Marking Scheme Mathematics Class X (2018)—set-1

Section A

1. Terminating decimal 1

2 4k2- 24k=0 ; k=0,6 1

3. Let A(x,y) ,then

x = 2, y =5 B= (2,5)

1

4. 3y + 5 -3y +1 =6: 5y +1-3y -5 =2y -4 2y-4 =6 y = 5

1

5. AD/ AB = DE /BC

DE = 3cm

1

6. secA – tanA = 1/p 1

Section B

7.

=> The system of equations has infinite many solutions.

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1

8. The prime factorization of 120

120 = 2 × 2 × 2 × 3 × 5

the prime factorization of 144

144 = 2 × 2 × 2 × 2 × 3 × 3

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5

LCM = 720

HCF = 24

1

1

9.

½

1/2

1

10.

½

½

½

½

11. Total number of cards in a deck = 52

No. of kings =4, no. of queens =4.

Total of kings and queens = 8

So, no. of cards excluding kings and queens = 52 – 8

= 44

i) P( neither king nor queen) =

ii) P( either king or queen) =

1

½

½

12. Probability(Blue ball)= 2*Probability(Red ball)

(x/(5+x))=2(5/(x+5))

=10

½

1

½

Section C

13. n,n+1,n+2 be three consecutive positive integers

We know that n is of the form 3q, 3q +1, 3q + 2

So we have the following cases

Case – I when n = 3q

In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3

Case - II When n = 3q + 1

n+1= 3q+2

n+2= 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not

divisible by 3

Case – III When n = 3q +2

n +1 = 3q +2 +1 = 3(q +1) is divisible by 3.

n+2 = 3q+2+2

but n and n+2 are not divisible by 3

Hence one of n, n + 1 and n + 2 is divisible by 3

1

1

1

14. Cos + Sin = 2 Cos

( Cos + Sin)2 = 2Cos

2

Cos2 + Sin

2+2Cos Sin = 2Cos

2

Cos2 - 2Cos Sin+ Sin

2 = 2Sin

2 2Sin

2 = 2 - 2Cos

2

(Cos - Sin)2 = 2Sin

2 1- Cos

2 = Sin

2 & 1 - Sin

2 = Cos

2

1

1

Cos - Sin = 2 Sin. OR

xcosθ - y sinθ = a

xsinθ + y cosθ = b

Squaring and adding

x2+y

2=a

2+b

2.

1

1

1

1

15. Given it is a parallelogram

=> diagonals bisect each other

=> AC = BD

now using mid pt. formula

[ (3+ x)/2 ; (3+7)/2 ] = [ (6+5)/2 ; (y+6)/2 ]

comparing and solving both sides

(since 'x' coordinates are equal and 'y' coordinates will be equal)

=> (3+ x) = 11 ; (y+6) = 10

=> x = 8 ; y = 4

OR

1

1

1

1

1

1

16. Let 50 % acids in the solution be x

Let 25 % of other solution be y

Total Volume in the mixture = x + y

x + y = 10 -------- (1)

100

50x +

100

25 y =

100

4010

2x + y = 16 -------- (2)

So x = 6 & y = 4

1

1

1

17. p (x) = 2 x

2 – 4 x + 5 + =

4

2

b

a

= 2

= 5

2

c

a

i) 2 +

2 = ( + )

2 – 2

Substitute, then we get, 2+

2 = -1

ii)

= 2

= 4/5

1

1

1

18.

2

1

1

19. Join OB We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. OAP = OBP = 90º Now,

OAP + APB + OBP + AOB = 360º [Angle sum property of quadrilaterals] implies 90º+ APB + 90º + AOB = 360º implies AOB = 360º - 180º - APB = 180º - APB ....(1) Now, in triangle OAB, OA is equal to OB as both are radii. implies OAB = OBA [In a triangle, angles opposite to equal sides are equal] Now, on applying angle sum property of triangles in AOB, we obtain

OAB + OBA + AOB = 180º implies 2 OAB + AOB = 180º implies 2 OAB + (180º - APB) = 180º [Using (1)] implies 2 OAB = APB Thus, the given result is proved

1

1

1

20. Total volume of 3 cubes = (33

+ 43

+ 53

) cm3

= (27 + 64 + 125)cm3

= 216 cm3

Now volume of new cube = 216 cm3

So edge of new cube = cube root of 216 cm3

= 6 cm

OR

Radius of cone = 3.5 cm, height of cone = 15.5 – 3.5 = 12 cm Slant height of

cone can be calculated as follows:

Curved surface area of cone can be calculated as follows:

Curved surface area of hemispherical portion can be calculated as follows:

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1

1

1

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1

21.

