MAHESH TUTORIALSSUBJECT : Maths

Chapter # 2, 3, 4, 5, 8, 9, 10, 12Model Answer Paper

Eng. Medium 10th GSEB Batch :

Test - Date: Time: 3 Hrs Marks : 100

PART - A1. (c) cubic2. (d) 43. (d) x - 34. (c) straight line5. (c) -36. (d) x - 97. (d) p(x) = 6x2 + 11x + 48. (b) 39. (c) x + 50

10. (d) x + 3 = y11. (c) unique solution12. (b) (-2, 1)13. (b) 214. (a) 10015. (d) 216. (c) 2x(3x - 5) + 1 = 3x(2x + 5) + 317. (c) D < 018. (c) 019. (d) -220. (b) 421. (c) 2

22. (c) -14

, -14

23. (a) (3, 0)24. (d) 1325. (b) (0, 2)26. (c) (2, -4)27. (a) 128. (a) 1229. (c) 3 : 230. (c) (6, 6)

31. (c)12

32. (c)269

33. (d)32

34. (d) 1

35. (c)43

SET - A

36. (b)1312

37. (c) 238. (c) tan2 39. (c) 2a40. (c) 60

41. (c) 10 342. (c) x > y

43. (b) 3 x44. (d) 34.645. (c) 2, 5, 8, 11, ...46. (d) 3647. (b) 80048. (c) 5149. (c) 1515050. (d) 100

PART - BSECTION - A

Solve the following sums : [2 Marks] 061. p(x) = x2 + 4x - 21

= x2 - 3x + 7x - 21= x(x - 3) + 7 (x - 3)= (x - 3) (x + 7)

To find zeros of p(x), let p(x) = 0 (x - 3) (x + 7) = 0 x - 3 = 0 or x + 7 = 0 x = 3 or x = (-7)Thus, 3 and (-7) are two zeros of p(x).

2. Here, + = 2 and = -3 + = 2

ba

=21

letb2 =

a1

= k, k 0

b = 2k, a = -k

Now, =ca

= -3

c = (-3) a c = (-3) (-k) c = 3kp(x) = ax2 + bx + c

= (-k)x2 + (2k)x + 3k= -kx2 + 2kx + 3k= k(-x2 + 2x + 3)= k(x2 - 2x - 3), k 0

3. m [a + (m – 1) d] = n [a + (n – 1) d] ma + (m2 – m) d = na + (n2 – n) d (m – n) a + (m2 – m – n2 + n) d = 0 (m – n) a + [(m – n) (m + n) – 1 (m – n) ] d = 0 (m – n) a + (m – n) (m + n – 1) d = 0 a + (m + n – 1) d = 0 [dividing by m - n (m n) Tm+n = 0

OR

3. Sn =n2

[2a + (n - 1)d]

S10 =10 [2 1 (10 – 1) 2]2

S10 = 5 [2 + 18] S10 = 5 × 20

= 100 S10 = 100

OR4. x + y = 7 ..... ..... ..... (i)

3x - y = 1 ..... ..... ..... (ii)From equation (i)y = 7 - xSubstituting y = 7 - x in equation (ii), we get,

3x - y = 1 3x - (7 - x) = 1

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3x - 7 + x = 1 4x - 7 = 1 4x = 1 + 7 4x = 8

x =84

x = 2Substituting x = 2 in equation (i), we get

2 + y = 7 y = 7 - 2 y = 5

{(x, y)} = {(2, 5)} is the solution of given pair of equations.

5. Let P(a, 0) be the point on X-axis at a distance 13 from A(11, 12). AP = 13 AP2 = 169 (a – 11)2 + (0 – 12)2 = 169 a2 – 22a + 121 + (–12)2 = 169 a2 – 22a + 121 + 144 – 169 = 0 a2 – 22a + 96 = 0 (a – 16) (a – 6) = 0 a – 16 = 0 (or) a – 6 = 0 a = 16 (or) a = 6 Co-ordinates of P are (16, 0) or (6, 0)

OR5. A(x1, y1) = A(3, – 6) and B(x2, y2) = B(– 2, – 1) are given points.

