Oswaal Maths

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Published by :

Karnataka Secondary Education Examination Board (KSEEB)

for SSLC 2015 Examination

Class Class

10

SAMPLE

QUESTION

PAPERS

Mathematics

solutions

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S O L U T I O N S

SAMPLE

QUESTION PAPER - 6

Self Assessment

______ ________________________________________________________________________________ Time

Time : : 3 3 Hours Hours 45 45 Minutes Minutes Maximum Maximum Marks Marks : : 8080 I.

I. Solutions of MSolutions of Multiple Choice ultiple Choice QuestionsQuestions 1. 1. (C) – 1(C) – 1 11 2. 2. (B) 6(B) 6 11 3. 3. (D) 0·3(D) 0·3 11 4. 4. (A) 30(A) 30 11 5. 5. (C)(C)aa = 1, = 1,bb = – 3 = – 3 11 6. 6. (B)(B) 11 7. 7. (D)(D) 11 8. 8. (C) (0, – 7)(C) (0, – 7) 11 II.

II. Solutions of Solutions of one mark one mark questionsquestions 9.

9.

(½ + ½ = 1) (½ + ½ = 1)

10.

10. nn(( A A∪∪BB) ) ==nn(( A A – –BB) +) +nn((BB – – A A) +) +nn(( A A∩∩BB))½½ = 7 + 11 + 4 = 7 + 11 + 4 = 22. = 22. ½½ 11. 11. f f (– 1) (– 1) = (1)= (1)33 + 2(1) + 2(1)22 – 3(1) + 4 – 3(1) + 4 = = 1 1 + + 2 2 – – 3 3 + + 44 = 4. = 4. 11 12.

12. StatemenStatement t :: The rectangle containedThe rectangle contained by any two sides of a triangle is by any two sides of a triangle is equalequal

to rectangle contained by altitude to rectangle contained by altitude drawn to the third side and the circum drawn to the third side and the circum diameter.

diameter. 11

13.

13. ∠∠ ABC ABC + +∠∠BOCBOC = 180° = 180° ½½

⇒

⇒∠∠ ABC ABC + + ∠∠ ABC ABC = 180° = 180°

⇒ ⇒ ∠∠ ABC ABC = 180° = 180° ⇒ ⇒ ∠∠ ABC ABC == × 180°× 180° = 72°. = 72°. ½½ 14.

14. Volume of double coneVolume of double cone = = 11 3 3 ππrr 2 2₍₍_h_h₊₊_H_H_{) units.}_{) units.} ₁₁

III. Solutions of two marks questions III. Solutions of two marks questions 15. 15. 12 12 = 2 × = 2 × 2 × 32 × 3 = 2 = 222 × 3, × 3, ½½ 15 15 = 3 × = 3 × 5 and 305 and 30 = 2 × 3 × 5 = 2 × 3 × 5 ½½ ∴ ∴ LCM LCM (12, (12, 15, 30) = 215, 30) = 222 × 3 × 5 × 3 × 5 = 60 = 60 ½½ HCF (12, HCF (12, 15, 30) 15, 30) = 3= 3 ½½

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S O L U T I O N S

SAMPLE

QUESTION PAPER - 6

Self Assessment

______ ________________________________________________________________________________ Time

Time : : 3 3 Hours Hours 45 45 Minutes Minutes Maximum Maximum Marks Marks : : 8080 I.

I. Solutions of MSolutions of Multiple Choice ultiple Choice QuestionsQuestions 1. 1. (C) – 1(C) – 1 11 2. 2. (B) 6(B) 6 11 3. 3. (D) 0·3(D) 0·3 11 4. 4. (A) 30(A) 30 11 5. 5. (C)(C)aa = 1, = 1,bb = – 3 = – 3 11 6. 6. (B)(B) 11 7. 7. (D)(D) 11 8. 8. (C) (0, – 7)(C) (0, – 7) 11 II.

II. Solutions of Solutions of one mark one mark questionsquestions 9.

9.

(½ + ½ = 1) (½ + ½ = 1)

10.

10. nn(( A A∪∪BB) ) ==nn(( A A – –BB) +) +nn((BB – – A A) +) +nn(( A A∩∩BB))½½ = 7 + 11 + 4 = 7 + 11 + 4 = 22. = 22. ½½ 11. 11. f f (– 1) (– 1) = (1)= (1)33 + 2(1) + 2(1)22 – 3(1) + 4 – 3(1) + 4 = = 1 1 + + 2 2 – – 3 3 + + 44 = 4. = 4. 11 12.

12. StatemenStatement t :: The rectangle containedThe rectangle contained by any two sides of a triangle is by any two sides of a triangle is equalequal

to rectangle contained by altitude to rectangle contained by altitude drawn to the third side and the circum drawn to the third side and the circum diameter.

diameter. 11

13.

13. ∠∠ ABC ABC + +∠∠BOCBOC = 180° = 180° ½½

⇒

⇒∠∠ ABC ABC + + ∠∠ ABC ABC = 180° = 180°

⇒ ⇒ ∠∠ ABC ABC = 180° = 180° ⇒ ⇒ ∠∠ ABC ABC == × 180°× 180° = 72°. = 72°. ½½ 14.

14. Volume of double coneVolume of double cone = = 11 3 3 ππrr 2 2₍₍_h_h₊₊_H_H_{) units.}_{) units.} ₁₁

III. Solutions of two marks questions III. Solutions of two marks questions 15. 15. 12 12 = 2 × = 2 × 2 × 32 × 3 = 2 = 222 × 3, × 3, ½½ 15 15 = 3 × = 3 × 5 and 305 and 30 = 2 × 3 × 5 = 2 × 3 × 5 ½½ ∴ ∴ LCM LCM (12, (12, 15, 30) = 215, 30) = 222 × 3 × 5 × 3 × 5 = 60 = 60 ½½ HCF (12, HCF (12, 15, 30) 15, 30) = 3= 3 ½½

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16.

16. 11½½

U =

U = Students appeared from examinationStudents appeared from examination M

M = = MathsMaths S

S = = ScienceScience

Percentage of students failed in both = 1% = Percentage of students failed in both = 1% =

½ ½ 17.

17. (i) Three-digit numbers less than 500(i) Three-digit numbers less than 500 = = 4 4 × × 4 4 × × 33 = 48 = 48 ½½ H H T T P P P P PP U U 4 4 1 1 44 11 33 11

(ii) Two-digit numbers less than 500 (ii) Two-digit numbers less than 500

= = 5 5 × × 4 4 = = 2020 ½½ T T P P PP U U 5 5 1 1 44 11

(iii) One-digit numbers less than 500 (iii) One-digit numbers less than 500

= 5 = 5 ½½ U U 5 5 1 1 P P ∴

∴ Total required numbers = 48 + 20 + 5 Total required numbers = 48 + 20 + 5 = 73. = 73. ½½ 18. 18. == == 11 = = ½½ ∴ ∴ xx = 121. = 121. ½½ 19. 19. ((3 13 18 2 8 2 1+ + 12 2 ))(( 550 0 −− 2277)) = = ((9 9 2 2 4 + + 4 3 3 5 ))((5 2 2 3 −−3 33)) ½½ = = 9 2 5 2 9 2 ((5 2 3 3 − − 3 3 4 3 ))++4 3 5 2 ((5 2 3 3−−3 3)) ½½ = = 445 5 2 ××2 2− −27 6 7 6 2++20 6 0 6 1−−12 2 3 ××3 ½½ = = = = ½½ 20. 20. 5 5 5 5 22 3 3 5 5 22 + + − − − − == 5 5 5 5 2 2 3 3 5 5 22 5 5 2 2 22 22 ( ( )) (( )) ( ( ) ) ( ( )) − − −− − − + + 1 1

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OSWAAL OSWAAL CBSE CBSE (CCE),(CCE),

Mathematics Class –10

= = ½½ = = as as 5 5 – – 4 4 = = 11 ½½ 21.

21. LetLetpp((xx) ) = = 22xx33 – – 55xx22 + +xx + +aa and and f f ((xx) ) ==axax33+ + 22xx22– – 33

Since

SinceRR₁₁ and andRR₂₂ are the r are the remainders whenemainders whenpp((xx)) and

and f f ((xx) are divided by () are divided by (xx – 2). – 2).

∴ ∴ RR₁₁ ==pp(2)(2) = = 2(2)2(2)33 – 5(2) – 5(2)22 + 2 + + 2 +aa = = 16 16 – – 20 20 + + 2 2 ++aa = – 2 + = – 2 +aa ½½ and and RR₂₂ == f f (2) =(2) =aa(2)(2)33 + 2(2) + 2(2)22 – 3 – 3 = 8 = 8aa + 8 – 3 + 8 – 3 = 8 = 8aa + 5 + 5 ½½ Given, Given, RR₁₁ = 2= 2RR₂₂ ⇒ ⇒ – – 2 2 ++aa = 2(8 = 2(8aa + 5) + 5) ½½ ⇒ ⇒ – – 2 2 ++aa = 16= 16aa + 10 + 10 ⇒ ⇒ 1515aa = – = – 1212 ∴ ∴ aa == ½½ OR OR For

For degreedegree qq((xx) = degree) = degreerr((xx), we have), we have degree

degree of of quotient quotient will will be be equal equal to to the the degreedegree of remainder when quotient is not constant. of remainder when quotient is not constant. Example : Example : p p((xx) ) = 7= 7xx33 + 2 + 2xx + 5, + 5, g g ((xx) = 7) = 7xx22 q q((xx) ) ==xx and andrr((xx) = 2) = 2xx + 5 + 5 11 Clearly, degree

Clearly, degreeqq((xx) = 1 = degree) = 1 = degree qq((xx)) ½½ Checking

Checking for for division division algorithm algorithm :: g

g ((xx))··qq((xx) +) +rr((xx) ) = 7= 7xx22 ( (xx) + (2) + (2xx + 5) + 5) = 7

= 7xx33 + 2 + 2xx + 5 = + 5 =pp((xx)) Thus,

Thus, division division algorithm algorithm is is satisfied.satisfied. ½½ 22. 22. Here,Here,aa = 1, = 1, bb = – 3, = – 3, cc = 1 = 1 m m + +nn == ½½ mn mn == ½½ ∴ ∴ == = = ½½

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OSWAAL OSWAAL CBSE CBSE (CCE),(CCE),

Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

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= = 9 – 2 = 7 ½ 23. L.H.S. = sec2θ + cosec2θ = ½ = sin cos cos · sin 2 2 2 2 θ θ θ θ + ½ = 1 2 2 cos θ · sin θ _½ = sec2θ· cosec2θ ½ = R.H.S. ½ 24. Point of division P ½ y-coordinate of P = 1 2 1 1 2 · y+ ( ) + − = 2 (given) ½ ⇒ = 2 ⇒ y = 2 × 3 + 2 = 8 ½

25. Radius of circle, OB = 13 cm

Draw OM⊥ AB.

