1291198421 Classx Math Samplepaper Saii 40

7/27/2019 1291198421 Classx Math Samplepaper Saii 40

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Summative Assesment-IITOPPER SAMPLE PAPER IV

MATHEMATICS

CLASS X

Time : 3- 31

2Hrs M.M: 80

GENERAL INSTRUCTIONS :

1. All questions are compulsory.

2. The question paper is divided into four sections

Section A : 10 questions (1 mark each)

Section B : 8 questions (2 marks each)

Section C : 10 questions (3 marks each)

Section D : 6 questions (4 marks each)

3. There is no overall choice. However, internal choice has been provided in 1

question of two marks, 3 questions of three marks and 2 questions of four marks

each.

4. Use of calculators is not allowed.

SECTION A

Q1. The values of k for which the quadratic equation 2x2-kx+k=0 has equal roots

is

(a) 0 only (b) 0,4 (c) 8 only (d) 0,8

Q2. In the given figure, the respective values of y and x are

(a) 60o and 30o

(b) 45o and 60o

(c) 60o and 45

(d) 30o and 45o

Q3. The next term of the given series 2 8 18 32 ............ would be

(a) 50 (b) 70 (c) 80 (d) 90


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Q4. To draw a pair of tangents to a circle which are inclined to each other at anangle of 35, it is required to draw tangents at the end points of those two

radii of the circle, the angle between which is(a) 105 (b) 70 (c) 140 (d) 145

Q5. If a die is thrown once the probability of getting a prime number is

(a)1

2(b)

1

3(c)

1

4(d)

1

5

Q6. The ordinate of a point is twice its abscissa. If its distance from the point

(4,3) is 10 , then the coordinates of the point are

(a) (1,2) or (3,5)

(b) (1,2) or (3,6)

(c) (2,1) or (6,3)

(d) (2,1) or (3,6)

Q7. The ratio in which the line segment joining A(3,4) and B(-2,1) is divided by

the y-axis is

(a) 1:2 (b) 2:3 (c) 3:2 (d) 2:5

Q8. The given figure shows a sector of a circle of radius 10.5 cm. The perimeter of

the sector is

(a) 28 cm (b) 30 cm (c) 32 cm (d) 34 cm

Q9. The radius of a cylindrical tank is 28m. If its capacity is equal to that of a

rectangular tank of size 28mx16mx11m then, its depth is(a) 2m (b) 3m (c) 4m (d) 5m

Q10. In given figure, a circle of radius 7.5cm is inscribed in a square, the remainingarea of the square is

(a) 48.91 cm sq.

(b) 48.375 cm sq.

(c) 46 cm sq.

(d) 52.32 cm sq.


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SECTION B

Q11. If the 3rd

and 6th

term of an AP are 7 and 13 respectively, find the 10th

term.Q12. In the given figure, XP and XQ are tangents from X to the circle with centre

O. R is a point on the circle. Prove that XA+AR=XB+BR.

Q13. A letter is chosen at random from the letters of the word ASSASSINATION.Find the probability that the letter chosen is a (i) vowel (ii) consonant

OR

Rekha tosses a one rupee coin three times and wins the game if she gets thesame outcome every time. What are her chances to win?

Q14. In given figure, ABCD is a square of side 14 cm and four congruent circles are

inscribed in it. Find the remaining area of the square.

Q15. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter

of the wheel.Q16. Determine the set of values of p for which the quadratic equation

px2+6x+1=0 has real roots.Q17. Find the condition that the point (x,y) may lie on the line joining (3,4) and (-

5,-6).

Q18. Two concentric circles are of radii 5cm and 3cm. Find the length of the chordof the larger circle which touches the smaller circle.

SECTION C

Q19. Solve :1 1 1 1

a b x a b x


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Q20. Find the area of the quadrilateral ABCD whose vertices are A(1,1), B(7,-

3),C(12,2) and D(7,21) respectively.

OR

One end of a line of length 10 units is at the point (-3,2). If the ordinate of the other

end is 10, then find the abscissa of other end.Q21. A boy standing on a horizontal plane finds a bird flying at a distance of 100m

from him at an elevation of 30o. A girl standing on the roof of a 20m highbuilding, finds the angle of elevation of the same bird to be 45o. the boy and

the girl are on the opposite sides of the bird. Find the distance of the bird

from the girl.OR

A tower is 50m high. Its shadow is x metres shorter when the suns altitude is

45o than when it was 300. Find the value of x.Q22. If the point (x,y) is equidistant from the points (a+b,b-a) and (a-b,a+b), then

prove that bx = ay.

