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MAHESH TUTORIALS I.C.S.E.

Model Answer Paper

Marks : 80

Time : 2½ hrs.

SUBJECT : MATHEMATICSICSE X

Exam No. : MT/ICSE/Semi Prelim I-Set-A-003

T17 I SP A003 Turn over

SECTION – A (40 Marks)

Attempt all questions from this Section.A.1(a) Sum deposited per month = ` 600

i.e., P = ` 600No. of months (n)= 20 monthsRate of interest = 10%Maturity value = ?

Interest earned = P ×( +1)2×12

n n × 100

r

I = 600 ×20(20 1)

2 12

×

10100

I =600 20 21 10

2 12 100

I = ` 1050The amount that Manish will get at the time of maturity

= (600 × 20) + 1050= 12000 + 1050

Maturity value = ` 13050 [3]

(b) n(S) = 100(i) Let A be the event that the bat is acceptable to Jai.

Jai accepts only good bats n(A) = 88

P(A) =n(A)n(S)

=88

100

P(A) =2225

(ii) Let B be the event that the bat is accepable to vijayVijay rejects only bats which have major defects

n(B) = 88 + 8 = 96

P(B) =n(B)n(S)

=96

100

P(B) =2425 [3]

Set A

Turn overT17 I SP A003

Set A

(c) A (4, 5)B (1, 2)C (4, 3)

and D (7, 6)

Slope of AB =2 – 51 – 4

=–3–3

Slope of AB = 1 .......... (i)

Slope of CD =3 – 64 – 7

=–3–3

Slope of CD = 1 .......... (ii)From (i) and (ii),Slope of AB = Slope of CD

AB || CD ........... (iii)

Slope of AD =5 – 64 – 7

=–1–3

Slope of AD =13

......... (iv)

Slope of BC =2 – 31 –4

=–1–3

Slope of BC =13

........... (v)

From (iv) and (v),Slope of AD = Slope of BC

AD || BC ........... (vi)In ABCD

AB || CD ......... [From (iii)]and AD || BC ......... [From (vi)]

ABCD is a parallelogram [A quadrilateral is said to be aparallelogram if its both pair ofopposite sides are parallel] [4]

A.2.(a) Diameter = 28 cm

radius (r) = 14 cmheight (h) = 21 cm

... 2 ...


Set A

(i) Curved surface area of the right circular cylinder = 2rh

= 2 ×227

× 14 × 21

Curved surface area = 1848 cm2

(ii) Total surface area of the right circular cylinder = 2r(r + h)

= 2 ×227

× 14 14 + 21

= 88(14 + 21) = 88(35)

Total surface area = 3080 cm2

(iii) Volume of the right circular cylinder = r2h

=227

× 14 × 14 × 21

Volume = 12936 cm3 [3]

(b) A =1 1

–2 0

; B =–2 1

1 1

3A – 2X = X – 2B3A + 2B = X + 2X

3X = 3A + 2B

3X = 31 1–2 0

+ 22 –11 1

3X =3 3

–6 0

+–4 2

2 2

3X =–3 4 3 2

–6+2 0 2

3X =7 1

–4 2

X =13

7 1–4 2

X =

7 13 34 2–3 3

[3]

(c) (i) C.P. of article for distributor = ` 6,000 Tax he pays= 12.5% of ` 6,000

=12510

×1

100 × 6,000

= ` 750

... 3 ...


Set A

S.P. of article for distributor = ` 7,500 Tax charged = 12.5% of ` 7,500

=12510

×1

100 × 7,500

=1875

2 = ` 937.50

VAT = (Tax charged) – (Tax paid)= ` 937.50 – ` 750= ` 187.50

(ii) C.P. of trader = ` 7,500 Tax he pays = 12.5% of ` 7,500

= ` 937.50S.P. of trader = ` 8,000

Tax charged = 12.5% of ` 8,000

=12510

×1

100 × 8,000

= ` 1,000 VAT = (Tax charged) – (Tax paid)

= ` 1,000 – ` 937.50= ` 62.50 [4]

A.3.(a) Let the required ratio be k : 1 and the point on the x-axis be (x , 0)

y = 2 1

1ky y

k

[Taking (2 , –3) (x1 , y1) and (5, 6) (x2 , y2)]

0 = × 6 + (-3)

1k

k 0 = 6k – 3

6k = 3

k =36

=12

m1 : m2 = 1 : 2Now,

x = 1 5 2 2

1 2

=5 4

3

=93

= 3 The ratio is 1 : 2 and the required point of intersection is (3 , 0) [3]

... 4 ...

