Page 1
CBSEX2017 EXAMINATION
CBSEXII2017 EXAMINATION
MATHEMATICS
Paper & Solution
Time: 3 Hrs. Max. Marks: 90
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections  A, B, C and D.
(iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C
contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each.
(iv) Use of calculators is not permitted.
SECTION – A
Question numbers 1 to 4 carry 1 mark each.
1. What is the common difference of an A.P. in which a 21  a 7 = 84 ?
Solution:
Given
a 21 – a 7 =84 ....................(1)
In an A.P a 1, a 2, a 3, a 4 …………
a n = a 1 + (n – 1)d d = common difference
a 21 = a 1 + 20 d ……………..(2)
a 7 = a 1 + 6d ……………….(3)
substituting (2) & (3) in (1)
a 1 + 20d – a 1 – 6d = 84
14d = 84
d = 6
? common difference = 6
2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is
60 , then find the length of OP.
Solution:
Page 2
CBSEX2017 EXAMINATION
CBSEXII2017 EXAMINATION
MATHEMATICS
Paper & Solution
Time: 3 Hrs. Max. Marks: 90
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections  A, B, C and D.
(iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C
contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each.
(iv) Use of calculators is not permitted.
SECTION – A
Question numbers 1 to 4 carry 1 mark each.
1. What is the common difference of an A.P. in which a 21  a 7 = 84 ?
Solution:
Given
a 21 – a 7 =84 ....................(1)
In an A.P a 1, a 2, a 3, a 4 …………
a n = a 1 + (n – 1)d d = common difference
a 21 = a 1 + 20 d ……………..(2)
a 7 = a 1 + 6d ……………….(3)
substituting (2) & (3) in (1)
a 1 + 20d – a 1 – 6d = 84
14d = 84
d = 6
? common difference = 6
2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is
60 , then find the length of OP.
Solution:
CBSEX2017 EXAMINATION
2 / 22
Given that 60 BPA ? = ?
OB = OA = a [radii]
PA = PB [length of tangents Equal]
OP = OP
PBO ?? and PAO ? are congruent. [By SSS test of congruency]
60
30
2
BPO OPA
?
? ? = ? = = ?
In a sin30
a
PBO
OP
? ? =
OP = 2a units
3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of
the sun ?
Solution:
Angle of equation of sun = GST ? = ?
Height of lower TG = 30m
Length of shadow GS = 10 3 m
TGS ? is a right angled triangle
30
tan
10 3
tan 3
60
? ? =
?=
? = ?
Page 3
CBSEX2017 EXAMINATION
CBSEXII2017 EXAMINATION
MATHEMATICS
Paper & Solution
Time: 3 Hrs. Max. Marks: 90
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections  A, B, C and D.
(iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C
contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each.
(iv) Use of calculators is not permitted.
SECTION – A
Question numbers 1 to 4 carry 1 mark each.
1. What is the common difference of an A.P. in which a 21  a 7 = 84 ?
Solution:
Given
a 21 – a 7 =84 ....................(1)
In an A.P a 1, a 2, a 3, a 4 …………
a n = a 1 + (n – 1)d d = common difference
a 21 = a 1 + 20 d ……………..(2)
a 7 = a 1 + 6d ……………….(3)
substituting (2) & (3) in (1)
a 1 + 20d – a 1 – 6d = 84
14d = 84
d = 6
? common difference = 6
2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is
60 , then find the length of OP.
Solution:
CBSEX2017 EXAMINATION
2 / 22
Given that 60 BPA ? = ?
OB = OA = a [radii]
PA = PB [length of tangents Equal]
OP = OP
PBO ?? and PAO ? are congruent. [By SSS test of congruency]
60
30
2
BPO OPA
?
? ? = ? = = ?
In a sin30
a
PBO
OP
? ? =
OP = 2a units
3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of
the sun ?
Solution:
Angle of equation of sun = GST ? = ?
Height of lower TG = 30m
Length of shadow GS = 10 3 m
TGS ? is a right angled triangle
30
tan
10 3
tan 3
60
? ? =
?=
? = ?
CBSEX2017 EXAMINATION
4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the
number of rotten apples in the heap ?
