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**SECTION - A**

**Questions 1 to 20 carry 1 mark each.****Q.1. If x = 2 + âˆš3, then value of is:****(a) 2 (b) 4 (c) âˆš6 (d) âˆš2****Ans.** (c) âˆš6

[âˆ´ x = 2 + âˆš3

â‡’

âˆ´

= 4 + 2 = 6

â‡’ **Q.2. If n(A) = 17, n(B) = 8 for two disjoint sets A and B, then n(A âˆª B) =(a) 9(b) 23(c) 25(d) None of theseAns.** Choice (c) is correct.

As A and B are disjoint sets.

âˆ´ n(A

(a) ordinate

(b) abscissa

(c) quadrant

(d) origin

Ans.

(a) (2, 0)

(b) (0, 3)

(c) (3, 0)

(d) (0, 2)

Ans.

2x + 3y = 6

2(0) + 3y = 6

0 + 3y = 6

y = 2

Hence, at the point (0, 2) the given linear equation cuts the y-axis.

(a) 60Â°

(b) 80Â°

(c) 120Â°

(d) 160Â°

âˆ´

â‡’ x + 20Â° + 2x - 20Â° + 60Â° = 180Â°

â‡’ 3x + 60Â° = 180Â°

â‡’ 3x = 120Â°

â‡’

Now, **âˆ **COD = 2x - 20Â° = 2 x 40Â° - 20Â° = 60Â°]**Q.6. The coefficient of x ^{50} in (1 + x)^{100} is** Choice (b) is correct.

(a) ^{100}C_{50} x^{50}

(b) ^{100}C_{50}

(c) x^{50}

(d) ^{n}C_{50}

Ans.

We know that (r + 1) term in the binomial expansion of (1 + x)

T

âˆ´ In the binomial expansion of (1 + x)

For coefficient of x

T

âˆ´ The coefficient of x

Ans.

âˆ´ k = o

(a) âˆš32 cm

(b) âˆš16 cm

(c) âˆš48 cm

(d) âˆš21 cm

Ans.

Area of isosceles Î” = 1/2 x Base x height

B = âˆš16

= 4 cm

By using Pythagoras theorem

(B)

(H)

H = âˆš32 cm

(a) 288 cm

Ans.

[âˆµ Dimensions of the cuboid formed by joining the cubes of side 4 cm are length (l) = 12 cm, breadth (b) = 4 cm and height (h) = 4 cm

âˆµ Surface area of the cuboid = 2 (lb + bh + hl)

= 2(12 x 4 + 4 x 4 + 4 x 12)

= 2(48 + 16 + 48) = 2 x 112 = 224 cm

(a) 1/52

(b) 13/52

(c) 4/52

(d) 1/6

Ans.

One card can be drawn in

â‡’ E

âˆ´

Q.11. A polynomial of degree one is called..........

Ans.

OR

Factorise 8x

Ans.

8x

= 8(x - 2)(x

Ans.

Ans.

Coefficient of variation =

OR

A frequency distribution in which each upper limit as well as lower limit are included, is called......

Ans.

OR

inclusive form.

Q.16. The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r.

Ans.

Volume of cone

= volume of hemisphere

Ans.

= y

= (y - 4)

Ans.

Ans.

âˆ´

Ans.

**SECTION - B**

**Question numbers 21 to 26 carry 2 marks each.****Q.21. Simplify: ****Ans. ****Q.22. In the expansion of which term is independent of x?****OR****If the different permutation of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?****Ans. **In the expansion of (x + a)^{n}, (r + l)th term is given by T_{r + 1} = ^{n}C_{r} x^{n - r} a^{r}

âˆ´ In the expansion of term is given by

â‡’

Suppose T_{r + 1} is independent of x.

âˆ´ 20 - 5r = 0 â‡’ 5r = 20 â‡’ r = 4

âˆ´ (r + 1)th = (4 + 1)th i.e., 5th term is the required term independent of x.**OR**

Total number of letters in the word EXAMINATION are 11.

Starting with A we have to find the number of arrangements of the remaining 10 letters, E, X, M, I, N, A, T, I, O, N, out of which there are 2Iâ€™s, 2Nâ€™s and the rest are all different.

The number of such arrangements

Clearly, we will have to start the next word with E (as in the dictionary)

So, the required numbers of words in the list before the first word starting with E are 907200.**Q.23. ABCD is a parallelogram and X is the mid-point of AB. If ar (Î”ADC) = 24 cm ^{2}, then find ar (AXCD).Ans. **Given, ABCD is a parallelogram and X is the mid-point of AS. Draw the diagonal AC.

We know that diagonal of a parallelogram divides it into two triangles of equal areas.

