]

Sol.4 (D)20m

[ Here, CE = 35 – 25 = 10m

and CAE = 30o

Now, AC = cosec 30o CE

AC = CE 2

= 10 2 = 20m

Sol.5 (B) ¾

[Here, sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Now, neither 3 heads nor 3 tails = { HHT, HTH, THH, HTT, THT, TTH}

Required probability = 6 = 3 ] 8 4 Sol.6 (C) 1

3

[Since (3, p) lies on the line 2x – 3y = 5

2(3) – 3p = 5

p q

x

– 3p = 5-6

p = -1 = 1 ] -3 3 Sol.7 (C) 7

[Distance covered in one round = 2 22 35 = 220 m

7

Now, number of rounds = 1540 = 7 ] 220

Sol.8 (C) 2 cm

[Here, Volume of cylinder = 1 1 16 = 16 cm3

Now, 12 4 r3 = 16 r3 = 16 3 =1

3 12 4

r = 1 cm

Thus, radius of the solid sphere is 1cm and diameter of the sphere is 2cm]

SECTION-B


Sol.9 2 - 5 + 2 = 0 x2 x

2 - 4 + 1 +2 = 0 2 1 - 2 -1 1 - 2 = 0 x2 x x x x x

1 - 2 2 - 1 = 0 1 - 2 = 0 or 2 - 1 = 0 x x x x

Or

(x+3)(x-1) = 3 x- 1 3

x2+2x-3 = 3x-1 x2 – x – 2 = 0

x2 –2x + x –2 = 0 x(x–2) +1(x–2) =0

(x –2 )(x+1) = 0 x–2=0 or x+1=0

x = 2 or x = –1

Sol.10 Let the given points be A(3, 5), B(m, 6) and C( 1 , 15) are collinear. 2 2

Since given points are collinear.

Area of triangle ABC formed by these points is zero.

1 3 6 – 15 +m 15 – 5 + 1 (5-6) = 0 2 2 2 2

18 – 45 +15m – 5m+ 5 – 3 = 0 36-45+5-6 + 15m -10m = 0 2 2 2 2 2

|-5 + 5m | = 0 5m = 5 2 2

m = 2

Sol.11 Here, radius (r) = 14 m and = 90°

Area of the field in which horse can graze

= r2 = 90o 22 14 14 360o 360o 7

= 1 22 14 14 = 154 m2 4 7

Sol.12 AB is a tangent to the inner circle at point C.

OC is the radius drawn at the point of contact.

OC AB OCA = OCB = 90°

Now, in rt. ed AOC and BOC

OA = OB = r

OC = OC [common]

OCA = OCB [proved above]

AOC BOC [by RHS congruency rule]

AC = BC [ c.p.c.t]

Hence, chord of a larger circle which is tangent to smaller circle is bisected at the point of

contact.

Sol.13 Here, first term (a) = 1 2

and eighth term a8 = 17

6

a + 7d = 17 1 + 7d = 17 6 2 6

7d = 17 - 1 = 14 d = 1 6 2 6 3

Now, a4 = a + 3d = 1 + 3 1 = 1 +1 = 3 2 3 2 2

Sol.14 Join PO. TO and TO

In PTO and PTO

TO = T’O = r

OP = PO (common)

PTO = PTO = 90o

PTO PT’O

(By RHS congruencey axiom)

PT = PT’ (C.P.C.T)

SECTION-C

Sol.15 Here, saving after one week is 15 and it increases every week by 5

So, given savings is an A.P. with first term 15 and common difference 5

(i) In one month, we have four weeks

a4=a+3d

= 15 + 3 x 5 = 15 + 15 = 30

Now, S4 = 4/3 (15+30) = 30

[Since Sn = n/2[a + 1 or an]

(ii) In one year, we have 52 weeks

S52 = -52[2a + (n-1)d] = 26[30 + 51 x 5] 2

= 26 x 285 = 7410

(iii) Savings for future makes individual self dependent.

Sol.16 4 - 3 = 5 .

X (2x +3)

4 - 3x = 5 . (4-3x)(2x + 3) =5x

X (2x +3)

-6x2 – x + 12 =5x

Or x2 + x -2 = 0

x(x + 2) –1(x+2 =0

x-1 = 0 or x +2 = 0

x = 1 or x = - 2

or

6x2 + 6x- 12 =0

x2 +2x-x-2 = 0

(X-1) (x+2) = 0

Let the two consecutive positive even integers be x and x + 2 according to statement

of the question, we have

X2 + (X + 2)2 = 340

X2 + X2 + 4x +4- 340 = 0

2x2 + 4x -336 = 0

X2 + 14x - 12x – 168 =0

(x-12) (X+ 14) =0

X – 12 = 0 or x + 14 = 0

X=12orX=-14 (rejecting , given number are positive)

Hence, the required integers are 12 and 14.

Sol.17 Given: A circle of radius 3 cm and a points P is 5 cm away from its centre.

Required: A pair of tangents.

Steps of construction:

1. Draw a circle C (0, r) with center O and

Radius 3 cm.

2. Take a point P, such that OP = 5 cm,

3. Draw AB, the perpendicular bisector of OP

and left it intersects OP in M.

4. With M as centre and PM or MO as radius,

draw another circle intersecting the given

circle in T and T.

5. Join PT and PT.

Thus, PT and PT are the required tangents from point P to the circle C(0, r).

Sol.18 (i) Number of kings = 4

Probability (a king) = 4 = 1

52 13

(ii) Number of kings of red suit = 2

Probability (a king) = _2_ = 1

52 26

Or

Total number of elementary events when two dice are thrown = 6 x 6 = 36

Number of favourable outcomes (difference of the numbers on the two dice is 2)

={(1,3), (2,4), (3,1), (3,5), (4,2), (4,6), (5,3), (6,4)}

=8

Required probability = 8 = 2

36 9

Sol.19 Let AB be the tower of height 24 m and ACB =30o

Distance covered by the man along the rope is CA.

Now, in rt. ed CBA

CA = cosec 30o

AB

CA = AB cosec 30o

= 24 x 2

= 48 m

Thus, distance covered by the man is 48 m.

Sol.20 Radius of two cones (r) =8 cm

Height of two cones (h) =15 cm

Slant height (l) = r2 + h2 = 64+225

= 289 = 17cm

Now, surface area of the shape so obtained

= 2(rl) = 2 22 8 17 7 = 854.86 cm2

Or

Radius of the conical and spherical portion (r) = 5 cm

Height of the conical portion (h) = 10- 5 = 5 cm

Now, volume of the ice-cream cone

= Vol. of conical portion

+ Vol. of hemispherical portion

= 1 r2h + 2 r3 = 1 r2 (h+2r) 3 3 3

= 1 22 5 5(5+10) = 392.86 cm3 3 7

Volume of the ice-cream = 1 – 1 392.86 = 5 392.86 = 327.38 cm3

6 6 Sol.21 Here, side (OA) of the square = 10 cm

Diagonal of the square = 102 cm

Now, area of the shaded region = Area of quadrant OPBQ – Area of square

= 90o 22 102 102 -10 10 360o 7

= 1 22 200-100 = 157.14 -100 4 7 = 57.14 cm2

Sol.22 Here, radii of the arcs with centres A, B, C and D = 7 cm.

AB = 18 cm, DC = 32 cm and distance between AB and

DC = 14 cm

Now, area of the shaded region = Area of trapezium ABCD

- Area of circle

with radius 7 cm

= 1(18+32)14 – 22 7 7 2 7 = 350 – 154 = 196 cm2

Sol.23 Radius of cylindrical neck = 2 = 1 cm

Height of cylindrical neck = 8 cm

Radius of spherical part = 4.25 cm

Total volume of water = Volume of spherical part

+ Volume of cylindrical neck

= 4 r3 + r2h = 4 3.14 (4.25)3 + 3.14 (1)28

3 3 = 321.39 + 25.12 = 346.51 cm3 = 347 cm3 (approx.)

Sol.24 Radius of hemispherical portion = Radius of cylindrical portion = 7 m

2 Total length of the tanker = 20 m

Length of cylindrical portion = 20 - 7 = 13 m

Now, Capacity (volume) of the oil tanker

= Volume of cylinder

+ 2 Volume of hemisphere

= r2h + 2 2 r3 = 22 7 7 13 +2 2 22 7 7 7 3 7 2 2 3 7 2 2 2 = 500.5 + 179.67 = 680.17 m3

SECTION-D


Sol.25 Here, radius of cone (r) = 3 m

Height of cone (h) = 7 m

Volume of cone = 1r2h = 1 22 3 3 7 = 66 m3 3 3 7 Height of the frustum = 10.5 – 7 = 3.5 m

Now, Volume of the lower portion (frustum of cone) = 1 22 3.5(32+22+32) 3 7

= 11(9+4+6) = 11 19 3 3 = 69.67 m3

Thus, Volume of the haystack = 66 + 69.67 = 135.67 m3

Here, r = 3 cm, R = 9 cm and height of the bucket (h) = 8 cm

Now, slant height (l) = h2 + (R-r)2 = 82 + 62 = 64 + 36 = 100 = 10 cm

(i) Since ABC AB’C’

AB = BC AB’ B’C’

h = 3 h+8 9

3h = h +8

2h = 8 h = 4

Thus, height of the cone is 8 + 4 i.e., 12 cm.

