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Class-IX CBSE
Latest Pattern Sample Paper {Mathematics}
Term-I Examination (SA I)
Time: 3hours
Max. Marks: 90
General Instructions
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into 4 sections A, B, C
and D. Section A comprises of 4 questions of 1 mark each, Section B comprises of
6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks
each and Section D comprises of 11 questions of 4 marks each.
(iii) There is no overall choice. However, internal choice has been provided in 1
question of 2 marks, 3 questions of 3 marks each and 2 questions of 4 marks
each. You have to attempt only one of the alternatives in all such questions.
(iv)Use of calculator is not permitted.
Section-A
1.
Let x be a rational number and y be an irrational number. It x + y necessarily an
irrational number?
Solution:
Yes, it is necessarily that sum of rational and irrational numbers is an irrational.
2.
If an isosceles right angled triangle has an area of 12 cm 2. Then find the length of its
base.
Solution:
Let equal sides of an isosceles triangle be x cm i.e. AB-BC= x cm.
Given, area of an isosceles triangle is 12 cm2.
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 Area = - x Base × Altitude
 12 
12
x  x
2
 24  x2
 x  24 [taking positive square root]
Hence, length of the base of a triangle is
3.
24 cm.
Find value of the polynomial p{x) = 5x + 3 – 4x2 at x = 2.
Solution:
Given,
p(x) = 5x + 3 – 4x2
At x=2, then p(2) = 5 × 2 + 3 – 4 × (2)2
= 10 + 3– 4 ×4
= 13 – 16 = – 3
Hence, the value of p(x) is – 3 at x = 2.
4.
Euclid divided his book 'Elements' into how many chapters?
Solution:
Euclid divided his book 'Elements' into 13 chapters.
Section-B
5.
The polynomials kx3 + 3x2 – 8 and 3x3 – 5x + k are divided by x + 2. If the remainder in
each case is the same, then find the value of k.
Solution:
Let p(x) = Kx3+ 3x2 – 8
and q(x) = 3x3 – 5x + k.
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When we divided p(x) and q(x) by x + 2, we get the remainder p(–2) and q(–2).
But according to the question, p(–2) = q(–2)
(–2)3 + 3(–2)2 – 8 = 3(–2)3 – 5(–2) + k
 – 8k +12 – 8 = – 24 +10 + k
 8k + 4 = –14 + k
 –9k = – 18
 k =2
0
6.
1

The angles of a triangle are (x – 40)°), (x – 20)° and  x  10  , find the angles of
2

triangle.
Or
In the given figure,  AOC and  BOC form a linear pair and a – b = 70°, find the values of
a and b.
Solution:
Let angles of a triangle be
 A = (x – 40)°,  S = (x – 20)°
0
1

and C   x  10 
2

We know that,  A +  B +  C = 180°
[by angle sum property of a triangle]
 x – 40° + x – 20° + – x –10° = 180°
1
 2x  x  700  1800
2
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 2x  x  2500
2

4x  x
 2500
2
 x
2500  2
5
 x = 1000
  A = x – 40° = 100° - 40° = 60°
 B = x – 20° = 100° – 20° = 80°
and  C =
1
1
x – 10° = × 100° – 10°= 50°-10° = 40°
2
2
Or
According to the question,
 AOC +  BOC = 180°
[linear pair axiom]
 a +b = 180°
...(i)
and
...(ii) [given]
a – b = 70°
On adding Eqs. (i) and (ii), we get
a + b =180°