Q23. Find the sum of the first 25 terms of an AP whose nth term is given by 3n+2.Q24. Draw a circle of any convinient radius. Draw a pair of tangents to the circle

such that they are inclined to each other at an angle of 90.

Q25. One card is drawn from a well shuffled pack of 52 cards. Calculate theprobability of getting

(i) A king or a queen

(ii) Neither a heart nor a red king

Q26. The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280.

The field is to be ploughed at the rate of Rs. 0.50 per m 2. Find the cost of

ploughing the field. (Take =22

7

)

OR

The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of

triangle as the centre, a circle is drawn with radius equal to half the length of

the side of the triangle. Find the area of the triangle not included in the

circles. (use =3.14 and 3 = 1.73205).

Q27. The ratio of the radii of two spheres is 1:2. The two spheres are meltedtogether to form a cylinder of height which is 12 times its radius . So what is

the ratio of the radius of the smaller sphere and the cylinder ?

Q28. If h,c and V respectively are the height, the curved surface area and the

volume of cone, then prove that 3Vh3

-c2

h2

+9V2

=0

SECTION D

Q29. The difference of two numbers is 5 and the difference of their reciprocals is

1

10. Find the two numbers.

OR

The difference of square of two natural numbers is 45. The square of the

smaller number is four times the larger number. Find the two numbers.


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Q30. The sum of four consecutive numbers in an AP is 32 and the ratio of the

product of the first and the last terms to the product of the two middle terms

is 7:15. Find the numbers.OR

If S1, S2 and S3 respectively be the sum of n, 2n and 3n terms of an AP, show

that S3=3(S2-S1).

Q31. From a window, h metres high above the ground, of a house in a street, theangles of elevation and depression of the top and the foot of another house

on the opposite side of the street are and respectively. Show that the

height of the opposite house is h(1+tan.cot) metres.Q32. The incircle of ABC touches the sides BC, CA and AB at D, E and F

respectively. Show that AF+BD+CE = AE+BF+CD=1

2(Perimeter of ABC)

Q33. Prove that the tangent of a circle is perpendicular to the radius at the point ofcontact.Q34. A building is in the form of a cylinder surmounted by a hemispherical dome.

The base diameter of the dome is equal to2

3of the total height of the

building. Find the height of the building, if it contains 671

21m3 of air.


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SOLUTIONS

Section A

Ans1. Option (d)2x2-kx+k=0

For equal roots, D=0b2-4ac=0

(-k)2-4(2)(k)=0

k2-8k=0k(k-8)=0

k=0 or k-8=0

k=0, 8 1mark

Ans2. Option (a)

In given figure, ABCO ABD + DBO =90 ABD + 30 =90 ABD = 60 ABD = BDO = y = 60 (alternate interior angles)Also, ABC = BCO = x = 30 (alternate interior angles)Thus, the value of y and x are 60 and 30 respectively. 1mark

Ans3. Option (a)

The given series can be written as 2 2 2 3 2 4 2 ............

So, the next term would be 5 2 , i.e., 50 1mark

Ans4. Option (d)The angle between a pair of tangents to a circle which are inclined to each

other at an angle is supplementary to the angle between the two radii of thecircle.Thus, the angle between the radii of the circle = 180-35=145 1mark

Ans5. Option (a)


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When a die is thrown once the outcomes are 1,2,3,4,5,6 out which 2,3 and 5are prime numbers.

Therefore, P(prime number) =3 1

6 2

1mark

Ans6. Option (b)

Let the coordinates of the point be P(x,2x). Let Q be the point (4,3).

PQ2=(4-x)2+(3-2x)2 = 1016+x2-8x+9+4x2-12x=10 5x2-20x+15=0 x2-4x+3=0 (x-3)(x-1)=0 x=1 or x=3So, 2x = 2 or 6

Hence, the coordinates of the required point are (1,2) or (3,6). 1mark

Ans7. Option (c)Let the line segment AB cut the y-axis at the point P(0,y).Let the ratio in which AB is divided by the point P be k:1.

0=2k 3 3

kk 1 2

Hence the required ratio is3

2: 1, i.e., 3:2. 1mark

Ans8. Option (c)

Required perimeter=2 r 2 22 10.5 60

2r360 7 360

+2(10.5)

= 11+21= 32cm 1mark

Ans9. Option (a)Let d be the depth of the cylindrical tank.

