icse board class x mathematics board paper 2013...
TRANSCRIPT
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
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
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)
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
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
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
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.
(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
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
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
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
(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
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.
(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
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
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
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)
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
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)
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