(ii) Volume of the bucket = 1h(R2+r2+Rr) 3

= 1 22 8(92+32+9 3) 3 7

= 1 22 8 117 = 980.57 cm3 3 7

(iii). Surface area of copper sheet= r2 + (r + R)l

= 22 9 22(3+9)10 7 7

= 198 + 2640 = 2838 = 405.43 cm2 7 7 7

Sol.26 Let the four terms in an A.P., be a-3d, a-d, a+d, a+3d

According to the statement of the question, we have

a-3d+ a-d + a + d + a + 3d = 20

4a = 20 a=5

Also, (a-3d)2+(a-d)2+(a+d)2+(ä+3d)2 = 120

(5-3d)2+(5-d)2+(5+d)2+(5+3d)2=120

25+9d2-30d+25+d2-10d+25+d2+10d+25+9d2+30d = 120

100 +20d2= 120

20d2 = 20 d = 1

Hence, the four terms are

5-3, 5-1, 5+1, 5+3 i.ë., 2,4,6,8

Or 5+3, 5+1, 5-1, 5-3 i.e., 8, 6, 4, 2

Sol.27 Let AB be the tower of height h m.

ACB= 30° and. ADB = 45o

Now, in rt.ed ADB, we have

AB = tan 45o AB = 1 DB DB

AB = DB = h m

In rt. ed CBA, we have

AB = tan 30o

CB

AB = CB tan 30o h = (CD + DB) 1

3

3h = CD + h CD = (3 – 1)h

Speed of the card = (3-1)h m/min 12 Time taken by the car to reach the tower after point D

= h = 12 = 12(3+1) = 6 2.732 min

( 3-1)h 3-1 2 12 = 16 minutes 24 secs.

Sol.28 Given: Let PA and PB be two tangents drawn from an external point P to the circle

with centre O.

To prove: APB + AOB = 180o

Proof: Because tangent is perpendicular to the

Radius of the circle at point of contact.

OA AP and OB BP

OAP = OBP = 90o

Now, OAP + APB + OBP + AOB = 360o

90 o + APB + 90 o + AOB = 360o

APB + AOB = 360o -180o = 180o

Sol.29 Let r be the radius of the circle.

OA = r

and OP = 2r [given]

[tangent is perpendicular to the radius through the point of contact]

In right OPA, we have

Sin (OPA+ = OA = r = 1 OPA = 30o OP 2r 2

Similarly, OPB = 30o

APB = 30o + 30o = 60o

Since PA = PB [ lengths of tangents from an external point are equal]

PAB = PBA

In APB, we have

APB + PAB + PBA = 180o

60o + 2PAB = 180o

PAB = 60o

PAB = 60o

Since all angle are 60o ABP is equilateral

Or

Since Tangent to any circle is perpendicular to circle at point of

contact.

TPO = TQO = 90o

In TPO and TQO, we have

OP = OQ [radii of a circle]

OT = OT [common side]

By RHS congruency, we have

TPO = TQO

1 = 2 [c.p.c.t.]

Again, in PTR and QTR

1 = 2 [proved above]

PT = QT [tangents from external point]

TR = TR [common]

By SAS congruency, we have

PTR = QTR

PRT = QRT [c.p.c.t.]

But PRT + QRT = 180o [linear pair]

Thus, PRT = QRT.= 90°

Also, . PR = QR [c.p.c.t.]

PQ and OT are right bisectors of each other.

Sol.30 Let Speed of be x km/h.

Speed of the boat upstream = (5-x) km/h

Speed of the boat downstream = )5+x) km/h

Time taken to go upstream = 5.25 5-x

Time taken to go upstream = 5.25 5-x

As per question, we have

5.25 - 5.25 = 1 5-x 5+x

5.25 5+x – 5+x = 1 (5-x)(5+x)

10.5x = 25-x2

x2 + 10.5x -25 = 0

or 2x2 + 21x -50 = 0

2x2 + 25x-4x-50 = 0

2(2x+25) -2(2x+25) = 0

(x-2)(2x+25) = 0

x-2 = 0 or 2x +25 = 0

x =2 or x= -12.5 (speed cannot be negative)

x = 2

Hence, speed of the stream is 2 km/h.

Sol.31 Here, number of triangles are 8 and number of squares are 10.

Total number of outcomes = 10 + 8 = 18

(i) Probability (lost card is a triangle) = 8 = 4 18 9

(ii) Probability (lost card is a square) = 10 = 5 18 9

(iii) Probability (lost card is a square of blue colour) = 6 = 1 18 3

(iv)Probability (lost card is a triangle of red colour) = 3 = 1 18 6 Sol.32 Let the coordinates of B and C be B(x, y) and C(p, q)

Now, D(2, -1) is the mid-point of AB

1+x = 2 and -4+Y = -1 2 2

x = 3 and y = 2

Thus coordinates of B are B (3, 2)

Also, E(0, - 1) is the mid-point of AC

1+P = 0 and -4+q = - 1

2 2

p=-1 and q = 2

Thus, the coordinates of C are C (-1, 2)

Ar(ΔABC) = 1 | 1 (2 - 2) + 3(2 + 4) – 1(-4 – 2)

2

= 1 | 0 + 18 +6 | = 1 × 24 = 12 sq. units

2 2

Sol.33 Let A (-1, 2) and C(3, 2) be the two opposite vertices of a square ABCD and let

coordinates of B be B(x, y).

Now, AB = BC

| AB |2 = |BC|2

(x + 1)2 + (y-2)2 =(3-x)2 + (2-y)2

X2 + 1 + 2 x+y2+4-4y= 9 +x2-6x+4+y2-4y

2x + 6x= 9 – 1

8x = 8 x =1

Also, AB2 BC2 = AC2

(x +1)2 + (y-2)2 + (3–x)2 + (2-y)2= (3+1)2 + (2-2)2

x2 +1 +2x +y2 +4-4y=9+x2-6x+4+y2-4y=16

2x2+2y2 – 4x – 8y =16 -18

x2+y2-2x-4y=-1

1+y2-2-4y+1=0 [Put x = 1]

y2-4y=0

y(y-4) = 0 y=0 or y =4

Thus, the point (x,y) is (1, 0 ) or (1,4)

Hence, the coordinates of b and d are b (1,0) and d (1,4).

Sol.34 Let number of marbles Amar had be x

Number of marbles with Gugu = 45- x

Both of them lost 5 marbles

Number of marbles left with Amar and Gugu be x – 5 and 45 –x -5 i.e.,

40 –x respectively

According to the statement of the question, we have

(x-5) (40-x) =124

-x2 + 45x-200 -124=0

x2-36x-9x +324 =0

(x-9) (x-36)= 0

x-9 = 0 or x-36 = 0

x=9 or x = 36

Thus, either Amar had 9 marbles and Gugu had 36 marbles or Amar had 36 marbles and

Gugu had 9 months.

Handover the purse containing money and some documents to the concerned teacher.

SAMPLE PAPER –2 (SA II)

MR SANDESH KUMAR BHAT KV 1 JAMMU

Mathematics

CLASS : X








SECTION – A

Question number 1 – 8 are multiple choice type questions carrying 1 mark each. For each

question four alternative choices have been given, of which only one is correct. You have to

select the correct choice.

Q1.If x = -2 is the root of quadratic equation x2-3x-a=0 then the value of a is

(a) -10 (b) 3 (c) 10 (d) -3

Q2. The missing term of the A.P:-3, □, 33, 48…………………is

(a) 12 (b) 15 (c) 18 (d) 21

Q3. If angle between two radii of a circle is 110° the angle between the tangents at the end of

the radii is

(a) 70° (b) 50° (c) 10° (d) 40°

Q4.From the given figure find the angle of elevation ‘A’

C

100 C

100

B 100 3 A

(a) 45° (b) 30° (c) 60° (d) 90°

Q5. If P (E) = 0.35 then the probability of not E is

(a) 0 (b) 0.45 (c) 0.65 (d) 0.53

Q6. The mid- point of a line segment joining the point A (-2, 8) and B (-6, -4) is

(a) (-4, -6) (b) (2, 6) (c) (4, 2) (d) (-4, 2)

Q7. If the perimeter and the area of a circle are numerically equal then the radius of the circle is

(a) 2 units (b) π units (c) 4 units (d) 7 units

Q8. Volume of two spheres is in ratio 64:27 the ratio of their radii is

(a) 3:4 (b) 4:3 (c) 9:16 (d) 16:9

SECTION – B

Question numbers 9 – 14 carry 2 marks each.