Let P(x, y) be the point on AB such thatAPPB

=32

mn

=32

P(x, y) divides AB from A in the ratiomn

=32 .

x =2 1mx + nx

m + n , y =2 1my + ny

m + n

x = 3 2 + 2 3

3 + 2

, y = 3 1 + 2 6

3 + 2

x = 0 , y = – 3 The coordinates of the point which divides AB in the ratio 3:2 from A are (0, – 3).

6. Let first number = x Second number = (27– x )The product of two numbers is 182 x (27– x ) = 182 27x – x2 = 182 –x2 + 27x – 182 0 x2 – 27x + 182 0 x2 – 13x –14x + 182 0 x (x –13) –14 (x – 13) 0 (x – 13) (x - 14) 0 x – 13 0 or x – 14 = 0 x = 13 or x = 14When x = 13,second number = 27 – x

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= 27 – 13= 14

When x = 14,second number = 27 – x

= 27 –14= 13

The two numbers are 13 and 14

7. cos (40 - ) sin (50 + ) +2 2

2 2cos 40 cos 50sin 40 sin 50

= sin [90 - (40 - )] - sin (50 + ) +2 2

2 2sin (90 40) sin (90 50)

sin 40 sin 50

= sin [90 - 40 + ] - sin (50 + ) +2 2

2 2sin 50 sin 40sin 40 sin 50

= sin (50 + ) - sin (50 + ) + 1= 0 + 1= 1

8. Here A is the kite

AC is the length of string of the kite.

ACB : angle made by the stringwith the horizontal.m ACB = 60

AB is the height of the kite.In ABC, m B = 90, m C = 60.

sin C =ABAC

sin 60 =AB100

32

=AB100

AB =100 3

2

AB = 50 3 AB = 50 1.73 AB = 86.5mHeight of the kite is 86.5m

SECTION - BSolve the following sums : [3 Marks] 12

9. Let the length of rectangle be x cmand it’s breadth be y cm.Length is twice it’s breadth x = 2y x - 2y = 0 ..... ..... ..... (i)It’s perimeter is 120 cm.

2(length + breadth) = 120 2(x + y) = 120 x + y = 60 ..... ..... ..... (ii)Subtracting equation (ii) from equation (i), we get,

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A

C B60

100

m

?

x - 2y = 0x + y = 60(-) (-) (-)

-3y = -60

y =603

y = 20Substituting y = 20 in equation (ii), we get,

x + 20 = 60x = 40

Length of rectangle is 40 cmand it’s breadth is 20 cm.

Area of rectangle = length breadth= 40 20= 800 sq. cm

Area of rectangle is 800 sq. cm.

10. L.H.S. : 2sec2 - sec4 - 2cosec2 + cosec4= 2 sec2 - 2cosec2 - sec4 + cosec4= 2 (sec2 - cosec2 ) - (sec4 - cosec4 )= 2(sec2 - cosec2 ) - (sec2 - cosec2 ) (sec2 + cosec2 )= (sec2 - cosec2 ) (2 - sec2 - cosec2 )= (1 + tan2 - 1 - cot2 ) (2 - 1 - tan2 - 1 - cot2 )

(sec2 = 1 + tan2 & cosec2 = 1 + cot2 )= (tan2 - cot2 ) (-tan2 - cot2 )= - (tan2 - cot2 ) ( tan2 + cot2 )= - [(tan2 )2 - (cot2 )2] ( ( a - b) (a + b) = a2 + b2)= - (tan4 - cot4 )= cot4 - tan4= R.H.S.

L.H.S. = R.H.S.OR

10. L.H.S. =sin cos 1sin cos 1

Dividing numerator and denominator by cos

=

sin cos 1cos

sin cos 1cos

=tan 1 sectan 1 sec

=(tan sec ) 1(tan sec ) 1

= 2 2

tan sec 1

tan sec sec tan

=tan sec 1

sec tan sec tan sec tan

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=tan sec 1

sec tan 1 sec tan

=tan sec 1

sec tan tan sec 1

=1

sec tan= R.H.S.

