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**SECTION - A**

**Question numbers 1 to 20 carry 1 mark each.****Q.1. If A and B are two sets, then A ∩ (A ****∩**** B) equals****(a) A ****(b) B ****(c) A ∩ B ****(d) φ****Ans. **Choice (c) is correct.

As A ∩ B ⊆ A, so A ∩ (A ∩ B) = A ∩ B**Q.2. Degree of the polynomial 4x ^{4} +0x^{3} + 0x^{5} + 5x + 7 is** (a)

(a) 4

(b) 5

(c) 3

(d) 7

OR

One of the factors of the expression

[x^{2} - y^{2} - z^{2} + 2yz + x + y - z] is:

(a) x - y + z + 1

(b) -x + y + z

(c) x + y - z + 1

(d) x - y - z + 1

Ans.

OR

(a)

∵ Given expression

= x

= x

= x

= [(x + y - z)(x - y + z)]+(x + y - z)

= [(x + y - z) (x - y + z) + 1]

= (x + y - z) (x - y + z + 1)

One factor of the expression is

(x - y + 2 + 1).

(b) I and II quadrant

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm

Ans.

AC = AC [common]

∠BAC = ∠DAC [AC is bisector of ∠BAD]

∠CBA = ∠CDA = 90°

∴ ΔAEC ≌ ΔADC [by AAS]

BC = DC [by CPCT]

⇒ (BC)

⇒ BC = 4 cm

∴ DC = BC = 4 cm

(a) 151

(b) 152

(c) 153

(d) None of these

Ans.

We have, (a

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

∴ Total number of terms in the expansion of (a - b)

Total number of terms in the expansion of (a — 3a

(a) 40°

(b) 45°

(c) 50°

(d) 60°

Ans.

ABCD is a rhombus such that ∠ACB = 40°.

We know that diagonals of a rhombus bisect each other at right angles.

In right triangle BOC, we have

∠OBC + ∠BOC + ∠BCO = 180°

(Angle sum property of triangle)

∠OBC =180° - (∠BOC + ∠BCO)

=180° - (90° + 40°) = 50°

∠DBC = ∠OBC = 50°

Now, ∠ADB = ∠DBC [Alt. int. angles]

∠ADB = 50°.

(a) 24 cm

(b) 30 cm

(c) 36 cm

(d) 48 cm

Ans.

= 6 + 30 = 36 cm

(a) 180°

(b) 360°

(c) 540°

(d) 720°

Ans.

(a) A rectangle is a square.

(b) A rectangle is not a square.

(c) A square is not a rectangle.

(d) A square is always a rectangle.

Ans.

Because, “A square is not a rectangle” is contradictory to the “A square is a rectangle”. Also because both the statements do not have the same truth.

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 1 cm

Ans.

Given: OD ⊥ AB and AB is a chord, OA = 5 cm, AB = 8 cm

We know that the perpendicular from the centre of a circle to the chord bisects the chord.

AB ⇒ AC = 4 cm

Now, In Δ AOC

(AO)

(5)

(OC)

OC = √9

OC = 3 cm

Now CD = OD - OC = 5 - 3 = 2 cm

Ans.

[∵ 101 x 105 = (100 + 1) (100 + 5) = 10000 + 500 + 100 + 5 = 10605]

Ans.

Ans.

There can be as many as numbers as there is permutation of four-digits 1, 3, 5, 7 taken all at a time.

∴ The required 4-digits numbers = 41 = 4 x 3 x 2 x 1 = 24

Ans.

Base = 4 cm, height =6 cm

The area of the triangle is 12 cm

Ans.

OR

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

Ans.

[since, angle subtended at the arc is half of the angle subtended at the centre]

Given, OA = 3 cm and OD = 2 cm

We know that diagonals of a parallelogram bisect each other.

∴ AC = 2 x OA = 6cm and BD = 2 x OD = 4cm

Ans.

= -3i + 2a + (1 - a)i + 5 = (2a + 5) + (-3 + 1 - a)i

= -(2a + 5) + (- 2 - a)i.

The complex number is real, so its imaginary part is zero.

∴ - 2 - a = 0 ⇒ a = - 2

Ans.

x = 10°

Smallest angle = 4x = 40°

Ans.

∴ Its area =

Now, new sides of the triangle are 2a units, 2b units and 2c units.

Thus, its area = =

Total increase in are =

Hence, percent increase = 300%

Ans.

