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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-B-003
T17 I SP A003 Turn over
SECTION – A (40 Marks)
Attempt all questions from this Section.A.1(a) Sum deposited per month (P) = ` 640
No. of months (n) = 54 monthsRate of interest (r) = 12%Maturity value = ?
Interest earned = P ×( +1)2×12
n n × 100
r
I = 640 ×54(54 1)
2 12
×
12100
I = 640 ×54 552 12
×12
100 I = ` 9504 Maturity value= (640 × 54) + 9504 Maturity value= ` 44,064 [3]
(b) Cards are marked as 1, 2, 3, 4, ....... 29, 30 Total possible outcomes = 30(i) A multiple of 4 or 6
4, 6, 8, 12, 16, 18, 20, 24, 28, 30i.e. number of favourable outcomes = 10
P (a multiple of 4 or 6) =1030
=13
(ii) Multiple of 3 and 5 are 15, 30
P (a multiple of 3 and 5) =2
30 =
115
(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30 Number of favourable outcomes = 14
P (a multiples of 3 or 5) =1430
=7
15 [3]
(c) Let A (–2, 4)B (4, 8)C (10, 7)
and D (11, –5)
Turn overT17 I SP A003
Set B
Let M be the mid-point of AB, By Mid-point formula,
M –2 + 4 4 + 8
,2 2
2 12
,2 2
(1, 6)Let N be the mid-point of CD
By Mid-point formula,
N 11 + 10 –5 + 7
,2 2
21 2
,2 2
21, 1
2Let P be the mid-point of AD
By Mid-point formula
P –2 + 11 4 + (–5)
,2 2
9 –1
,2 2
Let Q be the mid-point of BC By Mid-point formula
Q 4 + 10 8 + 7
,2 2
14 15
,2 2
157,
2 M (1, 6)
N
21, 1
2
P 9 –1
,2 2
and Q 15
7,2
Slope of QM = 2 1
2 1
y yx x
15 627 1
... 2 ...
(11, –5)
B
C
A
D(10, 7)
(–2, 4) (4, 8)
P
M
N
Q
Turn overT17 I SP A003
Set B
3
2 6
14
....(i)
Slope of QN = 2 1
2 1
y yx x
=
1512
21 72
=
13272
=137
....(ii)
Slope of NP = 2 1
2 1
y yx x
=
1 12
9 212 2
=
32122
=3
12
=14
....(iii)
Slope of PM = 2 1
2 1
y yx x
=
162
912
=
13272
=137
....(iv)
Slope of QM = Slope of NP [From (i) and (iii)]
... 3 ...
Turn overT17 I SP A003
Set B... 4 ...
QM || NP ....(v)Slope of QN = Slope of MP [From (ii) and (iv)]
QN || MP ....(vi) QNPM is a parallelogram [A quadrilateral is said to be a
parallelogram if its both pair of opposite sides are parallel] [4]
A.2.(a) Diameter of cylinder = 20 cm
its radius (r) = 10 cm(i) Curved surface area of the cylinder = 100 cm2
2rh = 100
2 ×227× 10 × h = 100
h =100×720×22
=5 7
22 =
3522
h = 1.6 The height of the cylinder is 1.6 cm
(ii) Volume of the cylinder = r2h
=227
× 10 × 10 × 1.6
=3520
7= 502.85= 502.9 cm3
Volume of the cylinder is 502.9 cm3 [3]
(b) A =–1 1
a b
A2 = A.A
=–1 1
a b
–1 1
a b
=
–1 × –1 + 1 × a –1 × 1 + 1 × b
a× –1 + b × a a × 1 + b × b
=
2
1 + a –1 + b
–a + ab a + b
Since A2 = I (Given)
2
1 + a –1 + b
–a + ab a + b
=
1 0
0 1
Turn overT17 I SP A003
Set B... 5 ...
