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ICSE Board
Class X Mathematics
Board Paper – 2013 Solution
Sol.1
(a) A+2X=2B+C
02
6222
20
04
04
23
02
62
2008
04462
02
62
28
422
02
62
28
422
26
1042
13
52
26
104
2
1
(b) P = Rs.4000, C.I. = Rs.1324, n = 3 years
Amount, A = P + C.I. = Rs.4000 + Rs.1324 =Rs. 5324
A=P
nr
1001
5324=4000
3
1001
r
1000
1331
4000
5324
1001
3
r
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33
10
11
1001
r
10
11
1001
r
10
11
10
11
100
r
10r %
(C) The observations in ascending order are:
11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47
Number of observations = 9 (odd)
thn
Median
2
1observation = 5 th observation
∴ x+4=24
x=20
Thus, the observations are:
11, 12, 14, 18, 24, 29, 32, 38, 47
9
473832292418141211 Mean
9
225
=25
Sol.2
(A) Let the number added be x.
xxxx 4320::15:6
x
x
x
x
43
20
15
6
xxxx 1520436
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22 1520300436258 xxxxxx
4214 x
3x
Thus, the required number which should be added is 3.
(b) Let p 142 23 bxaxxx
Given, 2x is a factor of p x
02Re pmainder
014222223
ba
0142416 ba
4a + 2b + 2 = 0
2a + b + 1 = 0 …. (1)
Given, when p(x) is divided by (x - 3), it leaves a remainder 52.
523 p
5214333223
ba
052143954 ba
01239 ba
043 ba …. (2)
Subtracting (1) from (2), we get
505 aa
From (1)
110110 bb
(C)
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Steps for calculation of mode.
(i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.
(ii) Mark the inner corner of adjacent rectangles as C and D.
(iii) Join AC and BD to intersect at K. From K, draw KL perpendicular to x-axis.
(iv) The value of L on x- axis represents the mode.
Mode = 13
Sol.3
(a) 0000 31sec59sin210cos.80cos3 ec
000000 31sec3190sin210cos1090cos3 ec
000 31sec,31cos210cos10sin3 ec
sin90cos,cos90sin 00
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1213 1sec.cos,1cos.sin c
523
(B) (i) ,ABDin
0180 ADBABDDAB
000 1807065 ADB
0000 456570180 ADB
000 904545, BDCADBADCNow
ADC is the angle of semi-circle so AC is a diameter of the circle.
(ii) ADBACB (angle subtended by the same segment)
045 ACB
(C)
(i) Radius AC 225723
22125
14425
169
=13 units
(ii) Let the coordinates of B be (x, y)
Using mid-point formula, we have
2
32
x
2
75
y
x 34 y 710
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7x 17y
Thus, the coordinates of points B are (-7, 17).
Sol.4
(a) x²-5x-10 = 0
Use quadratic formula: a = 1, b = -5, c = -10
a
acbbx
2
42
12
1014552
x
2
40255 x
2
655 x
2
06.85 x
2
06.3,
2
06.13 x
53.1.53.6 x
(b) (i) In ,DECABCand
090 DECABC (perpendiculars to BC)
DCEACB (Common)
ΔABC~ΔDEC (AA criterion)
(ii)Since ΔABC~ΔDEC
CD
AC
DE
AB
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CD
15
4
6
606 DC
106
60CD cm
(iii) It is known that the ratio of areas of two similar triangles is equal to the square of the ratio of
their corresponding sides.
4:916:364:6:: 2222 DEABDECarABCar
(c)
(i)
(ii) Co-ordinates of A' = (-6, -4)
Co-ordinates of B' = (0, -4)
(iii) ABA'B' is a parallelogram.
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(iv) AB = A'B' = 6 units
In ΔOBO'
OO' = 3 units
OB = 4 units
BO’ 2,
2 00 OB
22 34
25
= 5 units
Since BO' = 5 units
BA' = 10 units = AB'
Perimeter of ABA'B' = (6 + 10 + 6 + 10) units = 32 units
SECTION – B (40 MARKS)
Sol.5
(a) The given inequation is Rxxx
,6
1
3
11
23
3
11
23
xx
3
4
23
xx
6
1
3
11
2
x
3
4
6
32
xx and
3
4
6
1
2
x
3
4
6
5
x
6
81
2
x
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85 x 6
9
2
x
5
8x
6
18x
6.1x 3x
The solution set is {x: x R and 1.6 x < 3}
It can be represented on a number line as follows:
(b) Maturity amount = Rs 8088
Period (n) = 3 yrs = 36 months
Rate = 8% p.a
Let x be the monthly deposit.
S.I. =
100122
1 rnnp
xp 44.4100
8
24
3736
Total amount of maturity xxx 44.4044.436
Now,
40.44x=8088
200 x
Thus, the value of monthly installment is Rs 200
(c)
(i) No. of shares = Rs 50
Market value of one share = Rs 132
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Salman's investment = (132 x 50) = Rs 6600
(ii) Dividend on one share = 7.5% of Rs 100 = Rs 7.50
His annual income = 50 Rs7.50 = Rs375
(iii) Salman wants to increase his income by 150.
Income on one share = Rs 7.50
No. of extra shares he buys 2050.7
150
So.6
(a)
LHS
CosA
CosA
1
1
CosACosA
CosACosA
11
11
2
2
1
1
CosA
ACos
2
2
1 CosA
ASin
RHSCosA
SinA
1
(b) In cyclic quadrilateral ABCD,
0180 DB (opp. angles of cyclic quadrilateral are supplementary)
00 180100 ADC
080 ADC
Now, in ∆ACD,
0180 ADCCADACD
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40 000 18080 CAD
000 60120180 CAD
Now CADDCT (angles in the alternate segment)
∴ DCT = 60 0
(c)
On completing the given table, we get:
Principal for the month of Feb = Rs 4500
Principal for the month of March = Rs 4500
Principal for the month of April = Rs 4500
Principal for the month of May = Rs 6730
Principal for the month of June = Rs 1730
Principal for the month of July = Rs 7730
Total Principal for 1 month
= Rs (4500+4500+4500+6730+1730+7730) Rs = 29690
P = Rs 29690, ,12
1yearsT R=6%
∴ Interest 100
TRP
12100
1629690
= Rs 148.45
Sol. 7
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(a) The vertices of ABC are A (3, 5), B (7, 8) and C (1, -10).