Then, a + b + c = x - 2y + 2y - 3z + 3z - x = 0

∴ a

⇒ (x - 2y)

= 3(x - 2y)(2y - 3z)(3z - x)

**SECTION - B**

**Question numbers 21 to 26 carry 2 marks each.****Q.21. Let A = {9 ,10, 11, 12, 13} and let f : A → N be defined by f(n) = the highest prime factor of n. Find the range off.****OR****If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6} and D = {6, 7, 8, 9}, find:****(i) (B - C)'****(ii) (B ∩ D')****Ans. **A = {9, 10, 11, 12, 13}

The highest prime factor of 9 is 3, so f(9) = 3

The highest prime factor of 10 is 5, so f(10) = 5

The highest prime factor of 11 is 11, so f(11) = 11

The highest prime factor of 12 is 3, so f(12) = 3

The highest prime factor of 13 is 13, so f(13) = 13

∴ Range of f = {3, 5, 11, 13).**OR**

U= {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 4, 6, 8], C = {3, 4, 5, 6} and D= {6, 7, 8, 9}

B - C = { 2, 4, 6, 8} - {3, 4, 5, 6} = {2, 8}

(B - C)' = U - (B - C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 8) = {1, 3, 4, 5, 6, 7, 9}

D' = U - D = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {6, 7, 8, 9} = {1, 2, 3, 4, 5}

(B ∩ D') = {2, 4, 6, 8} ∩ {1, 2, 3, 4, 5} = {2, 4}**Q.22. State any two Euclid's axioms.Ans.** Euclid's axioms

(i) Things which are equal to the same thing are equal to one another.

(ii) If equals are added to equals, the wholes are equal.

How many straight lines can be drawn from A to C? State the Euclid Axiom which states the required result. Give one more Postulate.

Ans.

According to Euclid’s Postulate, “A straight line may be drawn from any point to any other point.” Another Postulate: “A circle may be described with any centre and any radius.”

OR

In the given figure, AE = AD and CE = BD.

Prove that ΔAEB ≌ ΔADC.

Ans.

∴ ∠DCB = ∠CBD = 40

[∵ angles opposite to equal sides are equal]

Also, ∠CBD + ∠BDC + ∠DCB = 180

∴ 40

⇒ ∠BDC = 180

As A, B, D and C are four points on a circle.

Therefore, ABDC forms a cyclic quadrilateral.

Hence, ∠A+ ∠D = 180°

[∵ in a cyclic quadrilateral, sum of two opposite angles is equal to two right angles i.e. 180°]

⇒ ∠A+ 100° = 180° [∵ ∠D = ∠BDC = 100°]

∴ ∠A = 80°

Given, AE = AD ...(i)

and CE = BD ...(ii)

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

AE+ CE = AD + BD ⇒ AC = AB

Now, in ΔAEB and ΔADC, we have

AE = AD [given]

AB = AC [proved above]

and ∠A = ∠A [common angle]

∴ ΔAEB ≌ ΔADC [by SAS congruence rule] Hence proved.

Ans.

where y litres of 100% pure milk is added to get 99% pure milk.

Now (800 x 80 + 100y) = (800 + y)99

or 100y - 99y = 800 x 99 - 800 x 80

or y = 800 x (99 - 80) = 800 x 19 = 15200

i.e., 15200 litres of 100% milk should be added to get 99% pure milk.

41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60

Find the median and the mode.

Ans.

n = 15 (odd)

Median =

= 8th observation

= 52

Mode = 52, Since it is repeating three times in a data.

** SECTION - C**

**Question numbers 27 to 32 carry 4 marks each.**

is not defined when 16 - x

i.e., when (4 - x)(4 + x) < 0

i.e ., when 4 - x < 0 and 4 + x > 0 Or i .e ., when 4 - x > 0 and 4 + x < 0

i .e ., when 4 < x and x > - 4 Or i. e . , when 4 > x and x < - 4

i. e . , when x > 4 and x > - 4 Or i.e., when x < 4 and x < - 4

i .e ., when x > 4 Or i.e., when x < - 4

⇒ x > 4 or x < - 4

Also for each real number x such that - 4 ≤ x ≤ 4,

So, domain of the given function = {x ∈ R : - 4 ≤ x ≤ 4}

Further

Clearly, x is not defined when (16 - y

But, (16 - y

Also,

∴ Range of the function = {y ∈ R: 0 ≤ y ≤ 4}

x + y = 7

At what points, does the graph cut the x-axis and the y-axis?

Ans.

When x = 0 then y = 7

When y = 0 then x = 7

When x = 3 then y = 4

Table for x + y = 7 is

Intercept on x-axis and y- axis = 7 units.

They intersect the x-axis at (7, 0) and the y-axis j at (0, 7). **Q.29. Awareness, amongst citizens for protection of environment and making pollution free air: Participatory role of residents in the welfare of the society. Resident Welfare Association of a colony Chandan Nagar decided to develop flower beds in the park of the colony and then plant Jasmine plants in the flower beds. The difference of three times the money spent on flower beds and two times the money spent on Jasmine plants was INR 6000. Write a linear equation in two variables for this above mention situation and draw its graph.Ans. **Let money spent on flower beds be INR x and that on Jasmine plants be INR y.

