ICSE Board Class IX Mathematics Paper 3 - Solution: SECTION - A (40 Marks) Q. 1
ICSE Board Class IX Mathematics Paper 3 - Solution: SECTION - A (40 Marks) Q. 1
ICSE Board Class IX Mathematics Paper 3 - Solution: SECTION - A (40 Marks) Q. 1
Class IX Mathematics
Paper 3 – Solution
Q. 1.
1 3
(a) P = Rs. 6000, R = 10% p.a., n = 1 years years
2 2
2n
R
A P1 Interest is compounded half yearly
2 100
3
10
6000 1
2 100
3
5
6000 1
100
6000 1.05
3
Rs. 6945.75
Amount = Rs. 6945.75
C.I. = 6945.75 – 6000 = Rs. 945.75
(b) We have
11 7 11 7 xy 77
11 7 11 7
11 77 77 7
x y 77
11 7
18 2 77
x y 77
4
9 1
77 x y 77
2 2
9 1
x ,y
2 2
1
(c)
2
Q. 2.
(a) In ΔDFC, DC2 = DF2 + FC2 [Pythagoras Theorem]
5 DF 4
2 2 2
52 42 DF2 DF2 25 16 DF 3
1 1
Area of DEC 4 4 3 8 3 12 cm2
2 2
FX = DX – DF = 9 – 3 = 6 cm
1 1
Area of trapezium CEBA = 4 4 6 6 6 20 6 60 cm2
2 2
Area of figure ABCDE = area of DEC + area of trapezium ECBA = 12 + 60 = 72 cm2
(b) Given that AB and CD are two chords of a circle with centre O, intersecting at a point
E. PQ is the diameter through E, such that ∠AEQ = ∠DEQ.
To prove that AB = CD.
Draw perpendiculars OL and OM on chords AB and CD respectively.
Now, m∠LOE = 180° – 90° – m∠LEO ... [Angle sum property of a triangle]
= 90° – m∠LEO
⇒ m∠LOE = 90° – m∠AEQ
⇒ m∠LOE = 90° – m∠DEQ
⇒ m∠LOE = 90° – m∠MEQ
⇒ ∠LOE = ∠MOE
In ΔOLE and ΔOME,
∠LEO = ∠MEO
∠LOE = ∠MOE
EO = EO
ΔOLE ≅ ΔOME
OL = OM
Therefore, cords AB and CD are equidistant from the centre.
3
(c)
1
8
2x 5
3
4 2
3
1
2x 5
4
1/3 2
81/3
1
2x 5
22/3 2
81/3
2 1
2 3 2
2x 5
2
2 1
2x 5
3 2
4x 1 15
7
x
2
Q. 3.
(a) Given, log x = a + b and log y = a – b
10x
log 2 = log 10x – log y2 [Using quotient law]
y
= log 10 + log x – 2 log y
= 1 + (a + b) – 2(a – b)
= 1 + a + b – 2a + 2b
= 1 – a + 3b
6. Join CB.
Q. 4.
(a) We can see that ΔABC is a right-angles triangle.
AB2 + BC2 = AC2 ....[By Pythagoras theorem]
152 + BC2 = 252
BC2 = 400
BC = 20 cm
Now BC = DB + CD
20 = DB + 7
DB = 13 cm
Again ADB is a right angled triangle.
AB2 + DB2 = AD2 ....[By Pythagoras theorem]
152 + 132 = 390
BC = 19.8 cm
In the right-angled CDE
ED2 + CE2 = CD2 ....[By Pythagoras theorem]
ED2 CD2 CE2 72 x2
5
In the right-angled AED
ED2 + AE2 = AD2 ....[By Pythagoras theorem]
ED2 AD2 AE2 19.82 25 x
2
(b)
tan cot
2
32
tan2 cot 2 2tan cot 9
1 cos
tan2 cot 2 2tan 9 cot
tan sin
tan2 cot 2 2 9
tan2 cot 2 7
6
Section – B (40 Marks)
Q. 5.
(a) Consider equation, x – 2y = 1 ….(1)
x 1
y
2
x 1 3 5
y 0 1 2
Since the lines intersect at (3, 1), therefore the solution is x = 3 and y = 1.