According to the given information,(28)2xd=28x16x11 d=2Thus, the depth of the cylindrical tank is 2 m. 1mark

Ans10.Option (b)Area of the circle = r2 = 3.14 (7.5)2= 176.625 cm2

Side of square = diameter of circle = 15cmArea of square = side2 = 152=225 cm2

Remaining area = area of square area of circle = 225 - 176.625=48.375cm2

1mark

SECTION B

Ans11.Here, t3=a + 2d = 7 and t6 = a + 5d = 131

2mark

Solving the two equations, we get,

a = 3; d = 21

2mark

Therefore, 10th term = a + 9d = 3 + 9 x 2 = 21 1mark

Ans12.Since the lengths of tangents from an exterior point to a circle are equal.


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Therefore, XP=XQ (from X) (i)AP=AR (from A) ..(ii)

BQ=BR (from B) ..(iii) 1mark

Now, XP=XQ XA+AP=XB+BQ

XA+AR=XB+BR (Using (ii) and (iii)) 1mark

Ans13.There are 13 letters in the word ASSASSINATION.

Total number of outcomes = 13

(i) There are 6 vowels in the word. (A,A,A,I,I,O)

Number of favourable outcomes = 6

P(vowel) =6

131mark

(ii) Number of consonants in the word = 13-6 =7

P(consonant) =7

131mark

OR

Tossing of a coin results in either a Head or a Tail.

After 3 tosses, the possible outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, which are 8 in number.

The outcomes which are in her favour to win are HHH or TTT, which are 2 in

number. 1 mark

Let E be the event that Rekha wins.

P(E) = Favourable outcomes 2 1Total outcomes 8 4

1 mark

Ans14.Area of square = 14cmx14cm=196 cm21

2mark

Diameter of each circle =14

2=7

Radius of each circle =7

2

Area of four circles = 4x r2

= 4 x

222 7

7 2

= 154 cm2 1mark

Remaining area = 196-154 = 42 cm21

2mark

Ans15.Let r be the radius of the wheel.Distance covered in 1 revolution = 2rDistance covered in 5000 revolution = 5000 2r = 11km 1mark

5000 22

7

(2r) =11 1000 metres

2r =7

10metres


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Hence diameter of the wheel =7

10m or 70 cm. 1mark

Ans16.The given equation will have real roots, if b2-4ac0.

(6)2-4(p)(1)0 1mark 36-4p0

364p

p9 1mark

Ans17.Since the point P(x,y) lies on the line joining A(3,4) and B(-5,-6), therefore,

P, A and B are collinear points.

So, area of triangle PAB = 01

2mark

0 = 1 2 3 2 3 1 3 1 21

x (y y ) x (y y ) x (y y )2

[x(4+6)+3(-6-y)-5(y-4)]=0 1mark10x-18-3y-5y+20=0

10x-8y+2=0

5x-4y+1=0, which is the required condition.1

2mark

Ans18.

Let PQ be the chord of the larger circle which touches the smaller circle at the

point L.Since PQ is tangent at the point L to the smaller circle with centre O.

Therefore, OL is perpendicular to PQ.Since PQ is a chord of the bigger circle and OL is perpendicular to PQ, OL

bisects PQ.

So, PQ=2PL 1mark

In right OPL,PL2=OP2-OL2

PL2= 25-9=16PL=4cmPQ=2PL= 2(4cm) =8cmHence, the length of the chord PQ is 8cm. 1mark

SECTION C

Ans19.


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1 1 1 1

a b x a b x

1 1 1 1

a b a b x x

a b x a b x1mark

ab a b x x

a ba b

ab a b x x

ax+bx+x2=-ab

x2+ax+bx+ab=0 1mark

x(x+a)+b(x+a)=0

(x+a)(x+b)=0

x+a=0 or x+b=0

x=-a,-b 1mark

Ans20. Area of quadrilateral ABCD = Area of ABC + Area of ACD

Area of triangle =1 2 3 2 3 1 3 1 2

1x (y y ) x (y y ) x (y y )

2

Area of ABC =1

2[1(-3-2) + 7(2-1) +12(1+3)]

=1

2[-5+7+48] = 25 sq. units 1mark

Area of ACD =1

2[1(2-21) + 12(21-1) +7(1-2)]

= 12

[-19+240-7] = 107 sq. units 1mark

Therefore, area of quadrilateral ABCD = 25 +107 = 132 sq. units. 1mark

OR

Let the abscissa of other end be x.Given, distance of A(-3,2) from B(x,10) is 10 units, i.e., AB =10

AB = 2 2

3 x 2 10 1mark

10 = 29 x 6x 6 4

Squaring both the sides, we get,

100 = x2 + 6x + 73

x2 + 6x -27 = 0 1mark

x2 + 9x 3x -27 = 0

x(x+9)-3(x+9) =0

(x+9)(x-3) =0

x = -9 or 3

Thus, the abscissa of other end is 9 or 3. 1mark

Ans21.