a  b  700
2a  2500
2500
 a
 1250
2
On putting the value of a in Eq. (i), we get
125° + b = 180°
 b = 180° – 125°
 b = 55°
7.
Express 15.7 12 in the form p/q.
Solution:
Let x = 15.7 12
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 x = 15.71212...
On multiplying both sides by 10, we get
10x = 157.1212
...(i)
On multiplying both sides by 100, we get
1000x = 157121212
...(ii)
On subtracting Eq (i) from Eq (ii), we get
1000x – 10x = 15712.1212...– 1571212...
 990x =15555
 x
15555
990
 x
3111
198
[dividing numerator and denominator by 5]
8.
Two points with coordinates (4, 3) and (4, – 2) lie on a line, parallel to which axis?
Solution:
As x-coordinate of both points is 4.
So, both points lie on the line x = 4 which is parallel to y-axis.
9.
Evaluate 103 × 107, without multiplying directly.
Solution:
103 × 107=(100 + 3)(100 + 7)
= (100)2 +100 × 7 + 3 ×100 + 3 × 7
= 10000+ 100(7+3)+21
= 10000 + 1000 + 21
= 11021
10. Find the factors of polynomial
4x2 + y2+4xy + 8x + 4y + 4
Solution:
4x2 + y2 + 4xy + 8x+4y+4
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= (2x)2 + (y)2 + 2(2x)(y) +2(2x)(2) + 2(2)(y) + 22
= (2x)2+(y)2+(2)2 + 2(2x)(y) + 2(2)(y) + 2(2x)(2)
= (2x + y + 2)2
[
(a + b + c)2 =a2 + b2 +c2 + 2ab + 2bc + 2ca]
Section-C
11. Evaluate
(i) (104)3
(ii) (999)3
Solution:
(i) We have, (104)3 =(100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
[
(a + B)3 = a3 + b3 +3ab (a + b)]
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
(ii) We have, (999)3 = (1000 – 1)3
= (1000)3 – (1)3 – 3 × 1000 × 1 × (1000–1)
[
(a – b)3 = a3 – (a – b)3 = a3 – b3 (a – b)]
= 1000000000 –1 – 3000 × 999
= 1000000000 – 1 – 2997000
= 997002999
12. Find the remainder when p(x) = x3 + 3x2 + 3x +1 is divided by 5 + 2x.
Or
Simplify
(x + y + z)2 – (x – y + z)2.
Solution:
By remainder theorem, when p(x) is divided by
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  5 
5 + 2x = 2  x      , then remainder is given by
  2 
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 5
p   .
 2
Now, p(x) = x3+3x2+3x+1
3
2
 5  5
 5
 5
 p         3    3    1
 2  2
 2
 2
=
125  25  15
 3    1
8
 4  2
=
125 75 15
  1
8
4 2
=
125  150  60  8
27

8
8
Hence, the required remainder is 
27
.
8
Or
(x + y + z)2 – (x – y + z)2
= [(x + y +z)2 – {x + (–y) + z}2]
= (x2 +y2 +z2 + 2xy + 2yz + 2zx) – [x2 + (–y)2 + (z)2 + 2(x) – (–y) + 2(–y) (z) + 2zx]
[
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (x2 + y2 + z2 + 2xy + 2yz + 2zx)
– (x2 + y2 + z2 – 2xy – 2yz + 2zx)
= x2 + y2 + z2 + 2xy + 2yz + 2zx – x2 – y2 – z2 + 2xy + 2yz – 2zx
= 4xy + 4yz
= 4y(x + z)
13. Evaluate
15
.
10  20  40  5  80
take
5 = 2.236 and 10 = 3.162)
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Or
If x = 9  4 5 , then find the value of
x
1
.
x
Solution:
Now,
10  20  40  5  80
= 10  2 5  2 10  5  4 5
= 3 10  3 5  3( 10  5)
15
10  20  40  5  80

=
15
5

3( 10  5)
10  5
=
15
10  5

10  5
10  5
[multiplying numerator and denominator by 10  50 ]
=
5( 10  5)
5( 10  5)

10  5
( 10)2  ( 5)2
[
=
a2 – b2 = (a – b) (a + b)]
5( 10  5)
5
= 10  5  3.162  2.236

10  3.162 


and 5  2.236
= 5.398
Or
Given x = 9 – 4 5

1
1

x 94 5
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94 5

94 5 94 5
[multiplying numerator and denominator by 9 + 4 5 ]
=
94 5
94 5
81  80
[
a  ba  b  a2  b2 ]
and x 
1
 9  4 5  9  4 5  18
x
..(i)
2
1 
1
1

Now,  x 

x


2
x.