∴ M bisect the chord AB. ½

∴ MB = AB

= × 24

= 12 cm ½

In rt.∆OMB, by Pythagoras theorem

OM2 = OB2 – MB2 ½ = 132 – 122 = 169 – 144 = 25 ∴ OM = 5 cm. ½ 26. Rough sketch : ½ Area of field ABCD = Ar (∆ APB) + Ar (trapezium PBCQ) + Ar (∆CQD) ½ = × 150 × 100 + (100 + 150) × 50 + × 100 × 150 ½ = 7500 + 6250 + 7500 = 21250 sq. m. ½ 27. l = 4 cm 2πR = 18 ⇒ R = = cm ½ 2πr = 6 ⇒ r = = cm ½

Curved surface area of frustum = πl(R + r) ½ =π × 4

= 4π ×

= 48 cm2. ½

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OSWAAL CBSE (CCE),

Mathematics Class –10

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OR Given, r₁ = 12 cm, h₁ = 20 cm, r₂ = 3 cm, h₂ = ? We know, = ⇒ = ⇒ h₂ = = 5 cm ½

∴ The cut to be made at 5 cm from the water.

∴ Height of frustum so obtained,

h = 20 – 5 = 15 cm ½

Volume of the frustum

= πh(r₁2 + r₂2 + r₁r₂) ½ = × 15(144 + 9 + 12 × 3)

= × 189

= 2970 cm3. ½

28. Given, N = 7, A = 10, R = 5 The graph is as follows :

2

29. Given, in∆LMN ,∠LMN =∠PNK = 46°

∠ M = a, PN = x or N = b, NK = c ½

∴ LM ||PN ½

(If corresponding angles are equal then lines are parallel)

∴ =

(corollary of Thales theorem) ½ =

∴ x = ½

30. Number of red balls = 6 Let number of blue balls = x

Total number of balls = 6 + x ½

P(red ball) = ½

P(blue ball) = ½

According to the question,

P(blue ball) = 2 × P(red ball) = 2 ×

⇒ x = 12 ½

Hence, number of blue balls is 12. IV. Solutions of three marks questions

31. Given, S_n = 3n2 + 4n ½ ∴ S_n_{– 1} = 3(n – 1)2 + 4(n – 1) = 3(n2 – 2n + 1) + 4n – 4 = 3n2 – 2n – 1 ½ nth term, T _n = S_n – S_n_{– 1} ½ = (3n2 + 4n) – (3n2 – 2n – 1) = 6n + 1 ½ ∴ T ₂₅ = 6 × 25 + 1 = 151 1 32. x f fx d x x d = − fd − − 2 2 32 2 64 10 100 200 37 5 185 5 25 125 42 6 252 0 0 0 47 5 235 5 255 125 52 2 104 10 100 200 20 840 650 2 n= fx fd = = Σ Σ 1 Arithmetic Mean = = 42 1 S.D. = ½ ∴ σ = = = 5·7 ½

33. The given equation is in the form Ax2 + Bx + C = 0, where A = b – c, B = c – a and C = a – b ½

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OSWAAL CBSE (CCE),

Mathematics Class –10

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OSWAAL Karnataka (SSLC),

Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

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For equal roots,

B2 – 4 AC = 0 ½

⇒ (c – a)2 – 4(b – c) (a – b) = 0 ½

⇒ c2 – 2ac + a2 – 4(ab – b2 – ca + cb) = 0 ½

⇒c2 – 2ac + a2 – 4ab + 4b2 + 4ca – 4bc= 0

⇒ c2 + a2 + 2ca – 4ab – 4bc + 4b2 = 0

⇒ (c + a)2 – 4b(a + c) + (2b)2 = 0 ½

⇒ [(c + a) – 2b]2 = 0⇒ (c + a) – 2b = 0

∴ c + a = 2b. ½

OR Let the cost price be`_x.

Selling price =`_18·75

∴ Loss = cost price – selling price

⇒ × x = x – 18·75 1 ⇒ x2 = 100x – 1875 ⇒ x2 – 100x + 1875 = 0 ⇒ x2 – 75x – 25x + 1875 = 0 1 ⇒ x(x – 75) – 25(x – 75) = 0 (x – 75)(x – 25) = 0 ∴ x – 75 = 0 or x – 25 = 0 ⇒ x = 75 or x = 25

The cost price is`_{75 or}`_25. ₁

34. Given,∆ ABC ~∆PQR

Since triangles are similar, therefore

= = = k (say) …(i) ½ ⇒ a = kp, b = kq, c = kr ½ ½ Now, = ½ = = = k …(ii) ½

From (i) and (ii), we have

= = = _· ½ OR Given : ∆ ABC ~∆PQR ar (∆ ABC) = ar (∆PQR) ½ To prove :∆ ABC≅ ∆PQR

Proof :We know that areas of similar triangles are proportional to the squares of the

corre-sponding sides. ½

∴ ₌ ₌ ₌ _½

⇒ 1 = = =

[·.· given, areas are equal] ½

⇒ AB2 = PQ2, BC2 = QR2 and CA2 = RP2

∴ AB = PQ, BC = QR and CA = RP ½ Hence,∆ ABC≅ ∆PQR [SSS criteria] ½ 35. Let the height of flagpost, PQ = x

Angle of elevation at C = 30°

When moved 6 m towards the post i.e., at B, angle of elevation = 30° + 15° = 45° In∆ AOB, tan 45° = ⇒ 1 = = OB = x 1 In∆OAC, tan 30° = ⇒ = ⇒ x + 6 = …(i) 1 = 6⇒ = 6

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OSWAAL CBSE (CCE),

Mathematics Class –10

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∴ x =

Hence, height of the post = m. 1 OR

Given, x sinθ = y cosθ

⇒ x = …(i) 1

and x sin3θ + y cos3θ = sinθ cosθ …(ii) Eliminating x from (i) and (ii), we get

.sin3θ + y cos3θ = sinθ cosθ ½

⇒ y cosθ sin2θ + y cos3θ = sinθ cosθ

y cosθ [sin2θ + cos2θ] = sinθ cosθ ⇒ y cosθ (1) = sinθ cosθ

⇒ y = sinθ …(iii) 1

Substituting this value of y in (i),

x = cosθ …(iv)

∴ Squaring (iii) and (iv) and then adding, we get

x2 + y2 = cos2θ + sin2θ = 1 ½ 36. Given : A and B are the centres of touching

circles. P is the point of contact. ½

To prove : A, B and P are collinear.

Construction :Draw the tangent XPY . Join AP

and BP. ½

Proof :Since radius drawn at the point of con-tact is perpendicular to the tangent.

∴ ∠ APX = 90° ½

and ∠BPX = 90° ½

⇒ ∠ APX =∠BPX = 90°

⇒ AP and BP lies on the same line. ½ Hence, A, B and P are collinear. ½

OR

Given : O is the centre of the circle. P is an external point. PA and PB are the two tangents drawn from an external point P. ½

To prove :∠ AOB +∠ APB = 180° ½ Proof :In quadrilateral PAOB,

∠PAO +∠ AOB +∠OBP +∠ APB = 360° ½ But ∠PAO = 90° and∠OBP = 90° ½

∴ 90° +∠ AOB + 90° +∠ APB = 360° ½

⇒ ∠ AOB +∠ APB = 360° – (90° + 90°) = 360° – 180°

= 180°. ½

V. Solutions of four marks questions 37. Three numbers are in G.P.

Let numbers are a, ar.

Product of numbers = × a × ar

⇒ 343 = a3⇒a =

∴ a = 7 1

Sum of the three numbers.

⇒ 57 = + a + ar 57r = a + ar + ar2 [Multiply by r] 57r = 7 + 7r + 7r2 ½ 57r – 7r = 7r2 + 7 ∴ 7r2 – 50r + 7 = 0 7r2 – 49r – r + 7 = 0 7r(r – 7) – 1(r – 7) = 0 1 (r – 7) (7r – 1) = 0 r – 7 = 0, 7r = 1 r = 7 or r = ½ Numbers are a, ar 7, 7 × 7 i.e., 1, 7, 49. 1

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OSWAAL CBSE (CCE),

Mathematics Class –10

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Mathematics Class –10

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OR

Let r be a common ratio of the G.P. a, b, c, d then b = ar, c = ar2 and d = ar3. ½

L.H.S. = (b – c)2 + (c – a)2 + (d – b)2

= (ar – ar2)2 + (ar2 – a)2 + (ar3 – ar)3 1 = a2r2 (1 – r)2 + a2 (r2 – 1)2 + a2r2 (r2 – 1)2 = a2[r2(r – 2r + r2) + (r4 – 2r2 + 1) + r2(r4 – 2r2 + 1)] 1 = a2[r2 – 2r3 + r4 + r4 – 2r2 + 1 + r6 – 2r4 + r2] = a2 (r6 – 2r3 + 1) = a2 (1 – r3)2 1 = (a – ar3)2 = (a – d)2 = R.H.S. ½ 38. y = x2 + 2

The value of y for corresponding values of x are given in the following table :

x y

− 2 −1 0 1 2 6 3 2 3 6

2

The curve i.e., parabola does not intersect the x-axis.