Q9. Find the value of K for which the equation 2x2+Kx+3=0 has two equal roots.

Q10. Find the roots of the quadratic equation 6x2-x-2=0 by factorization method.

Q11. Find the 10th term of the A.P:- 2, 7, 12………………………………

Q12. Two concentric circles of radii 5cm and 3cm. Find the length of the chord of the larger

circle which touches the smaller circle.

Q13. Find a relation between x and y such that the point (x, y) is equidistant from the point

(3, 6) and (-3, 4)

OR

Find the points on the x- axis which is equidistant from (2, -5) and (-2, 9)

Q14. In what ratio does the point (-4, 6) divide the line segment joining the points A (-6, 10) and

B (3, -8).

SECTION – C

Question numbers 15 – 24 carry 3 marks each.

Q15. The sum of the 4th and 8th term of an A.P is 24 and the sum of the 6th term and 10th term is

44. Find the first three terms of the A.P .

Q16. Prove that the tangent at any point of a circle is perpendicular to the radius through the

point of contact.

Q17. Draw a pair of tangents to the circle of radius 5cm which are inclined to each other at an

angle of 60°.

Q18. The shadow of a tower standing on a level ground is found to be 40m longer when sun’s

altitude is 30° than when it is 60°. Find the height of the tower.

Q19. A child has a block in the shape of a cube with one word written on each face as.

(i) The block (cube) is thrown. What is the probability of getting word ‘to’?

(ii) Use all words of the block (cube) to make a meaningful value base sentence

Q20. Find the value of k for which the points (7, -2), (5, 1) and (3, k) are collinear.

Q21. Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) taken in order are the vertices of a

square.

Q22. A toy is in the form of cone of radius 3.5cm mounted on a hemisphere of same radius.

The total height of the toy is 15.5cm. Find the total surface area of the toy.

Q23. (i) Find the area of the shaded region given in figure.

(ii) Write the four values given in the figure. Which value you consider important in

student’s life and Why?

learn to serve go come to

14cm

14cm 14cm

Q24.In a fig. a square OABC is inscribed in a quadrant OPBQ. If OA=20cm. Find the area of the

shaded region.

SECTION – D

Question numbers 25 to 34 carry 4 marks each

Q25.A plane left 30 minutes later than the schedule time and in order to reach its destination

1500km away in time. It has to increase its speed by 250km/h from its usual speed. Find its

usual speed.

Q26. Sum of the areas of two squares is 468m2. If the difference of their perimeter is 24m. Find

the sides of the two Squares.

Q27. A sum of Rs 700 is to be used for giving seven cash prizes to students of a school for their

overall academics performance. If each prize is Rs 20 less than its preceding prize. Find the

value of each prizes.

Q28. Prove that the length of tangents drawn from an external point to a circle is equal.

Q29. Construct a triangle similar to a given triangle ABC with its sides equal to ¾ of the corresponding sides of the triangle ABC (i.e., of scale factor

Hard

work Obedi

ent

Toler

ance Punct

uality

yyyy

.Q30. A straight highway leads to the foot of a tower. A man standing at the top of the tower

observes a car at an angle of depression of 30˚ which is approaching the foot of the tower with

a uniform speed. Six seconds later the angle of depression of a car is found to be 60˚. Find the

time taken by the car to reach the foo of the tower from this point.

Q31. 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.

(iii) A spade.

(iv) A black face card.

Q32. A car has two wipers which do not overlap. Each wiper has blade of length 25cm sweeping

through an angle of 115˚. Find the total area cleaned at each sweep of the blades.

Q33. Water in a canal 6m wide 15m deep is flowing with a speed of 10km/h. how much area

will it irrigate in 30 minutes? If 8cm of standing water is needed.

Q34. An open metal bucket is in theshape of a frustum of a cone, mounted on a hollow cylindrical base

made of the same metallic sheet (see Fig. 13.23). The diameters of the two circular ends of the bucket are

45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm.

Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle

of the bucket. Also, find the volume of water the bucket can hold. (Take = 22/7)


Section A

1. X2-3x-a=0

Put X= -2

(-2)2-(3)(-2)-a=0

4+6-a=0

10-a=0

a=10

2. Let missing term be x

-3,x,33,48,………an are in A.P

There for a2-a1 = a3- a2

So, x-(-3)=33-x

=> x+3=33-x

=> 2x= 33-3

=> x=

Q 3.

In quad. . In Quad. In POQT . ⎳P+⎳Q +⎳T+ ⎳ O=360 900 + 900 +⎳T + 0o = 360o ⎳T= 70o

4.In C

0

⎳ 00 B A

5.P ( E) = 0.35,

As we know that P( not E) + P ( E) = 1 P( not E) = 1- P ( E)

= 1-0.35 So, = 0.65

Q6. Let P(x,y) be a mid point Then by mid point formula.

X=

Y=

Q7. Let ‘r’ is the radius of circle Area of circle = Perimeter of circle Πr2= 2Π r r =2 units Q8. Let ‘r’ be the radius of small sphere. And volume is ‘v’ Therefore v= 4/3πr3 Let ‘R’be the radius of largest sphere and Volume is ‘V’ V= 4/3πR3

Now=

=

Section B

Q.9 As we know that For equal and real roots b2-4ac=0 k2 -24=0 k2=24

=> k= ± 24 or ±2 6

Q.10 As 6x2 +3x-4x-2=0 => 3x (2x+1)-2(2x+1)=0 => (3x-2)(2x+1)=0 => X=3/2 or x=-1/2 Q11. It is given that a=2, n=10 and d=a2-a1

7-2=5 And An=a+(n-1)d

An =a+9d 2+9(5)

=> 2+45=47

Q.12 As AB is a tangent of the inner circle

From fig. BD2 +OD2= OB2 BD2+9 = 25 BD2= 25-9 BD2= 16

= 4 cm Q.13 Let C (x,y) be a point. And it is given that A (3, 6) &B (-3, 4)

AC2=CB2 [(x-3)2+(y-6)2]= [(x+3)2+(y-4)2] 12x+4y-20=0 or 3x+y-5=0

OR A (2,-5) B(-2,9) Let P(x, 0) is the point on x –axis So, PA=PB By Distance formula

√( ) + ( + ) √( + ) + ( )

√( ) + = √( + ) + X2+4-4x +25 =x2 +4 +4x+81

So, x=-7 Q.14 Let line segment A (-6,10) and B (3,-8) is divided by (-4,6) in k:1 => -4=3k-6/k+1 => -4k-4=3k-6 => -4k-3k=-6+4 => -7k=-2 => k=2/7 So, k:1 = -2:7 Or Let ratio = k :1

By using y coordinate of [

]

We get 6=10-8k/k+1 6k+6=10-8k

=> 6k+8k=10-6 14k=4 K=2/7

Therefore K: 1=2:7 result

SECTION C Q.15 Let ‘a’ be the first term and ‘d’ is common difference So accordind to statement

a4+a8=24 and a6+a10=44 2a+10d=24 ____________(1) 2a+14d=44 ____________(ii)

By subtraction (i) from (ii) we get, d =5, and put the value of ‘d’ in (i) we get a=-13 . There fore A.P is : -13,-8,-3 ……………………………….. Q.16 Solution:Take a point Q on XY other than P and join OQ The point Q must lie outside the circle. Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is,

OQ > OP.

Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. Q.17 . In Quad. In AOBP . ⎳ +⎳B +⎳P+ ⎳ O=360 ⎳APB + ⎳ AOB= 180 ⎳ AOB=120 Construction of tangents 1.Draw a circle with radius =5cm 2. Draw angle ⎳ AOB=120 at O. 3. At A and B draw perpendiculars intersecting each other at P. h Q.18 Let height of tower AB =h and AD = x tan 30 =h/(x+40)

1/ =h/(x+40) tan 60°=h/x 3 = h/x h= 3x

3× 3x=x+40 2x=40 x=20 h=2 3m Q.19 P (E) = favorable outcomes/total outcomes

P ( )

=

Come to learn go to serve

Q.20 A (7,-2) B (5, 1) C (3,3k) => Condition of collinear points => Ar. ABC=0 => 1/2[x1(y2-y3) +x2(y3-y1) +x3(y1-y2)]=0 => [7(1-3k) +5(3k+2) +3(-2-1)]=0 => 7-21k+15k+10-9 =0 => 8-6k=0 Therefore K=4/3 D(-4,4) C(-1,-1) Q.21 diagonals of a square are equal.