1 + a = 1 a = 1 – 1 = 0
– 1 + b = 0 b = 1 a = 0; b = 1 [3]
(c) (i) Marked price of article = ` 500S.P. of wholesaler or C.P. of retailer at 20% discount
= 500 ×
100 – 20100
= 500 ×80
100 = ` 400
Sales tax of 12.5% charged by wholesaler
=12.5100
× 400 = ` 50
S.P. of retailer or C.P. of customer at marked price = ` 500
Sales tax charged by retailer =12.5100
× 500 = ` 62.50
Also, sales tax paid by retailer = Sales tax chargedby wholesaler = ` 50
(i) Price paid by the customer = 500 + 62.50= ` 562.50
(ii) VAT paid by the retailer = Tax charged by retailer – Tax paid by retailer
= 62.50 – 50= ` 12.50 [4]
A.3.(a) Let the ratio be k : 1
Let A (–1 , 4) (x1 , y1)
B (4 , –1) (x2 , y2)
Now, x =1 2 2 1
1 2
++
m x m xm m
1 =(4) + 1 (–1)
+ 1k
k
1 =4 – 1
+ 1kk
k + 1 = 4k – 1 2 = 4k – k 2 = 3k
k =23
m1 : m2 = 2 : 3
A(–1, 4)(x1, y1)
P(1, a)
B(4, –1)(x2, y2)
k 1
Turn overT17 I SP A003
Set B... 6 ...
Now, a =1 2 2 1
1 2
++
m y m ym m
a =2(–1) + 3 (4)
2+ 3
a =–2 +12
5
a =105
a = 2 [3]
(b) P(a, b) is reflected in origin, M0(a, b) = (–a, –b)
Now (–a, –b) is reflected in y-axis to P My(–a, b) = (a, –b)
But P = (4, 6) (given) (a, –b) = (4, 6)
a = 4 and –b = 6i.e. b = –6Hence a = 4 and b = –6 [3]
(c) Let the two parts be ` x and ` (50,760 – x)For 1st part :-Sum invested = xRate of Dividend = 8%
Face Value = ` 100Discount = ` 8
Market Value = ` 100 – ` 8= ` 92
No. of shares =Sum investedM.V. of 1share
=92x
Income = No. of shares × Rate of Dividend × face value
=92x
×8
100 × 100 = `
223x
For 2nd part :-
Sum invested = 50, 760 – x
Rate of Dividend = 9%Face Value = ` 100
Premium = ` 8
Market Value = ` 100 + ` 8= ` 108
Turn overT17 I SP A003
Set B... 7 ...
No. of shares =Sum investedM.V. of 1share
=50,760 –
108x
Income = No. of shares × Rate of Dividend × face value
=50760 –
108x
×9
100 × 100
=50760 –
12x
Income from both the investment are equal
223x
=50760 –
12x
2x × 12 = 23(50760 – x)
24x = 1167480 – 23x 24x + 23 x = 1167480 47x = 1167480
x =1167480
47 x = 24840
and (50760 – x) = 50760 – 24840= 25920
The two parts are ` 24,840 and ` 25,920 [4]
A.4.(a) Mean x = 62.8
Class f Mid-value x fx0 – 20 5 10 5020 – 40 f1 30 30 f1
40 – 60 10 50 50060 – 80 f2 70 70f2
80 – 100 7 90 630100 – 120 8 110 880
f = 30 fx = 30f1 + f1 + f2 +70f2 + 2060
f = 50 30 + f1 + f2 = 50 f1 + f2 = 20 ....(i)
x =fxf
Turn overT17 I SP A003
Set B... 8 ...