∴ There is no real value of x for x2 + 2 = 0. Hence, there are no real roots. 1 39. Given : ABC is right angled at B and D is the

mid-point of BC. ½

BD = DC = BC ½

Sample Question Papers

(SA-2)

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In∆ ABD, AD2 = AB2 + BD2

(Pythagoras Theorem) … ½ In∆ ABC, AC2 = AB2 + BC2

(Pythagoras theorem) …(ii) ½ From eqn. (i),

AD2 = AB2 +

(D is the mid-point of BC) ½

⇒ 4 AD2 = 4 AB2 + BC2

⇒ BC2 = 4 AD2 – 4 AB2 …(iii) ½ Using this in eqn. (ii),

AC2 = AB2 + 4 AD2 – 4 AB2 ½ AC2 = 4 AD2 – 3 AB2. Proved. ½

40. C₁ = R = 5·5 cm C₂ = r = 3·5 cm d = 5·5 + 3·5 = 9 cm C₃ = R – r = 5·5 – 3·5 = 2 cm 1 Steps of construction : (a) Draw AB = 9 cm.

(b) Construct circles C₁ and C₂ at A and B, with radius 5·5 cm and 3·5 cm respectively. ½

(c) Draw C₃ with centre A and radius R – r

= 2 cm. ½

(d) Construct perpendicular bisector to AB bisecting AB at M.

(e) With M as centre and radius MA, draw C₄. ½ (f) Join AM and AN to intersect C₃ at M and N

and produce them to meet the circle C₁ at P

and R. ½

(g) Join BM and BN .

(h) With the radius MB with P and R as centres, draw arcs to intersect C₂ at Q and S.

(j) Join PQ and RS. ½

∴PQ and RS are D.C.T.

By measurement :PQ = RS = 8·7 cm. ½

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OSWAAL CBSE (CCE),

Mathematics Class –10

SAMPLE

QUESTION PAPER - 7

Self Assessment

___________________________________________

Time : 2 Hours 45 Minutes Maximum Marks : 80

III. Solutions of two marks questions

15. If the number is divisible by 35, 56 and 91, then it is the LCM of these numbers.

By prime factorization,

35 = 5 × 7, 56 = 23 × 7 and 91 = 7 × 13

∴ L.C.M. (35, 56, 91) = 23 × 5 × 7 × 13

= 3640 1

The least number divisible by 35, 56 and 91 is 3640.

Since it leaves a remainder 7, the required number is

3640 + 7 = 3647. 1

16. Two sets A and B are disjoint sets when

A∩B =φ. ½

A = {1, 2, 3, 4} and B = {4, 5, 6, 7} ½

A∩B = {4}≠ φ ½

∴ A and B are not disjoint sets. ½

17. nC₂ = 10

⇒ = 10 ½

⇒ n2 – n – 20 = 0

⇒ (n – 5) (n + 4) = 0 ½

⇒n = 5 or n = – 4, which is not possible. ½

∴ n = 5. ½

18. The 3 particular books can be arranged in 3P₃

ways. ½

3 particular books as one and remaining 4 books can be arranged in5P₅ ways. ½ Total number of ways

=3P₃ ×5P₅ ½

= (3 × 2 × 1) × (5 × 4 × 3 × 2 × 1) = 6 × 120

= 720 ½

I. Solutions of Multiple Choice Questions

1. (C) 24th 1 2. (D) a haptagon 1 3. (C) 0≤r≤ 1 1 4. (B) 0·7 1 5. (B) x3 1 6. (D) 1 7. (B) II quadrant 1 8. (C) (0, 0) 1

II. Solutions of one mark questions 9. Euclid’s division lemma : Given

positive integers a and b there exist unique integers q and r satisfying a = bq + r, 0≤r < b.

10. A∩ (B∪C) = ( A∩B)∪ ( A∩C). 1 11. Since the curve cut the x-axis at 3

points. ½

∴ Number of zeroes of f (x) = 3. ½ 12. Two triangles are said to be similar if :

(i) their corresponding sides are

proportional and ½

(ii) their corresponding angles are

equal. ½

13. There two tangents are parallel. 1 14. Volume of sphere I Volume of sphere II = = ½ ⇒ = ⇒ ½

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OSWAAL CBSE (CCE),

Mathematics Class –10

19. = a ab ab b 3 3 _· ₊3 _· 3 1 = ½ = ½ 20. x = ⇒ = ½ ⇒ = 1 2 6 5 2 6 5 2 6 5 + × − − ½ = = ½ ∴ x + = ½ 21. Let f (x) = x3 + x2 – 3x + 5 and g (x) = x – 1 By synthetic division, 1 1 1 – 3 5 1 2 – 1 1 2 – 1 4 ∴ Quotient, q(x) = x2 + 2x – 1 1 and remainder, r(x) = 4 1 OR Here, p(x) = ax2 + bx + c

Since zeroes are reciprocal to each other, then letα and be the zeroes of p(x). 1

∴ Product of zeroes,

α × =

⇒ 1 = or a = c, 1

Which is the required condition.

22. Here, a = 49, b = – k , c = – 81 ½ Let one root is m, then other root is – m. ½

∴ Sum of roots = ½ ⇒ m + (– m) = ⇒ 0 = ⇒ k = 0 ½ 23. L.H.S. = = ½ = = ½ = cos2θ – (1 – cos2θ) ½ = 2 cos2θ – 1 = R.H.S. ½ 24. For line I, slope m₁ =

= = 1 ½

For line II, slope m₂ =

= ½

Since two lines are perpendicualr, then

m₁m₂ = – 1 ½ ∴ 1 × = – 1 ⇒ 1 = – m – 2 ⇒ m = – 1 – 2 = – 3 ½ 25. Steps of construction :

(a) Draw a line segment O₁O₂ = 5 cm.

(b) With O₁ and O₂ as centres draw two circles C₁ and C₂ of same radii 3 cm.

1 + 1

26. Scale 25 m = 1 cm

1

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OSWAAL CBSE (CCE),

Mathematics Class –10

AP = 100 m = 4 cm AQ = 150 m = 6 cm AR = 200 m = 8 cm AD = 300 m = 12 cm QE = 50 m = 2 cm QB = 75 m = 3 cm RC = 100 m = 4 cm 1 27. Given, r = = 8 cm, h = 15 cm l = = = 17 cm ½ T.S.A. = 2πrh +πr2 +πrl ½ =πr(2h + r + l) = × 8[30 + 8 + 17] ½ = × 8 × 55 = 1383 cm2 (Approx.) ½ OR l = ( )4 15 2 2 2

+ 

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



_



Curved surface area = 2πrh +πrl

=πr(2h + l) ½

= [4·8 + 8·5]

= × 13·3

= 313·5 m2 ½

∴ Cost of canvas used at `_50/m2

= 313·5 × 50

=`_15675. _½

28. F = 6, E = 12, V = 8 ½

Euler’s formula for polyhedra

F + V = E + 2 ½

⇒ 6 + 8 = 12 + 2 ½

⇒ 14 = 14, which is true

Hence, Euler’s formula is verified. ½

29. AB2 = AC· AM (Corollary) ½

⇒ AB2 = AC·16 MC

⇒ AB = 4 AC MC_· _{…(i) ½}

Again BC2 = AC· MC (Corollary)

⇒ BC = AC MC_· _{…(ii) ½}

AB = 4BC ½

30. If three coins are tossed together, then the sample space is

S = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }

∴ n(S) = 8 ½

(i) An event of getting atmost two heads is A = {HHT , HTH , THH , HTT , THT , TTH , TTT }

⇒ n( A) = 7 ½

∴ P( A) = = _½

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

(ii) An event of getting all heads is B = {HHH }⇒n(B) = 1

∴ P(B) = = ½

IV. Solutions of three marks questions 31. Let the three terms in H.P. are a, b, c.

∴ a = 2c, (given) …(i) ½

Now, b = H.M. between a and c = 20, (given)½

⇒ = 20 ⇒ = 20 ½ ⇒ = 20⇒ 4 3 c = 20 ∴ c = = 15 ½ From (i), a = 2 × 15 = 30 ½

Hence, three terms are 30, 20 and 15. ½ 32. C.I. mid-point 2 f x fx d x x = d fd = − − − − − 2 2 20 40 30 60 30 78 48 2304 4608 40 600 50 350 50 78 28 784 5488 60 80 70 840 70 78 8 64 768 80 100 9 7 12 19 − − − − − − = = 00 1710 90 78 12 144 2736 100 120 5 110 550 110 78 32 1024 5120 − − − − − = = Totall n= 45 _Σfx= 3510 _Σfd2=18720 1 Now, mean = = = 78 ½ Variance, σ2 = = = 416 ½ S.D., σ = = 20·396 ½

This means, each score on an average deviates from mean value, 78 by 30·396. ½ 33. Let the length of the smaller square = x m

and length of the larger square = y cm According to the question,

y2 – 2x2 = 14 …(i) ½ and 2 y2 + 3x2 = 203 …(ii) ½

From (i), y2 = 14 + 2x2 ½

From (ii), 2(14 + 2x2) + 3x2 = 203 ½ ⇒ 28 + 4x2 + 3x2 = 203 ⇒ 7x2 = 203 – 28 ⇒ x2 = = 25 ∴ x = 5 cm ½

From (i), y2 = 2x2 + 14 = 2(5)2 + 14 = 64

∴ y = 8 cm. ½

Hence, the length of the sides of squares are 5 cm and 8 cm.