x
x
x

[
(a – b)2 = a2 + b2 – 2ab]
= x
1
2
x
= 18 – 2 = 16

[from Eq. (i)]
1
 x 4
x
x
14. Simplify
 21  32 
 22 34 


7/2
 22  33 
  3 5 
 2 3 
5/2
.
Or
If 2x = 3y = 6–z, then prove that
1 1 1
   0.
x y z
Solution:
 21  32 
 22  34 


7/2
 32  34 
= 2
1
2  2 
 22  33 
 3
5 
2  3 
7/2
5/2
 33  35 
 3
2
2  2 
 m 1 
 a  am 


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7/2
5/2
 36 
= 3
2 
 38 
 5 
2 
3 
=
2 
 25 
 8 
3 
[
a m  1/ am ]
36  7/2 25 5/2
= 37/2  8  5/2
2
3
[
a 
6
3
7/2
7/2
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[ a m  a n  a m n ]
5/2
m
n
 a mn ]
321 225/2
= 21/2  20
2
3
= 3
2120
 2
25/221/2
 am
m n 
 an  a 


= 3  2  3  4  12
1
To prove,
2
1 1 1
  0
x y z
Let 2x = 3y = 6-z = k
Thus, 2= k1/x, 3 = k1/y = and 6 = k–1/z
Now, 2 × 3 = 6
 k1/x  k1/y  k 1/z
[put 2 = k1/x , 3  k1/y , 6  k 1/z ]
 k1/x1/y  k 1/z [ am  an  amn ]
On comparing the exponent, we get
1 1
1
 