B(5 , 6)(x2 , y2)

P(x , 0)

A(2 , –3)(x1 , y1)

21


Set A

(b) (i) Let co-ordinates of P be (x, y).Under reflection in x-axis, P becomes (–4, 5)

Mx(x, y) = (x, –y)Hence, (x, –y) = (–4, 5) x = –4and –y = 5i.e. y = –5Hence P is (–4, –5)(ii) Image of P under reflection in y-axis is

My(–4, –5) = (4, –5) [3]

(c) No. of shares = 60Nominal Value = ` 100

Premium = 60%

Market Value = ` 100 +60

100 × 100 = ` 160

Sale Proceeds = 160 × 60Sale proceeds = ` 9600

Firm 2 :Nominal Value = Face Value = ` 50

Discount = 4%

Market Value = 50 –4

100 × 50 = ` 48

Rate of Dividend = 18%Sales Proceeds of 1st firm becomes sum invested for the 2nd firm

No. of shares =960048

= 200

No. of shares= 200Dividend = No. of shares × Rate of Dividend × Face Value

= 200 ×18100

× 50

Dividend = ` 1800 [4]

A.4.(a)

Mean = 2117

(Given)

... 5 ...

Classinterval

0–1010–2020–3030–4040–50

Frequency (f )

82231f2

Mid - Value(x)

515253545

f x

4033077535f90

f = 63 + f fx = 1235 + 35 f


Set A

=148

7

Mean =fxf

148

7=

1235 3563

ff

148 (63 + f) = 7 (1235 + 35 f ) 9324 + 148 f = 8645 + 245 f 9324 – 8645 = 245 f – 148 f 679 = 97 f

f =67997

f = 7 [3]

(b) SPHERE : Radius (rs) = 15 cmCONE : Radius (rc) = 2.5 cm

Height (hc) = 8 cmLet ‘N’ cones be recastVolume of Sphere = N × Volume of cone

N =Volume of SphereVolume of Cone

N =

3

2

4 r3

1 r3

s

c ch

N =4 15 15 152.5 2.5 8

N = 270 270 cones are recast [3]

(c) In ABC and EBD,BD = 5 cm, BE = 6 cm, EC = 4 cmBC = BE + EC = 6 + 4 = 10 cm

Now, ACB = EDB (Given)ABC = DBE (Common angle)

ABC ~ EBD (AA postulate)

(i) ABBE =

BCBD (Corresponding sides of similar

triangles are proportional)

AB6 =

105

AB =10×6

5 AB = 12 cm

... 6 ...

D

B

A

CE


Set A

(ii) Since, ABC ~ EBD

Area of ABCArea of EBD

2

2

BCBD

2

2

105

10025

Area of ABC = 4 (Area of EBD) = 4(9) = 36 cm2

Area of ABC = 36 cm2 [3]

SECTION – II (40 Marks)Attempt any four questions from this Section.

A.5.(a) Let A (0, 2) (x1, y1)

B (–3, –1) (x2, y2)

Slope of AB =2 1

2 1

– –

y yx x

=–1 –2–3 –0

=–3–3

Slope of AB = 1Now,line AB is parallel to the line passing through (–1, 5) and (4, a)

Slope of line passing through (–1, 5) and (4, a) = Slope of AB = 1

1 = – 5

4 – (–1)a

1 = – 55

a

5 = a – 5 5 + 5 = a a = 10 [3]

(b) Face value of 1 share = ` 100Market value of 1 share = ` 85

Annual income = ` 1800Rate of dividend = 12%

Income = Number of shares × Rate of dividend × Face value

... 7 ...