So. ar (parallelogram AECD) = 2 x ar (Î”ADC)

= 2 x 24 = 48 cm

Since, CX is the median of Î”ABC.

We know that a median of a triangle divides it into two triangles of equal areas.

Now, ar(AXCD) = ar(parallelogram ABCD) - ar (Î”BCX)

= 48 - 12 = 36 cm

Ans.

or 4x = 90Â° + 10Â° = 100Â°

or

or, x = 25Â°

Angles are 60Â° and 30Â°.

OR

Ten observations 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 are written in an ascending order. The median of the data is 24. Find the value of x.

Ans.

Sum of last 8 observations = 8 x 20 = 160

Sum of 15 observations = 15 x 19 = 285

Thus, 8th observation ~ 144 + 160 = 285 = 304 - 285 = 19

Hence, 8th observation is 19.

OR

Given data is in ascending order and n = 10 (even number)

âˆ´ Median = Average of

= Average of 5

Also, median = 24

âˆ´

â‡’ 3x - 12 = 48

â‡’ 3x = 60

â‡’ x = 20

Ans.

...(i)

If the line (1) cuts the x-axis, then by putting y = 0 in (1), we have x = h

âˆ´ The coordinates of the point A are (h, 0).

Similarly, the line (1) cuts the y-axis, then by putting x = 0 in (1), we have y = k

âˆ´ The coordinates of the point B are (0, k).

If P(a, b) is the mid-point of AB, then

â‡’ h = 2a and k = 2b

Substituting the values of h and k in (1), we get the required equation of the line AB is

â‡’ **OR**

Let the given points A(4, 7, 8), B(2, 3, 4), C (- 1 , - 2 , 1) and D{ 1, 2, 5) be vertices of a quadrilateral.

The coordinates of the mid-point of diagonal AC are

The coordinates of the mid-point of diagonal BD are

Since, the mid-points of the diagonals AC and BD are same, so the diagonals bisect each other. We know that if the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Hence, ABCD is a parallelogram.

**SECTION - C**

**Question 27 to 34 carry 3 marks each.****Q.27. Observe the graph of the equation f(x) in two variables. Based on graph, answer the following questions:****(i) Name the type of graph.(ii) Write the coordinates of A, B and C.(iii) Name the triangle formed by the line and coordinate axes. Also, name of the type of triangle.Ans.** (i) It is a linear graph i.e., it is a line.

(ii) Coordinates of A, B and C are A(4, 0), B (0, 3) and C

(iii) Î”AOB, since âˆ AOB = 90Â°.

âˆ´ It is a right-angled triangle.

Ans.

In this case the boy may borrow Chemistry Part II. So, he has to select now two books out of the remaining 7 books of his interest This can be done in

In this case the boy does not want to borrow Chemistry Part II. So, he has to select three books from the remaining 6 books. This can be done in

Hence the required number of ways

Ans.

BD and CE are two medians.

To prove BD = CE

...(i)

[âˆµ angles opposite to equal sides are equal]

...(ii) [dividing by 2 both sides]

Thus, in Î”DCB and Î”EBC, we get

DC = EB [from Eq. (ii)]

CB = BC [common side]

and âˆ DCB = âˆ EBC [from Eq. (i)]

OR

Sides of a triangle ABC are in the ratio 12 :17:25 and its perimeter is 540 cm. Find its area.

Ans.

c = 42 - 28 = 14 m

Given: sides of triangle are in the ratio 12:17:25

âˆ´ Sides are 12x, 17x and 25x respectively.

Perimeter = 540 cm

12x + 17x + 25x =540

54x = 540

x = 10

Now

s = 27x = 27 x 10 = 270

OR

In the given fig., lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE.

Ans.

â‡’

â‡’

â‡’

Now, in Î”BDE, we have

â‡’

â‡’

â‡’

â‡’ x

â‡’ x = 30

âˆ AOC and âˆ BOD are vertically opposite angles.

âˆ´ âˆ AOC = âˆ BOD

â‡’ [given]

Now, [given]

â‡’ [using (i)]

â‡’

â‡’

Also, [linear pair axiom]

â‡’

â‡’

â‡’

Now, âˆ COE + reflex âˆ COE = 360Â° [angles at a point]

â‡’ 110

â‡’ reflex âˆ COE = 360Â° - 110Â°

â‡’ reflex âˆ COE = 250Â°

Ans.

For a, b âˆˆ A, a - 2b = 0 â‡’ a/2

Ans.