x y
z

1 1 1
  0
x y z
15. In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that
AC = BD and AC BD.
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Solution:
Given In the figure, AO = OB, OC = OD
To prove AC = BD and AC
BD
Proof In  AOC and  BOD, we have AO = BO
[
O is the mid-point of AB]
 AOC =  BOD [vertical opposite angles]
and CO = DO
[
 AOC   BOD
[by SAS congruence rule]
Then,
and
O is the mid-point of CD]
AC = BD
[by CPCT]
 CAO =  DBO
[by CPCT]
  CAB =  DBA
...(i)
But  CAB and  DBA are alternate interior angles formed when transversal AB
intersects CA at A and DB at B.
 AC
BD
Hence, AC = BD and AC
BD.
Hence proved.
16. Write down Euclid's five postulates.
Solution:
First A straight line may be drawn from any one point to any other point.
Second A terminated line can be produced indefinitely.
Third A circle can be drawn with any centre and any radius.
Fourth All right angles are equal to one another.
Fifth For every line L and for every point P not lying on L, there exists a unique line M
passing through P and parallel to L.
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17. In the given figure, if ,AB CD, EF  CD and  GED = 126°, then find  AGE,  GEF and
 FGE.
Solution:
Given, AB CD, GE is a transversal line, then
 AGE =  GED
  AGE =126°
Now,
[alternate interior angles]
 GEF =  GED –  FED
  GEF = 126° – 90°
[
 FED = 90°]
  GEF =36°
Also,  FGE +  AGE = 180°
[linear pair axiom]
  FGE +126° =180°
  FGE = 180°– 126° = 54°
Hence,  AGE =126°,  GEF =36°
and
 FGE = 54°.
18. Prove that two distinct lines cannot have more than one point in common.
Solution:
Given Two distinct lines l1 and l2
To prove Lines l1 and l2 have only one point in common.
Proof Suppose, lines l1 and 12 intersects at two distinct points, say P and Q.
Then, line l1 contains points P and Q.
Also, line 12 contains points P and Q.
So, two lines l1 and l2 pass through two distinct points P and Q.
But only one (unique) line can pass through two distinct points.
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So, our assumption that two lines can pass through two distinct points is wrong.
Hence, two distinct lines cannot have more than one point in common.
19. If the sides of a triangle are produced in order, then prove that the sum of the exterior
angles so formed is equal to four right angles.
Let ABC be a triangle whose sides AB, BC and CA are produced in order, forming
exterior  CBF,  ACD and  BAE.
In  ABC, we have
 CBF =  1 +  3
...(i)
[exterior angle is equal to the sum of opposite interior angles]
Similarly,
and
 ACD =  1 +  2
 BAE =  2 +  3
...(ii)
...(iii)
On adding Eqs. (i), (ii) and (iii), we get  CBF +  ACD +  BAE =2[  1 +  2 +  3]
= 2 × 180° = 4 × 90°
[by angle sum property of a triangle is 180°]
  CBF +  ACD +  BAE = 4 right angles
Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so
formed is equal to four right angles.
20. Suppose E and F are the mid-points of the sides AB and AC of  ABC. CE and BF are
produced to X and Y respectively, so that EX = CE and FY = BF.AX and AY are joined. Find
in figure, a triangle congruent to  AEX and demonstrate the congruency. Prove that XAY
is a straight Line.
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Given E and F are the mid-points of the sides AB and AC of AABC, CE and BF are
produced to X and Y, respectively such that EX = CE and FY = BF.
To prove (i)  AEX =  BFC
(ii) XAY is a straight line.
Construction Join AX and AY.
Proof In  AEX and  BEC, we have
AE = BE
[
E is the mid-point of AB]
 AEX =  BEC
[vertically opposite angles]
and EX = EC
[given]
 AEX  BEC
[by SAS congruence rule]
  XAE =  CBE
or
[
[by CPCT]
 XAB =  CBA
 XAE =  XAB and  CBE =  CBA]
But  XAB and  CBA are alternate interior angles formed when a transversal AB meets
XA at A and BC at B.
 XA BC ...(i)
Similarly, it can be proved that
AFY  CFB and AY BC
...(ii)
From Eqs. (i) and (ii), we get
BC XA and BC Ay
Hence, XAY is a straight line.
Hence proved.
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Section-D
21. In the given figure, if AB = AC, then prove that AF > AE.
In the given figure,  B <  A and  C <  D. Prove that AD< BC.
Solution:
Given in the Figure, AB = AC
To prove AF > AE
Proof hi  ABC, we have
AC = AB
[given]
 1 = 2
...(i)
[angles opposite to equal sides are equal]
In  DBE, whose side BE is extended to A, we have
5 > 1
...(ii)
[exterior angle > each opposite interior angle]
From Eqs. (i) and (ii), we get
5 > 2
...(iii)
Now,  3 =  4
...(iv)
[vertical opposite angles]
Considering  FDC, whose side DC is extended to B, we have
2 > 3
...(v)
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From Eqs. (iv) and (v), we get
2 >4
...(vi)
From Eqs. (iii) and (vi), we get
5 > 4
 AF > AE
Or
[sides opposite to greater angle is longer]
Hence,
AF > AE
Or
Given In the figure  B <  A and  C <  D
To prove AD < BC
Proof In  ABO, we have
 B < A
[given]
 AO < BO .
..(i)
[sides opposite to smaller angle is shorter]
In  COD, we have
C < D
 OD < OC
...(ii)
[sides opposite to smaller angle is shorter]
On adding Eqs. (i) and (ii), we get
AO + OD < BO + OC
 AD < BC
Hence,
AD < BC
22. Draw the quadrilateral with vertices (– 4,4), (–6,0), (–4, –4)and (–2, 0). Name the type of
quadrilateral and find its area.
Solution:
Firstly, plot the points A(–4,4), B(–6,0),C(–4, –4) and D(–2,0) on a graph paper and join
all these points.
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We obtained quadrilateral is a rhombus because its all sides are equal.
i.e.,
AB = BC = CD = DA
and diagonals are not equal.
Now, area of a rhombus =
=
d1  4



and d2  8 
1
 d1  d2
2
1
48
2
= 16 sq units
Hence, are a of quadrilateral is 16 sq units.