OR

Let one odd positive number be x. then other will be = x + 2 According to the question,

x2 + (x + 2)2 = 130 1 ⇒ x2 + x2 + 4 + 4x = 130 ⇒ 2x2 + 4x + 4 = 130 ⇒ 2x2 + 4x – 126 = 0 ⇒ x2 + 2x – 63 = 0 ⇒ x2 + 9x – 7x – 63 = 0 ½ ⇒ x(x + 9) – 7(x + 9) = 0 ⇒ (x – 7)(x + 9) = 0 ½ ⇒ x – 7 = 0 or x + 9 = 0 ⇒ x = 7 or x = – 9 (not possible) So, x = 7

Hence, odd positive numbers are 7, 9. 1 34. Given :In∆ ABC and∆DEF,

∠BAC =∠EDF

∠ ABC =∠DEF

and ∠ ACB =∠DFE

To prove :

=

Construction : Mark X on AB and Y on AC, such that AX = DE and AY = DF, Join XY . 1 Proof :In∆ AXY and∆DEF,

∠ A =∠D [Given]

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OSWAAL CBSE (CCE),

Mathematics Class –10

AX = DE, [By construction] AY = DF [By construction]

∴ ∆ AXY =∆DEF [SAS-criteria]

∴ XY = EF ½

and ∠ AXY =∠DEF [Congruent triangles]

⇒ ∠ AXY =∠DEF =∠ ABC [Given] Since ∠ AXY =∠ ABC [... ∠ AXY and∠ ABC are corresponding angles]

∴ XY ||BC ½

⇒ = ½

[Basic proportionality theorem and corollary]

⇒ = Proved. ½

OR

In the figure, ABCD is a rhombus.

∴ AB = BC = CD = AD AD||BC and AB||DC ½ In∆ ADQ and∆PCQ, AD||PC ∴ = ½ ⇒ = [·.· AD = BC] ½ ⇒ = + 1 ½ = + 1 [·.· DC = BC] ½ ⇒ 1 = ∴ = Proved. ½

35. Let the height of the cliff be x from the ground.

∴ OD = x – 24 In∆OCD, tan 45° = = 1 ⇒ = 1⇒ y = x – 24 1 In∆OAB, tan 60° x y = ⇒ y = 1 ∴ x – 24 = _½ ⇒ = 24 ⇒ = 24⇒x = ½

Hence, height of the cliff, x = m. OR

= 4

⇒ cos ( sin ) cos ( sin )

( sin )( sin ) θ θ θ θ θ θ 1 1 1 1 + + + − − = 4 1 ⇒

cos cos sin cos cos sin sin θ θ θ θ θ θ θ + + − − 1 2 = 4 ½ ⇒ = 4 ⇒ = 4 1 ⇒ cosθ = = = cos 60° ∴ θ = 60°. ½

36. Let AB be a diameter of a given circle and let CD and EF be the tangent lines drawn to the circle at A and B respectively.

Since tangent at a point to a circle is per-pendicular to the radius through the point.

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

∴ AB⊥CD and AB⊥EF 1

∴ ∠CAB = 90° and∠ ABF = 90° ½

∠CAB =∠ ABF ½

∠CAB and∠ ABF are alternate int. angles.

∴ CD||EF 1

OR (a) In∆ ABC, AB = BC

∴ ∠C =∠ A Also, ∠ A +∠B +∠C = 180° ∠C + 68° +∠C = 180° 2∠C = 112° ⇒ ∠C = 56° ∴ ∠ ACB = 56° 1 (b) In the figure, ∠ AOB = 2∠ ACB = 2 × 56° = 112° 1

(c) In the quadrilateral OADB,

∠ A =∠B = 90°

(·.· AD and BD are tangents to the circle)

∴ ∠ AOB and∠ ADB = 180° 112° +∠ ADB = 180°

∴ ∠ ADB = 180° – 112°

= 68° 1

V. Solutions of four marks questions

37. Let the three terms of A.P. are a – r, a and a + r. Sum of these terms = 15, given

⇒ a – r + a + a + r = 15 3a = 15

∴ a = = 5 1

The sum of squares of extremes = 58

∴ (a – r)2 + (a + r)2 = 58 (5 – r)2 + (5 + r)2 = 58 (25 + r2 – 10r) + (25 + r2 + 10r) = 58 1 25 + r2 + 25 + r2 = 58 2r2 = 58 – 50 r2 = = 4 ⇒ r = ± 2 1

The terms of AP in order : a – r, a, a + r i.e., 5 – 2, 5, 5 + 2 or 5 + 2, 5, 5 – 2

i.e., 3, 5, 7 or 7, 5, 3. 1

OR

Let, given series be 4 + 12 + 16 + … …(i) Here, a = 4, r = = 3 =

⇒ Series (i) is geometric series. ½

S_n = , if r > 1 ½ ∴ S₃ = = = 52 1 and S₆ = = = 1456 1 Now, = ∴ S₃ : S₆ = 1 : 28. 1

38. x2 – x – 2 = 0⇒x2 – (x + 2) = 0 split the equation y = x2 and y = 2 + x

Steps :

(a) Prepare the table for corresponding values of x and y satisfying the equation y = x2.

x y x y 0 1 1 2 2 0 1 1 4 4 0 0 1 1 1 1 2 4 2 4 − − − − ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) 1

(b) Prepare the table for corresponding values of x and y satisfying the equation y = 2 + x.

x y x y 0 1 2 1 2 2 3 4 1 0 0 2 1 3 2 4 1 1 2 0 − − − − ( , ) ( , ) ( , ) ( , ) ( , ) ( , ) 1

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OSWAAL CBSE (CCE),

Mathematics Class –10

(c) Choose the scale of

x-axis, 1 cm = 1 unit y-axis, 1 cm = 1 unit

(d) Plot the points (0, 0); (1, 1); (– 1, 1); (2, 4) and (– 2, 4) on the graph sheet.

(e) Join the points by a smooth curve.

(f) Plot the points (0, 2); (1, 3), (2, 4); (– 1, 1) and (– 2, 0) on the graph sheet.

(g) Join the points to get a straight line.

(h) From the intersecting points (– 1, 1) and (2, 4) of curve and the line, draw the perpendiculars

to the x-axis.

1

(i) Perpendiculars meet the x-axis at the points A(– 1, 0) and B(2, 0).

∴ Roots of the equation x2– x – 2 = 0 are x = – 1

and x = 2. 1

39. Given :In∆ ABC,∠BAC = 90°

To prove : BC2 = AB2 + AC2 ½ Construction :Draw AD⊥BC

Proof :In∆ ABC and∆DBA,

∠BAC =∠BDA = 90° ∠ ABC =∠ ABD ∴ ∆ ABC|| ∆DBA ½ (AA-similarity criteria) ⇒ = ⇒ BC·BD = AB2 …(i) ½

In∆_{ABC and}∆_DAC ,

∠BAC =∠ ADC = 90°

∠ ACB =∠ ACD

∴ ∆ ABC|| ∆DAC ½

⇒ =

⇒ BC·DC = AC2 …(ii) ½

By adding (i) and (ii), we get BC·BD + BC·DC = AB2 + AC2

⇒ BC(BD + DC) = AB2 + AC2 ½

⇒ BC·BC = AB2 + AC2,

(as BD + DC = BC)

∴ BC2 = AB2 + AC2 ½

Hence, in a right-angled triangle, square of hypotenuse is equal to the sum of the square

of the other two sides. ½

40. C₁ = R = 3 cm, C₂ = r = 2 cm C₃ = R + r = 5 cm

Distance of the centres,

d = (3 + 2 + 3) cm

= 8 cm 1

Steps of construction :

(a) With ‘ A’ as centre, draw circle C₁ with radius 3 cm. With ‘B’ as centre, draw circle C₂ with radius 2 cm. With ‘ A’ as centre draw circle

C₃ with radius 5 cm. ½

1

(b) Construct the tangents BM to circle C₃ from the external point B. Join AM. ½ (c) Let AM intersect circle C₁ at P.

(d) From B, draw AP||BR. Join PR.

∴PR is the transverse common tangent. ½

By measurement, PR = 7 cm. ½

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

SAMPLE

QUESTION PAPER - 8

Self Assessment

___________________________________________

1. (B) A, G, H are in G.P. 1 2. (D) 3 1 3. (D) 1 4. (D) Coefficient of variation 1 5. (C) 0 1 6. (C) 1 7. (B) 12 1 8. (D) x = 2, y = 6 1

II. Solutions of one mark questions 9. We know that

L.C.M. (a, b) × H.C.F. (a, b) = a × b ½ Here a = 36, b = 32 and L.C.M. (36, 32) = 288

∴ H.C.F. (36, 32) =

=

= 4. ½

10. B – A = {1, 2, 3}

∴ (B – A)′= U – (B – A) = {0, 4, 5, 6, 7, 8, 9} ½ 11. Given, product of zeroes = 3

∴ = ½

⇒ 3k – 6 = k ⇒ 2k = 6⇒k = 3. ½ 12. If a straight line is drawn parallel to one side

of a triangle, then it divides the other two sides

proportionally. 1

13. Since M is mid point of AB, then

AM = AB = × 16 = 8 cm and ∠OMA = 90° ½ ∴ OM2 = OA2 – AM2 = (10)2 – (8)2 = 100 – 64 = 36 ⇒ OM = 6 cm. ½

14. T.S.A. of right circular hollow cylinder = 2π(R + r) (h + R – r).

III. Solutions of two marks questions 15. Let is a rational number say r.

∴ = r

⇒ ½

Squaring both sides, we get 2 =

⇒ = ½

∴ =

which is a contradiction as right hand side is a rational number while 5 is an

irrational number. ½

Hence is an irrational number.

