Math Past Year Paper with Solution - 2017, Class 10

# 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:
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:
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:
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:

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
```

## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests

## FAQs on Math Past Year Paper with Solution - 2017, Class 10 - Mathematics (Maths) Class 10

 1. What is the format of the Class 10 Math Past Year Paper with Solution for 2017?
Ans. The Class 10 Math Past Year Paper with Solution for 2017 is a document that includes the question paper from the 2017 exam along with detailed solutions for each question.
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Ans. You can access the Class 10 Math Past Year Paper with Solution for 2017 by searching for it online. Many educational websites and platforms provide these papers for free or for a nominal fee.
 3. What topics are covered in the Class 10 Math Past Year Paper with Solution for 2017?
Ans. The Class 10 Math Past Year Paper with Solution for 2017 covers a wide range of topics in mathematics, including algebra, geometry, trigonometry, statistics, and probability. It is designed to test the understanding and application of concepts taught in Class 10.
 4. How can solving the Class 10 Math Past Year Paper with Solution for 2017 help in exam preparation?
Ans. Solving the Class 10 Math Past Year Paper with Solution for 2017 can be extremely beneficial for exam preparation. It gives you an idea of the exam pattern, the types of questions asked, and the level of difficulty. By practicing these papers, you can improve your speed, accuracy, and problem-solving skills.
 5. Are the solutions provided in the Class 10 Math Past Year Paper with Solution for 2017 explained in detail?
Ans. Yes, the solutions provided in the Class 10 Math Past Year Paper with Solution for 2017 are explained in detail. Each question is solved step by step, with explanations and reasoning given for each step. This helps students understand the concepts and techniques used to solve the problems effectively.

## Mathematics (Maths) Class 10

116 videos|420 docs|77 tests

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