Now n = 51. S0, n/ 2 = 51/ 2 =25.5 this observation lies in the class 145 – 150

Then,

L (the lower limit) = 145

cf (the cumulative frequency of the class preceding 145 – 150) = 11

f (the frequency of the median class 145 – 150) = 18

h (the class size) = 5

Using the formula, median = l+( )×h, we have

= 145 + 72.5/ 18 = 149.03

So, the median height of the girls is 149.03 cm

This means that the height of about 50% of the girls in less than this height, and 50% are taller than this height.

1

½

½

1

22. Let BC = a, A = Ar( ABC)

A = x a x AB

AB = (1)

ADB ABC (AA)

…………………(2)

From ABC

AB2 + BC

2 = AC

2

= AC2

AC=

1

1

From 1 & 2

BD =

OR

Area of triangle ACB with base AB = (1/2)AB*CD = (1/2)cp

Area of triangle ACB with base AC = (1/2)AC*BC = (1/2)ba

Both areas are same. Therefore, on equating, we get,

c = ab/p ... (1)

Now, using Pythagoras theorem in triangle ABC, we have

AB2 = AC

2+BC

2

c2 = b

2+a

2

a2b

2/ p

2 = b

2+a

2 [Using (1)]

1/ p2= (b

2+a

2)/ a

2b

2

Hence, 1/ p2= 1/a

2+1/b

2

1

1

1

1

SECTION –D

23. Given a5 = 30

That is (a + 4d) = 30 …. (1)

Also given a12 = 45

(a + 11d) = 45 ….. (2)

Subtract (1) from (2), we get

a + 11d = 45

a + 4d = 30

--------------------

7d = 15

d = 15/7

Put d = 15/7 in (1), we get a = 150/7

Sum of n terms of AP is given by Sn = (n/2)[2a + (n – 1)d]

S20 = (20/2)[2(150/7) + (20 – 1)(15/7)]

= 10[(300/7) + (285/7)]

= 10[585/7] = 5850/7

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1

1

1

24.

Given: A right triangle ABC right angled at B.

To prove: AC2

= AB2 + BC

2

Construction: Draw BD AC

Proof: In ADB and ABC

ADB = ABC (each 90°)

BAD = CAB (common)

ADB ~ ABC (By AA similarity criterion)

Now,

(corresponding sides are proportional)

AB2= AD × AC …(i)

Similarly BDC ~ ABC

BC2 = CD × AC …(ii)

Adding (1) and (2)

AB2 + BC

2= AD × AC + CD × AC

AB2 + BC

2= AC × (AD + CD)

AC2

= AB2 + BC

2

, Hence Proved

OR

Basic Proportionality Theorem (Thales theorem): If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio. Given : In ∆ABC , DE || BC and intersects AB in D and AC in E. Prove that : AD / DB = AE / EC

Construction : Join BC,CD and draw EF ┴ BA and DG ┴ CA.

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1

1

1

1

½

½

2

25.

Given height of the cone = 30 cm Let the small cone is cut off at a height ‘h’ from the top Let the radius of big cone be r1 and small cone be r2. Volume of the big cone, From the figure, ΔACD ~ ΔAOB [AA similarity criterion]

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1

Thus at a height 20 cm above the base a small cone is cut.

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1

26.

Let the height of the cloud from the surface of the lake be CE = h metres Hence the reflection of cloud in the lake F will be at the same distance ‘h’ metres from the surface of the lake.

1

Let AD = BE = x metres Also we have AB = 60 m

⇒ CD = (h – 60) m In right ΔADC, θ = 30°

Thus the height of the cloud from the surface of the lake is 90 metres.

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1

1

27. Let the speed of the fast train be x km/hr

Let the speed of the slow train be (x- 10) km/hr

Hence the time taken by the fast train is 600/x

Time taken by slow train is 600/(x-10)

Therefore (600/x-10) - (600/x) = 3

On reducing we get,

3x2-30x-6000 = 0 or x

2 -10x -2000 = 0

On factorising we get x = 50 or -40

The negative value is not acceptable

Hence x = 50

That is the speed of the fast train is 50km/hr and the speed of the slow train is

40km/hr

OR Let side of 1st square=x cm. ∴ area of 1st square=x²cm²

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1

given, sum of areas =400cm² ∴ area of 2nd square=(400-x²)cm² and side of 2nd square=√[(20-x)² i.e.20-x .......(1) Difference of perimeters=16cm. then- 4x-4(20-x)=16 (assuming that 1st square has larger side) x=12 hence - side of 1st square = 12cm ; side of 2nd square=20-12=8cm. [from (1)]

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1

1

28.

2

1

VALUE: CHARITY towards old age people 1

29. Sec = x +

x4

1

Sec2 =( x +

x4

1)

2 (Sec

2= 1 + Tan

2)

Tan2 = ( x +

x4

1)

2-1

Tan2 = ( x -

x4

1)

2

Tan = + x -

x4

1

Sec + Tan = x + x4

1+ x -

x4

1

= 2x or x2

1

1

1

1

1

30.

OR

2

1

1

1

1

1

1