½

M

A

T

H

E

M

A

T

I

C

S

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OSWAAL CBSE (CCE),

Mathematics Class –10

16. (i)

1 (ii)

From (i) and (ii), 1

A∩ (B∪C) = ( A∩B)∪ ( A∪C)

17. Two bowlers out of 5 can be chosen in 5C₂ ways. ½ Remaining 9 players out 17 – 5 = 12 can be

chosen in12C₉ ways. ½

∴Total number of ways

=5C₂ ×12C₉ =5C₂ ×12C₃ ½ = = 10 × 220 = 2200. ½ 18. nP₂ = 20nP₁ ⇒ n(n – 1) = 20 × n 1 ⇒ n – 1 = 20 ⇒ n = 21. 1

19. L.C.M. of orders 3 and 4 surds = 12 ½

∴ = _½ and = 4 3× 3 3 1227 = ½ ∴ = = ½ 20. = 5 5 3 3 5 3 5 2 3 2 ( ) ( ) ( ) ( ) + − − − 1 = ½ = = 4 ½ 21. Let p(x) = = = = 1 p(x) is zero, when = 0 or = 0 i.e., when x = or x =

∴ Required zeroes are and · 1

OR

Let p(x) = 4x4 + 2x3 – 3x2 + 8x + 5a

By factor theorem, (x + 2) is a factor of p(x) if

p(– 2) = 0 1 ∴ p(– 2) = 0 ⇒ 4(– 2)4 + 2(– 2)3 – 3(– 2)2 + 8(– 2) + 5a = 0 ⇒ 64 – 16 – 12 – 16 + 5a = 0 ⇒ 20 + 5a = 0 ∴ a = = – 4. 1 22. Here, m = and n = ∴ m + n = = 4 and mn = ( 2 3 2 3+ )( − ) = = 4 – 3 = 1 1 ∴ Required equation is x2 – (m + n)x + mn = 0 x2 – 4x + 1 = 0. 1 23. 4 cos θ = 11 sinθ ⇒ cosθ = sinθ ½ ∴ = ½ = ½ = · ½ 24. d = AB =

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Mathematics Class –10

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Mathematics Class –10

= _½ = BC = = 9 2 +( −1)2 = 81 1 + = 82 ½ CA = = 8 2 ₊82 = ½

Since AB = BC, the triangle is isosceles. ½ 25. Given, OP⊥ AB and OQ⊥CD

⇒OP bisects AB and OQ bisects CD

⇒ AP = AB and CQ = CD ½

Join OA and OC. In∆OPA and∆OQC,

OP = OQ (Given)

OA = OC (Radii of a same circle)

∠P =∠Q (each is a rt. angle)

∴ ∆OPA ≅ ∆OQC (RHS property) ½

⇒ AP = CQ (C.P.C.T.) ½ ⇒ AB = CD⇒ AB = CD. ½ 26. Given, Scale 20 m = 1 cm AD = 140 metre = = 7 cm AP = 50 metre = = 2·5 cm QE = 80 metre = = 4 cm PB = 40 metre = = 2 cm AQ = 100 metre = = 5 cm AR = 120 metre = = 6 cm RC = 60 metre = = 3 cm 1 27. Volume of hemispherical tank

= m3

= m3 1

Volume to be emptied = m3 = 99000

28 litres ½

∴ Required time = seconds

= 990 seconds

= 16·5 minutes ½ OR

Diameter of hemisphere

= Side of cubical box

⇒ 2R = 7 cm

∴ R = cm ½

S.A. of solid = S.A. of the cube – area of base of hemisphere + CSA of

hemisphere ½

= 6l2 –πR2 + 2πR2

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OSWAAL CBSE (CCE),

Mathematics Class –10

= 6l2 +πR2 ½

= 6 × 49 +

= 332·5 cm2. ½

28. A network is traversable if it contains :

(i) two odd nodes and any number of even

nodes, or 1

(ii) all even nodes. 1

29. (i) In∆ ABC and∆ AMP,

∠ A =∠ A (Common)

∠ ABC =∠ AMP= 90° (Given)

∴ by AA-criteria,

∆ ABC ~∆ AMP 1 (ii) Since∆ ABC ~∆ AMP (Proved)

∴ Corresponding sides are proportional.

⇒ = 1

30. The total number of mangoes after mixing = 9 + 30 = 39

∴ n(S) = 39 ½

(a) An event of selecting a good mango, n( A) = 30

P( A) = ½

(b) An event of selecting a rotten mango, n(B) = 9

∴ P(B) = _· ₁

IV. Solutions of three marks questions 31. In∆ ABC, let

∠ A = 30°

Since∠ A,∠B and∠C are in A.P.

∴ ∠B = 30° + d and∠C = 30° + 2d ½ But∠ A +∠B +∠C = 180° ½ ⇒ 30° + 30° + d + 30° + 2d = 180° ½ ⇒ 3d = 90° ⇒ d = 30° ½ ∴ ∠B = 30° + d = 30° + 30° = 60° and ∠C = 30° + 2d = 30° + 2 × 30° = 90° ½

∴ ∆ ABC is a right angled triangle. ½ 32. For English : Given,

= 56 andσ_{= 5·75}

C.V. = × 100

= × 100

= 10·27 1

C.V. for English = 10·27

Similarly, C.V. for mathematics

= × 100

= 8·56 ½

and C.V. for science

= × 100

= 9·68 ½

On comparison, the co-efficient of variation (C.V.) in mathematics is the least.

Hence, the performance in Mathematics is more

consistent. 1

33. Let the number be x. According to the question,

3x2 – 4x = 15 ⇒ 3x2 – 4x – 15 = 0 1 ⇒ 3x2 – 9x + 5x – 15 = 0 ⇒ 3x(x – 3) + 5(x – 3) = 0 ⇒ (x – 3)(3x + 5) = 0 1 ⇒ x – 3 = 0 or 3x + 5 = 0 ⇒ x = 3 or x = (not possible)

Hence, the whole number = 3. 1

OR

Here, a = 1, b = 1, c = – (a + 2)(a + 1)

x = ½

=

= ½

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

= −1 4 12 9 2 2 ± a + a+ = ½ = ½ ∴ x = − 1 2 3 2 + ( a+ ) = = a + 1 ½ and x = = = – (a + 1) ½

34. Given : ABC is a triangle in which DE||BC.

To prove : = ½

Construction :Draw DN ⊥ AE and EM⊥ AD.

Join BE and CD. ½ Proof : = = …(i) ½ and area area ( ) ( ) ∆ ∆ ADE DEB = = …(ii) ½

∆DEB and∆DEC lies on the same base DE and between the same parallel lines DE and BC. ½

∴ area (∆DEB) = area (∆DEC) ½ Hence from (i) and (ii), we have

= ½ OR In∆PQR,∠PQR = 90° and QP⊥PR ∴ PQ2 = PR × PD (corollary) But PD = 4DR (given) ∴ PQ2 = PR × 4DR ⇒ PR = PQ DR 2 4 …(i) 1 Again, QR2 = PR × DR (corollary) ⇒ PR = …(ii) 1

From equations (i) and (ii), we get =

⇒ PQ2 = 4QR2

⇒ PQ = 2QR. Proved. 1

35. Let BD is the height of the tree. It is broken at A such that

AD = AC = x, say

In∆ ABC, ∠ ACB = 60°,∠B = 90° ∴ tan 60° = ⇒ AB = m 1 By Pythagoras theorem, AC2 = AB2 + BC2 x2 = = 1200 + 400

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OSWAAL CBSE (CCE),

Mathematics Class –10

= 1600

⇒ x = 40 1

Height of the tree

= BD = AB + x =

= m. 1

OR L.H.S. =

= tan _tansec (sec _sec tan )

θ θ θ θ θ θ + + − − − 2 2 1 [·.· 1 = sec2θ – tan2θ] 1 = (tan sec ) [(sec tan ) (sec tan ) ]

tan sec θ θ θ θ θ θ θ θ + + + − − − ₁ ½ = (tanθ + sec θ) ½ = (tanθ + sec θ) × ½ = = R.H.S. ½

36. Given :Circles with centres X and Y touch each other externally at P.

Construction : Join AP, BP, PC and PD. ½ Proof :Since AB is the diameter.

i.e., ∠ APB = 90° (Angle in a semi-circle) …(i) ½

Again, CD is the diameter.

∴ ∠CPD = 90°

(Angle in a semi-circle) …(i) ½ From (i) and (ii),

∠ APB =∠CPD = 90° ½

⇒ Vertically opposite angles are equal.

⇒ APD and BPC are two intersecting straight

lines. ½

⇒BPC is a straight line.

Hence, B, P, C are collinear. ½

OR Given, AB = 10 cm

and AO = OB = OQ (Radii of a circle)

= AB = × 10 = 5 cm ½ ∴ OC = AC – OA = 6 – 5 = 1 cm

Let PC = PQ = x (Radii of a circle)

∴ OP = OQ – PQ

= (5 – x) cm ½

Since OB is a tangent at C to a smaller circle.

∴ PC ⊥CO

In∆PCO, by Pythagoras theorem, OP2 = OC2 + PC2 (5 – x)2 = (1)2 + x2 1 25 + x2 – 10x = 1 + x2 x2 – 10x – x2 = 1 – 25 x = = 2·4

∴ Radius of the smaller circle = x = 2·4 cm. 1 V. Solutions of four marks questions

37. Given, b = G.M. of a and c = …(i) ½ x = A.M. of a and b = …(ii) ½ y = A.M. of b and c = …(iii) ½

From (i), b2 = ac⇒c = ½

Now, ₌ _½

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

= ½ = ½ = ½ OR

Let the three terms of G.P. be a, ar. ½ Given, × a × ar = 216

⇒ a3 = (6)3 ⇒ a = 6 ½ and = 156 ½ ⇒ a2 = 156 ⇒ 36 = 156 ⇒ r2 + r + 1 = r = r ⇒ 3r2 + 3r + 3 – 13r = 0 ⇒ 3r2 – 10r + 3 = 0 1 ⇒ 3r2 – 9r – r + 3 = 0 ⇒ 3r(r – 3) – 1(r – 3) = 0 ⇒ (r – 3) (3r – 1) = 0 ⇒ r = 3 or r = 1

∴ Required terms are 6, 6 × 3 i.e., 2, 6, 18. ½ 38. y = x2 + x – 6

The values of y for corresponding values of x are given in the following table :

x 0 1 2 –1 –2 –3

y – 6 – 4 0 – 6 – 4 0

1

The curve is a parabola which intersect the x-axis at the points A(– 3, 0) and B(2, 0).

