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**SECTION - A**

**Questions 1 to 20 carry 1 mark each.****Q.1. The total number of terms in the expansion of (a ^{2} + 2ab + b^{2})^{101} is:**

(b) 203

(c) 204

(d) None of these

Ans.

We have, (a

We know that the total number of terms in the expansion of (x + a)

âˆ´ Total number of terms in the expansion of (a + b)

Total number of terms in the expansion of (a

(a) 0

(b) 1

(c) 2

(d) 3

OR

Which of the following expressions is a polynomial?

(a) 5x

(b) x + 2/x

(c) x

(d) None of these

Ans.

OR

(a)

(a) abscissa is 6 (b) ordinate is 6 (c) abscissa is - 6 (d) ordinate is-6

Ans.

(a) parallelogram (b) rhombus (c) rectangle (d) trapezium

Ans.

(a) ( 5 , 0 , 0 )

(b) (0,7,0)

(c) (0, 0, 10)

(d) None of these

Ans.

(a) 40Â° (b) 50Â° (c) 80Â° (d) 130Â°

Ans.

Given that AB = AC and âˆ B = 50Â°

In Î”ABC, AB = AC (given)

âˆ C = âˆ B (angles opposite to equal sides are equal)

âˆ C = 50

[âˆµ If a triangle and parallelogram are on the same base and between the same parallel, the area of the triangle is equal to half of the parallelograms.

(a) 360Â° (b) 450Â° (c) 540Â° (d) 180Â°

Ans.

= (5 - 2) x 180Â° = 540Â°

(a) 6

(b) 4

(c) 8

(d) 2

Ans.

If the die is thrown once more, then the probability that it shows 5 is:

(a) 9/50

(b) 3/40

(c) 4/25

(d) 7/25

OR

A coin is tossed 200 times. The head appears 79 times. The probability of a tail is

(a) 79/200

(b) 121/200

(c) 1

(d) 0

Ans.

P (getting 5) = 150/1000 = 3/20

(b)

Q.11. If (x + 1) is a factor of p(x) = x

Ans.

[âˆµ p(-l) = 0

â‡’ (-1)

â‡’ 1 - 2 + p = 0

â‡’ P = 1]

Ans.

**OR**

a/b

**Q.14. In the given figure, âˆ ABC = 69Â°, âˆ ACB = 31Â°, find âˆ BDC.Ans.** âˆ BAC = 180Â°-(69Â°+ 31Â°)

= 80Â°

âˆ BDC =âˆ BAC = 80Â°

(Angles in same segment are equal)

Ans.

OR

Evaluate (0.2)

Ans.

On squaring both sides, we get

Let a = 0.2, b = -0.3 and c = 0.1

Then, a + b + c = 0.2 - 0.3 + 0.1 = 0

We know that,

a

âˆ´ (0.2)

= -0.018

Ans.

We know that n(A âˆ© B) = n(A) - n(A - B) = 115 - 47 = 68

and n( A âˆª B) = n(A) + n(B) - n(A âˆ© B) = 115 + 326 - 68 = 373

OR

An equilateral triangle is an acute angled triangle

Ans.

OR

True.

In an equilateral triangle, all the angles are 60Â°.

OR

If the volume of a cube is 3âˆš3 a

Ans.

6 x (side)

(side)

side = 11

Hence, the length of the edge of cube is 11 cm.

Let x be edge of a cube.

â‡’

â‡’ x = âˆš3 a

Total surface area =

Ans.

[by Heronâ€™s formula]

**SECTION - B**

**Question numbers 21 to 26 carry 2 marks each.****Q.21. What is the value of ?****Ans. ****Q.22. Express in the form of P/q where p and q are integers and q â‰ 0.****OR****Write the following in decimal form and say what kind of decimal expansion each has?****(i) 49/100 (ii) 2/5****Ans. **Let,

x = 0.6666.......... ...(i)

Multiplying by 10 on both the sides, we get,

10x = 6.6666 ...(ii)

From (ii) - (i), we get

9x = 6.0

Or, **OR**

(i) 49/100= 0.49, so it has terminating decimal expansion.

(ii) so it has no terminating decimal expansion.**Q.23. Write Euclidâ€™s fifth postulate. Does Euclidâ€™s fifth postulate imply the existence of parallel lines? Explain.Ans.** If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

Yes.