Solution:
Probability of selecting rotten apple =
Number of rotten apples
Total number of apples
No. of rotten apples
0.18
900
?=
No. of rotten apples = 900 0.18 162 ?=
SECTION B
Question numbers 5 to 10 carry 2 marks each.
5. Find the value of p, for which one root of the quadratic equation px
2
 14x + 8 = 0 is 6 times the other.
Solution:
Given Quadriatic Equation Px
2
– 14x + 8 = 0
Also, one root is 6 times the other
Lets say one root = x
Second root = 6x
From the equation : Sum of the roots =
14
P
+
Product of roots =
8
P
2
2
2
14
6.
2
8
6
28
6
6 4 8
3
xx
P
x
P
x
P
PP
PP
P
? + =
=
?=
??
?=
??
??
?
=
=
Page 4
CBSEX2017 EXAMINATION
CBSEXII2017 EXAMINATION
MATHEMATICS
Paper & Solution
Time: 3 Hrs. Max. Marks: 90
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections  A, B, C and D.
(iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C
contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each.
(iv) Use of calculators is not permitted.
SECTION – A
Question numbers 1 to 4 carry 1 mark each.
1. What is the common difference of an A.P. in which a 21  a 7 = 84 ?
Solution:
Given
a 21 – a 7 =84 ....................(1)
In an A.P a 1, a 2, a 3, a 4 …………
a n = a 1 + (n – 1)d d = common difference
a 21 = a 1 + 20 d ……………..(2)
a 7 = a 1 + 6d ……………….(3)
substituting (2) & (3) in (1)
a 1 + 20d – a 1 – 6d = 84
14d = 84
d = 6
? common difference = 6
2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is
60 , then find the length of OP.
Solution:
CBSEX2017 EXAMINATION
2 / 22
Given that 60 BPA ? = ?
OB = OA = a [radii]
PA = PB [length of tangents Equal]
OP = OP
PBO ?? and PAO ? are congruent. [By SSS test of congruency]
60
30
2
BPO OPA
?
? ? = ? = = ?
In a sin30
a
PBO
OP
? ? =
OP = 2a units
3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of
the sun ?
Solution:
Angle of equation of sun = GST ? = ?
Height of lower TG = 30m
Length of shadow GS = 10 3 m
TGS ? is a right angled triangle
30
tan
10 3
tan 3
60
? ? =
?=
? = ?
CBSEX2017 EXAMINATION
4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the
number of rotten apples in the heap ?
Solution:
Probability of selecting rotten apple =
Number of rotten apples
Total number of apples
No. of rotten apples
0.18
900
?=
No. of rotten apples = 900 0.18 162 ?=
SECTION B
Question numbers 5 to 10 carry 2 marks each.
5. Find the value of p, for which one root of the quadratic equation px
2
 14x + 8 = 0 is 6 times the other.
Solution:
Given Quadriatic Equation Px
2
– 14x + 8 = 0
Also, one root is 6 times the other
Lets say one root = x
Second root = 6x
From the equation : Sum of the roots =
14
P
+
Product of roots =
8
P
2
2
2
14
6.
2
8
6
28
6
6 4 8
3
xx
P
x
P
x
P
PP
PP
P
? + =
=
?=
??
?=
??
??
?
=
=
CBSEX2017 EXAMINATION
4 / 22
6. Which term of the progression
113
20,19 ,18 ,17 ,....
4 2 4
is the first negative term ?
Solution:
Given progression 20,
113
19 ,18 ,17 ,.......
4 2 4
This is an Arithmetic progression because
Common difference (d) =
1 1 1
19 20 18 19 ..............
4 2 4
 =  =
3
4
d

=
Any n
th
term a n = 20 +
3 83 3
( 1)
44
n
n
 ??
=
??
??
Any term an < 0 when 83 < 3n
This is valid for n = 28 and 28
th
term will be the first negative term.
7. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.
Solution:
Need to prove that
BAP ABP ? = ?
