SET A

Class XMathematics

Marking Scheme 2018-19

Time allowed: 3 hours Max. Marks: 80

Q.No. Section-A Marks

1. √ = 5 ½(4-1)2 + (k-0)2

9 + k2 = 25⟹ k2 = 16⟹ k = ± 4 ½

2. D = b2 – 4ac = 42 – 4 x 2 x (-7) = 16 + 56 = 72 > 0 ½Hence roots are real and unequal. ½[OR]Putting x = -2 in the quadratic equation,LHS = 3x2 + 13x + 14

1= 3 (-2)2 + 13 (-2) + 14= 12 – 26 + 14 = 0 = RHS

Hence x = -2 is a solution.3. 2cot2θ + 2 = 2 (cot2θ + 1) = 2cosec2θ = 2/sin2θ ½

= 2 / (⅓)2 = 2 x 9 = 18 ½[OR]

cot B = cot (90o-A) (∵A+B = 90o) ½= tan A = ¾ (∵cot(90-θ) = tanθ) ½

4. a18 – a14 = 32 ⟹ (a+17d) – (a+13d) = 32 ½⟹ 17d – 13d = 32⟹ d = 32/4 = 8 ½

SET A

5. ΔABC ~ ΔDEF ⟹ AB = BC ½DE EF⟹ AB = 8 (∵DE = 2AB)

2AB EF⟹ 1 = 82 EF∴EF = 16 cm ½

6. 619 = 619 ½325 52 x 13The denominator is not of the form 2m x 5n, where m and n are non-negativeintegers.

Hence 619 will have a non-terminating repeating decimal expansion.½325

Section-B

7. 1001 = 385 x 2 + 231

385= 231 x 1 + 154

231= 154 x 1 + 77

154= 77 x 2 + 0

As remainder is zero and divisor is 77, HCF = 77

[OR]Let us assume, to the contrary that 5 - √3 is irrational.

5 - √3 = p , where p & q are co-prime and q≠0 q

√3 = 5q - pq

5q - p is rational = √3 also becomes rational which is a contradiction.q

Hence 5 – √3 is irrational.

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SET A

8. a = 18, d = -2, Sn = 0

n [36 + (n-1)(-2)] = 0 12⟹ n(36 - 2n + 2) = 0⟹ n(38 - 2n) = 0 ⟹ n = 19 1

[OR]nth term from the end = l – (n-1)d ½7th term from the end = 213 – (7-1)(2) 1

= 213 – 12 = 201 ½

9. Let the ratio be k:1

-3 = -2k -5 ½k+1

-3k – 3 = -2k – 5⟹ k = 2 ∵Ratio is 2:1

½P = 2 x 3 – 4 = 213 3

10. Total no. of prime numbers from 1 to 25 = 9 1

∴Non-prime numbers from 1 to 25 = 25 – 9 = 16 ½⟹ P(non-prime numbers)= 16 ½

25

11. Sample space is {HH, HT, TH, TT}

Total no. of outcomes = 4 1

Favourable outcomes are {HT, TH, TT}

No. of favourable outcomes = 3Required probability = 3 1

4

12. a1 = 1

a2 8b1 = 1b2 8c1 = 32 1

c2 25a1 = b1 ≠ c1 ½a2 b2 c2The given system does not represent a pair of coincident lines ½

SET A

Section-C

13. Let the HCF of the two numbers be x. Then LCM = 14xx + 14x = 600 ⟹ 15x = 600 ⟹ x=40 1HCF=40 , LCM=560 1HCF x LCM = Product of two numbers560 x 40 = 280 x second numbersecond number = 560 x 40/ 280 = 80 1