Set A

1800 = No. of shares ×12

100 × 100

No. of shares =1800 10012 100

= 150

Sum invested = No. of shares × Market value= 150 × 85

Sum invested = ` 12,750

% Return =Total income

Total investment × 100

=1,80012,750 × 100

= 14.12 % The percentage return on this investment is 14.12% [3]

(c)

PB = 3.6 cm [4]

A.6.(a) In DFE and DHG

D = D (Common angle)DFE = DHG (each 90º)

DFE ~ DHG (By AA postulate)

DFDH

=DEDG

(sides of similar triangles are proportional)

... 8 ...

X

M

A

l

B C

P

G

H DE

F

12

8 cm


Set A

128

=4 23 –1x

x 12 (3x – 1) = 8 (4x + 2) 36x – 12 = 32x + 16 36x – 32x = 16 + 12 4x = 28

x =284

x = 7DG = 3x – 1

DG = 3 (7) – 1 DG = 21 – 1

DG = 20 unitsAnd, DE = 4x + 2 DE = 4 (7) + 2 DE = 28 + 2 DE = 30 units [3]

(b) Let G be the centroid of PQR G (2 , –5) (x , y)

Let P (x1 , y1)Q (–6 , 5) (x2 , y2)

and R (11 , 8) (x3 , y3) By centroid formula,

2 = 1 6 113 x

6 = x1 – 6 + 11 6 + 6 – 11 = x1

x1 = 1

–5 = 1 5 83 y

–15 = y1 + 13 y1 = –15 –13 y1 = –28 P (x1, y1) (1, –28) [3]

(c) (i)2 13 4

X =76

Of the form, A2 × 2 Xm × n = B2 × 1

Multiplication is possible Number of columns of A = Number of rows of X m = 2.

... 9 ...

P (x1 , y1)

R (11 , 8)

Q (–6 , 5)

G (2, –5)


Set A

Further, Order of B = (No. of rows of A) × (No. of columns of X) 2 × 1 = 2 × n n = 1 Order of matrix X is 2 × 1

(ii) Let X =2 1

xy

2 13 4

xy =

76

2 1 2 1

2 73 4 6

x yx y

By equality of matrices,2x + y = 7 ...........(ii)–3x + 4y = 6 ............(ii)

Multiplying equation (i) by 4, we get,8x + 4y = 28 .............(iii)Subtracting equation (ii) from equation (iii), we get,

8 4 283 4 6( ) ( ) ( )

11 22

x yx y

x

2211x

x = 2 ...........(iv)From (i) and (iv),

2(2) + y = 7 4 + y = 7 y = 7 – 4 y = 3

X =

23 [4]

A.7.(a) Number of total outcomes in a single throw of two dice = 6 × 6 = 36

(i) Possible doublets are (1,1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) Number of favourable outcomes = 6

P (a doublet) =636

=16

(ii) An odd number on one die and a number less than or equal to 4 onthe other die.

Case (a) : Odd number on 1st die and a number less than or equal to4 on 2nd die(1,1), (1,2), (1,3), (1,4)

... 10 ...


Set A

(3,1), (3,2), (3,3), (3,4)(5,1), (5,2), (5,3), (5,4)

Case (b) : A number less than or equal to 4 on 1st die and odd numberon 2nd die(1,5)(2,1), (2,3), (2,5)(3,5)(4,1), (4,3), (4,5)

Number of favourable outcomes = 20

Required Probability =2036

=59 [3]

(b) Internal diameter = 21 cm

Internal radius, r =212

= 10.5 cm

External diameter = 28 cm

External radius, R =282

= 14 cm

(i) Internal curved surface =12

(4r2)

= 2r2 = 2 ×227

×212

×212

= 693 cm2

Internal curved surface area = 693 cm2

(ii) External curved surface =12

(4R2)

= 2R2 = 2 ×227

× 14 × 14

= 1232 cm2

External curved surface = 1232 cm2 [3]

(c) Let the time of Recurring Deposit be n months.

The equivalent principle for one month = `300 ×( +1)

2n n

Interest on it = ` 300×( +1)

2n n

×12 ×1

100×12

= `3 ( +1)

2n n

300×n+3 ( +1)

2n n

= 8,100

600n + 3n2 + 3n = 16,200 3n2 + 603n – 16200 = 0 n2 + 201n – 5400 = 0 ( n + 215) ( n – 24) = 0 n = –215 or n = 24

... 11 ...