23. If x = 2  5
1
2
  2  5 
1
2

 

1
2
and y  2  5  2  5 , then evaluate x2 + y2.
Solution:

Given, x  2  5

and y  2  5

 
1/2
1/2
 2 5

  2 5

1/2

1/2
Now,  x  y   [(2x  5)1/2  (2  5)1/2 + (2  5)1/2  (2  5)]2
2
= [2(2  5)1/2 ]2
1
2
2
= 4[2  5]
[
a 
m
n
 a mn ]
= 4(2  5)  8  4 5
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and 2xy  2 [(2  5)1/2  (2  5)1/2 ] × [(2  5)1/2  (2  5)1/2 ]
= 2[{(2  5)1/2 }2  {(2  5)1/2 }2 ]
[
a  b a  b  a2  b2 ]
= 2[2  5  2  5]  2  4
=8
  x  y   2xy  8  4 5  8
2
 x2  y 2  2xy  2xy  4 5
[
(a + b)2 = a2 + b2 + 2ab]
 x2 + y2 = 4 5
24. In a class, a teacher conducted a small quiz to solve a question on blackboard. She needs
two students and a prize will be given to the students who solve the question first. For
this purpose she choose a boy and a girl. The problem is that in the given figure, AB CD.
Find the values of x, y and z.
Which of these values is depicted by the teacher in this question?
(i) Social value
(ii) Freedom
(iii) Truth value
(iv) Gender equality
Solution:
In ACO, AO  AC
Then,
[given]
 ACO =  AOC
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[angles opposite to equal sides are equal]
In  ACO +  AOC +  CAO =180°
[ by angle sum property of a triangle is 180° ]
  ACO +  ACO + 74°=180°
[
 AOC =  ACO]
  ACO =180° – 74°
  ACO = 106°
1060
 530
 ACO 
2
i.e.,  ACO =  AOC = 53°
Again,
 AOC +  COP = 180°
[linear pair axiom]
 53° + x = 180°
[
 AOC = 53°]
 x = 180° – 53°
 x = 127° =  POC
In A=  POC,  POC +  OCP +  CPO = 1800
[by angle sum property of a triangle is 1800]
 127°+15°+ y =180°
 142° + y =180°
 y = 180° – 142°
= 380
Since, AB
CD and AP is a transversal line.
Then, y=z [alternate interior angles]
z = 38°
Hence, x = 127°, y = 38° and z = 38°
Value depicted by the teacher is gender equality.
25. If in two right triangles, the hypotenuse and one side of one triangle are equal to the
hypotenuse and one side of the other triangle, then prove that the two triangles are
congruent.
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Solution:
Given Two right angled  ABC and  DEF in which  B =  E = 90°, AC = DF and
BC = EF
To prove  ABC   DEF
Construction Produce DE to G such that GE = AB.
Join GF.
Proof In  ABC and  GEF, we have
BC = EF
[given]
 ABC =  GEF = 90°
[by construction]
AB = GE
[by construction]
 ABC  GEF
[by SAS congruence rule]
Then,  A =  G
and AC = OF
But
[by CPCT ]
...(i)
AC = DP
[by CPCT]
[given]
 GF = DF
 D = G
[angles opposite to equal sides are equal]
 A = D
Now, in  ABC and  DEF,
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A = D
[ from Eq. (i)]
B = E
and
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[given]
Remaining  C = Remaining  F
Now, in  ABC and  DEF, we have
BC = EF
[given]
C = F
and AC = DE
[given]
 ABC   DEF
[by SAS congruence rule]
Hence proved.
26. Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a)
Solution:
To prove,
(a + b + c)3 –a3 –b3 –c3 = 3(a + b)(b + c)(c + a)
LHS = [(a + b + c)– a3]– [b3 + c3]
= [(a + b + c) – a][(a + b + c)2 + a(a + b + c) + a2] – (b + c) (b2 + c2 – bc]
[
x3 – y3 = (x – y) (x2 + y2 + xy)
and x3 + y3 =(x + y)(x2 +y 2–xy]
= (b + c)[a2 + b2 +c2 + 2ab + 2bc +2ca + a2 + ab + ac + a2] –[(b + c) (b2 + c2 – bc)
= (b + c)[3a2 + b2 +c2 + 3ab +2bc +3ca – b2 – c2 + bc]
= (b + c) [3a2 + 3ab + 3bc + 3ca]
= 3(b + c) [a2 + ab + bc + ca]
= 3(b + c) [a(a + b) + c(b + a)]
= 3(a + b) (b + c) (c + a) = RHS
Hence proved.
27. Simplify
4
2187
3/7