AC = √( + ) + ( + )

BD = √( + ) + ( ) AB2+ BC2 = AC2

(√( ) + ( ) ) + (√( + ) + ( + ) ) = ( )

+ By the converse of Pythagoras

⎳B = 900

Therefore quad. ABCD is a square. Q.22 For cone r=3.5 cm, h=12 cm l= 3.52+122 =12.5cm2 Total surface area=πrl+2πr2 =22/7+3.5*12.5+2*22/7*3.5 =214.5cm2 Q.23 Side of square = 14 cm, Therefore area = side× side= 14 × 14 = 196 cm2 Diameter of circle = 7 cm Radius = 3.5 cm Area of 4 circles = 4×πr2= 4×22/7 × 3.5 ×3.5 = 154 cm2

There fore area of shaded portion = area of sq.-area of 4 circles= 196-154= 42 cm2 All the four values are important in students life because these are bases for successful life.

Q.24 Join OB , In right angled triangle OAB => OB2= OA2 + AB2= (20)2 + (20)2 => OB2 = 800

=> OB = 00 = 0

Area of sector=

=

( 0 )

= 628 cm2

Area of shaded region = Area of sector – area of square = 628- 400

= 228 cm2

Section D Let original speed = x km/h Increase speed =(x+250) km/h Acc. to question 1500/x-1500/(x+250)=1/2 X2+250x=750000 X2+250x-750000=0 X2+1000x-750x-750000 =0 X(x+1000)-750(x+1000)=0 (x+1000)(x-750)=0 So, X= -1000, 750 Rejecting –ve speed of 1000 km/h Therefore speed = 750 km/h Q.26 Let side of smaller square =x Perimeter= 4x Perimeter of larger square =24+4x Side=6+x ATQ x2+ (6+x)2= 468 x2+ 36+x2 +12x -468=0

x2+ 6x – 216 =0 x2+ 18x-12x – 216 =0 x(x+18)-12(x+18)= 0

(x+18)(x-12) =0 x=-18 or 12 Reject -18 Sides 12m and 18m Q.27 Let the first prize be of Rs x The next prize will be Rs (x-20),(x-40) x,(x-20),(x-40)……......... a=x d=-20 Sn=n/2[2a+ (n-1) d] S7=7/2[2a-120] =700 a=160 The prizes are Rs160, Rs140, Rs120, Rs100, Rs80, Rs60, Rs40 Q.28 Given: a circle ,with centre o. in which PQ, and PR are two tangents. To prove : PQ= PR Construction : Join OP ,OQ and OR . Proof : We are given a circle with centre O, a point P lying outside the circle and two tangents

PQ, PR on the circle from P . We are required to prove that PQ = PR. For this, we join OP, OQ and OR. Then ∠ OQP and ∠ ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP, OQ = OR (Radii of the same circle) OP = OP (Common) Therefore, Δ OQP ≅ Δ ORP (RHS) This gives PQ = PR (CPCT)

Q.29

1 Steps of Construction 1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. 2. Locate 4 (the greater of 3 and 4 in 3/4) points B1, B2, B3 and B4 on BX so that BB1 = B1B2 = B2B3 = B3B4. 3. Join B4C and draw a line through B3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to B4C to intersect BC at C′. 4. Draw a line through C/ parallel to the line CA to intersect BA at A/

Then, triangle A’BC' is the required triangle. Q.30 Let speed of the car=x m/sec Let time taken to reach from D to A =v sec Distance CD=6x Distance AD=V x m tan 30°=h/6x+vx tan 60°=h/vx (v+6)x / 3=vx 3 v=3 Time taken 3 seconds Q.31 P (E) =favorable outcomes/total outcomes P (E) =2/52=1/26 Sample space 4 kings + 4 queen + 4 jacks P (fcard) 12/52=3/13 Favorable outcomes of spade=13 P (spade)=13/52=1/4 Black cards=03+03=6 P(E)=6/52=3/26

Q.32

Length of the blade = radius=25 cm Sector angle=115° Area swept by one blade=𝛉𝛑r2/360 So, Area swept by two wipers=(2×115×22×625)/360×7 =158125/126 cm2 = 1254.96 cm2 Q.33 Depth=1.5m Width=6m Volume=l×b×h Speed=10km/h =10,000m/h Volume of water flowing in 1 Hour = Volume of cuboid of dimensions 10,000 m × 6m× 1.5 m Now volume of water = l×b×h =10,000×6×1.5 m3 = 90,000 m3 Volume of water flowing in canal in 30 minutes=45000m3 Let area of the field being irrigated= x m2 Area of field × height of water= 45000 m3

Area of field ×8/100=45000 => Area of field =5625900m2 Q.34 The total height of the bucket = 40 cm, which includes the height of the base. So, the height of the frustum of the cone = (40 – 6) cm = 34 cm.

Therefore, the slant height of the frustum, l = √ + ( )

where r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm.

So, l = √ + ( ) = 35.44 cm The area of metallic sheet used = curved surface area of frustum of cone + area of circular base + curved surface area of cylinder = [π × 35.44 (22.5 + 12.5) + π × (12.5)2

+ 2π × 12.5 × 6] cm2

= 22/7 (1240.4 156.25 150) cm2

= 4860.9 cm2

1

SAMPLE PAPER –3 (SA II) MRS.KIRAN WANGNOO

Mathematics

CLASS : X








SECTION -A

Question numbers 1 to 8 carry 1 mark each. For each of the questions 1-8, four alternative choices have been provided of which only one is correct. You have to select the correct choice. Q.1. Which of the following equations has the sum of its roots as 3?

(A) x2 + 3x – 5 = 0 (B) -x2 + 3x + 3 = 0

(C) 2x2 -

x (D) 3x2 - 3x – 3 = 0

Q.2. The sum of first five multiples of 3 is:

(A) 45 (B) 65

(C) 75 (D) 90

Q.3. If radii of the two concentric circles are 15 cm and 17 cm, then the length

of each chord of one circle which is tangent to other is

(A) 8 cm (B) 16 cm

(C) 30 cm (D) 17 cm

Q.4. In given fig. PQ and PR are tangents to the circle with

centre O such that QPR = 500,

then OQR is equal to:

(A) 25° (B) 30o

(C) 40° (D) 50o

Q

P O

R

50o

2

Q.5. To draw a pair of tangents to a circle which are inclined to each other at an angle of 1000, it

is required to draw tangents at end points of those two radii of the circle, the angle between

tangents should be:

(A) 100° (B) 50°

(C) 800 (D) 200°

Q.6. The height of a cone is 60 cm. A small cone is cut off at the top' by a

plane parallel to the base and its volume is

the volume of original cone. The

height from the base at which the section is made is: (A) 15 cm (B) 30 cm

(C) 45 cm (D) 20 cm

Q.7. A pole 6 m high casts a shadow 23 m long on the ground, then the sun's elevation is:

(A)60° (B) 450

(C) 300 (D) 90o

Q.8. Which of the following cannot be the probability of an event?

(A) 1/5 (B) 0.3

(C) 4% (D) 5/4

SECTION- B


Q.9. Two tangents making an angle of 120° with each other, are drawn to a circle of radius 6 cm,

then find the length of each tangent .

Q.10. If the circumference of a circle is equal to the perimeter of a square then taking

=

find the ratio of their areas .

Q.11. Find the roots of the following quadratic equation:

x2 - x –

= 0

Q.12 lf the numbers x-2, 4x-1 and 5x+ 2 are in A.P. Find the value of x.

Q.13. The tangents PA and PB are drawn from an external point P

to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.

Q.14. In given fig., a circle of radius 7 cm is inscribed in a square

Find the area of the shaded region

6m

23

3√

A

B C

A

P

B

O

3

SECTION—C


Q.15. How many spherical lead shots each having diameter 3 cm can be made from a cuboidal

lead solid of dimensions 9 cm x 11 cm x 12 cm?

Q.16. Point P (5, 3) is one of the two points of trisection of the line segment joining the points

A(7,- 2) and B (1, - 5) near to A. Find the coordinates of the other point of trisection.

Q.17 Show that the point P (- 4, 2) lies on the line segment joining the points A (-4, 6) and

B(-4,-6).

Q.18. Two dice are thrown at the same time Find the probability of getting Indifferent numbers

on both dice.

Or

A coin is tossed two times. Find the probability of getting atmost One head.

Q.19. Find the roots of the equation

+

=

x

, 5

Or

A natural number, when increased by 12, becomes 160 of its reciprocal. Find the number.

Q.20. Find the sum of integers between 100 and 200 that are divisible by 9.

Q.21. In given figure two tangents PQ and PR are drawn to a circle with

centre O from an external point P. Prove that QPR = 2OQR.

Or

Prove that the parallelogram circumscribing a circle is a rhombus .

Q.22. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60°.