Now, according to Euclidâ€™s fifth postulate, when line â€˜nâ€™ falls on line 'l' and â€˜m â€™ such that âˆ 1 + âˆ 2 < 180

We find that the lines which are not according to Euclidâ€™s fifth postulate i.e., âˆ 1 + âˆ 2 < 180

Ans.

Here,

and

âˆ´

Ans.

âˆ´ Total number of elementary events =

There are 4 cards of king, 12 cards of hearts other than king and 12 cards of red colour other than heart and kings. Out of 28 cards, one card can be drawn in

âˆ´ Favourable number of elementary events =

Required probability =

OR

Find the length of the longest rod that can be placed in a room 12 cm long, 9 m broad and 8 m high.

Ans.

But surface area = 4Ï€r

Volume = 539/3 or 179.67 cm

**SECTION - C**

**Question 27 to 34 carry 3 marks each.****Q.27. Calculate mean and mean deviation about median for the following data:**

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |

Frequency | 6 | 7 | 15 | 16 | 4 | 2 |

**OrA committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man? (b) one man? (c) two men?**

Here, N= 50.

âˆ´ N/2 = 50/2 = 25, so median class is 20 - 30.

From the table: l = 20, h= 10, f= 15, cf= 13

Using the formula:

Thus, mean deviation about median

= M.D. (Median)

Calculation of Mean:

Let the assumed mean a = 35

âˆ´ Mean is given by

The total number of persons = 2 men + 2 women = 4

Out of these four persons, two persons can be selected in

(a) No man in the committee of two persons

= There will be two women in the commits

Out of 2 women, two can be selected in

âˆ´

(b) One man in the committee of two persons means that there is one man and one woman in the committee.

One man out of 2 men can be selected in ^{2}C_{1} ways and 1 woman out of 2 women can be selected in ^{2}C_{1} ways. Together they can be selected in ^{2}C_{1} x ^{2}C_{1} ways.

âˆ´

(c) Two men can be selected out of 2 men in^{ 2}C_{2} way.

âˆ´ **Q.28. Factorize 2x ^{2} + 3âˆš5 x + 5.ORFactorize x^{3} - 2x^{2} - x + 2Ans.**

x

= x

= (x -2 )(x

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

OR

Two chords PQ and QR of a circle are equal. Prove that the centre of the circle lies on the angle bisector of âˆ PQR.

Ans.

To Prove: OO' is perpendicular bisector of the chord PQ.

Const. : Join OP, OQ, O'P and O'Q

Proof: In Î”OPO' and Î”OQO'

OP = OQ [radii of same circle]

O'P = O'Q [radii of same circle]

OO' = OO' [common]

â‡’ Î”OPO' = Î”OQO' [by SSS congruence axiom]

â‡’ âˆ POM = âˆ QOM [c.p.c.t,]

Now, in Î”POM and Î”QOM

OP = OQ [radii of same circle]

âˆ POM = âˆ QOM [proved above]

OM = OM [common]

â‡’ Î”POM â‰… Î”QOM [by SAS congruence axiom]

PM = QM and

âˆ PMO = âˆ QMO [c.p.c.t.]

Also, âˆ PMO + âˆ QMO'= 180Â° [a linear pair]

â‡’ âˆ PMO = âˆ QMO

= 90Â°

Hence, OO' is the perpendicular bisector of the chord PQ.

Given: Two equal chords PQ and QR of a circle C(O, r).

To Prove: Centre O of the circle lies on the bisector of âˆ PQR.

Const. : Join PR, draw bisector QX of âˆ PQR and let it intersects PR in M.

Proof: In Î”PQM and Î”RQM

PQ = RQ [given]

âˆ PQM = âˆ RQM [by construction]

QM = QM [common]

â‡’ Î”PQM = Î”RQM [by SAS congruence axiom]

â‡’ PM = RM

and âˆ PMQ = âˆ RMQ [c.p.c.t.]

But âˆ PMQ + âˆ RMQ = 180

â‡’ âˆ PMQ = âˆ RMQ = 90Â°

â‡’ QM is the perpendicular bisector of chord PR.

â‡’ QM passes through the centre O.

[âˆµ perpendicular bisector of a chord always passes through the centre]

Hence, the centre of the circle lies on the angle bisector of âˆ PQR.

Ans.