62.8 = 1 130 + 70 + 206050
f f
3140 = 30f1 + 70f2 + 2060 30f1 + 70f2 = 3140 – 2060 30f1 + 70f2 = 1080 3f1 + 7f2 = 108 ....(ii)
Multiplying (i) by 3 and subtracting (ii) from3f1 + 3f2 = 603f1 + 7f2 = 108
– – ––4f2 = –48
f2 =–48–4
= 12
Put this value in (ii), f1 + 12 = 20
f1 = 20 – 12= 8
Hence f1 = 8, f2 = 12 [3]
(b) CONE :Radius (rc) = 5 cmHeight (hc) = 8 cmSPHERE :Radius (rs) = 0.5 cmLet ‘N’ spheres be formedVolume of Cone = N × Volume of Sphere
N =Volume of Cone
Volume of Sphere
N =
2
3
1 r34 r3
c c
s
h
N =5 5 8
4 0.5 0.5 0.5
N = 400 Number of spheres formed is 400 [3]
(c) Given : PQR in which XY || QRPX : XQ = 1 : 3 and QR = 9 cmand ar (PXY) = x cm2
To find : (i) Length XY(ii) ar(PQR)(iii) ar(trapezium XQRY)
Proof : (i) XY || QR PXY = PQR (Corresponding angles)Also, P = P (Common angle) PQR ~ PXY (AA postulate)
R
P
X
Q
Y
Turn overT17 I SP A003
Set B... 9 ...
PQPX
=QRXY
.....(1) (Corresponding sides of similar
triangles are proportional)
NowPXXQ =
13
XQPX
=31
XQ + PX
PX=
3 +11
PQPX
=41
...... (2)
From (1) and (2)
41
=QRXY
41
=9
XY
XY =94
XY = 2.25 cm
(ii) Further
ar ΔPXY
ar ΔPQR =2
2
XYQR
ar ΔPQRx
=2
14
ar ΔPQRx
=1
16 ar (PQR) = 16x cm2
(ii) Area (trap. XQRY) = ar (PQR) – ar (PXY)= 16x – x= 15x cm2
Area (trap. XQRY) = 15x cm2 [4]
SECTION – II (40 Marks)Attempt any four questions from this Section.
A.5.(a) Let A (–4, –2) (x1, y1)
B (2, –3) (x2, y2)
Slope of AB =2 1
2 1
– –
y yx x
=–3 – (–2)2 – (–4)
Turn overT17 I SP A003
Set B... 10 ...
=–3 + 22 + 4
=–16
Now,line AB is perpendicular to the line passing through (a, 5) and (2, –1)
Slope of line passing through (a, 5) and (2, –1) =–1
Slope of AB
=–1–16
= 6
6 =–1 – 52 –a
6(2 – a) = –6 12 – 6a = –6 – 6a = –18 6a = 18
a =186
a = 3 [3]
(b) (i) Face value of 1 share = ` 80Premium = 30% of ` 80
= `30
100 × 80 = ` 24
Market value of 1 share = ` 80 + ` 24 = ` 104 Market value of 150 shares = ` (150 × 104) = ` 15,600
Market value of 150 shares is ` 15,600(ii) Dividend on 1 share = 18% of ` 80
= `18100
× 80 = `1440100
Dividend on 150 shares = (14.40 × 150)= ` 2,160
Dividend on 150 shares is ` 2,160
(iii) % return =2,160
15,600 × 100
=18012
= 13.846 = 13.85%
% return = 13.85% [3]
Turn overT17 I SP A003
Set B... 11 ...
(c) Construction :
Steps :For triangle :1. Draw AB = 7.0 cm.2. Construct MBA = 60º.3. Mark a point C on ray BM, such that BA = 8.0 cm.4. Draw AC.
ABC is the required triangle.
For point P :1. Draw perpendicular bisector of BC.2. Draw angle bisector of ABC.3. Mark point of intersection of perpendicular bisector and angle
bisector as P.4. P is the required point.
(i) P is equidistant from B and C as it lies on perpendicularbisector of BC.
(ii) P is equidistant from AB and BC as it lies on anglebisector of ABC.
(iii) PB = 4.6 cm [4]
A.6.(a) Given : In the figure,
(i) AB and DE are perpendiculars to BC.(ii) AB = 9 cm, DE = 3 cm, AC = 24 cm.