11. Let A(x1, y1) = A(0, 0), B(x2, y2) = B(1, 2), C(x3, y3) = C(2, 5) and D(x4, y4) = D(1, 4)

ABC =12 |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

=12 |0(2 – 5) + 1(5 – 0 ) + 2 (0 – 2)|

=12 |0 + 5 – 4|

=12 |1|

=12 sq. units

ACD=12 |x1 (y3 – y4) + x3 (y4 – y1) + x4 (y1 – y3)|

=12 |0(5 – 4) + 2 (4 – 0) + 1 (0 – 5)|

=12 |0 + 8 – 5|

=32 sq. units

Now,Area of ABCD = ABC + ACD

=12 +

32

= 2Area of ABCD is 2 sq.units.

12. Here, AB is the height of the tree.

BC is the width of the river.A person moves from point C to point D.Hence CD = 20mm ACB = 60 and m ADB = 30.In ABC, m B = 90, m C = 60.

tan C =ABBC

3 =ABBC

AB = 3 BC .........(i)

In ABD, mB = 90, mD = 30.

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300 600

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D 20 m ?C B

A

D(1, 4) C(2, 5)

A(0, 0) B(1, 2)

tan D =ABBD

tan 30 =AB

BC CD ( D -C - B)

13

=AB

BC 20 ( CD = 20m)

13

=3 BC

BC 20 ( from (i))

BC + 20 = 3 3 BC BC + 20 = 3BC 3BC - BC = 20 2BC = 20

BC =202

BC = 10mNow, from (i)

AB = 3 BC= 1.73 10

AB = 17.3mThe height of the tree is 17.3m and width of the river is 10m.

SECTION - CSolve the following sums : [4 Marks] 12

13. Let the usual speed of cylist be x km/ hr

Then the time taken to cover 35 km distance at usual speed =35x

hours

If there is an increase of 2 km/ hr in the speed,The new increased speed = (x + 2)km/ hr

Then the time taken to cover 35km distance at the increased speed =35

x +2 hours.

The cylist reaches the destination 2 hours earlier.

35

x +2 =35x

– 2

Multiplying the equation by x(x + 2), 35x = 35(x + 2) – 2x (x + 2) 35x = 35x + 70 – 2x2 – 4x 35x– 35x – 70 + 2x2 + 4x = 0 2x2 + 4x – 70 = 0 2(x2 + 2x – 35) = 0 x2 + 2x – 35 = 0 x2 + 7x – 5x – 35 = 0 x(x+ 7) – 5(x+ 7) = 0 (x + 7) (x - 5) = 0 x+ 7 = 0 or x – 5 = 0 x = – 7 or x = 5

Speed cannot be negative. x – 7

x = 5 The speed of cyclist is 5 km / hr

OR13. In ABC, let mB = 90 .

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Let AB = x be the shortest sideHypotenuse is 2 less than 3 times its shortest side.

AC = 3x – 2Remaining side is BC.Remaining side is 2 more than twice the shortest side.

BC = 2x + 2Now AC2 = AB2 + BC2 (Pythagoras theorem)

(3x – 2)2 = (x)2 + (2x + 2)2 9x2 – 12x + 4 = x2 + 4x2 + 8x + 4 9x2 – 5x2 – 12x – 8x + 4 – 4 = 0 4x2 – 20x = 0 4x(x – 5) = 0 4x = 0 or x – 5 = 0 x = 0 or x = 5

Length of side cannot be zero. x 0 x = 5 AB = x = 5 units

BC = 2x + 2= 2 (5) + 2= 10 + 2= 12 units

Now Area of ABC =12

× base × altitude

=12

× BC × AB

=12

× 12 × 5

= 30 Thus, area of right angled triangle is 30 sq. units.

