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**SECTION - A**

**Questions 1 to 20 carry 1 mark each.**

**Q.1. The value of when **

(b) 31.62

(c) 0.0316

(d) 316.2

(a) 0

(b) 1

(c) 2

(d) âˆš2

Ans:

(a) lies on Y-axis

(b) lies on X-axis

(c) in first quadrart

(d) in second quadrant

Ans:

(5,0) is a point on lies 2x + ky - 10k

âˆ´ 10 = 10k â‡’ k = 1

(a) 60Â°

(b) 80Â°

(c) 90Â°

(d) 100Â°

Ans:

[âˆµ Let âˆ p = 4x and âˆ q - 5x

Now, âˆ p + âˆ q = 180Â° [linear, pair axiom]

â‡’ 4x + 5x = 180

â‡’ 9x = 180

x = 20

âˆ´ âˆ r = âˆ p = 4x = 4 x 20

(a) 6 and 3

(b) 3 and 6

(c) 3 and 9

(d) 6 and 9

Ans:

Suppose A and B has m and n elements respectively.

âˆ´ Number of subsets of set A = 2

and number of subsets of set B = 2

According to given condition, we have

2

â‡’ 2

â‡’ 2

â‡’ 2

âˆ´ n= 3 and m - n= 3

â‡’ n = 3 and m = 3 + 3 = 6

(a) X-axis

(b) Y-axis

(c) Both of them

(d) None of these

Ans:

(a) 0.5

(b) 0.6

(c) 0.7

(d) 0.8

Ans:

Explanation: Total no. of people = 642

No. of people with high school certificate = 514

P (person with high school certificate) =

(a) halved

(b) doubled

(c) same

(d) four times

Ans:

[âˆ´ Let r and h be radius and height of the cylinder, then C.S.A. = 2Ï€rh Now, radius is doubled and height is halved

âˆ´ New radius = 2r and new height =h/2

New C.S.A. - 2Ï€ x 2r x h/2 = 2Ï€rh]

Ans:

**Questions 11 to 15 carry one mark each.**

**Q.11. Degree of the zero polynomial is .........****Ans:** The degree of zero polynomial is not defined.**Q.12. ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm ^{2}, then ar (ABC) = 24 cm^{2} ** False.

Ans:

Now, ar (ABCD) = ar (AXCD) + ar (XBC) ...(1)

Diagonal AC of a parallelogram divides it into two triangles of equal area.

ar (ABCD) = 2ar (ABC) ...(2)

Again, X is the mid-point of AB, So

...(3)

[Median divides the triangle in two triangles of equal area]

From Eq (1)

[Using (1), (2) and (3)]

Ans:

Required line is y - 3 = tan 60Â°(x - 2)

â‡’

Ans:

**Questions 16 to 20 carry one mark each**

**Q.16. Find the value of x from the adjacent figure.Ans:** 2x + 3x + 5x = 180Â°

10x = 180Â°

x = 18Â°

Ans.

Now, each side of the triangle is doubled.

âˆ´ New dimensions of the triangle are 2a, 2b, 2c

âˆ´ Î”

Hence, the percentage increase in area of a triangle if its every side is doubled, is 300%

Ans:

If the line (1) passes through the mid-point of the line joining the points (0,-4) and (8,0), i.e., through the

Hence, the slope of a line passes through the origin is -1/2.

Show that the lines m and n are parallel.

Ans:

60Â° + âˆ 2 = 180Â°

â‡’ âˆ 2 = 120Â°

We have, âˆ 6 = 120Â° â‡’ âˆ 2 = âˆ 6 = 120Â°

which are corresponding angles.

Hence, m || n

OR

Evaluate : 249

Ans:

OR

249

= (249 +248) (249 - 248)

= 497 x 1

= 497

**SECTION - B**

**Questions 21 to 26 carry two marks each**

**Q.21. Simplify ****Or****Simplify ****Ans:**

Or

= 3-8 x 6 + 15 x 2 + 15

= 3-48+30+15 = 48-48 =0**Q.22. A and B are mutually exclusive events of an experiment. If P(A) = 0.4 and P(B) = 0.5, find Ans: **We have

[ âˆ´ A and B are mutually exclusive]

(i) First, draw a line segment SC = 6 cm.