To find : ADSoln. In ABC and DEC,
ABC = DEC, (Each 90º)C = C (Common angle)
ABC ~ DEC (By AA postulate)
ABDE
=ACDC
(Sides of similar triangles are proportional)
BA
P
C
M
7cm
60º
Turn overT17 I SP A003
Set B... 12 ...
93
=24DC
DC =24 3
9
DC = 8 cmAD = AC – DC
= 24 – 8= 16
AC = 16 cm [3]
(b) A (5 , x)B (–4 , 3)C (y , – 2)Let O be the centroid of ABC
The centroid is the origin O (0 , 0) By centroid formula,
0 =5 4
3 y
0 = 5 – 4 + y 0 = 1 + y y = –1
0 =3 23
x
0 = x + 1
x = – 1 [3]
(c)
8 – 2
1 4 . X =
12
10
Order of matrix
12
10 is 2 × 1
Order of
8 – 2
1 4 is 2 × 2
Let order of matix X be a × b
i.e. 2×2
8 – 2
1 4 × Xa×b = 2×1
12
10(1st matrix) (2nd matrix) (resulting matrix)Since, product of matrices is possiblel, only where the number ofcolumns in the first matrix is equal to the number of rows in thesecond.
a = 2Also, the number of columns of product (resulting) matrix is equal to
B CE
D
A
?
9 cm
3 cm
24 cm
Turn overT17 I SP A003
Set B... 13 ...
the number of rows of 2nd matrix. b = 1 Order of matrix X = a × b = 2 × 1
Let X =
xy
8 – 2
1 4
xy =
12
10
8 – 2
4
x y x y =
12
10
8x – 2y = 12 ...(i) x + 4y = 10 ...(ii)
Multiplying (i) by 2 adding (ii) to it, 16x – 4y = 24 x + 4y = 10
17x = 34
x =3417
x = 2Putting this value in (i);
8x – 2y = 12 8 × 2 – 2y = 12 –2y = 12 – 16 –2y = –4
y =–4–2
= 2
X =
xy =
2
2
Hence, (i) 2 × 1 (ii)
22 [4]
A.7.(a) Area of square = 142
n(S) = 196Side of square=2 × diameter of 1 circle
14 = 2 × d
d =142
d = 7 cm
r =2d
r =72
cm
Turn overT17 I SP A003
Set B... 14 ...
Area of 4 circles = 4 × r2
= 4 ×227
×72
×72
= 154 cm2
Area of shaded region = Area of square – Area of 4 circles= 196 – 154= 42 cm2
Let A be the event that the die falls on the shaded region n(A) = 42
P(A) =n(A)n(S)
=42
196
P(A) =3
14 [3]
(b) Internal diameter = 21 cm
Internal radius, r =212
= 10.5 cm
External diameter = 28 cm
External radius, R =282
= 14 cm
Volume of the material of the vessel =12
×43(R3 – r3)
=23
×227
33 2114 –
2
=23
×227
92612744 –8
=23
×227
21952 – 92618
=23
×227
×12691
8
=11×1813
3 × 2
=19943
6 = 3323.83 cm3
Volume of the material of the vessel = 3323.83 cm3 [3]
(c) Sum deposited per month (P) = ` 400Rate of interest (r) = 8%Maturity value = ` 16,176
Turn overT17 I SP A003
Set B... 15 ...
Calculate : Period i.e., no. of months (n)
Maturity value = (P × n) + P ×( +1)2×12
n n × 100
r
Maturity value = (400 × n) + P ×( 1)2 12
n n
× 100r
16,176 = 400n + 400 ×( 1)2 12
n n
×8
100
16,176 = 400n +4 ( 1)
3n n
16,176 =21200 4 4
3n n n
16,176 × 3 = 4 (n2 + 301n)
n2 + 301n =16,176 3
4
n2 + 301n = 12,132 n2 + 301n – 12,132 = 0 n2 + 337 n – 36n – 12,132 = 0 n(n + 337) – 36 (n + 337) = 0 (n – 36) (n + 337) = 0 n – 36 = 0 or n + 337 = 0 n = 36 or n = – 337
No. of months can not be negative, hencen = – 337 is not applicable value.
no. of months = 36 Period of R.D. = 3 years. [4]
A.8.