7
(b) In 15 days A and B together can do a piece of work.
1
Therefore, in 1 day they do work
15
Let us assume that A takes x days and B takes y days to do the work alone.
1
So A’s one day’s work =
x
1
B’s one day’s work =
y
1 3 1
.
x 2 y
3x 2y 0
2y 3x
3x
y ....(i)
2
1 1 1
Also,
x y 15
1 2 1
x 3x 15
32 1
3x 15
3x 75
x 25
3 25
y 37.5
2
Hence, A will do the work alone in 25 days and B will do it alone 37 and half days.
(c) Let there be n sides of the polygon. Then, each interior angle is of measure
2n 4
n 90
2n 4
90 108
n
(2n 4) 90 108n
180n 360 108n
180n 108n 360
72n 360
n 5
Hence the given polygon has 5 sides.
8
Q. 6.
5600 14 1
(a) (i) Interest for first year = Rs. 784
100
(ii) Amount at the end of the first year = 5600 + 784 = Rs. 6384
(iii) Interest for the second year =
6384 14 1
Rs. 893.76 Rs. 894 to the nearest rupee
100
(b)
(i) Since, the point P lies on the x-axis, its ordinate is 0.
(ii) Since, the point Q lies on the y-axis, its abscissa is 0.
(iii) The co-ordinates of P and Q are (–12, 0) and (0, –16) respectively.
9
Q. 7.
(a)
3n 1 9n 1
3n(n 1) 3n 1 n 1
3
n 1
n 1 n 1
3
3n(n 1) 9n 1
3n 1 3
n 1 n 1
n(n 1)
3 3
n 1
3
3n 1 3
n2 1
3
n2 n n 1
2
3
3n 1
3
n2 1
2 2n 2
3n n 3
2 1 (n2 n)(2n 2)
3n 1 n
2 1 n2 n 2n 2
3n 1 n
32
1
32
1
9
(b)
1
Area of an isosceles = b 4a2 b2
4
(where b is the base and a is the length of equal sides)
Given, b = 8 cm and area =12 cm2
1
8 4a2 82 12
4
4a2 82 6
4a2 82 36
4a2 100
a2 25
a 5 cm
Perimeter = 2a + b = 2 5 + 8 = 18 cm
10
(c) Given AC = CD
To prove: BC < CD
Proof: In ACD,
mACD = 180 – 70 = 110 [Linear pair]
70
CAD = ADC = 35 [Angles opposite to equal sides are equal]
2
In ABC,
mBAC = 70 – 35 = 35 [BAC = BAD – CAD]
mABC = 180 – (70 + 35) [Sum of all s of a is 180]
= 75
BAC < ABC
BC < AC
So, BC < CD [Since AC = CD]
Q. 8.
(a)
2 mn
cos =
mn
Now,
sin2 cos2 1
sin2 1 cos2
2
2 mn
1
m n
4mn
1
m n 2
m n 4mn
2
m n
2
m2 n2 2mn 4mn
m n
2
m2 n2 2mn
m n
2
m n
2
m n
2
2
mn
mn
mn
sin
mn 11
(b) Mean = 20
Number of terms = 5
Total sum = 20 5 = 100
Let the excluded number be x.
Then,
100 x 23
4
100 – x = 23 4 = 92
x=8
Hence, the excluded number is 8.
(c) Let the side of each of the three equal cubes be 'a' cm.
Surface area of one cube = 6a2 cm2
Therefore, sum of surface areas of the three cubes = 3 × 6a2 = 18a2 cm2
Now,
Length of the new cuboid = 3a cm
Breadth of the new cuboid = a cm
Height of the new cuboid = a cm
Total surface area of the new cuboid = 2[(3a × a) + (a × a) + (a × 3a)]
= 2[3a2 + a2 + 3a2]
= 2[7a2]
= 14a2 cm2
Thus, the required ratio of T.S.A. of the new cuboid to that of the sum of the S.A. of the
3 cubes = 14a2 : 18a2 = 7 : 9.