...(i)

Now, correct value of âˆ‘x

â‡’ Correct value of âˆ‘x

= 1990/50 = 39.8

Number of girls | 2 | 1 | 0 |

Number of families | 475 | 814 | 211 |

**Find the probability that a family chosen at random, having(i) 2 girls (ii) 1 girl (iii) no girlAns.**

**SECTION - D**

**Questions 35 to 40 carry 4 marks each****Q.35. If find the value of 3ya ^{2} - 4xa + 3y.**

Squaring both sides of eqn. (i), we have

Ans.

âˆ´ Function is defined for all real values of x as 1 + x

Hence, domain of the function is R.

âˆ´

â‡’

âˆ´ Range of f(x) = [0,1)

OR

On a busy road, following data was observed about cars passing through it and number of occupants

Find the chance that it has

(i) exactly 5 occupants (ii) more than 2 occupants (iii) less than 5 occupants (iv) greater than 5 occupants

Ans.

Number of times 3 tails appeared = 34

Number of times 2 tails appeared = 55

Number of times 1 tail appeared = 72

Number of times 0 tail appeared =19

(i) P (getting 3 tails)

(ii) P (getting 2 tails)

Total number of cars = 100

âˆ´ The total numbers of trials = 100

(i) The numbers of trial, exactly 5 occupants = 5

âˆ´ P(exactly 5 occupants) = 5/100 = 1/20

(ii) The numbers of trial, more than 2 occupants

= 23+17+5 = 45

âˆ´ P(more than 2 occupants) = 45/100 = 9/20

(iii) The numbers of trial, less than 5 occupants

= 29+26+23+17 = 95

âˆ´ P(less than 5 occupants) = 95/100 = 19/20

(iv) The number of trials, greater than 5 occupants = 0

âˆ´ P (greater than 5 occupants) = 0/100 = 0

(a) (102)

OR

Find the value of 'a' if remainder is same when polynomial p(x) = x

Ans.(a)

[âˆµ (a + b)

(100 + 2)

102

= 10,00,000 + 8 + 61200

102

= (100)

[a

104 x 96 = 10000 - 16 = 9984

OR

p(x) = x

p(x) leave the same remainder when divided by (x + 2) and (x + 1).

OR

Diagonals PR and QS of quadrilateral PQRS intersect at T, such that PT = TR. If PS = QR, show that: ar(Î”PTS) = ar(Î”RTQ).

Ans.

i.e., ar(Î”ABD) = ar(Î”CDB)

and ar(Î”ABC) = ar(Î”ACD)

To Prove: ABCD is a parallelogram

Proof: Diagonal AC divides the quad. ABCD into two triangles of equal area

âˆ´ ar(Î”ABC) = ar(Î”ACD) ...(i)

Also, ar(Î”ABC) + ar(Î”ACD) = ar(quad. ABCD)

â‡’ ar(Î”ABC) + ar(Î”ABC) = ar(quad. ABCD) [using (i)]

â‡’ ...(ii)

Again, diagonal BD of the quad. ABCD divides it into two triangles of equal area

âˆ´ ar(Î”ABD) = ar(Î”BCD) (iii)

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

â‡’ ar(Î”ABD) + ar(Î”ABD) = ar(quad. ABCD) [using (iii)]

â‡’ ...(iv)

From (ii) and (iv), we obtain

ar(Î”ABC) = ar(Î”ABD)

Given : In quad. PQRS, diagonals PR and QS intersect at T.

To prove: ar(Î”PTS) = ar(Î”RTQ)

Construction: From P and R, draw perpendiculars PE âŠ¥ SQ and RF âŠ¥ SQ.

Proof: In Î”PET and Î”RTF,

TP = TR [given]

âˆ PTE = âˆ RTF [vert. opp. angles]

âˆ PET = âˆ RFT = 90

Î”PET â‰… Î”RFT [AAS congruency]

â‡’ ar(Î”PET) = ar(Î”RFT)...(i) [congruent figures have equal area]

and PE = RF [c.p.c.t.]

Now, in Î”PSE and Î”RQF,

âˆ PES = âˆ RFQ [right angle]

PE = RF [proved above]

PS = QR [given]

So, Î”PSE â‰Œ Î”RQF [RHS congruency]

â‡’ ar(Î”APSE) = ar(Î”RQF) ...(ii)

Adding (i) and (ii), we have

ar(Î”PET) + ar(Î”PSE) = ar(Î”RFT) + ar(Î”RQF)

â‡’ ar(Î”PTS) = ar(Î”RTQ)

Ans.

(i) As the number of terms in the expansion of

Since 13 is an odd number, then there is only one middle term viz., term, i.e., T7 term.

Now, [Putting r= 6 in (1)]

(ii) Let T

Then,

The term will be independent of x, if the power of x is zero, i.e., 24 - 3r = 0 â‡’ r = 8

Substituting r = 8 in (1), we get

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