(a) A =
1 0
2 1 B =
2 3
–1 0 A2 + AB + B2 = A.A + A.B + B.B
=
1 0
2 1
1 0
2 1 +
1 0
2 1
2 3
–1 0 +
2 3
–1 0
2 3
–1 0
=
1 + 0 0 + 0
2 + 2 0 + 1 +
2 + 0 3 + 0
4 – 1 6 + 0 +4 – 3 6 + 0
–2 + 0 – 3 + 0
=
1 0
4 1 +
2 3
3 6 +
1 6
–2 – 3
=
1 + 2 + 1 0 + 3 + 6
4 + 3 – 2 1 + 6 – 3
A2 + AB + B2 =
4 95 4 [3]
Turn overT17 I SP A003
Set B... 16 ...
(b) Sum Invested = ` 3072Rate of Dividend = 5%
Face Value = ` 10Market Value = ` 16
(i) His annual Income :
No. of shares =Sum Invested
Market Value of 1 share
=307216
= 192 No. of shares = 192
Annual Income = No. of Shares × Rate of Dividend× Face Value
= 192 ×5
100 × 10
= ` 96 Annual Income = ` 96(ii) His Percentage income on his investment :
Rate of dividend × F.V = Income % × Market Value 5 × 10 = Income % × 16
Income % =5 ×10
16Percentage income = 3.125% [3]
(c) P = (4 , 2)Q = (–1 , 5)O = (–3 , 2)
Let R = (h , k) andS = (x , y)
PQRS is a parallelogram(Given) Diagonals of PQRS bisect each other Point O is the midpoint of diagonals PR & QS.
For PR :By mid-point formula,
– 3 =4
2 h
– 6 = 4 + h – 6 –4 = h h = – 10
2 =2
2k
4 = 2 + k k = 4 – 2 = 2
P (4 , 2) Q (–1, 5)
S(x, y) R (h , k)
O (–3 , 2)
Turn overT17 I SP A003
Set B... 17 ...
R = ( –10 , 2)For QS :By mid-point Formula,
–3 =12
x
–6 = –1 + x –6 +1 = x x = – 5
2 =5
2y
4 = y + 5 y = – 1 S = (–5 , –1) [4]
A.9.(a) Scale factor, m = 0.8
LMN ~ LMN
L MLM
=M NMN
=L NLN
= m ...(i)
(i)M NMN
= m (From (i))
M N
8
= 0.8.
M N = 0.8 × 8 M N = 6.4 cm
(ii)L MLM
= m (From (i))
5.4LM
= 0.8
LM =5.40.8
LM =548
LM = 6.75 cm [3]
(b) Sum deposited per month (P) = ` 300No.of months (n) = 24 monthsMaturity value = ` 7725Rate of interest (r) = ?
Maturity value = (P × n) + P ×( +1)2×12
n n × 100
r
Maturity value = ` (300 × 24) + P ×24(24 1)
2 12
× 100
r
Turn overT17 I SP A003
Set B... 18 ...
7725 = 7200 +300 24(25)
2 12
× 100
r
7725 – 7200 = 75 r525 = 75 r
r =52575
r = 7% Rate of interest = 7% [3]
(c)
Mode = 53 [4]
A.10.(a) x-intercept = 4 units
Corresponding point on x-axis = (4, 0)The other given point = (2, 3)
Slope m =3 – 02 – 4
=3–2
=3–2
Equation of line through (2, 3) and slope
–32 is
y – y1 = m (x – x1)
y – 3 =–32
(x – 2)
2y – 6 = –3x + 63x + 2y = 12 [3]
20
2
8
4
10
6
12
30 40 50 60 70 800
Mod
e
Class-intervals
Freq
uenc
y
BA
K
L
D
C
X
Y
Turn overT17 I SP A003
Set B... 19 ...
