Set A MAHESH TUTORIALS...
Transcript of Set A MAHESH TUTORIALS...
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
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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
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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
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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]
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B(5 , 6)(x2 , y2)
P(x , 0)
A(2 , –3)(x1 , y1)
21
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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)
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Classinterval
0–1010–2020–3030–4040–50
Frequency (f )
82231f2
Mid - Value(x)
515253545
f x
4033077535f90
f = 63 + f fx = 1235 + 35 f
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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
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D
B
A
CE
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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
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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)
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X
M
A
l
B C
P
G
H DE
F
12
8 cm
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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.
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P (x1 , y1)
R (11 , 8)
Q (–6 , 5)
G (2, –5)
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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)
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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
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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
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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 ...
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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]
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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
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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
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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]
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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
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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
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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)
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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
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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]