âˆ´ âˆ 4 = âˆ 1 ...(i) [alternate interior angles]

and âˆ 5 = âˆ 2 ...(ii) [corresponding angles]

On adding Eqs. (i) and (ii), we get

[adding âˆ 3 both sides]

âˆ´

Hence proved.

OR

Find the square root of - 7 - 24i.

Ans.

Now,

...(1)

Squaring both sides, we have

Comparing real and imaginary parts on both sides, we have

x

and 2xy = - 24 â‡’ 4x

Now, (x

Taking square roots of both sides, we have

x

Adding and subtracting (2) and (4), we get

Since xy < 0, therefore x and y are of opposite sign.

âˆ´ x = 3, y = -4 or x = -3, y = 4

Putting the values of x and y in (1), we get

Ans.

Let PQRS be a rectangle

In Î”SPQ and Î”RQP,

SP = RQ. (Opp. Sides of rectangle are equal)

âˆ SPQ = âˆ RQP = 90Â°

(Each angle of rectangle is right angle)

PQ = QP (common)

âˆ´

Ans.

Length of rectangle = 6 units

Breadth of rectangle = 4 units

Area of rectangle = 6 x 4 = 24 sq. units

Now, point of intersection of diagonals is P(4, 1).

OR

The slant height and base diameter of a conical tomb are 25 m and 14 m, respectively. Find the cost of white-washing its curved surface area at the rate of â‚¹210 per 100 m

Ans.

Area of each triangular piece

[by Heronâ€™s formula]

Now, there are 4 triangular pieces of one colour and 4 triangular pieces of other colour.

Hence, total area of cloth of each colour.

Given, diameter of conical tomb = 14 m

and slant height of conical tomb (l) = 25 m

**SECTION - D**

**Questions 35 to 40 carry 4 marks each****Q.35. If prove that: ****ORProve that: cos ^{2}A + cos^{2} **

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’ ...(i)

and ...(ii)

Now,

[using (i) and (ii)]

= R.H.S.

We know that

Ans.

= 1

Ans.

Frequency density or adjusted frequency for a class =

Here, the minimum class size = 10 - 0 = 10

Let us represent weekly pocket money along x-axis and corresponding adjusted frequencies along y-axis on a suitable scale, the required histogram is as given below:

Ans.

In horizontal axis, take 1 block is equal to 10 units and in vertical axis, take 1 block is equal to 5 units.

(i) Draw the base line segment BC = 4.6 cm.

(ii) At the point B, make âˆ XBC = 45Â°.

(iii) Now, cut a line segment BD equal to AB + AC= 8.2 cm from the ray BX.

(iv) Join DC.

(v) Draw perpendicular bisector MN of CD which meets BX at A.

(vi) On joining AC, we net the required Î”ARC.

Here, 31 is the (m + 2)th term. i.e..

31 = 1 + [(m+2) - 1]d

â‡’ 31 = 1 + (m + 1)d

â‡’

â‡’

â‡’

â‡’

â‡’ [âˆµ a = 1]

â‡’ 14 = m

Hence, the value of m is 14.

Since, p, q, r are in G.P., therefore

q

The roots of the given equation

px

are given by

[using (1)]

â‡’ ...(2)

From (2), it is clear that both the roots of equation px

It is given that the equations px

âˆ´

â‡’ dq

â‡’ dpr - lepq + fp

â‡’ p[dr - 2eq +fp]= 0

â‡’ dr - 2eq +fp = 0 [âˆµ p â‰ 0]

â‡’ [Dividing both sides by pr]

â‡’ [using (1)]

â‡’

â‡’

â‡’

17,2, 7,27, 5,14,18,10,24,25,48,10,8, 7,10,28,25. Find median and mode of the series.

OR

Find the mean salary of 60 workers of a factory from the following table.

Salary (Rs.) | No. of Workers |

3000 | 16 |

4000 | 12 |

5000 | 10 |

6000 | 8 |

7000 | 6 |

8000 | 4 |

9000 | 3 |

10000 | 1 |

TOTAL | 60 |

**Ans.** Arrange the data in ascending order.

2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 48

N = 16, which in even

median = 12

Mode = The most frequently occurring observation

= 10 i.e., 3 times occurring 10

âˆ´ Mode = 10**OR**

= 5083.33

âˆ´ Mean salary of 60 workers = â‚¹ 5083.33

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