(b) Cost Price of 15 identical articles = ` 840
Cost Price of 1 article =84015
= ` 56
Cost Price of 6 articles = 6 × Cost Price of 1 article Cost Price of 6 articles = 6 × 56 = ` 336
Selling Price of 1 article = ` 65 Selling Price of 6 articles = 6 × Selling Price of 1 article
Selling Price of 6 articles = 6 × 65 = ` 390Rate of Sales Tax = 8%VAT paid by shopkeeper against the sale of 6 articles = ?
C.P. Tax 8% S.P. Tax 8% VAT
Shopkeeper ` 336 ` 26.88 ` 390 ` 31.20 ` 4.32
Tax on Cost Price = Tax paidTax on Selling Price = Tax charged
Tax paid =8
100 × 336 = ` 26.88
Tax charged =8
100 × 390 = ` 31.20
VAT paid by Shopkeeper = Tax charged – Tax paid VAT paid by Shopkeeper = 31.20 – 26.88
VAT paid by Shopkeeper = ` 4.32 [3]
(c) Total Surface area of the cuboid = 2(lb + bh + lh)= 2(42 × 30 + 30 × 20 + 42 × 20)= 2(1260 + 600 + 840)= 5400 cm2
l2 = r2 + h2
=
2142
+ 242
= 72 + 242
= 49 + 576 l2 = 625 l = 25 cm
Curved surface area of the cone = rl
=227
×142
× 25
= 22 × 25= 550 cm2
Area of the circular base = r2
=227
×142
×142
= 11 × 14 = 154 cm2
Turn overT17 I SP A003
Set B... 20 ...
Surface area of the remaining solid = Total surface area of the cuboid – Area of circular base + curved surface area of the cone= 5400 + 550 – 154 = 5796 cm2
Surface area of remaining solid is 5796 cm2 [4]
A.11.(a) (a) Line x = 0 is the equation of y-axis
My(–4, 3) = (4, 3)
Q is (4, 3)
(b) Q (4, 3) is reflected in liney = 0 i.e. x-axis get image R
Mx(x, y) = (x, –y)
R = (4, –3)
(c)
Plot the points P(–4, 3), Q(4, 3) and R(4, –3).Join PQ, QR and PR.
The figure PQR is right angled triangle.(d) PQ = 4 + 4 = 8 units
QR = 3 + 3 = 6 units
Area PQR =12
× PQ × QR
=12
× 8 × 6 = 24 sq. units [4]
2
3
1
-2
-3
-11 2 3 4-5 -4 -3 -2 -1 -0
-4
-5
4
5Y
XX 5
Y
R(4, –3)
Q(4, 3)P(–4, 3)
Turn overT17 I SP A003
Set B... 21 ...
(b) Height (in cm) No. of Cumulativestudents frequency c.f.
135 – 140 0 0140 – 145 12 12145 – 150 20 32150 – 155 30 62155 – 160 38 100160 – 165 24 124165 – 170 16 140170 – 175 12 152175 – 180 8 160
20
40
60
80
100
160
1400 X
(140, 0)
120
140
145 150 155157.5
160 165 170 175 180
(175,152)
(180,160)
(145, 12)
(155,62)
Scale : On X axis : 2 cm = 5 cm height On Y axis : 2 cm = 20 students
(170,140)
(165,124)
(160,100)
(150, 32)No.
ofst
ude
nts
Height
B
P
Q
R
Turn overT17 I SP A003
Set B... 22 ...
(i) Median :N = 160
N2
=160
2 = 80
N = (157.5, 0)
Median height = 157.5 cm
(ii)N4
=160
4 = 4
Q1 = 151 cm
3N4
=1603 ×
4= 3 × 40
= 120
Q3 = 163 cm
Inter quartile range = Q3 – Q1
= 163 – 151
= 12 cm
(iii) Number of students whose height is less than 172 cm = 146
Number of students whose height is above 172 cm = 160 – 146 = 14 [6]