(ii) Taking S and C as centre and radius 6 cm, draw two arcs.

(iii) The arc of step (ii) cut any point A join the lines. AB and AC.

Thus, Î”ABC is an equilateral triangle.

Ans:

âˆ´ Hypotenuse (AC) is the longest side of Î”ABC.

âˆ´ âˆ B = 90Â° and AB = BC

or (Î” ABC) = 200 cm

BC

By using Pythagoras theorem

AC

=BC

AC=20 x 1.414= 28.280 cm

AC = 28.28 cm

Ans.

Ans:

Let the point P divide the line joining Q and R in the ratio k:1.

The coordinates of the point P are

But the coordinates of P are given to be (a, b, 3).

From last equation of (1), we have

Putting value of k in (1), we have

â‡’ a = 4 and b = 2

Or

Ans:

PA = PB = PC.

Since, PA = PB

â‡’ PA

â‡’ 4y + 4z = 7

â‡’ ...(1)

Now,

PB - PC

PB

â‡’ ...(2)

Subtracting (2) from (1), we get

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

Hence, the required point in yz-plane is

**SECTION - C**

**Questions 27 to 34 carry 3 marks each**

**Q.27. The sides of a quadrilateral ABCD are AB = 13 cm , BC = 16 cm, CD = 20 cm and DA = 5 cm. If BD = 12 cm, find the area of the quadrilateral using Heronâ€™s formula.Ans:** Here, given quadrilateral ABCD is divided into two triangles, Î”ABD and Î”BCD.

For Î”ABD,

Let a = 13 cm, b = 12 cm and c = 5 cm

For Î”BCD,

Let a = 16 cm, b = 20 cm, c = 12 cm

âˆ´

= 8 x 4 x 3 = 96 cm

Now, ar(quad. ABCD) = ar(Î”ABD) + ar(Î”BCD)

= 30 +96 =126 cm

(x - h)

Since the circle passes through (2, -2) and (3,4), we have

(2 - h)

and (3 - h)

From (3) and (4), we have

(2 - h)

â‡’ (2 - h)

â‡’ (2 - h + 3 - h)(2 - h - 3 + h) ={4 - k + (- 2 - k)}{4 - k - (- 2 - k)}

â‡’ (5 - 2h)(- 1) = (2 - 2k)(6)

â‡’ -5 + 2h = 12 - 12k

â‡’ 2h + 12k =17 ...(4)

Also, since the centre (h, k) lies on the line 2x + 2y = 7, we have

2h + 2k = 7 ...(5)

Subtracting (5) from (4), we get

10k = 10

â‡’ k=1 ...(6)

Substituting k = 1 in (5), we get

2h +2 = 7

â‡’ 2h = 5

â‡’ h=5/2 ...(7)

â‡’

â‡’

â‡’ ...(8)

From (1), (6), (7) and (8), we get the required equation of the circle as

Or**Find the equation of the circle touching positive y-axis at a distance of 4 units from the origin and cutting intercepts of 6 units on positive x-axis.****Ans:** As shown in the figure, let AB - 6 and OC - 4 as given in the question.

Let equation of the circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 ...(1)

At A or B,y = 0. Then

x^{2} + 2gx + c = 0

Let coordinates of A be (x_{1},0) and B be (x_{2},0).

Then x_{1} + x_{2} = - 2g and x_{1}.x_{2} = c

Also AB - x_{2} - x_{1} = 6

Now,

or 4g^{2} = 36 + 4c

or g^{2} = 9 + c ...(2)

Further, at C we have x = 0

Then from (1), y^{2} + 2fy + c = 0 ...(3)

Since equation (3) has real equal roots, we have

(2f)^{2}-4(1)c = 0 [D = B^{2} - 4AC = 0]

or 4f^{2} - 4c = 0

or f^{2} = c ...(4)

At C,y = 4. From (3), we get

4^{2} + 2f(4) + c = 0

or 16 + 8f+ c = 0

or c = - 16 - 8f ...(5)

From (4) and (5), we have

f^{2} = - 16 - 8f

or

f^{2} + 8f + 16 = 0

or (f + 4)2 = 0 â‡’ f = - 4 â‡’ c = 16 [From (4)]

Now from (2) g^{2} = 9 + 16 = 25 â‡’ g = Â± 5

Putting values of g = Â± 5, f = - 4 and c = 16 in (1), we get equations of circles as

x^{2} +y^{2} + 10x - 8y + 16 = 0**Q.29. Find three irrational numbers between 5/7 and 9/11.OrAns:** By Song divsion, we have

âˆ´

Also,

âˆ´

So, we can consider three irrational numbers between which are non-terminating non-repeating decimals as follows

0.72072007200072...