Then construct a triangle whose sides are ¾ time the corresponding sides of

ABC.

Q.23. In given fig., OABC is a square inscribed in a quadrant OPBQ. If OA = 20 cm, find the area

of shaded region. [Use = 3.14]

4

Q.24. A hemispherical depression is cut out from one face of a cubical wooden block such that

the diameter l of the hemisphere is equal to the edge of the cube.

Determine the surface area of the remaining solid.

Or

A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of

uniform thickness. Find the thickness of the wire.

SECTION—D


Q.25. A tower stands vertically on the ground. From a point on the ground which is 20 m away

from the foot of the tower, the angle of elevation of the top of the tower is found to be 600

Find the height of the tower.

Q.26. Prove that the points A (4, 3), B (6,4), C (5, -6) and D (3, -7) in that order are the vertices of

a parallelogram.

Q.27. The points A2, 9), B(a, 5), C(5, 5) are the vertices of a triangle ABC

right-angled at B. Find the value of 'a' and hence the area of ABC.

Q.28. Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the

box. Find the probability that the card which is selected has a number which is a perfect square

Q.29. A train travels at a certain average, speed for a distance àf 63 km and then travels a

distance of 72 km at an average speed. of 6 km/h more than its original speed. If it

takes 3 hours to complete the total journey, what is its .original average speed?

Or

Find two consecutive odd positive integers, sum of whose squares is 290.

Q.30. A sum of Rs 400 is to be used to give seven cash prizes to students of a school for their

overall academic performance. If each prize is Rs 40 less than the preceding price, find the

value of each of the prize.

Q.31. Prove that the lengths of tangents drawn from an external point to a circle

are equal.

Q.32. A well of diameter 3 m and 14 m deep is dug. The earth, taken out of it, has

been evenly spread all around it in the shape of a circular ring of width 4 m

to form an embankment. Find the height of the embankment. Comment

5

on the importance of water in our daily life.

Or

21 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of water filled in the box.

Q.33. The slant height of the frustum of a cone is 4 cm and the

circumferences of its circular ends are 18 cm and 6 cm. Find curved

surface area the frustum.

Q.34. From a point on the ground, the angles of elevation of the bottom and top of a

transmission tower fixed at the top of a 20 m high building are 45° and 60°

respectively. Find height of the tower..how transmission towers are harmful to us?

6

SOLUTIONS SAMPLE PAPER –2 (SA II)

ANSWERS

SECTION -A

Question numbers 1 to 8 carry 1 mark each. For each of the questions 1-8, four alternative choices

have been provided of which only one is correct. You have to select the correct choice.

Ans.1

Sol. (B)3

[ Sum of the roots = -b = -3 = 3 a -1

Ans.2

Sol. (A) 45 [ Required sum = 3 + 6+ 9 + 12 + 15 = 45]

Ans.3

Sol. (B) 16 cm

[ OA2 = AD2 + OD2

172 = AD2 + 152

289-225 AD2 = AD2 =64

AD=8cm

AB = 2AD = 16 cm]

Ans.4

Sol. (A) 25°

[ QOR = 180o – 50o = 130o

OQR = ORQ [ Angle opposite to equal side of a Triangle]

2OQR = 180o – 130o = 50°

OQR = 250 ]

Ans.5

Sol. (C) 800 [ Angle between the radii 180° - 100° = 80]

Ans.6 Sol. (C) 45 cm

[ ABE ACD .

h = r

60 R

Q

P O

R

50o

7

1 r2h = 1 R2 60

3 64

r2h = 1 60 R2

64

h = 60 R2

64 r

h = 60 602

64 h2

h3 = 603 = 153

43

H = l5cm

Thus, height from the base = 60 - 15 = 45 cm]

Ans.7 Sol. (A) 600

[ tan = 6 = 3 = 3

23 3

tan = tan 60°

=60° ]

Ans8

Sol. (D) 5/4 [ Probability of an event 1 in any case]

SECTION-B

Ans.9

Sol. (D) 23 cm

[ In rt. PQO

PQ = Cot 600 = 1

OQ 3

PQ = OQ 1 = 6

3 3

= 6 3 = 23 cm

3 3 Ans.10 Sol

2r = 4 x Side

r = Side

2

6m

23

m

A

B C

8

Area of circle : Area of square = r2 : r 2

2

= 4:

= 4 : 22

7

= 14:11

Ans.11

Sol. Given equation is 2 x2 - x – 3 = 0 5 5

2x2 – 5x -3 = 0

2x2 - 6x + x - 3 = 0

2x (x-3) + 1 (x-3)= 0

(2x+1)(x-3)=0

X = 3, X = -½

Ans.12

Sol. x-2, 4x-l and 5x+2 are in A.P.

(4x-1)- (x-2) = (5x+2) - (4x-1)

4x-1-x+2 = 5x+2- 4x+ 1

2x=2

x=1

Ans.13

Sol. Since angle between the radius and the tangents at the

Point of contact is 90o

PAO + PBO = 90° + 90° = 180°

or APB + AOB = 180°

= AOBP is a cyclic quadrilateral

Opposite angles of a quadrilateral are supplementary

Ans.14

Sol. Here, Side of the square = Diameter of the circle

= 2 x 7 = 14 cm

Area of the shaded region = Area of the square - Area of the circle

= (14)2 – (22/7)

=196 – 154

= 42cm2

A

P

B

O

9

SECTION-C

Ans.15

Sol. Let number of spherical lead shots be n

n x (4) r3 = L x B x H 3

n x 4 x 22 x 3 x 3 x 3 = 9 x 11 x 12 3 7 2 2 2

9 x 11 x 12 x 3 x 7 x 2 x 2 x 2 4 x 22 x 3 x 3 x 3

n = 84

Hence, the required spherical lead shots is 84.

Ans.16

Sol. AP = PQ = QB

Q is the mid-point of PB.

Coordinates of Q are

Q 5+1 , -3 -5 2 2

Q (3,-4)

Ans.17

Sol. The abscissa (x-coordinate) of the points P, A and B is -4

Points P, A and B lie on the line x = 4

Hence, P A and B are collinear.

Ans.18

Sol. Number of outcomes of the sample space when two dice are thrown = 6x 6=36

Number of outcomes of getting same number on both

dice = 6 [(1, 1), (2, 2), (3, 3),(4, 4), (5, 5), (6, 6)]

Number of favorable outcomes (different numbers on both dice) = 36 -6 = 30

Required probability =. -36 = 5 36 6

Or

All possible outcomes when a coin is tossed twice = HH, HT, TH, UF

Favourable outcomes (atmost one head) TT, HT, TH

Required probability = 3

4

A

(7,-2)

P

(5,-3) B

(1,-5)

Q

10

.

Ans.19.

Sol. The given equation is 1 + 1 = 1

2x-3 x-5

x – 5 + 2x - 3 = 1

(2x-3)(x-5)

3x-8 = (2x-3)(x-5)

3x - 8 = 2x2 - 13x + 15

2x2 – 16 x + 23 = 0

x = 16 ± √( ) 2 x 2

x = 16 ± √ 4

x = 16 ± √ = 16 ± 6 √ = 4 ± 3 √ 4 4 2

OR

Let the number be x

According to the statement of the question

x + 12 = 160

x2 + 12x – 160 = 0

x2+20x-8x-160 = 0

(x-8)(x+20) = 0

= x = 8 or x = -20

(Rejecting - ve value since x is the natural number) Hence, the number is 8.

Ans.20.

Sol. Integers divisible by 9 between 100 and 200 are 108, 117, 126, 135, ……198

an = 198

a + (n-1) d = 198

108 + (n-1)9 = 198

(n-1)9 = 90

(n-1) = 10

n = 11

11

Now Sn = n (a + an)

2

S11 = 11 (108 + 198) = 11 x 306 = 1683

2 2

Ans.21.

Sol. Join ‘OR’

Now. QOR + QPR = 180o

QOR = 180o - QPR ... (i)

Also. OQR = ORQ ….(ii)

[s opposite to equal sides of a Triangle]

OQR + ORQ + QOR = 180o

QOR = 180°-OQR - ORQ

= 180o - 2OQR .... (iii) [using (ii)]

Now, from (i) and (iii), we have

180o - QPR 180o - 2OQR

QPR = 2OQR

Or

Given: A parallelogram ABCD, circumscribes a circle.

To prove: ABCD is a rhombus i.e.,

AB = BC = CD = DA.

Proof: Since ABCD is a parallelogram.

AB = DC and BC = AD ... (i)

AP and AS are two tangents from an external point A to the circle.