Coordinates of the mid-point D of BC are
2,
2
2121 yyxx
2
)10(8,
2
17
2
2,
2
8
1,4
Slope of AD12
12
xx
yy
34
51
61
6
Now, the equation of median is given by:
11 xxmyy
365 xy
1865 xy
0236 yx
(b) Since, the shopkeeper sells the article for Rs 1500 and charges sales-tax at the
rate of 12%.
Tax charged by the shopkeeper = 12% of Rs 1500 = Rs 180
VAT = Tax charges - Tax paid
Rs 36 = Rs 180 - Tax paid
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Tax paid = Rs 144
If the shopkeeper buys the article Rs x
Tax on it = 12% on Rs x = Rs 144
x Rs 144 12
100Rs 1200
Thus, the price (inclusive of tax) paid by the shopkeeper = Rs 1200 + Rs 144 = Rs 1344.
(c)
(i) In ∆ AEC,
Tan 30EC
AE0
x
y
3
1
)1.......(3
xy
In ∆ DBA,
Cot 60EC
AE0
603
1 x
mx 64.343203
60
Therefore, the horizontal distance between AB and CD = 34.64 m.
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(ii) Substituting the value of x in (1),
3
320 y m20
Height of the lamp post = CD = (60 - 20) m = 40 m
Sol. 8
(a)
yy
xx
4
3
1
2=
12
5
yy
xx
42
32=
12
5
1532 xxx
and 21242 yyy
(b) Sphere : R = 15 cm
Cone : r = 2.5 cm, h = 8 cm
Let the number of cones recasted be n
n Volume of one cone = Volume of solid sphere
32
3
4
3
1Rhrn
3215485.2 xn
85.25.2
1515154
n
270n
Thus, 270 cones were recasted.
(c) x 2 + (p - 3) x + p = 0
Here, A = 1, B = (p - 3), C = p
Since, the roots are real and equal, D = 0
042 acb
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01432
pp
04692 ppp
09102 pp
0)9)(1( pp
1 p or 9p
Sol. 9
(a) Area of the quadrant OACB =
2
4
1r
5.35.37
22
4
1
2625.9 cm
Area of the triangle OAD =
heightbase2
1
25.32
1
2,5.3 cm
Shaded Area = Area of quadrant OACB - area of triangle OAD area of triangle OAD
= 9.625 – 3.5 cm 2
= 6.125 cm 2
(b) Number of white balls in the bag = 30
Let the number of black balls in the box be x
Total number of balls = x + 30
P (drawing a black ball) 30
x
x
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P (drawing a white ball) 30
30
x
It is given that:
P (drawing a black ball) x5
2 P(drawing a white ball)
30
30
5
2
30
xx
x
30
12
30
xx
x
12 x
Therefore, number of black balls in the box is 12.
(c)
Here, A = 55, h = 10
Mean hf
fdA
1050
2755
4.555
6.49
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Sol. 10
(i) Steps of constructions:
(a) Draw a line segment BC = 6 cm.
(b) At B, draw a ray BX making an angle of 1200 with BC.
(c) From point B cut an arc of radius 3.5 cm to meet ray BX at C
(d) Join AC
ABC is the required triangle
(ii)
(a) Bisect BC and draw a circle with BC as diameter.
(b) Draw perpendicular bisectors of AB. Let the two bisectors meet the ray of angle
bisector of ABC at point P. P is equidistant from AB and BC
(iii) On measuring BCP = 30 0
(b)
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602
120n
(i) Through marks 60, draw a line segment parallel to x-axis which meets the curve
at A. From A, draw a line perpendicular to x-axis meeting at B.
Median = 43
(ii) Through marks 75, draw a line segment parallel to y-axis which meets the curve
at D. From D, draw a line perpendicular to y-axis which meets y-axis at 110.
Number of students getting more than 75% = 120 - 110 = 10 students
(iii) Through marks 40, draw a line segment parallel to y-axis which meets the curve
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at C. From C, draw a line perpendicular to y-axis which meets y-axis at 52.
Number of students who did not pass = 52.
Sol. 11
(a) Let the coordinates of A and B be (x, 0) and (0, y) respectively.
Given P divides AB is the ratio 2:3,
Using section formula, we have:
32
3023
x
32
0324
y
5
33
x
5
24
y
x315 20=2y
5x 10y
Thus, the coordinates of A and B are (-5, 0) and (0, 10) respectively
(b) 8
17
2
12
4
x
x
Using componendo and dividendo,
817
817
21
214
24
xx
xx
9
25
1
122
22
x
x
2
2
22
3
5
1
1
x
x
3
5
1
12
2
x
x
35
35
11
1122
22
xx
xx (Using componendo and dividend)
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2
8
2
2 2
x
42 x
2 x
(c) Original cost of each book = Rs x
Number of books brought for Rs x
960960
In 2 nd case:
The cost of each book = (x - 8)
Number of books bought for Rs 8
960960
x
From the given information, we have:
4960
8
960
xx
48
8960960960
xx
xx
19204
8960)8(
xx
0192082 xx
04048 xx
48 x or 40
But x can't be negative,
x = 48
Thus, the original cost of each book is Rs 48