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1mark

Let O be the position of the bird, B be the position of the boy, G be theposition of the girl and FG be the building at which the girl is standing.BO= 100m , FG= 20m

In OLB,

OL

BO= sin30

OL 1

100 2 OL =50m 1mark

OM = OL-ML = OL-FG = 50-20 = 30m

In OMG,

oOM 1sin 45 OG 30 2 m 42.3 m

OG 2

Thus, the distance of the bird from the girl is 42.3 m. 1mark

OR

1mark

Let PQ be the tower and let PA and PB be its shadows when the altitudes of

the sun are 45o and 30o respectively. Then, PAQ = 45o, PBQ = 30o, PQ =50m, AB =x m.

In APQ,

cot 45 =AP

PQ


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1 =AP

AP 50m50

1mark

In BPQ,

cot 30 =BP

PQ

3 =x 50

50

x= 50( 3 1 ) 1mark

Ans22.Let P(x,y), Q(a + b, b a) and R(a b, a + b) be the given points.It is given that PQ = PR

PQ2 = PR21

2mark

{x (a + b)}2 +{y (b a)}2 = {x (a - b)}2 +{y (a + b)}2

1mark x2 2 x (a + b) + (a + b)2 + y2 2y (b a) + (b-a)2

= x2 + (a b)2 2x (a - b) + y2 2y (a + b) + (a + b)2

-2 x (a + b) 2 y (b a) = -2x (a b) 2 y (a + b) -ax - bx - by + ay = -ax + bx - ay - by

2 bx = 2 ay

bx = ay 11

2marks

Ans23.a1=3(1)+2=5a2=3(2)+2=8

a3=3(3)+2=11The AP with nth term as 3n+2 is 5,8,11,14,

Here, a=5 and d=3 11

2marks

S25=25

2[2x5+(25-1)3] =

25

2[10+72] =1025 1

1

2marks

Ans24. The steps of construction are as follows:1. Draw a circle of any convinient radius and mark its centre as O.

2. Draw one of its radius OA and draw a line perpendicular to OA at A.

3. Draw an angle of 90 at O such that it intersects the circle at B.4. Draw a line perpendicular to OB at B and let both perpendiculars intersect

at point P.Thus, PA and PB are the required tangents. 1mark


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2 marks

Ans25.Total number of outcomes =52

(i) Number of kings=4Number of queens = 4

P(king or queen) = 4 4 8 2

52 52 13

1

1

2marks

(ii) Number of hearts =13

Number of red kings =2 (out of these 1 ia a heart)

P(neither a heart nor a red king) = 113 2 1 38 19

52 52 26

1

1

2marks

Ans26.For Rs. 24 , the length of fencing =1m

For Rs. 5280, the length of fencing =1

24 5280 = 220 metres. 1mark

Circumference of the field = 220 m

2r =220

2x22

7 r =220

220 7r 35m

44

Area of the field = r2 = (35)2 =1225 m2 1markCost of ploughing = Rs. 0.50 per m2

Total cost of ploughing the field = Rs. 1225 x 0.50

= Rs.1225 22 1

175 11 Rs.19257 2

1mark

OR


14/18

Let ABC be the equilateral triangle whose each side is of length a.

3

4 a

2

= 17320.5

1.73205

4 a

2

= 17320.5 a2

= 4 10000 ora=200 1mark

Radius of each circle = 100cm1

2mark

Area of the required region = area of equilateral area of 3 sectors

= 17320.5 360

360 (100)2 = 17320.5

1

2 3.14 10000

= 17320.5 1.57 x 10000 = 17320.5 -15700 = 1620.5 cm2 11

2marks

Ans27. Let the radius of the smaller sphere be r , so the radius of the larger sphere

is 2r

Let the base radius of the cylinder be RSo, height of the cylinder=12R...given 1mark

ATQ,

3 34 4r + (2r)

3 3= 2R (12R) 1mark

312r = 312R r=R The ratio of the radii is 1:1 1mark

Ans28.Curved surface area of a cone , c= rl

Volume of a cone , V =1

3

r2h 1mark

L.H.S.= 3Vh3-c2h2+9V2

= 3h3 21

r h3

-(rl)2h2+9

2

21 r h3

= 2r2h4- 2r2h2l2 +9 2 4 21

r h9

1mark

= 2r2h4 - 2r2h2(r2+h2) + 2r4h2 (Since, l2=r2+h2)= 2r2h4 - 2r4h2 - 2r2h4 + 2r4h2 =0 = R.H.S. 1mark

SECTION D

Ans29.Let the numbers be x and x-5.Given,

1 1 1 1 1

x 5 x 10 x 5 x

11

2marks

x x 5 1

x 5 x 10

(x-5)x=50

x2-5x-50=0 1mark

(x-10)(x+5)=0

x=10 or x=-51

2mark


15/18

When x=10, x-5=5When x=-5, x-5=-5-5 = -10

Thus the required numbers are 10 and 5 or -5 and -10. 1mark

OR

Let x and y be the two natural numbers and x>y.