AP = AS ... (ii)

[ Tangents drawn from an external point to the circle are equal]

Similarly, we have

BP=BQ

CR=CQ …(iv)

and DR = DS …(v)

Adding (ii), (iii), (iv) and (v), we have

(AP + BP) + (CR + DR) = (AS + DS) ± (BQ + CQ)

12

AB+CD =AD+BC

AB + AB = AD + AD [using (i)]

2AB=2AD

AB = AD

i.e., adjacent sides of the parallelogram are equal. Thus, all the sides are equal.

Hence, ABCD is a rhombus. .

Ans.22. Sol. Steps of Construction:

1 Draw a line segment BC = 6 cm

2. At B, construct an angle 60° such that BA = 5 cm..

3. Join AC, so ABC is the given triangle.

4. Through B, construct an acute angle CBX, (<90o).

5. Mark four points B1, B2, B3 and B4, such that

BB1=B1B2=B2B3=B3B4

6. Join B4C.

7. Through B3, draw B3C' || B4C, intersecting BC in C'.

8. Through C', draw C'A' || CA, intersecting BA in A'.

9. Hence, A'BC' is the required triangle.

Ans.23.

Sol. Here, OA = 20 cm and OABC is a square,

OA =AB = BC = CO =20 cm

OB = OA2 + AB2 [by Pythagoras Theorem]

= 202 + 202=

= 202 cm

Now, area of the shaded region

= Area of quadrant OPBQ - Area of square OABC

= 90o 3.14 202 202-20 20 360o

[ OB=r=202 cm]

= ¼ 3.14 400 2-400

13

= 628-400

= 228cm2

Ans.24.

Sol. Edge of the cube = l

Radius of the hemisphere = l/ 2

Surface area of the remaining solid

= S.A. of cube - S.A. of the top of hemisphere

+ C.S.A. of hemisphere

= 6l2 - l/2 l/2 + 2 l/2 l/2

= 6l2 + 12

4

= ¼ l2 (24+) sq. units

Or

Volume of the wire = Volume of the copper rod

r2 1800 = ½ ½ 8

r2 = 8 = 1

4 1800 900

r = 1/30 cm

Thickness of the wire = Diameter of wire

= 2 1 = 1 cm

30 15

SECTION-D

Ans.25.

Sol. Let us assume the AB be the tower and C is a point 20 m away from the ground Angle of elevation of the top of the tower is 60o

In rt.CBA,

AB = tan 60o

CB

AB = 3

20

AB = 203 m

Hence, the height of the tower is 203 m.

14

Ans.26.

Sol. Here, AB = (6-4)2 + (4-3)2 = 4 +1 = 5 units

CD = (3-5)2 + (-7-6)2 = 4 +1 = 5 units

AB = CD

Again, BC = (5-6)2 + (-6-4)2 = 1 +100 = 101 units

AD = (3-5)2 + (-7-6)2 = 1 +100 = 101 units

BC = AD

Now, in quadrilateral of ABCD both pair of opposite sides are equal.

Hence, it is a parallelogram.

Ans.27.Sol.

Given ABC is right-angled at B.

By Pythagoras theorem, we have

AB2 + BC2 = AC2

(a - 2)2 + (5- 9)2 + (5-a)2 + (5- 5)2- (5-2)2 + (5-9)2

a2+ 4-4a + 16 + 25 + a2-10a = 9 + 16

2a2-14a-20 = 0

a2-7a-10 = 0

(a-5)(a-2) = 0

a = 5 or a = 2

Rejecting a = 5, BC reduces to zero.

Thus, a = 2

Area of ABC = ½ AB BC

= ½ 4 3

= 6 sq. units

Ans.28.

Sol. Total number of cards in the box = 100

Favorable outcomes (perfect squares) are 4, 9,16,25,36,49,64,81,100

15

Required probability = 9

100

SECTION—D


Ans.29.

Sol. Let the average speed be x km/h;


63 + 72 = 3

x x+6

63(x+6)+72x = 3

x(x+6)

63x+378+72x = 3

x2 + 6x . .

135x + 378 = 3x2 + 18x

3x2-117x-378=0 .

x2-39x-126 = 0

(x-42)(x+3) = 0

x = 42 or x = -3 (rejecting -ve value because speed cannot be -ve)

Hence, original average speed is 42 km/h. .

Or

Let two consecutive odd positive integers be x, x + 2.


x2 + (x + 2)2 = 290

x2+x2+4+4x = 290 .

2x2+4x – 286 = 0

x2 + 2x - 143 = 0

x2 + 13x-11x - 143 = 0

(x+13)(x-11) = 0 . . .

x = -13 or x = 11 .

16

(rejecting –ve value because x is an odd +ve integer)

Numbers are 11 and 13.

Ans.30.

Sol. Total amount of seven prizes = 1400

Let the value of first prize be x

According to given statement, the seven prizes are

x, x - 40, x - 80, x - 120, ..., x -240

Now, x-40-x = -40

x - 80 - x + 40= - 40, which is constant.

Thus, it is an A.P. with first term (a) as x and common difference (d) as - 40.

Sn =

{2a+ (n-1) d}

1400 =

{2x+(7-1)(-40)}

400 = 2x – 240}

2x = 640

x = 320

Hence, the amount of each prize (in ) is 320, 320 - 40, 320 - 80, 320 -120, 320- 160, 320-200, 320-

240 i.e., 320, 280, 240, 200. 160, 120, 80.

Ans.31.

Sol. Given: A circle C (O, r) with centre O. Through the external point P tangents PT and PT' are drawn.

To prove: PT = PT' T

Const.: Join PO, TO and T'O

Proof: In PTO and PT'O, we have

TO =T'O = r

hypt. PO = hypt. PO [common]

PTO = PT'O = 90o

PTO PT'O [by RHS cong. rule]

PT = PT'. .. [c.p.c.t.]

17

Ans.32.

Sol. Here, Radius of the well = 3/2 m

Depth of the well = 14 m

Width of the embankment = 4 m

Radius of the embankment = 1.5 + 4 = 5.5 m

Let 'h' be the height of the embankment.

Volume of the embankment = Volume of the well (cylinder)

(5.52 - 1.52) h = (1.5)2 14

(30.25- 2.25) h = (2.25 14) 4 m

28 h = 31.5

h = 31.5

28

h = 1.125m

Or

Radius of sphere = 2 cm

Volume of 21 spheres = 21 4 22 222

3 7

= 704cm3

Volume of cuboid = 16 x 8 x 8 = 1024 cm3

Volume of water = 1024 - 704 = 320 cm3

Ans.33.

Sol. Slant height of the frustum of a cone (l) = 4 cm

Circumference of top end = 18 cm

2R = 18

2

R = 18

R= 18 7 = 63 cm

2 22 22 and circumference of bottom end = 6 cm

2r = 6

2

r = 6

r = 6 7 = 21 cm

2 22 22

Curved surface area = l (R + r)

=

4

+

=

4

= 48 cm2

18

Ans.34.

Sol. Let AB be the transmission to fixed on the top of the building of height 20 m. Let AB = h m

and .P be a point on ground, such that BPC = 45°, APC = 60°.

In rt. PCB, C = 90° A

BC = tan 45o

PC

20 = 1 PC = 20m

In rt. PCA, C = 90°

AC = tan 60o

PC

h+20 = 3 h+20 = 20 3

20

h = 20 3 -20 = 20 (3 -1)

Hence, the height of the tower is 20 (3 -1)m.

1

SAMPLE PAPER –4 (SAII) MR AMIT. KV NANGALBHUR

Mathematics

CLASS : X




The question paper consists of 34 questions divided into 4 sections, A,B,C and D. Section – A comprises of 8 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each and Section- D comprises of 10 questions of 4 marks each.

Question numbers 1 to 8 in Section –A multiple choice questions where you are to select one correct option out of the given four.

There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions.

Use of calculator is not permitted.

Section-A Question number 1 to 10 carry 1 mark each.

1) The discriminant of the quadratic equation .3√3 x2 + 10x + √3 = 0 is

A) 30 B) 36 C) 64 D) 100

2) If a, 2 are three consecutive terms of an A.P., then the value of a.

A) 3/5 B) 5/3 C) 4/5 D) 7/5

3) In figure below, ΔABC is circumscribing a circle. he length of BC is

A) 10cm B)7cm C) 4cm D) 3cm

4) HCF and LCM of two numbers are 12 and 36 respectively, then product of these two

numbers is:

A) 36 B)432 C) 342 D) 12

5) Find the perimeter of the given figure, where AED is a semi-circle and ABCD is a rectangle.

A) 76cm B)68cm c) 22cm D)54cm

2

6) To draw a pair of tangents to a circle which are inclined to each other at an angle of

1000 . it is required to draw tangents at end points of those two radii of the circle, the

angle between which should be

A) 1000 B) 2000 C) 500 D) 800

7) Two Tangents making an angle of 1200 are drawn to a circle of radius 6cm , then the

length of each tangent is equal to

A) 2√3 cm B) √2 cm C) 6√3 cm D) √3cm

8) A pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

A) 600 B) 900 C) 300 D) 450

. Section-B Question number 9 to 14 carry 2 marks each

9) Solve by factorisation : √7 y2 – 6y - 13√7 = 0.