∴ The roots of equation x2 + x – 6 = 0 are

x = – 3 and x = 2. 1

39. Let AD = x

then CA = 2 AD = 2x

and BD = 3 AD = 3x ½

In∆ ADC, ∠ ADC = 90°

So, by Pythagoras theorem AC2 = AD2 + CD2 ⇒ (2x)2 = x2 + CD2 ⇒ CD2 = (2x)2 – x2 = 4x2 – x2 = 3x2 ⇒ CD = 3 _x2 ₊_x 3 ₁ In∆BDC, ∠D = 90° By Pythagoras theorem, BC2 = BD2 + CD2 = = 9x2 + 3x2 = 12x2 BC = ₁

Also, AB = AD + DB

= x + 3x = 4x

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OSWAAL CBSE (CCE),

Mathematics Class –10

Now, AC2 + BC2 =

= 4x2 + 12x2 = 16x2

= AB2 1

∴ By converse of Pythagoras theorem.,

∆ ABC is right angle triangle, right angled at C.

Hence, ∠BCA = 90°. Proved. ½

40. Distance between the centres, d = 2 cm C₁ = R = 5 cm C₂ = r = 3 cm C₃ = R – r = 1·5 cm 1 Steps of construction :

(a) With ‘ A’ as centre draw circle C₁. With ‘B’ as centre draw circle C₂ and with ‘ A’ as centre,

draw circle C₃. 1

1 (b) Construct the tangents BN and BM to circle

C₃ from the external point B. Join AN and AM.

(c) Let AN produce meet the circle C₁ at P and

AM produce at R. 1

(d) With centres P and R and radius PB draw two arcs to intersect C₂ at Q and S. Join PQ and RS.

∴PQ and RS are the direct common tangents. 1

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

SAMPLE

QUESTION PAPER -9

Self Assessment

___________________________________________

1. (D) 15 1

2. (A) 24 1

3. (C) Determining the boiling point of water 1

4. (A) are equal to one another 1

5. (D) (B) and (C) are correct 1

6. (D) 2 1

7. (A) 14 1

8. (C) 1

II. Solution of one mark questions

9. = ½

=

which is a irrational number. ½

10. A = {1, 2, 3} (·.· x∈N )

and B = {3, 1}

(·.· 3 y = 9⇒ y = 3 and 2 y = 2⇒ y = 1) ½

∴ A∪B = {1, 2, 3}∪ {1, 3} = {1, 2, 3} = A ½ 11. Dividend = Divisor × Quotient + Remainder 1

12. BD2 = AD × CD ½ = 16 × 4 = 64 ∴ BD = 4 cm ½ 13. ·.· OQ ⊥ AB ⇒ ∠OQB = 90° ½ In∆OPQ, OP = OQ ⇒ ∠OPQ =∠OQP = 90° – 65° = 25° ½ 14. Volume =πr2h ½ = × 7 × 7 × 10 = 1540 cm3. ½

III. Solutions of two marks questions 15. 344 = 2 × 2 × 2 × 43 = 23 × 43 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5 ½ ∴ H.C.F. = (344, 60) = 22 = 4 ½

But L.C.M. (a, b) × H.C.F. (a, b) = a × b ½ ∴ L.C.M. (344, 60) = = = 5160. ½ 16. 1

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Mathematics Class –10

21. Given, p(x) = x3 – 1, g (x) = x – 1

)

x x x x x x x x x x x − − − − − − − − − − 1 1 1 1 1 1 0 3 2 3 2 2 2 + + + + + ( ) ( ) ( ) ( ) ( ) ( ) ∴ Quotient, q(x) = x2 + x + 1 and remainder, r(x) = 0 1

Verification :We know that p(x) = g (x)·q(x) + r(x) Now, g (x)·q(x) + r(x)

= (x – 1)(x2 + x + 1) + 0 = x3 + x2 + x – x2 – x – 1 = x3 – 1 = p(x)

Thus, Division Algorithm is verified. 1 OR

Let p(x) = x3 – 3x2 + ax – 10

By factor theorem, if (x – 5) is a factor of x3– 3x2 + ax – 10, then p(5) = 0. ∴ p(5) = 0 ⇒ (5)3 – 3(5)2 + a(5) – 10 = 0 1 ⇒ 125 – 75 + 5a – 10 = 0 ⇒ 40 + 5a = 0 ∴ a = = – 8 1 22. Given, B = ⇒ a2 = ∴ a = ± = ± 2 1 Now, if B = then a = ± 2 = ± 2 = ± 2 × 4 = ± 8. 1 n( A) = 35, n(B) = 20 + 10 = 30 n( A) only = 35 – 10 = 25 ∴ n( A∪B) = 25 + 10 + 20 = 55 ½ 17. Teachers = 6, Doctors = 4 No. of Committees :

Possibilities Teachers Doctors No. of ways

(i) 6 4 15 4 2 6 15 6 C ₌ C ₌ _× ₌₉₉₀ 20 4 20 4 80 15 1 15 1 15 185 6 3 4 3 6 2 4 4 (ii) (iii) Total C C C C = = × = = = × = Total = 185 committees 2 18. (n + 1) ! = 12 × (n – 1) ! ⇒ (n + 1)n × (n – 1) ! = 12 × (n – 1) ! ½ ⇒ (n + 1)n = 12, as (n – 1) ! ≠_{0 ½} ⇒ (n + 1)n = 4 × 3 = (3 + 1) = 3 ½ ∴ n = 3 ½

19. Let a = 21/3 and b = 2 – 1/3, then

a3 = 2 and b3 = 2 – 1 = ½ ∴ a3 + b3 = 2 + = ⇒ (a + b) (a2 + b2 – ab) = ½ ⇒ (21/3 + 2 – 1/3) [(21/3)2 + (2 – 1/3)2 – 21/3 × 2 – 1/3)] = ⇒ (21/3 + 2 – 1/3) (22/3 + 2 – 2/3 – 1) = ½

Since is a rational number, so R.F. of (21/3 + 2 – 1/3) is (22/3 + 2 – 2/3 – 1). ½ 20. (3 2 2 3 2 3 4 2+ )( − ) = 3 2 2 3 4 2 2 3 2 3 4 2( − )+ ( − ) 1 = 6 6 12 2 4 3 8 6₋ _× ₊ _× ₋ = = 1

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Mathematics Class –10

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OSWAAL CBSE (CCE),

Mathematics Class –10

23. = ½ = ½ = ½ = tan A cot B ½

24. Let the required ratio be m : n. Given, (x₁, y₁) = (– 3, 10), (x₂, y₂) = (6, – 8) and (x, y) = (– 1, k ) ½ By section formula, x = ½ ⇒ – 1 = ½ ⇒ – m – n = 6m – 3n ⇒ – m – 6m = – 3n + n ⇒ – 7m = – 2n ⇒ = ∴ m : n = 2 : 7. ½ 25. From construction, ½ ∠ ACB = 90° =∠ ADB ½ ∠ AEB = 90° =∠ AFB ½

Hence, angles in semi-circle are right angle.½

26. Scale 20 m = 1 cm AP = 50 m = = 2·5 cm PB = 100 m = = 5 cm AQ = 80 m = = 4 cm 1 QE = 60 m = = 3 cm AR = 120 m = = 6 cm RC = 70 m = = 3·5 cm AD = 200 m = = 10 cm 1

27. Curved surface area of frustum of a cone

=πl(r₁ + r₂) 1

= × 5(10 + 4) ½

= × 5 × 14

= 220 cm2. ½

OR

Given, for melted cone, h = 3·6 cm, r = 1·6 cm For recast cone,

R = 1·2 cm, H = ? ½

According to the question,

Vol. of melted cone = Vol. of recasted cone

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OSWAAL CBSE (CCE),

Mathematics Class –10

⇒ πr2h = πR2H ½ ⇒ (1·6)2 × (3·6) = (1·2)2 × H ½ ⇒ H = = = _½ = 6·4 cm

Hence, height of recasted cone = 6·4 cm.

28. N = 8, A = 8, R = 6 1

Euler’s formula for a graph

N + R = A + 2 ½

4 + 6 = 8 + 2

10 = 10, which is true ½ Hence, Euler’s formula is verified by given network.

29. In∆ AEB and∆CED,

∠ ABE =∠EDC (Given)

∠ AEB =∠CED (V.O.A.)

∴ by AA-criterion,

∆ AEB ~∆CED ½

⇒ = ½

⇒ = (Given, CD = 4 AB)

⇒ DE = = 4BE ½

Now, BD = BE + ED = BE + 4BE

∴ BD = 5BE ½

30. Here, n(S) = 36

(i) Let A be the event of getting an even sum.

∴ A = {(1, 1), (1, 3), (1, 5), (2, 2) (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

⇒ n( A) = 18

∴ P( A) = , 1

(ii) Let B be the event of getting a total of 7.