Q. 9.
(a) Given: In quadrilateral ABCD; AD = BC. P, Q, R, S are the mid-points of AB, BD, CD and
AC respectively.
To Prove: PQRS is a rhombus.
1
Proof: In ACD, RS||AD and RS AD ....(i)
2
[Line joining the mid-points of the two sides of triangle is
parallel and half of the third side.]
Similarly,
1
In ABD, PQ||AD and PQ AD ....(ii)
2
1
In BCD, QR||BC and QR BC ....(iii)
2
1
In ABC, SP||BC and SP BC ....(iv)
2
As AD = BC [Given]
RS = PQ = QR = SP and RS||PQ and QR||SP [From (i), (ii), (iii) and (iv)]
Hence PQRS is a rhombus. 12
(b) Given that we have to construct a grouped frequency distribution table of class size 5.
So, the class intervals will be as 0 – 5, 5 – 10, 10 – 15, 15 –20, and so on.
Required grouped frequency distribution table is as follows:
Distance (in km) Tally marks Number of engineers
0–5 5
5 – 10 11
10 –15 11
15 – 20 9
20 – 25 1
25 – 30 1
30 – 35 2
Total 40
(c)
log x (8x 3) log x 4 2
8x 3
log x 2
4
8x 3 2
x
4
8x 3 4x 2
4x 2 8x 3 0
4x 2 6x 2x 3 0
2x(2x 3) 1(2x 3) 0
(2x 3)(2x 1) 0
2x 3 0 or 2x 1 0
3 1
x or x
2 2
13
Q. 10.
(a)
Let us assume, on the contrary that 5 is a rational number.
a
Therefore, we can find two integers a, b (b ≠ 0) such that 5
b
Where a and b are co-prime integers.
a
5
b
a 5b
a2 5b2
Therefore, a2 is divisible by 5 then a is also divisible by 5.
So a = 5k, for some integer k.
Now,a2 (5k)2 5(5k 2 ) 5b2
b2 5k 2
This means that b2 is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
So our assumption that 5 is rational is wrong.
(b)
1 1
tan 1 tan 2 2 3
tan 1 2
1 tan 1 tan 2 1 1
1
2 3
32 5
5 6
tan 1 2 6 6 1
1 6 1 6 5
1
6 6
tan 1 2 1 tan45
1 2 45
14
x2 1
(c) 4
x
x2 1 4x
x2 4x 1 0 ....(i)
On dividing equation (i) by x, we have
1
x4 0
x
1
x 4 ....(ii)
x
On cubing equation (ii) both sides, we have
3
1
x + x = 4
3
1 1 1
x3 + 3 + 3× x × x+ = 64
x x x
1
x3 + 3 + 3 × 4 = 64
x
1
x3 + 3 = 64 12
x
1
x3 + 3 = 52
x
2 1
2x3 + 3 = 2 x3 + 3 = 2 × 52 = 104
x x
Q. 11.
(a) 4a3b – 44a2b + 112b
4ab a2 11a 28
4ab a2 7a 4a 28
4aba(a 7) 4(a 7)
4ab a 7 a 4
15
(b) Construction: Draw TM ⊥ QS
1 1
Area of RQS QS RN 35 20 350 cm2
2 2
Now, QS QM MS
35 25 MS
MS 10 cm
In STM,
MS2 TM2 ST2
TM2 ST2 MS2 (26)2 (10)2 676 100 576
TM 24 cm PQ
1 1
Area of trapezium PQST (PT QS) PQ (25 35) 24 720 cm2
2 2
Thus, area of given figure Area of RQS Area of trapezium PQST
350 cm2 720 cm2
1070 cm2
16