14.x5

-y3

=4

15Multiplying the equation by 15

3x - 5y = 4 ..... ..... ..... (i)

x2 -

y9

=7

18Multiplying the equation by 18, we get,

9x - 2y = 7 ..... ..... ..... (ii)

Multiplying the equation (i) by 3, we get, 9x - 15y = 12 ..... ..... ..... (iii)

9x - 2y = 7 9x - 15y = 12(-) (+) (-)

13y = -5

y =5

13

Substituting y =5

13

in equation (ii), we get,

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A

B C

33 - 2x

2x + 2

x

A

B C

33 - 2xx

9x - 25

13

= 7

9x +1013

= 7

9x = 7 -1013

9x =91 10

13

9x =8113

x =81

13 9

x =9

13

{(x, y)} =9 5,

13 13

is the solution of given pair of equations.

15. Here, distance between two rungs = 25cm. Here, distance between the first and the last rung is 2.5m = 250cm.

No. of rungs =25025 + 1

= 11 Assuming that there is a uniform increase in the length of staircase, we have

a = T1 = 40 and T11 = 60

d = m nT Tm n

= 11 1T T11 1

=60 40

10

d =2010

d = 2 The length of wood required can be obtained by finding S11

Sn =n2 (2a + (n - 1)d)

S11 =112 (2 40 + (11 - 1) 2)

=112 (80 + 20)

=112 100

= 11 50= 550

The length of wood required in the rungs is 550 cm = 5.5m

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SECTION - DSolve the following sums : [5 Marks] 10

16. Case 1 : Triangles are acute angled triangles.

A

P

B QC RL N

Given : Correspondence ABCPQR of ABC and PQR is a similarity.

To prove :ABCPQR =

2

2

ABPQ =

2

2

BCQR =

2

2

ACPR

Proof : Draw altitudes AL and PN .The correspondence ABC PQR is a similarity. B Q

andABPQ =

BCQR =

ACPR

gives2

2

ABPQ =

2

2

BCQR =

2

2

ACPR

In ABL and PQN,B QALB PNQ (right angles)

The correspondence ABL PQN is a similarity.

ABPQ =

ALPN

ALPN

=ABPQ =

BCQR

Now, area of a triangle =12 Base Altitude

ABCPQR =

1212

BC . ALQR . PN =

BCQR

ALPN

=BCQR

BCQR =

2

2

BCQR

ABCPQR =

2

2

ABPQ =

2

2

BCQR =

2

2

ACPR

OR16. Prove that : In a triangle, if the square of a side is equal to the sum of the square s

of other two sides, then the angle opposite to the first side is a right angle.Eg : If BC2 = AB2 + AC2 in ABC, then A (opposite to BC ) is a right angle.Data : BC2 = AB2 + AC2 in ABC.To prove : A is right angle.

Proof : Let OX be any ray.

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We can construct OY such that OY OX .

Let M OY such that OM = AC.

Let N OX such that ON = AB.

Draw MN .

MON is a right angled triangle, as OM ON .

MON is right angle. MN is the hypotenuse. According to pythagoras Theorem.

MN2 = OM2 + ON2 = AC2 + AB2 [OM = AC and ON = AB]But AB2 + AC2 = BC2

MN2 = BC2

MN = BC In ABC and ONM consider the correspondence ABC ONM, we have

AB ON

AC OM

BC MN The correspondence ABC ONM is a congruence.So, ABC ONM A Obut O in ONM is a right angle by construction.A is a right angle.

17. Data : (O, 3) is given. Point P is in the exterior of the circle suchthat OP = 7 cm.

To Construct : To draw tangents to (O, 3) through P.

O MP

Q

R

Steps of construction :i. (O, 3) is constructed and P is chosen such that OP = 7 cm.

ii. OP is drawn.

iii. Construct perpendicular bisector of OP and obtain the midpoint M.

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C

A B

M

O N X

Y

iv. Draw (M, OM) is constructed intersecting (O, 3) at Q and R.

v. DrawPQ and

PR .

Thus,PQ and

PR are the required tangents.

**** Best of Luck ****

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