10) Find a point on x-axis which is equidistant from the points (-2,5) and (2, -).3

11) Determine the ratio in which the point P(m, 6) divides the join of A( -4, 3) and B (2,8).

12) Spherical ball of diameter 21 cm is melted and recasted into cubes , each of side 1

cm . find the number of cubes thus formed.

13) A die is thrown once , find the probability of getting

(i) a prime number (ii) a number divisible by 2

14) All cards of ace, jack and queen are removed from the deck of playing cards. One

card is drawn at random from the remaining cards. Find the probability that the card

drawn is :(a) a face card (b) not a face card

Section-C Question number 15 to 24 carry 3 marks each

15) Determine the AP whose 3rd term is 16 and when fifth term is subtracted from 7th

term, we get 12.

16) Construct a triangle similar to a given ∆ ABC in which AB =4cm , BC =6cm and

∟ABC = 600 , such that each side of the new triangle is ¾ of the corresponding

sides of given ∆ABC.(Steps of Constructions not required)

3

17) A

F E

B D C

In the figure above , the incircle of ABC touches the sides BC, CA and AB at D, E and F

respectively. If AB = AC , prove that BD = CD.

18)

E

A 2.8 cm B 1.4cm C

F

In the fig. , find the perimeter of the shaded region where ADC, AEB and BFC are semi-circles on

the diameter AC, AB, and BC respectively.

19) Largest sphere is carved out of the cube of side 7cm . find volume of the sphere?

20)The curved surface area of the cone is 12320 cm2. If the radius of the base is 56cm, find its

height.

21) The volume of the right circular cylinder of height 7cm is 567 cm3. Find its curved surface area.

22)An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the

chimney from his eye is 450. What is height of the chimney?

23)Find the value of ‘p’ for which the points (-5, 1) , (1,p) and (4, -2) are collinear.

OR

If the coordinates of the mid-points of triangle are (1,2) , (0,-1) and (2,-1). Find the coordinates of its

vertices.

24) Find the roots of quadratic equation : a2b2x2 + b2x – a2x – 1 = 0

4

Section-D

Question number 25 to 34 carry 4 marks each

25) The product of Tanay’s age five years ago and his age after 10 years is 16. Find his

present age.

26) Find the sum of the first 31 terms of an AP. Whose nth term is given by 3 + 2n/3.

27) OABC is a rhombus whose three vertices A,B and C lie on a circle with centre O. If

the radius of the circle is 10cm, find the area of the rhombus.

28) Prove that ‘ the lengths of the tangents drawn from an external point to a circle are

equal. Use this theorem for the following:

If all sides of the parallelogram touches a circle, show that it is a rhombus.

29) From top of a hill the angles of depression of two consecutive kilometre stones due

east are found to be 300 and 600. Find the height of the hill.

30) A school has 5 houses A, B , C , D and E. In a class having 23 students, 4 are from

house A , 8 from house B , 5 from house C , 2 from house D and rest are from E . 1

student is selected at random to be a class monitor.

A) Find the probability that the selected student is not from B, C and E.

B) find the probability that the student is from house E.

C) what moral values the student of this class must share?

31) Find the ratio in which the line 2x + y = 4 divides the join of A(2, -2) and B(3,7). Also

find the coordinates of the point of their intersection.

32) A bucket made up of metal sheet is in the form of a frustum of a cone of the height

16cm with radii of its lower and upper ends as 8cm and 20cm respectively. Find the

cost of the bucket, if the cost of the metal sheet is Rs. 15 per 100 cm2.

33) A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all

around to a width of 5m to form an embarkment. Find the height of the embankment.

34) Sum of the areas of the two squares is 468 m2. If the difference of their perimeters is

24m, find the sides of the two squares.

5


SECTION –A

1) C) 64

Explanation: - Here a =3√3, b = 10, c = √3

D = b2 – 4ac

= 102 – 4 X 3√3 X√3

= 100 – 36

= 64

2) D) 7/5

Explanation:- as 4/5, a, 2 are in AP therefore the common difference is same

i.e a- 4/5 = 2 – a

2a = 2 + 4/5

2a = 14/5

a = 7/5

3) A) 10cm

Explanation :- AQ = 4cm ( Tangents from same external point to same circle)

AC = AQ + QC

or 11 = 4 + QC

QC = 7cm

PC = QC = 7cm ( Tangents from same external point to same circle)

BP = 3cm ( Tangents from same external point to same circle)

BC = PC + BP

= 7 + 3

= 10cm

4) B) 432

Explanation :- Product of two numbers = HCF X LCM

= 12 x 36

= 432

5) A) 76cm

Explanation :- Required perimeter = AB + BC + CD + r [ here r = ½ X BC]

= 20 + 14 + 20 + 22/7 X 7

= 54 +22

=76 cm

6) D) 800

Explanation : As angles between tangents and at centre are supplementary

Required angle = 1800 -1000

= 800

7) A) 2√3 cm

Explanation :-

B

A O

6

∟OAB = 120/2 [ the line joining the external point to the centre bisect the angle between the

tangents at external point]

Therefore ∟OAB = 600

Now in AOB

Tan 600 = OB / AB

√3 = 6 / AB ( As tan 600 = √3)

AB = 6/√3

= 6√3/(√3 X√3)

= 6√3/ 3

= 2√3 cm A

8) A) 600

Explanation:- Pole

x

B C

Shadow

Let the inclination = x

Therefore tan x = AC/BC

Tan x = 6/2√3

= 6√3/(2√3X√3)

= 6√3/(2X3)

= √3

= tan 600

Section –B

9) √7y2 – 6y - 13√7 = 0

Or √7y2 – 13y + 7y - 13√7 =0

Or Y( √7y – 13) +√7 ( √7y – 13) =0

Or (√7y – 13) ( y + √7) = 0

Or (√7y – 13) = 0 or ( y + √7) = 0 A(-2,5)

Y = 13/√7 or y = - √7

10) Let the point on X axis be P(a,0) P(a,0) X-axis

According to question

AP = PB

√[(a+2)2 + (-5)2] = √[( a-2)2 + 32]

√[ a2 + 4 + 4a +25] = √[ a2 +4 – 4a + 9] B (2, -.3)

a2 + 4a -a2 + 4a =13-29

8a = -16

7

a = -2 (Ans)

11)

A(-4,3) K P(m,6) 1 B(2,8)

Let required ratio = k : 1

Using section formula ( on y coordinate) {[mx1 + nx2]/[m+n] , [my1 + ny2]/[m+n]}

6 = (8k +3)/(k+1)

Or 6(k+1) = 8k + 3

Or 6k + 6 = 8k +3

Or 6k – 8k = 3 – 6

-2k = -3

K = 3/2

Therefore reqired ratio = 3:2

12) Let the number of cubes formed = N

N X vol of cube = vol of sphere

N X 13 = 4/3 X

N = 4/3 X 22/7 X 21/2 X21/2 X21/2

N = 38808/8

N = 4851

13) Total outcomes = 6[ 1,2,3,4,5,6]

i) Total prime numbers =3[2,3,5]

P(E) = 3/6

= ½

ii) Total numbers divisible by 2 =3[ 2,4,6]

P(E) = 3/6

= ½

14) Number of cards removed = 12 [4A +4J +4Q]

Remaining cards = 52 – 12 =40

a) Number of face cards = 4 [kings]

P( E) = 4/40 = 1/10

b) Probability[not a face card] = 1 – P( E)

= 1 – 1/10

= 9 /10

SECTION – C

15) Given a3 = 16 i.e a + 2d = 16-------------------(1)

Also a7 –a5 = 12

a + 6d – ( a + 4d) =12

a +6d – a – 4d = 12

2d = 12

d = 6

Put d = 6 in (1)

8

a + 2X6 = 16

a + 12 = 16

a = 4

therefore required AP : 4 , 10 , 16 , 22 ……………..

Q16

16) Given AB = AC

Therefore AF + FB = AE + EC { from fig.)