Thus, the required equation as per statement of the question is:

3x - 2y = 6000

Table of solution is:

Required graph is shown alongside.

Ans.

(i) First, draw PQ = 8.2 cm.

(ii) At point P, construct ∠RPQ = 60° and at point Q, construct ∠SQP = 45°.

(iii) Draw bisector of ∠RPQ and ∠SQP, which intersect each other at point A.

(iv) Join AP and AQ.

(v) Draw perpendicular bisector of AP which intersects PQ at point B and also draw perpendicular bisector of AQ which intersects PQ at point C.

(vi) Join AB and AC.

Thus, we get the required ΔABC.

Or

Let 5 be the sum, P the product and R the sum of the reciprocals of n terms of a G.P. Prove that P

Ans.

= Cash down payment at the time of purchase + Sum of the balance in annual

instalments of INR 500 plus 12% interest on the unpaid amount ...(1)

Total cost of the used tractor = INR 12000

Balance cost of the used tractor = INR 12000 - INR 6000 (cash paid at the time of purchase) = INR 6000

Balance payment made at the end of 1 year by the farmer

Balance payment made at the end of 2nd year by the farmer

Balance payment made at the end of 3rd year by the farmer

and so on.

Thus, the payments form an A.P. whose first term, a = INR 1220 and common difference, d= -60 (1160 -1220) and n = be the number of instalments obtained by dividing INR 6000 by INR 500.

Therefore, the A.P. is

1220, 1160, 1100, 1040, 980, 920, 860, 800, 740, 680, 620, 560

Now, from (1), we get

Total cost of a used tractor will be =

= INR [6000 + 6( 1780)]

= INR [6000 + 10680]

= INR 16680

Let a, ar, ar

S = Sum of terms of a G.P.

⇒ S = a + ar+ ar

⇒

⇒ ...(i)

P = Product of n terms of a G.P.

⇒

...(2)

and R = Sum of reciprocals of n terms of a G.P.

⇒

⇒ ...(3)

Multiplying (2) and (3), we get

= S

OR

In the given figure, C and D are points on the semicircle with AB as diameter. If ∠BAD = 75° and ∠DBC = 40°, find ∠ABD, ∠ACB and ∠BDC.

Ans.

Given: OD ⊥ BC

Proof: In ΔOBD and ΔOCD

OB = OC (radii)

OD = OD (Common)

∠ODB = ∠ODC (90°)

ΔOBD ≌ ΔOCD (RHS congruence)

∠BOD = ∠COD (c.p.c.t)

But ∠BOC = 2∠BOD

2 ∠BAC = 2 ∠BOD

(Angle subtended by an arc of a circle at the centre is twice the one subtended at the circumcentre)

or ∠BOD = ∠BAC

Since AB is the diameter.

∠ADB = 90° (angle in the semi circle)

In ΔADB

∠ABD + ∠ADB +∠BAD = 180° (angle sum property)

∠ABD + 90° +75° = 180°

∠ABD = 15°

Also, ∠ACB = 90° (angle in the semi circle)

∠BCD + ∠BAD = 180°(opp. ∠s of circlic quad. are supplementary)

∠BCD + 75° = 180°

∠BCD = 180° -75° = 105°

In ΔBDC

∠BDC +∠BCD + ∠DBC = 180° (angle sum property)

∠BDC + 105° + 40°= 180°

∠BDC = 180° - 145°

= 35°

Ans.

AB + BC + CA = 10 cm,

AB = AC and AM ⊥ BC, AM = 3 cm

Required: To construct ΔABC.

Steps of construction:

1. Draw any line segment XY = 10cm.

2. Draw ‘p’ the perpendicular bisector of XY and let it intersect XY in M.

3. Cut-off AM = 3 cm.

4. Join AX and AY.

5. Draw the perpendicular bisectors of XA and YA and let they intersect XY in B and C.

6. Join AB and AC.

Thus, ΔABC is the required triangle.

Ans.