Given, x2-y2=45 (i) 1mark

y2=4x .(ii)1

2mark

Using (ii) in (i), we get,

x2-4x-45=01

2mark

(x-9)(x+5)=0

x=9 or x=-5 1mark

Rejecting x=-5 because x is a natural number.Therefore, x=9

From (ii),

y2=4x = 4x9=36y = 6

But y is a natural number. Therefore, y=-6 is rejected.

Thus, y=6.Hence the required natural numbers are 9 and 6. 1mark

Ans30.Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.

So, a-3d+a-d+a+d+a+3d=32

4a=32

a=8 11

2marks

Also,

a 3d a 3d 7

a d a d 15

1mark

or, 15(a2-9d2)=7(a2-d2)or, 15a2-135d2=7a2-7d2

or, 8a2-128d2=0

or, d2=4 or d=2 1markSo, when a=8 and d=2, the numbers are 2,6,10,14.

When a=8,d=-2 the numbers are 14,10,6,2. 12

mark

OR

Here, S1 =n

2[2a+(n-1)d]

S2 =2n

2[2a+(2n-1)d]

S3 =3n

2[2a+(3n-1)d] 1

1

2marks


16/18

S2-S1= 2an+n(2n-1)d an n n 1 d

2

=an +[(2n-1) n 1

2

]nd

= an + n 3n 1 d

2

1

1

2marks

3(S2-S1)=3an + 3n 3n 1 d

2

= S3 1mark

Ans31.

1mark

Let B be the window of a house AB and let CD be the other house. Then, AB=

= EC =h metres.Let CD = H metres. Then, ED= (H-h) m

In BED,

cot =

BE

ED

BE = (H-h) cot .. (a) 1mark

In ACB,

AC

AB= cot

AC=h.cot . (b) 1mark

But BE=AC(H-h) cot = hcot

H= h cot cot

cot

H = h(1+tan cot) m 1mark

Hence proved.

Ans32.


17/18

1

2mark

Since lengths of the tangents drawn from an external point to a circle are

equal.Therefore, AF=AE (From A)(i)

BD=BF (From B) (ii)

CE=CD (From C) ..(iii) 1markAdding equations (i), (ii) and (iii), we get

AF+BD+CE = AE+BF+CD1

2mark

Now, Perimeter of ABC = AB+BC+CA

Perimeter of ABC = (AF+FB)+(BD+CD)+(EC+AE) 1mark= (AF+AE)+(BD+BF)+(EC+CD)

= 2(AF+BD+CE)

AF+BD+CE =1

2(Perimeter of ABC)

Hence, AF+BD+CE = AE+BF+CD =1

2(Perimeter of ABC) 1mark

Ans33. Given: A circle C (O, r) and a tangent AB at a point P.

To Prove: OP is perpendicular to AB.Construction: Take any point Q, other than P, on the tangent AB. Join OQ.

2marks

Since, Q is a point on the tangent AB, other than the point of contact P, so Qwill be outside the circle.

Let OQ intersect the circle at R.Then, OQ=OR+RQ

OQ>OR OQ>OP (OR=OP=radius)

Thus, OP


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But, among all the line segments, joining the point O to a point on AB, theshortest one is the perpendicular from O on AB.

Hence, OP is perpendicular to AB. 2marks

Ans34.Let the radius of the hemispherical dome be r metres and the total height of

the building be h metres.

Since the base diameter of the dome is equal to2

3of the total height, 2r =

2h

3.

This implies r =h

3. Let H metres be the height of the cylindrical portion.

Therefore, H= h h3

= 2h3

metres. 1mark

Volume of the air inside the building

= volume of air inside the dome+ volume of the air inside the cylinder

=

3 2

32 h h 2h 8 h3 3 3 3 81

cu.meters 2marks

Volume of the air inside the building is 671

21

m3.therefore,8

81

h3 =1408

21

This gives h=6m 1mark

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1291198421 Classx Math Samplepaper Saii 40