AF – AE + FB = EC

AF – AF + FB = EC { AF =AE , lengths of two tangents from same external point }

FB = EC { lengths of two tangents from same external point }

BD = DC { lengths of two tangents from same external point }

17) Required perimeter = perimeter(semi-circle ADC) + perimeter(semi-circle AEB) +

perimeter(semi-circle BFC)

= X 4.2/2 + X 2.8/2 + X 1.4/2

=6.6 + 4.4 + 2.2

= 13.2 cm

9

18) Diameter of the largest sphere = side of the cube

Therefore diameter = 7 cm

Volume = 4/3 X

= 4/3 X 22/7 X7/2 X 7/2 X 7/2

=539 / 3 cm3

19) Given curved surface area = 12320 cm3

X rl = 12320

X 56X l = 12320

Or l = 12320 / ( X 56)

= 12320 X7 /(22X56)

l = 70 cm

Now h2 =r2 + l2

Or h = √[ l2 – r2]

= √[ 702 – 562]

= √[ 4900 – 3136]

= √1764

= 42 cm

20) Volume = 567 cm3 [given]

Therefore

r2h = 567

h = 567 / r2

h = 567/ 7 X 7 [ radius = 7cm]

h = 81/7 cm

curved surface area = 2 rh

= 2 X 22/7 X 7 X 81/7 A

= 3564 / 7 cm 2

21)

x

D 450

B

1.5m

E

28.5m C

Let the height of the chimney = x + 1.5

In ADB

Tan 450 = x / DB

1 = x / 28.5 { EC = DB}

X = 28.5cm

Therefore height of the chimney = 28.5 + 1.5

= 30 m(Ans)

10

22) Given that points A( 5,1) , B ( 1,p) and C (4,-2) are collinear

Therefore area ABC = 0

i.e 1/2[ x1(y2-y3) + x2(y3-y1) + x3(y1-y2) ] = 0

5(p +2) +1 (-2-1) +4(1-p) =0

5p +10 -3+ 4 -4p =0

P+11 = 0

P = -11(Ans)

OR

A(x1, y1)

D(1,2) E(2.-1)

B(x2, y2) F(0,-1) C (x3, y3)

As D(1,2) is the mid point of AB, By mid point formula

X= (x1+x2)/2 and y= (y1+y2)/2

Therefore 1 = (x1+x2)/2 and 2 =(y1+y2)/2

X1 + x2 = 2 ---------------- --------------------------(1)

And y1+y2 = 4---------------------- --------------------------(2)

Similarly, x2 + x3 =0----------------------------------------(3)

Y2+y3 = -2----------------- ------------------------(4)

x1+x3 = 4-------------------------------------------(5)

y1+y3 = -2------------------------------------------(6)

Add (1), (3) and (5)

2(X1 + x2+ x3) =6

X1 + x2+ x3 =3------------------------------------------------(7)

Put (1) in (7)

2 + x3 = 3

X3 =1

Similarly Putting (3) in (7)

X1 +0 = 3

X1 =3

Similarly Putting (5) in (7)

X2 +4 = 3

X2 =-1

Similarly y1 = 2, y2 = 2 and y3 = -4

11

23) a2b2x2 + b2x – a2x – 1. = 0

b2x(a2x +1) -1 (a2x +1) =0

a2x +1=0 or b2x -1 =0

x = -1/a2 or x = 1/b2

section -D

24) let Tanay’s present age = x years

therefore Tanay’s age 5 year’s ago = (x-5) years

and Tanay’s age after 10 years = ( x + 10) years


(X – 5) ( X +10) =16

X2 + 10x -5x -50 =16

X2 + 5x – 66 =0

X2 + 11x -6x – 66 =0

X(x +11) – 6 (x + 11) =0

x-6 =0 or x +11 = 0

x =6 or x = -11

x = -11 rejected ( as age cannot be negative)

therefore Tanay’s present age = 6 years

25) Given a n= 3 + 2n/3

Put n= 1 , 2 , 3 ……

a =11/3 , a2 =13/3, a3 = 15/3 and so on

therefore a =11/3 and d = 13/3 – 11/3 = 2/3

Sn = n/2[ 2a + (n-1)d]

= 31/2 [ 2 X11/3 + 30X 2/3]

= 31/2[ 22/3 + 60/3]

= 31/2 X 82/3

=1271/3 (Ans)

26) A

B O

C

As its given that OABC is a rhombus

Therefore OA = AB = BC = CO = Radius =10cm[ As the sides of rhombus are equal]

More over OB =10cm~

12

Area OABC = Area OAB + area OBC

= √3/4 X 102 +√3/4 X 102 { area of an equilateral triangle = √3/4 X Side2 }

=50√3 cm2

27) Given:- A circle with center O. Two tangents PT and PQ are drawn from external point P.

To Prove :- PT = PQ

Construction :- Join TO, OP and OQ

T

O

P

Q

Proof:- In ∆POT and ∆POQ

∟OTP = ∟OQP [ Each 900 as radius and tangents are perpendicular to each other]

OP = OP [Common hypotenuse]

OT = OQ [ Radii of the same circle]

Therefore ∆POT =᷈ ∆POQ [ by RHS]

PT = PQ ( c.p.c.t)

A

E F

B

H D

G

C

Given :- A parallelogram ABCD such that its sides touches a circle at E,F,G and H as shown in fig.

Let AF=AE =a, BE = BH = b, CH = CG =c , DG =DF = d

Now AB = CD { opposite sides of the parallelogram}

Therefore AE+EB = DG +GC

a + b = d + c ------------------------(1)

again BC = AD

13

BH +HC = AF +FD

b +c = a + d --------------------------------------(2)

adding (1) and (2)

a +2b +c = a +2d +c

b = d ---------------------------------(3)

But AB= a +c

= a + d ( using 3)

=AD Therefore ABCD is a rhombus because if adjacent sides of parallelogram are equal

then it’s a 3rhombus.

28)

A

300

600

h

D 600 C 300 B

h/√3 1km

Let AD is hill of height h km and C and B are two kilometre stones

Therefore CB = 1 km

In ADC

Tan 600 = h/DC

√3 = h / DC

DC = h /√3--------------------(1)

In ADB

Tan 300 = h / (DC + 1)

1/√3 = h / (DC +1)

14

DC+1 = h√3

1 = h√3 - h /√3 [ from 1]

1 = 2h/√3

h = √3/2 km

29)

A) Probability that the student is not from houses B,C and D = No. of students in A and D/

Total No. of students in class

= (4 + 2)/23

= 6/23

B) Number of students in house E = 23 – (4 + 8 + 5 + 2)

= 23 – 19

= 4

the probability that the student is from house E= No. of students in E/ Total No. of

students in class

= 4 /23

C) Cooperation

31)

K 1

A(2,-2) B(3,7)

Let the required ratio be k:1

Therefore the point of intersection is given by { (3k+2)/(k+1) , (7k- 2 )/ (k +1) }

As this point lie on the line 2x + y = 4 therefore it must satisfy the equation

i.e 2(3k+2)/(k+1) + (7k- 2 )/ (k +1) =4

or 6k +4 +7k – 2 = 4(k+1) ( By taking LCm and cross multiplication)

or 6k + 4 + 7k – 2 = 4k + 4

or 13k – 4k = 4 -2

or 9k = 2

K = 9/2

15

Therefore the required ratio is 9:2

For point of intersection

[( 3 X 9/2 + 2)/ 9/2 +1 , ( 7X 9/2 – 2)/ 9/2 +1 ]

( 31/11 , 59/11)

D) Here r = 8cm, R=20cm , h = 16cm

Slant height , I = √[ h2 + ( R – r)2]

= √[ 162 + ( 20 – 8)2]

= 20 cm

Total surface area = ( )

= 3.14[ 20(20+8) + 8 X8]

=1956.36 cm2

Cost of metal used = 1959.36 X 0.15

= Rs. 293.90.

E) Diameter of the well = 10m , height/depth = 14m

Inner radius of the embankment = 5m

Outter radius of the embankment = 10m

As the mud dug out of well is used to make the embankment

Therefore the vol. of well = vol of embankment

= ( ) ( )

= (10X10 -5X5)

14X25 = h (100- 25)

14 X25 = h X 75

(14 X25) /75 = h

14/3m = h

F) Let the side of larger square be x m and y m

Therefore the area of larger square = x2

And the area of smaller square = y2

Perimeter of larger square = 4x and of smaller square = 4y


Difference of perimeters = 24

4x – 4y = 24

Or x- y = 6

Or x = 6 + y

According to second condition

Sum of areas = 468

X2 + y2 = 468

(6 + y)2 + y2 = 468

36 + y2 + 12y + y2 = 468

16

2y2 + 12y -432 =0

Y2 + 6y -216 = 0

Y2+ 18y -12y – 216 =0

Y (y + 18) -12(y + 18) =0

(y +18) (y -12) =0

Y +18 =0 or y – 12 =0

Y = -18 or y = 12

Y = -18 rejected because side cannot be negative

Therefore side of smaller square = 12m

And side of larger square = 18m

Mathematics CLASS : X - K.V. No. 3 Amritsar · Mathematics CLASS : X ... However, internal choice...

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Transcript of Mathematics CLASS : X - K.V. No. 3 Amritsar · Mathematics CLASS : X ... However, internal choice...