∴ B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

⇒ n(B) = 6

∴ P(B) = , 1

IV. Solutions of three marks questions 31. Let 3 terms of G.P. are a, ar.

∴ + a + ar = 7 …(i)

and × a × ar = 8 …(ii) ½

(ii)⇒ a3 = 23⇒a = 2 ½ From (i), + 2 + 2r = 7 ½ ⇒ 2 = 7 – 2 ⇒ 2 = 5 ⇒ 2r2 – 5r + 2 = 0 ⇒ (r – 2) (2r – 1) = 0 ⇒ r = 2 or r = ½

When r = 2, three terms of G.P. are

2, 2 × 2 i.e., 1, 2, 4 ½ When r = , three terms of G.P. are

2 1 2/ , 2, 2 × i.e., 4, 2, 1. ½ 32. Scores x d x x d x d = = = − − − − 2 2 40 8 64 36 12 144 64 16 256 48 0 0 52 4 16 240 48 Σ Σ 00 1

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OSWAAL CBSE (CCE),

Mathematics Class –10

Here n = 5 Mean = ½ σ = 1 ∴ C.V. = σ x ×100= × = 9 79 48 100 20 39 · · _½

33. Let time taken by tap of larger diameter = x hrs. and time taken by tap of smaller diameter = (x + 2) hrs.

Since both the water taps together can fill a tank in 2 hrs. i.e., in hrs.

∴ In 1 hr. both fill the portion of tank =

⇒ = 1 = ⇒ 35(2x + 2) = 12x(x + 2) ⇒ 12x2 – 46x – 70 = 0 ⇒ 6x2 – 23x – 35 = 0 or 6x2 – 30x + 7x – 35 = 0 or 6x(x – 5) + 7(x – 5) = 0 ⇒ (6x + 7) (x – 5) = 0 1 ⇒ x = or x = 5 ⇒ Rejecting x =

(because time cannot be negative)

∴ x = 5 hrs.

Hence, smaller tap can fill the tank in 5 + 2 = 7

hrs and larger tap in 5 hrs. 1

OR

Let the age of son = x years and the age of father = y years

Condition I : y = 2x2 …(i) ½ After 8 years,

son’s age = x + 8 and father’s age = y + 8 Condition II :

⇒ y + 8 = 3(x + 8) + 4

⇒ y = 3x + 24 + 4 – 8

⇒ y = 3x + 20 …(ii) ½

2x2 = 3x + 20 ½ ⇒ 2x2 – 3x – 20 = 0 ⇒ 2x2 – 8x + 5x – 20 = 0 ½ ⇒ 2x(x – 4) + 5(x – 4) = 0 ⇒ (x – 4) (2x + 5) = 0 ⇒ x = 4 or x = (not possible) ∴ x = 4 ½

Hence, age of son = 4 years

and age of father, y = 2x2 = 2(4)2 = 32 years. ½ 34. Given :In∆ ABC in which P and Q are points on AB and AC respectively such that PQ|| BC and AD is the median, cutting BC at E.

To prove : PE = EQ 1 Proof :In∆ ABD, PE ||BD (... PQ||BC, given) ∴ = …(i) 1 (Thales theorem) In∆ ADC, EQ ||DC (·.· PQ||BC, given)

∴ By Thales theorem,

= …(ii)

From (i) and (ii), we have =

But BD = DC (·.· AD is the median)

∴ PE = EQ. 1

OR

Given :∆ ABC and∆BDC are on the same base BC.

To prove : ₌ ½

Construction :Draw AE⊥BC and DF⊥BC ½

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OSWAAL CBSE (CCE),

Mathematics Class –10

Proof :In∆ AOE and∆DOF

∠ AOE =∠DOF (Vertically opposite angles)

∠ AEO =∠DFO = 90° (Construction)

∴ ∆ AOE ~∆DOF (AA-Criteria)

⇒ = 1

Now, = ½

= _· Proved. ½

35. Let height of aeroplane, PC = h km

Distance between two stones A and B = 1 km Let AC = x km⇒BC = (1 – x) km In∆PCA, tanα = ⇒ x = 1 In∆PCB, tanβ = ⇒ 1 – x = ₁ ⇒ = ⇒ 1 = ⇒ 1 = h ⇒ h = 1 OR L.H.S. = = ½ = ½

= tan 2 _θ₊cot 2_θ₊₂ tan cot_θ _θ

as tan a cotθ = 1 1

= ½

= tanθ + cotθ = R.H.S. ½ 36. Since tangents from an exterior point to a circle

are equal in length.

∴ BP = BQ (tangents drawn from B) …(i) CP = CR (tangents drawn from C) …(ii) and AQ = AR (tangents drawn from A)…(iii)1

From (iii), we have AQ = AR

⇒ AB + BQ = AC + CR

⇒ AB + BP = AC + CP (iv) 1

Now,

Perimeter of∆ ABC

= AB + BC + AC

= AB + (BP + PC) + AC = ( AB + BP) + ( AC + PC)

= 2( AB + BP), using (iv) = 2( AB + BQ), using (i) = 2 AQ

∴ AQ = (Perimeter of∆_ABC) ₁

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Mathematics Class –10

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Mathematics Class –10

OR

Three circles with centres A, B and C touch each other externally.

Let r₁, r₂ and r₃ are the radius of circles C₁, C₂

and C₃ respectively. ½ Let AB = 7 cm ⇒ r₁ + r₂ = 7 …(i) BC = 8 cm ⇒ r₂ + r₃ = 8 …(ii) and CA = 9 ⇒ r₃ + r₁ = 9 …(iii) 1

Adding all there, we get

2(r₁ + r₂ + r₃) = 7 + 8 + 9 = 24

⇒ r₁ + r₂ + r₃ = 12 …(iv) ½ Now apply [(iv) – (i)], we get

⇒ r₃ = 12 – 7

= 5 cm Apply [(iv) – (ii)], we get

⇒ r₁ = 12 – 8

= 4 cm Apply [(iv) – (iii)], we get

⇒ r₂ = 12 – 9

= 3 cm

So, radii of the circles are 5 cm, 4 cm and 3 cm. 1 V. Solutions of four marks questions

37. T _n = a + (n – 1)d = …(i) ½

T _m = a + (m – 1)d = …(ii) ½

Apply [(i) – (ii)], we get

(n – m)d = =

⇒ α = 1

(i)⇒a + (n – 1) =

⇒ a = =

= = 1

∴ Sum of first mn terms

= ½

=

= (mn + 1). ½

OR (i) Here, Arithmetic series is

1800 + 1750 + 1700 + … a = 1800, d = T ₂ – T ₁ = 1750 – 1800 = – 50, n = 12, S₁₂ = ? ½ S_n = [2a + (n – 1)d] ½ ∴ S₁₂ = [2 × 1800 + (12 – 1)(– 50)] ½ = 6[3600 + 11(– 50)] = 6[3600 – 550] = 6 × 3050 = 18300 1

∴ Total amount paid in 12 instalments =`_18,300.

(ii) Extra amount paid

=`_{18,300 –} `_{15,000 =}`_3,300 ₁

38. y = – x2 + 8x – 16

The values of y for corresponding values of x are given in the following table :

1

2

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OSWAAL CBSE (CCE),

Mathematics Class –10

The curve i.e., parabola intersects the x-axis at only one point A(4, 0).

∴ The roots of the equation are 4 and 4. 1 39. In right∆ ABD, by Pythagoras theorem

AB2 = AD2 + BD2 …(i) ½ In right∆ ADC, by Pythagoras theorem

AC2 = AD2 + DC2 …(ii) ½

Subtracting (ii) from (i), we get

∴ AB2 – AC2 = BD2 – DC2 ½ = – DC2, Given, BD = CD AB2 – AC2 = ½ Since = ½ ⇒ = ⇒ CD = BC ½ ∴ AB2 – AC2 = ½ =

Thus 2 AB2 + BC2 = 2 AC2. ½ 40. Steps of construction :

(a) C₁ and C₂ are two concentric circles with common centre O and radii 2 cm and 4 cm

respectively. ½

(b) Take point P from O at a distance 8 cm such

that OP = 8 cm. ½

(c) M is mid-point of OP. ½

(d) With M as centre and OM as radius, draw a circle to cut C₁ and C₂ at A, B, C and D respectively. Join PA, PB, PC and PD. 1½ (e) PA and PB are tangents to C₁. ½ (f) PC, PD and tangents to C₂. ½

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Mathematics Class –10

SAMPLE

QUESTION PAPER -10

Self Assessment

___________________________________________

∴ C.S.A. of a cone

=πrl ½

= × 7 × 10

= 220 cm2. ½

III. Solutions of two marks questions 15. 3825 = 3 × 3 × 5 × 5 × 17

= 32 × 52 × 17. 1

1 16. To prove,

A∪ (B∩C) = ( A∪B)∩ ( A∩C) B∩C = {2, 4, 6, 8, 10} ∩ {1, 3, 5, 7, 9, 11}

= { }

A∪ (B∩C) = {1, 2, 3, 5, 7, 11} ∪ { } = {1, 2, 3, 5, 7, 11} …(i) 1 Again ( A∪B)∩ ( A∪C)

= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11}

∩ {1, 2, 3, 5, 7, 9, 11} = {1, 2, 3, 5, 7, 11} …(ii) From (i) and (ii), we have

A∪ (B∩C) = ( A∪B)∩ ( A∪C) 1 I. Solutions of Multiple Choice Questions

1. (C) 200 1 2. (C) 1 3. (B) 40% 1 4. (A) × 100 1 5. (B) – 4 1 6. (D) cot 0° 1 7. (C) y = 3x 1 8. (D) None of these 1

II. Solutions of one mark questions 9. We know that

first number × second number = HCF × LCM ½

∴ HCF × LCM = 50 × 95 = 4750 ½

10. n( A′) = n(U) – n( A) ½

= 28 – 12

= 16. ½

11. The graph of f (x) intersect the x-axis at two

points. ½

∴ Number of real zeroes of f (x) = 2. ½ 12. If a straight line divides two sides of a triangle

proportionally, then the straight line is parallel

to the third side. 1

13. Since AB ||PT and∠PAB = 60° ½

∴ ∠ APT =∠PAB = 60° ½ 14. r = = 7 cm, l = 10 cm

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17. nC_r = 1 n_C n –r = = =nC_r 1 18. 9P₃ + 3 ×9P₂ = ½ = ½ = = ½ =10P₃. ½ 19. = = = 1 = = = 1 20. R.F. of denominator ( 3 6+ )_is ₍₆− ₃₎ ∴ ₌ ½ = ½ = _½ = = ½ 21. Let p(x) = ax3 + 3x2 – 13 and g (x) = 2x3 – 4x + a

By remainder theorem, the two remainders are p(3) and g (3).