AB is the chord
We know that OA = OB (radii)
90 OBP OAP ? = ? = ?
Join OP and OP = OP
By RHS congruency
Page 5
CBSEX2017 EXAMINATION
CBSEXII2017 EXAMINATION
MATHEMATICS
Paper & Solution
Time: 3 Hrs. Max. Marks: 90
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into four sections  A, B, C and D.
(iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks each, Section C
contains 10 questions of 3 marks each and Section D contains 11 questions of 4 marks each.
(iv) Use of calculators is not permitted.
SECTION – A
Question numbers 1 to 4 carry 1 mark each.
1. What is the common difference of an A.P. in which a 21  a 7 = 84 ?
Solution:
Given
a 21 – a 7 =84 ....................(1)
In an A.P a 1, a 2, a 3, a 4 …………
a n = a 1 + (n – 1)d d = common difference
a 21 = a 1 + 20 d ……………..(2)
a 7 = a 1 + 6d ……………….(3)
substituting (2) & (3) in (1)
a 1 + 20d – a 1 – 6d = 84
14d = 84
d = 6
? common difference = 6
2. If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is
60 , then find the length of OP.
Solution:
CBSEX2017 EXAMINATION
2 / 22
Given that 60 BPA ? = ?
OB = OA = a [radii]
PA = PB [length of tangents Equal]
OP = OP
PBO ?? and PAO ? are congruent. [By SSS test of congruency]
60
30
2
BPO OPA
?
? ? = ? = = ?
In a sin30
a
PBO
OP
? ? =
OP = 2a units
3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the angle of elevation of
the sun ?
Solution:
Angle of equation of sun = GST ? = ?
Height of lower TG = 30m
Length of shadow GS = 10 3 m
TGS ? is a right angled triangle
30
tan
10 3
tan 3
60
? ? =
?=
? = ?
CBSEX2017 EXAMINATION
4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0·18. What is the
number of rotten apples in the heap ?
Solution:
Probability of selecting rotten apple =
Number of rotten apples
Total number of apples
No. of rotten apples
0.18
900
?=
No. of rotten apples = 900 0.18 162 ?=
SECTION B
Question numbers 5 to 10 carry 2 marks each.
5. Find the value of p, for which one root of the quadratic equation px
2
 14x + 8 = 0 is 6 times the other.
Solution:
Given Quadriatic Equation Px
2
– 14x + 8 = 0
Also, one root is 6 times the other
Lets say one root = x
Second root = 6x
From the equation : Sum of the roots =
14
P
+
Product of roots =
8
P
2
2
2
14
6.
2
8
6
28
6
6 4 8
3
xx
P
x
P
x
P
PP
PP
P
? + =
=
?=
??
?=
??
??
?
=
=
CBSEX2017 EXAMINATION
4 / 22
6. Which term of the progression
113
20,19 ,18 ,17 ,....
4 2 4
is the first negative term ?
Solution:
Given progression 20,
113
19 ,18 ,17 ,.......
4 2 4
This is an Arithmetic progression because
Common difference (d) =
1 1 1
19 20 18 19 ..............
4 2 4
 =  =
3
4
d

=
Any n
th
term a n = 20 +
3 83 3
( 1)
44
n
n
 ??
=
??
??
Any term an < 0 when 83 < 3n
This is valid for n = 28 and 28
th
term will be the first negative term.
7. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.
Solution:
Need to prove that
BAP ABP ? = ?
AB is the chord
We know that OA = OB (radii)
90 OBP OAP ? = ? = ?
Join OP and OP = OP
By RHS congruency
CBSEX2017 EXAMINATION
OBP OAP ? ? ?
By CPCT BP AP ?=
In ABP BP AP ?=
Angles opposite to equal sides are equal
BAP ABP ? ? = ?
Hence proved.
8. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA
Solution:
ABCD is the Quadrilateral
Circle touches the sides at P, Q, R,S points
For the circle AS & AP are tangents
...............(1) AS AP ?=
In the similar way
BP = BQ……..(2)
CQ = CR…………(3)
RD = DS …………..(4)
Now AB + CD = AP + PB + CR + RD
BC + AD = BQ + QC + DS + AS
Using (1), (2), (3), (4) in above equation BC + AD = BP + CR + RD + AP
? AB + CD = BC + AD
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