5
256
1/4

2
1331 
2
1/3
.
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Solution:
4
2187
=
=
4
3 
7
 3

3/7
4
3

3/7
5
256
5
4 
4
5

1/4
 4
1
1/4



2
1331 
2
1/3
2
[11 ]2  1 /3
3
2
11
[
2
1
m
n
 amn ]
 m 1 
a  a m 


= 4  3  5   4   2  11 
3
a 
2
= 4 × 27 – 5 × 4 + 2 × 121
= 108 – 20 + 242
= 350 – 20 = 330
1
1
28. Factroise a3  a2b  ab2  b3 .
3
27
Factorise x3 
1
 14.
x3
Solution:
1
1
a3  a2b  ab2  b3
3
27
2
3
1 
1  1 
=  a   3 a  b   3 a   b    b 
3 
3  3 
3
 
2
3
 1 
= a  b
3 

[
(a – b)3 = a3 – b3 – 3ab (a – b)]
 1  1  1 
=  a  b  a  b  a  b 
3 
3 
3 

1
1
Hence, a3  a2b  ab2  b3 =
3
27
 1  1  1 
 a  b  a  b  a  b 
3 
3 
 3 
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OR
3
1
1
3
 1
 1
x  3  14  x3  3  8  6 = x3       2  3 x      2
x
x
 x
 x
3
 1 
=  x   2
 x 
[
 2  1 2

2
 1  1
x




2

x




2


2
x












 

 x
 x  x


x3 + y3 + z3 – 3xyz = (x + y + z)
(x2 + y2 + z2 – xy – yz – zx)]
1 
1
2


=  x   2   x2  2  4  1   2x  =
x 
x
x


1  2 1
2


 x   2   x  2  5   2x 
x 
x
x


29. Find the area of the cyclic quadrilateral AB CD by using Brahmagupta's formula, in
which AB = 9cm, BC = 12 cm, CD = 12 cm and DA = 15 cm.
Solution:
According to the given informations, the rough sketch of the cyclic quadrilateral ABCD
will be as shown alongside.
Let a = AB = 9cm, t = BC = 12cm,
C = CD = 12cm and d = AD=15cm
 s
=
a bcd
2
9  12  12  15 48
=
= 24 cm
2
2
 Area of cyclic quadrilateral
=
 s  a s  bs  c s  d 
[by Brahmagupta’s formula]
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24  924  1224  1224  15
= 151212 9 = 3  5  12  12  3  3
= 12  3 3  5 = 36 15 cm2
Hence, area of cyclic quadrilateral is 36 15 cm2 .
30. What is the maximum number of digits in the repeating block of digits in the decimal
expansion of
1
? Perform the division to determine your answer.
17
Solution:
By long division method, we get
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0.0588235294117647058
17 100
85
150
136
140
136
40
34
60
51
90
85
50
34
160
153
70
68
20
17
30
17
130
119
110
102
80
68
120
119
100
85
150
136
14
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The remainders start repeating after 16 divisions.
1
 0.0588235294117647
17
Hence, the maximum number of digits in the repeating block of digits in the decimal
expansion of
1
is 16.
17
31. A point O is taken inside an equilateral four sided figure ABCD such that its distances
from the angular points D and B are equal. Show that AO and OC are in the same straight
line.
Solution:
Let AB
CD and transversal XY cuts AB and CD at P and Q, respectively. Let PR and QR be
the bisectors of  BPQ and  DQP respectively, meet at R.
AB
CD and XPQY is the transversal. Then,
 BPQ +  DQP = 180°
[cointerior angles]

1
BPQ  DQP  1800
2
[dividing both sides by 2]
 1  2  900
...(i)





1

BPQ  1
2

1
DQP 2 

2
In PRQ , we have
 RPQ +  PQR +  PRQ = 180°
[ by angle sum property of a triangle is 180°]
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  1 +  2 +  3 = 180°
 90° +  3 = 180°
[from Eq. (i),  1 +  2 = 90°]
  3 = 180° – 90° = 90°
Hence,  PRQ = 90°
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