By the given condition, p(3) = g (3) ⇒ a(3)3 + 3(3)2 – 13 = 2(3)3 – 4(3) + a 1 ⇒ 27a + 27 – 13 = 54 – 12 + a ⇒ 27a + 14 = 42 + a ⇒ 26a = 28 ∴ a = · 1 OR Let p(x) = xn – 1

In order to show that (x – 1) is a factor of (xn – 1), it is sufficient to show that p(1) = 0 1

∴ p(1) = 1n – 1 = 1 – 1 = 0

∴ (x – 1) is a factor of ( xn – 1). 1 22. Here a = 1, b = p, c = q

Let one root be m, then other root is 3 m. ½

∴ sum of roots = ⇒ m + 3m = p 1 ⇒ 4m = – p ⇒ m = – p 4 ½

and product of roots =

⇒ m × 3m = ½ ⇒ = q ∴ 3p2 = 16q. ½ 23. cosθ = ½ By Pythagoras theorem, AB2 = AC2 – BC2 = (13)2 – (5)2 = 169 – 25 = 144 ∴ AB = 12. ½

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Now, sinθ = ½ and tanθ = ½ 24. AD is median. ∴D is mid-point of BC. ½ Co-ordinates of D = = (– 2, 3) ½ Now, AD = ½ =

∴ Length of median through A = units. ½

25. ∠ ADB =∠ ACB = 70° ½

(Angles in the same segment of a circle)

In∆ ABD,∠DAB +∠DBA +∠ ADB = 180° ½

⇒ 60° +∠DBA + 70° = 180° ½ ⇒ ∠DBA = 180° – (60° + 70°) = 180° – 130° = 50° ½ 26. Given, Scale 25 m = 1 cm AF = 25 m = = 1 cm FB = 50 m = = 2 cm AH = 50 m = = 2 cm HE = 50 m = = 3 cm AG = 75 m = = 3 cm GC = 75 m = = 3 m AD = 100 m = = 4 cm. 1 27. Given, h = 20 cm External radius R = 12·5 cm Internal radius r = 11·5 cm TSA of the pipe

= 2πh(R + r) + 2π(R2 – r2) 1 = 2π(R + r) (h + R – r) = 2 × (12·5 + 11·5) (20 + 12·5 – 11·5)½ = × 44 × 24 × 21 = 3168 cm2. ½ OR

For cone, r₁ = 5 cm and h = 20 cm Let r₂ be the radius of sphere, then

Vol. of the recasted sphere

= Vol. of the metallic cone

πr₂3 = πr₁2h 1

⇒ 4r₂3 = r₁2h

⇒ r₂3 =

∴ r₂ = 5 cm. 1

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OSWAAL CBSE (CCE),

Mathematics Class –10

28. For hexahedron,

F = 6, V = 8, E = 12 1

Euler’s formula for solid

F + V = E + 2 ½

6 + 8 = 12 + 2

⇒ 14 = 14, which is true ½

Hence, Euler formula is verified. 29.

In ∆KLM and ∆ XYZ, corresponding vertices are

M↔Z, L↔Y , K ↔ X 1 Corresponding sides are

KL↔ XY , KM↔ XZ, ML↔ZY ½ Corresponding ratios are

= · ½

30. There are 64 squares in a chess out of which 3 are to be selected, so

n(S) =64C₃ =

= 64 × 21 × 31 ½

On a chess board, there will be 32 black and 32 white squares. Let A be the event of selecting 2 black and 1 white or 1 black and 2 white,

n( A) =32C₂ ×32C₁ +32C₁ ×32C₂ = 2 ×32C₂ ×32C₁

= 2 × × 32

= 32 × 31 × 32 ½

∴ P( A) = · 1

IV. Solutions of three marks questions 31. Here, n = 25, which is odd

So, middle term = term

= term = 13th term ½ ∴ T ₁₃ = 12 (Given) ⇒ a + (13 – 1)d = 12 ⇒ a + 12d = 20 …(i) 1 ∴ Sum, S₂₅ = [2a + (25 – 1)d] ½ = [2a + 24d] = 25[a + 12d] ½ = 25[20] = 500 ½

32. (a) The yield of rice

= × 8100

= 900 tons ½

The yield of ragi = × 8100

= 2250 tons ½

The yield sugarcane

= × 8100

= 1800 tons

The yield of others = × 8100 ½

= 3150 tons ½

(b) From pie-chart increase in degree of ragi over rice

= 100° – 40° = 60° ½

∴ % increase = × 100

= 16·66% ½

33. Let the three consecutive numbers be x, x + 1, x + 2. As per given condition,

x2 + (x + 1)2 + (x + 2)2 = 149 ½ ⇒ x2 + x2 + 2x + 1 + x2 + 4x + 4 = 149 ½ ⇒ 3x2 + 6x + 5 – 149 = 0 ⇒ 3x2 + 6x – 144 = 0 ⇒ x2 + 2x – 48 = 0 ½ ⇒ x2 + 8x – 6x – 48 = 0 ⇒ x(x + 8) – 6(x + 8) = 0 ½

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Mathematics Class –10

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Mathematics Class –10

⇒ (x + 8) (x – 6) = 0 ⇒ x = – 8 or x = 6 ½ Since x ≠ – 8 ∴ x = 6 x + 1 = 7, x + 2 = 8 ½ Hence, the three numbers are 6, 7, 8.

OR

Let altitude of the triangle be x cm, then it’s

base is (x + 4) cm. ½

Area of triangle = × base × altitude ½

⇒ 48 = × (x + 4) × x ½ ⇒ 96 = x2 + 4x ⇒ x2 + 4x – 96 = 0 ⇒ x2 + 12x – 8x – 96 = 0 ½ ⇒x(x + 12) – 8(x + 12) = 0 ⇒ (x + 12) (x – 8) = 0 ½ ⇒ x = 8 or x = – 12 (not possible) ∴ Altitude = 8 cm. ½

34. Proof :In∆ ABC, D, E and F are the mid-points of AB, BC and AC respectively.

FE joins the mid-points of AC and BC.

∴FE|| AB and FE = AB

(∴ Mid-point theorem) Similarly, DF||BC and DF = BC

and DE|| AC and DE = AC 1

In∆ ABC and∆EFD,

∠ ABC =∠EFD

(·.· opp. angles of paralellogram BEFD)

∠BCA =∠FDE

(·.· opp. angles of parallelogram ECFD)

∴ by AA-criteria,

∆ ABC ~∆EFD 1

= =

Hence, Area of ∆DEF : Area of∆ ABC = 1 : 4. 1 OR

In∆ ADE, EF ||CD

By B.P.T., = …(i) 1

In∆ ABC, DE ||BC

By B.P.T., = …(ii) 1

From equations (i) and (ii), we have =

⇒ AD2 = AF × AB. 1

35. Let the distance of the cloud P from the observation S is y.

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OSWAAL CBSE (CCE),

Mathematics Class –10

Height of cloud above lake = PT = H

R be the reflection of P, then TR = H ½ In∆PQS, tanα =

⇒ H – h = x tanα …(i) ½

In∆SQR, tanβ =

⇒ H + h = x tanβ …(ii) ½

Apply [(ii) – (i)], we get

2h = x tanβ – x tanα ½ ∴ x = Now in∆PQS, secα = ½ ⇒ y = x secα = · ½ OR

Given, sinθ – cosθ =

⇒ (sinθ – cosθ)2 =

⇒ sin2θ + cos2θ – 2 sinθ cosθ = ½

⇒ 1 – 2 sinθ cosθ =

½

⇒ 2 sinθ cosθ = ½

Now, (sin θ + cosθ)2

= sin2θ + cos2θ + 2 sinθ cosθ ½

= 1 + 2 sinθ cosθ ½ = 1 + 3 4 7 4 = _½ 36. Let ∠BPQ = x°

In∆PBQ, BP = BQ (Radii of a same circle)

⇒ ∠BQP =∠BPQ(Angles opposite to equal sides of a triangle)

∴ ∠BQP = x° …(i) 1

Similarly, in∆PAR AP = AR

⇒ ∠ ARP =∠ APR

∴ ∠ ARP = x° …(ii) 1

∠BQP =∠ ARP

⇒ Corresponding angles are equal. ½

∴ AR ||BQ. Hence proved. ½

OR

Given, AB ||CD

∴ ∠ ABC =∠BCD (Alternate angles) ½ But, again ∠ ABC = 55°

∴ ∠BCD = 55° ½

Again ∠BOD = 2∠BCD

(central angle is twice the inscribed angle)

∴ ∠BOD = 2 × 55° = 110° 1

Since the tangents at B and D intersect each other at point P.

∴ ∠BOD +∠BPD = 180° ½

⇒ 110° +∠BPD = 180°

∴ ∠BPD = 180° – 110° = 70°. ½

V. Solutions of four marks questions

37. Let the common ratio of G.P. is r and first term is a, then G.P. is

a, ar, ar2, ar3, ar4, … According to first condition,

a × ar × ar2 × ar3 × ar4 = 1

⇒ a5r10 = 1

⇒ ar2 = 1 …(i) 1

36 |

OSWAAL CBSE (CCE),