14. √3x²-8x+4√3=0

√3x²-6x-2x+4√3=0 1

√3x(x-2√3)-2(x-2√3)=0

(√3x-2)(x-2√3)=0

x=2/√3,2√31

we know sum of roots = - x coefficient /x² coefficient

2/√3+2√3=8/√3

8/√3=8/√3 hence verified

we know that product of roots is constant term /x² coefficient

(2/√3)(2√3)=4√3/√3

4=4 hence verified 1

15. ∠A+∠B+∠C=180°

x+3x+y=180°

4x+y=180°,3y-5x=300 1Operating,3×(4x+y)-(3y-5x)=3×180-30°

12x+5x=540°-30°

17x=510°

x=30° 1

∠A=30°

∴∠B=3x=3×30=90°1

Hence it is a right angled

SET A

16. Let co-ordinates of A be (x,y)

Co-ordinates of the centre = ( x+1/2, y+4/2) = (2,-3) 1⟹x+1/2=2 , y+4/2 = -3 1x=3, y=-10co-ordinates of A are (3, -10) 1

[OR]

For collinearity, area of triangle formed = 0

1/2(x(2-0) + 1(0-y) + 7(y-2))=0 1

1/2(2x-y+7y-14)=0 1

2x+6y-14=0

x+3y=7 is the required equation. 1

17. Consider LHS 1 – sinθ 1 + sinθ

Multiply and divide with (1 – sinθ)

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= RHS

[OR]Given L.H.S = (sinA + cosecA)2 +(cosA + secA)2

= sin2A+cosec2A+2sinAcosecA+cos2A+sec2A+2cosAsecA 1

= 1+cosec2A+2+sec2A+2 1= 1+1+cot2A+2+1+tan2A+2

= 7+cot2A+tan2A = RHS 1

SET A

18.

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19. Draw the perpendicular from A to BC. Let it intersect BC at E. Then in triangle AEB we have:

AB2 = AE2 + BE2 ½In triangle AED we have:

AD2 = AE2 + DE2 ½Subtracting gives:

AB2 - AD2 = BE2 - DE2

½= (BE + DE)(BE - DE)

Now, the perpendicular to the unequal side in an isosceles triangle is also thebisector of the unequal side.

This can be proved by considering triangles AEB and AEC

AB = AC

angle AEB = angle AEC, AE = AE

By RHS criterion, the two triangles are congruent.

So BE = CE1BE + DE = BD and BE - DE = CD

We get: (BE+DE)*(BE-DE) = BD*CD.

Thus: AB2 - AD2 = BD*CD½

SET A

[OR]AD is median. BD = CD.

In right angled triangle AEB, AB2 = AE2 + BE2

but BE = BD - DE,

therefore, AB2 = AE2 + (BD - DE)2, 1

AB2 = AE2 + BD2 - 2BD*DE + DE2 (eq.1)

Similarly, in the right angled triangle AEC,

AC2 = AE2 + CE2

but CE = CD + DE,

therefore, AC2 = AE2 + (CD + DE)2 1

AC2 = AE2 + CD2 + 2CD*DE + DE2 (eq. 2)Adding equations 1 and 2 we get,

AB2+AC2 = (AE2+BD2 - 2BD*DE + DE2) + (AE2+ CD2 + 2CD*DE +DE2 )

= 2AE2 + 2DE2 + BD2 + CD2 - 2BD*DE + 2CD*DE= 2AE2 + 2DE2 + BD2 + BD2 - 2BD*DE + 2BD*DE= 2AE2 + 2DE2 + 2BD2

= 2(AE2 + DE2) + 2BD2

AE2 + DE2 = AD2 (triangle AED is right angled ) 1AB2 + AC2 = 2AD2 + 2BD2 = 2(AD2 + BD2)

20. Sol:

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½

1

Area of segment APBA = Area of sector APBO- Area ofAOB=154-98=56cm2 ½

SET A

21. Here the length of the rectangular reservoir = 120m, Breadth = 75m

Rise in water level in 18 hours = 2.4m

Volume of water in reservoir = l x b x h = 120 x 75 x 2.4 = 21600m3 1

Let the speed of the water flowing be x km/hr.