On putting x = 2 in Eq. (i), we get

f(2) = (2)

= 5 - 10 = - 5

On putting x = -1 in Eq. (i), we get

f(-1) = (-1)

On putting x = 1/3 in Eq. (i), we get

**SECTION - D**

**Question numbers 33 to 36 carry 6 marks each.****Q.35. In a town of 10.000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% of families buy B and C and 4% buy A and C. If 2% of families buy all the three newspapers, find:(a) the number of families which buy newspaper A only.(b) the number of families which buy none of A, B and C.Ans.** Total families in a town = 10,000 = U(say)

n(A) = Number of families buy newspaper A

= 40% of 10,000

= 4,000

n(B) = Number of families buy newspaper B

= 20% of 10,000

= 2,000

n(C) = Number of families buy newspaper C

= 10% of 10,000

= 1,000

n(A ∩ B) = Number of families buy both newspapers A and B

= 5% of 10,000

= 500

n(B ∩ C) = Number of families buy both newspapers B and C

= 3% of 10,000

= 300

n(A ∩ C) = Number of families buy both newspapers A and C

= 4% of 10,000

= 400

n(A ∩ B ∩ C) = Number of families buy all the newspaper A, and B and C

= 2% of 10,000

= 200

(a) The number of families which buy newspaper A only is

n(A ∩ B' ∩ C)

= n(A) - n[A ∩ (B ∪ C)']

= n(A) - n(A ∩ (B ∪ C)]

=n(A) -n[(A ∩ B) ∪(A ∩ C)] [P ∩ (Q ∪ R)= (P ∩ Q) ∪ (P ∩ R)]

= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B) ∩ (A ∩ Q)] [n (P ∪ Q)] = n(P) + n( Q) - n(P ∩ Q)]

= n( A) -n ( A ∩ B ) - n( A ∩ C) + n(A ∩ B ∩ C)

= 4,000 - 500 - 400 + 200

=3,300

OR

Factorise x

Ans.

and q(x) = x

Put, x - 3 = 0 or x = 3 in p(x) and q(x)

p(3) = a(3)

= 27a + 36 + 9 - 4

= 27a + 41

q(3) = (3)

= 27 - 12 + a

= 15 + a

According to the question,

p(3) = q(3)

⇒ 27 a + 41 = 15 + a

⇒ 27 a - a = 15 - 41

⇒ 26a = -26

⇒ a = -1.

Let p(x) = x

p(-1) = 0 ⇒ (x + 1) is a factor

OR

In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that

Ans.

To Prove:

Proof: We know that,

∠ABC + ∠CBD = 180° [linear pair]

⇒

⇒ [∵ BO is the bisector of ∠CBD]

⇒ ...(i)

Similarly, ∠ACB + ∠BCE = 180° [linear pair]

⇒

⇒ [∵ CO is the bisector of ∠BCE]

⇒ ...(ii)

In ΔOBC,

[sum of ∠s of a Δ is 180°]

⇒

⇒ [using (i) and (ii)]

⇒

⇒

∴

⇒

⇒

⇒

⇒

⇒

In ΔPQR

...(i) [ext. ∠ = sum of int. opp. ∠s]

Again, in ΔTQR

...(ii) [ext. ∠ = sum of int. opp. ∠s]

Now, QT and RT are the bisectors of ∠PQR and ∠PRS respectively.

⇒ ...(iii)

and ...(iv)

From (i), we have

⇒ ...(v) [using (iii) and (iv)]

From (ii) and (v), we have

⇒

(i) the lateral or curved surface area of a closed cylindrical petrol storage tank, which is 4.2 m in diameter and 4.5 m in high.

(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank?

Ans.

and height, h = 4.5m

Curved surface area of a closed cylindrical petrol storage tank =

= 44 x 0.3 x 4.5 = 59.4 m

(ii) Now, total surface area of cylindrical tank

= Curved surface area + 2πr

Let x m

Since, 1/12 of the actual steel used was wasted. Therefore, the area of the steel, which was actually used for making the tank

Hence, actually steel used in making a tank is 95.04 m

In the expansion of (x + a)

...(1)

...(2)

...(3)

Dividing (2) by (1), we have

...(4)

Dividing (3) by (2), we have

...(5)

Dividing (5) by (4), we get

⇒ 4n - 8 = 3n - 3 ⇒ n = 5

Putting the value of n in (1) and (2), we get

5 x

⇒ 10x

Dividing (7) by square of (6) to eliminate a, we get

⇒ x = 2 ...(8)

Putting the value of x in (6), we get

5(2)

Hence, x = 2, a = 3 and k = 5.

[Expanding by binomial theorem]

Thus, the coefficients of the first three terms of respectively.

It is given that the sum of the coefficients of the first three terms in the expansion

∴

⇒

⇒ 2 - 6m + 9m

⇒ 9m

⇒ 3m

⇒ reject it, m being a natural number

Now,

We need the term containing x

Thus, the required term is

Draw a histogram to represent the data above.

Ans.

The lower limit of 18 - 25 = 18

The upper limit of 10 - 17 = 17

The difference = 18 - 17 = 1

∴ Half the difference = 1/2 = 0.5

The histogram of given data is shown below

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