0.73073007300073...

0.74074007400074...

or

Ans.

To Trove : AX || CY

Construction: Join XY

Proof: ABCD is a parallelogram

(i) Write the points whose abscissa is 0.

Ans:

(ii) G(5, 0), l(-2, o), O(0, 0)

(iii) D(-5, 1) and H(-5, -3).

= 3 cosec

Ans:

In Î”ABC, AB is the' shortest side.

[âˆµ angle opposite to the shortest sice is smaller]

In Î”ACD, CD is the longest side.

[âˆ´ angle opposite to the longest side is longer]

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

Now, draw the diagonal BD in Î”ABD, AB is the shortest side.

...(iii)

[âˆµ angle opposite to the shortest side is smaller]

In A BCD, CD is the longest side.

[angle opposite to the longest side is longer]

On adding Eqs. (iii) and (iv), we get

From the above data, what is the probability of getting a sum :

(i) More than 10.

(ii) Betweeen 8 and 12.

OR

Fifty seeds were selected at random from each 5 bags seeds and were kept under standardized conditions favourable to germination. After few days, the number of seeds which had germinated in each collection were counted and recorded as follows

What is the probability of germination of

(i) More than 40 seeds in a bag

(ii) 49 seeds in a bag

(iii) More than 35 seeds in a bag

(ii) Favourable events n(E) = 8 < n(E)<12

n(E) = 53 + 46 + 28 = 127

Probability of getting a sum between 8 and 12

Or

Total number of bags = 5

(i) P(more than 40 seeds in a bag) = 2/5 = 0.4

**SECTION - D**

**Questions 35 to 40 carry 4 marks each**

**Q.35. Simplify: OrAns:** Here

and

Or

Ans.

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

or**Prove that : cos ^{2} x + cos^{2}** We have

Ans:

Or

A kite in the shape of a square with each diagonal 36 cm and having a tail in the shape of an isosceles triangle of base 10 cm and equal side 6 cm, is made of three different shades as shown in the adjoining figure. How much paper of each shade has been used in it? [Given,

Ans.

Join the diagonal BD.

In ABCD, we have

(BD)

= (12)

â‡’ BD = 13m [taking positive square root]

âˆ´ Î”BCD is a right angled triangle.

Now, area of quadrilateral ABCD

= Area of Î”BCD + Area of Î”ABD

= 30+35.4=65.4 m

Hence, the area of park is 655 m

Or

One shaded portion is Î”ABD.

âˆµ Kite is the shape of square, so diagonals are equal and bisect each other

Second shaded portion is Î”BCD.

and area of Î”BCD = area of is Î”ABD = 324 cm

Third shaded portion is Î”CEF in which

CE = CF = 6 cm and EF = 10 cm.

Here, a = 6 cm, b = 6 cm and c = 10 cm

âˆ´

= 5x3.31 = 16.55 cm

Ans:

and x - y - 2 = 0 Table for line x + y - 6 = 0 or y = 6 - x is

Table for line x-y-2 = 0 ox y = x-2 is

Ans:

âˆµ 2Ï€r =17.6

Now, areas of two hemispherical domes = 2Ï€r

Cost of painting at the rate of Rs 10 per cm

= Rs 10 x 837.76 = Rs 8377.60

Out of these

I is repeated four times

S is repeated four times

P is repeated two times

and rest are different.

Thus, total number of distinct permutations

If four Iâ€™s always come together, then we have to arrange 7 (11 - 4) other letters + 1 group of four Iâ€™s (IIII) = 8 letters, that can be done in

Hence, the total number of distinct permutations of the letters in MISSISSIPPI such that four Iâ€™s do not come together

= 33810

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