Distance travelled in 18 hours = 18x [s = d/t]

Now, Volume of water in reservoir = Volume of water flows out of pipe 1 ∴21600 = 20 x 20 x 18x ⟹ x = 21600 x 100 ⟹ x = 30000 m/hr

10000 2 x 2 x 18

∴ Speed of water = 30 km/hr 1

[OR]

Given: Diameter of each solid sphere = 6cm ⟹ r1 = 3cm

1Diameter of beaker =18cm ⟹ r2 = 9cm, h = 40cm

1n(4/3)πr1

3 = πr22h

∴ n = r2 2h = 9×9×40×3 = 90 solid spheres1

(4/3)r1 3 4×3×3×3

22.N = 600⟹ Median = N th term

2= 600 = 300th term

12

Median class = 1000 – 2000

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Section-D

23. Let speed of stream = x km/h

∴ Downstream speed = (15+x) km/h

∴ Upstream speed = (15-x) km/h

∴ 30/(15+x) + 30/(15-x) = 4 1/2

⟹ 900/225-x² = 9/2

⟹ 225 - x² = 200

⟹ x² = 25⟹ x = 5

∴ Speed of stream = 5 km/h

[OR]

SET A

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SET A

24. In an AP, nth term is given as an = a+ (n-1)d

n = 21

sum of last three terms i.e. a19, a20, a21:

a + 18d + a + 19d + a + 20d = 237⟹ 3a + 57d = 237

⟹ a + 19d = 79 (eq. 1) 1

sum of three middle terms i.e. a10,a11 and a12:

a + 9d + a + 10d + a + 11d = 129⟹ 3a + 30d = 129a + 10d = 43 (eq. 2) 1

Subtract both equations:

a + 19d – a – 10d = 79 – 43⟹ 9d = 36 ⟹ d = 36/9 = 4 1

Put the value of d, into eq 1 or eq2 ½So, First term a =3

common difference d= 4. The A.P is 3, 7, 11, 15,…… ½

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SET A

26.

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SET A

27. Let AB be the building of height 7 m and EC be the height of tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°

EC = DE + CD

also, CD = AB = 7 m. and BC = AD

In right ΔABC,

tan 45° = AB/BC

⇒ 1= 7/BC

⇒ BC = 7 m = AD

also,

In right ΔADE,

tan 60° = DE/AD

⇒ √3 = DE/7

⇒ DE = 7√3 m

Height of the tower = EC = DE + CD

=(7√3 + 7) m = 7(√3+1) m

[OR]In DBC,

tan 60o = h / x⇒ √3 = h / x.

⇒ h = √3x -----(i)

In ΔDAC,

tan 30o = h / (40 + x)

⇒ 1 / √3 = h / (40 + x) ⇒ (40 + x) = √3h -----(ii) Substituting h = √3x in equation (ii)

⇒ (40 + x) = √3x√3x ⇒ (40 + x) = 3x

2x = 40x = 20.width of the river = 20 mh = √3x = 20√3.Height of the tree is 20√3 m

There fore ,width of the river = 20 meters

h = √3x = 20√3.

Hence, the height of the tree is 20√3

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SET A

28.

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MEDIAN MARKS = 24 1

[OR]

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SET A

29. Capacity of glass = Volume of frustum of cone

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30. m + n = tanθ + sinθ + tanθ – sinθ = 2tanθ ½m - n = tanθ + sinθ – tanθ + sinθ = 2sinθ ½mn = (tanθ + sinθ) (tanθ - sinθ) = tan²θ - sin²θ ½m²-n² = (m + n) (m - n) = 2tanθ.2sinθ ½

= 4sinθtanθ-------- (1)

4√mn = 4√(tan²θ - sin²θ)

= 4√(sin²θ/cos²θ-sin²θ) ½= 4√sin²θ(1/cos²θ-1)

= 4sinθ√(1-cos²θ)/cos²θ) ½= 4sinθ/cosθ√sin²θ [∵ sin²θ+cos²θ=1]

= 4sinθtanθ-----------(2) ½From (1) and (2), m² - n² = 4√mn ½