Set A

n = 24 [n = –225 is not possible]Hence, Required time = 24 months = 2 years [4]

A.8.

(a) A2 – 5A + 7I =3 1 3 11 2 1 2

– 53 11 2

+ 71 00 1

=3 3 1 1 3 1 1 21 3 2 1 1 1 2 2

–

15 5 7 05 10 0 7

=8 5 15 5 7 05 3 5 10 0 7

=8 15 7 5 5 05 5 0 3 10 7

A2 – 5A + 7I =

0 00 0 [3]

(b) Sum invested = ` 11200Rate of dividend = 6%Face value = ` 100Market value = ` 140Annual dividend = ?

No.of shares =Sum Invested

Market value of 1 share

=11200140

No. of shares = 80Dividend = No. of shares × rate of dividend × face value

= 80 ×6

100 × 100

= 480

Annual dividend = ` 480Rate of dividend × face value = Income % × market value

6 × 100 = Income % × 40

Income % =6 ×100

40 Income% = 15 [3]

(c) Let A (–5, 2)B (3, – 6)C (7, 4)

AD is the Median D is the mid-point of BC By mid-point formula,

... 12 ...

A (–5 , 2)

E (1 , 3)F (–1 , –2)

B(3 , –6) D

(5 , –1)

C(7 , 4)

O


Set A

D =1 2 1 2,

2 2

x x y y

=3 7 –6 4

,2 2

=10 2

,2 2

D = (5 , –1)BE is the Median

E is the Mid-point of AC By mid-point Formula,

E =1 2 1 2,

2 2

x x y y

=5 7 2 4

,2 2

=2 6

,2 2

E = (1 , 3)CF is the Median

By mid-point formula,

F =1 2 1 2,

2 2

x x y y

=5 3 2 6

,2 2

=2 4

,2 2

F = (–1 , –2)Now,

AD = 2 2

2 1 2 1x x y y

= 2 25 –5 1 2

= 2 25 5 3

= 2 210 3

= 100 9

AD = 109 units

BE = 2 2

2 1 2 1x x y y

... 13 ...


Set A

= 2 23 1 6 3

= 2 22 9

= 4 81

BE = 85 units

CF = 2 22 1 2 1 x x y y

= 2 27 1 4 2

= 2 27 1 4 2

= 228 6

= 64 36

= 100 CF = 10 units

The length of all its medians are as follows :AD = 109 units

BE = 85 unitsCF = 10 units [4]

A.9.(a) (i) Length of the ship = 4 × 200

= 800m

Length of the ship = 800 m

(ii) Area of the deck of the model =160000200 200 m2

= 4m2

Area of the deck of the model = 4 m2

(iii) Volume of the model = 200liter

=200

1000m3

= 0.2m3

Volume of the ship = 0.2 × (200)3 m3

=2

10× 8000000

= 1600000m3

Volume of the ship = 1600000 m3 [3]

... 14 ...


Set A

(b) Sum deposited per month = ` 140No. of months (n) = 48 monthsMaturity value = ` 8092Rate of interest = r = ?

Maturity value = (P × n) + P ×( +1)2×12

n n × 100

r

8092 = ` (140 × 48) + 140 ×48(48 1)

2 12

×100

r

8092 – 6720 = 140 ×48(48 1)

2 12

× 100

r

1372 =14 48 49

2 12 10

r

r =1372 1014 2 49

r = 10% Rate of interest = 10% [3]

(c)

Mode = 23 [4]

A.10.(a) Equation of given line is

4x + 5y = 6

... 15 ...

1

0

2

3

4

5

7

6

8

9

10

10

X

C

A

Y

20 30 40 50 60 70 80Class Internal

FRE

QU

EN

CY

MODE

D

B

L

K

Scale :X-axis : 1 cm = 10 unitsY-axis : 1 cm = 1 units


Set A

5y = –4x + 6

y =–45

x +65

Comparing it with y = mx + c

Slope, m =–45

Slope of r line =–

1–45

=54

.

