Math Past Year Paper with Solution - 2017, Class 10 | Mathematics (Maths) Class 10 PDF Download

 Page 1


CBSE-X-2017 EXAMINATION 
CBSE-XII-2017 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


CBSE-X-2017 EXAMINATION 
CBSE-XII-2017 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: 
 
 
CBSE-X-2017 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


CBSE-X-2017 EXAMINATION 
CBSE-XII-2017 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: 
 
 
CBSE-X-2017 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
? ? =
?=
? = ?
  
CBSE-X-2017 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


CBSE-X-2017 EXAMINATION 
CBSE-XII-2017 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: 
 
 
CBSE-X-2017 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
? ? =
?=
? = ?
  
CBSE-X-2017 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
? + =
=
?=
??
?=
??
??
?
=
=
  
 
CBSE-X-2017 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


CBSE-X-2017 EXAMINATION 
CBSE-XII-2017 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: 
 
 
CBSE-X-2017 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
? ? =
?=
? = ?
  
CBSE-X-2017 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
? + =
=
?=
??
?=
??
??
?
=
=
  
 
CBSE-X-2017 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 
CBSE-X-2017 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