Equation of line through (–2, 1) having slope54

is

y – y1 = m (x – x1)

y – 1 =54

[x – (–2)]

4y – 4 = 5x + 10 0 = 5x + 10 – 4y + 4 5x – 4y + 14 = 0 [3]

(b) Marked price of article = ` 5,000S.P. of manufacturer at 25% discount on M.P.

= 5,000 ×

100 – 25100

= 5,000 ×75

100= ` 3,750

C.P. of wholesaler = ` 3,750S.P. of wholesaler at 15% discount on M.P.

= 5,000 ×

100 – 15100

= 5,000 ×85

100= ` 4,250

C.P. of retailer = ` 4,250S.P. of retailer at M.P. = ` 5,000

Sales tax paid by wholesaler = 3,750 ×8

100= ` 300

Sales tax changed by wholesaler = 4,250 ×8

100= ` 340

VAT paid by wholesaler = Tax charged – Tax paid= ` 340 – ` 300= ` 40 [3]

... 16 ...


Set A

(c) Total height of the toy = 60 cmHeight of the cone = 24 cm

Height of the cylinder = 60 – 24= 36 cm

Radius of the base of the cone = 10 cmSince radius of the base of the cone is twicethat of the cylinder,

Radius of base of cylinder =102

= 5 cm

Curved surface area of the cylinder = 2rh= 2 × × 5 × 36= 360cm2

Area of circular base of cylinder = r2

= × 5 × 5= 25cm2

Now for cone,l2 = r2 + h2

l2 = 102 + 242

l2 = 100 + 576 l2 = 676 l = 26 cm

Total surface area of the cone = r(r + l)= × 10(10 + 26)= × 10(36)= 360cm2

Total surface area of the toy = Total surface area of the cone – Area of the circular surface of the cylinder + Curved surface area of the cylinder+ Area of the circular base of the cylinder

= 360 – 25+ 360 + 25= 720

= 720 × 3.14

= 2260.8

Total surface area of the toy is 2260.8 cm2 [4]

... 17 ...

60 cm

24 cm

10 cm


Set A

A.11.(a) (i) Plot the points A (3, 5), B (–2, –4)

Take 1 cm = 1 unit

(ii) Since A is the image of A when reflected in the x-axis A is (3, –5)(iii) Let B1 be the image of B reflected in the y-axis B is (2, –4)Now B is the image of B1, in the origin B is (–2, 4)(iv) Clearly AABB is an isosceles trapezium.(v) (4, 0) and (–3, 0) are invariant points under reflection in the x-axis. [4]

(b) Weight frequency (f) c.f.35 – 40 0 040 – 45 5 545 – 50 17 2250 – 55 22 4455 – 60 45 8960 – 65 51 14065 – 70 31 17170 – 75 20 19175 – 80 9 200

Total = 200

... 18 ...

2

3

1

-2

-3

-11 2 3 4-5 -4 -3 -2 -1 -0

-4

-5

4

5Y

XX 5

YA (3, –5)

B (2, –4)1B(–2, –4)

B (–2, 4)

A(3, 5)


Set A

(i) Mark point P at 55 kg. PC x-axis from C, draw CQ y-axis. OQ = 44 Students weighing more than 55 kg = 200 – 44 = 156

% students =156200

× 100

= 78%

... 19 ...

10

40 45 50 55 60 65 70 75 80

20

30

40

50

60

80

70

90

100

110

130

150

170

190

120

140

160

180

200

0

A

B

Q

P

D

RE

F

G

H

C

Y

Weight ( in kg)

No.

ofSt

ude

nts

Scale : On X axis:1 cm = 5kgScale : On Y axis:1 cm = 10 Students

(80, 200)

(75, 191)

(70, 171)

(65, 140)

(55, 44)

(50, 22)(50, 22)

(40, 0) (45, 5)X

(60, 89)

(55, 44)

T


Set A

... 20 ...

(ii) 30% students are30

100 × 200 = 60

The heaviest 30% are represented at point R. Draw RE || x-axisand ET x-axis

OT = 65 Heaviest 30% students have weight above 65 kg.(iii) Standard weight = 55.70 kg

(a) Under weight students = 46 (approx)(b) over weight students = 154 (approx) [6]

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