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Page 1
1
SAMPLE QUESTION PAPER
MARKING SCHEME
SUBJECT: MATHEMATICS- STANDARD
CLASS X
SECTION - A
1 (c) 35
1
2 (b) x
2
–(p+1)x +p=0
1
3 (b) 2/3
1
4 (d) 2
1
5 (c) (2,-1)
1
6 (d) 2:3
1
7 (b) tan 30°
1
8 (b) 2
1
9
(c) x=
????
?? +??
1
10 (c) 8cm
1
11 (d) 3v3cm
1
12 (d) 9p cm
2
1
13 (c) 96 cm
2
1
14 (b) 12
1
15 (d) 7000
1
16 (b) 25
1
17 (c) 11/36
1
18 (a) 1/3
1
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct
explanation of assertion (A)
1
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
Page 2
1
SAMPLE QUESTION PAPER
MARKING SCHEME
SUBJECT: MATHEMATICS- STANDARD
CLASS X
SECTION - A
1 (c) 35
1
2 (b) x
2
–(p+1)x +p=0
1
3 (b) 2/3
1
4 (d) 2
1
5 (c) (2,-1)
1
6 (d) 2:3
1
7 (b) tan 30°
1
8 (b) 2
1
9
(c) x=
????
?? +??
1
10 (c) 8cm
1
11 (d) 3v3cm
1
12 (d) 9p cm
2
1
13 (c) 96 cm
2
1
14 (b) 12
1
15 (d) 7000
1
16 (b) 25
1
17 (c) 11/36
1
18 (a) 1/3
1
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct
explanation of assertion (A)
1
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
2
SECTION – B
21 Adding the two equations and dividing by 10, we get : x+y = 10
Subtracting the two equations and dividing by -2, we get : x-y =1
Solving these two new equations, we get, x = 11/2
y = 9/2
½
½
½
½
22 In ?ABC,
?1 = ?2
? AB = BD ………………………(i)
Given,
AD/AE = AC/BD
Using equation (i), we get
AD/AE = AC/AB ……………….(ii)
In ?BAE and ?CAD, by equation (ii),
AC/AB = AD/AE
?A= ?A (common)
? ?BAE ~ ?CAD [By SAS similarity criterion]
½
½
½
½
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent)
?AOB = 105° (By angle sum property of a triangle)
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the
angle subtended by the arc at the centre)
½
½
1
24
We know that, in 60 minutes, the tip of minute hand moves 360°
In 1 minute, it will move =360°/60 = 6°
? From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?
of 210° and radius of 6 cm
=
210
360
x p x 6
2
=
7
12
x
22
7
x 6 x 6
=66cm
2
OR
Let the measure of ?A, ?B, ?C and ?D be ?1, ?2, ?3 and ?4 respectively
Required area = Area of sector with centre A + Area of sector with centre B
+ Area of sector with centre C + Area of sector with centre D
½
½
½
½
½
Page 3
1
SAMPLE QUESTION PAPER
MARKING SCHEME
SUBJECT: MATHEMATICS- STANDARD
CLASS X
SECTION - A
1 (c) 35
1
2 (b) x
2
–(p+1)x +p=0
1
3 (b) 2/3
1
4 (d) 2
1
5 (c) (2,-1)
1
6 (d) 2:3
1
7 (b) tan 30°
1
8 (b) 2
1
9
(c) x=
????
?? +??
1
10 (c) 8cm
1
11 (d) 3v3cm
1
12 (d) 9p cm
2
1
13 (c) 96 cm
2
1
14 (b) 12
1
15 (d) 7000
1
16 (b) 25
1
17 (c) 11/36
1
18 (a) 1/3
1
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct
explanation of assertion (A)
1
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
2
SECTION – B
21 Adding the two equations and dividing by 10, we get : x+y = 10
Subtracting the two equations and dividing by -2, we get : x-y =1
Solving these two new equations, we get, x = 11/2
y = 9/2
½
½
½
½
22 In ?ABC,
?1 = ?2
? AB = BD ………………………(i)
Given,
AD/AE = AC/BD
Using equation (i), we get
AD/AE = AC/AB ……………….(ii)
In ?BAE and ?CAD, by equation (ii),
AC/AB = AD/AE
?A= ?A (common)
? ?BAE ~ ?CAD [By SAS similarity criterion]
½
½
½
½
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent)
?AOB = 105° (By angle sum property of a triangle)
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the
angle subtended by the arc at the centre)
½
½
1
24
We know that, in 60 minutes, the tip of minute hand moves 360°
In 1 minute, it will move =360°/60 = 6°
? From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?
of 210° and radius of 6 cm
=
210
360
x p x 6
2
=
7
12
x
22
7
x 6 x 6
=66cm
2
OR
Let the measure of ?A, ?B, ?C and ?D be ?1, ?2, ?3 and ?4 respectively
Required area = Area of sector with centre A + Area of sector with centre B
+ Area of sector with centre C + Area of sector with centre D
½
½
½
½
½
3
=
?? 1
360
x p x 7
2
+
?? 2
360
x p x 7
2
+
?? 3
360
x p x 7
2
+
?? 4
360
x p x 7
2
=
(?? 1 + ?? 2 + ?? 3 + ?? 4)
360
x p x 7
2
=
(?????? )
360
x
????
7
x 7x 7 ( By angle sum property of a triangle)
= 154 cm
2
½
½
½
25
sin(A+B) =1 = sin 90, so A+B = 90……………….(i)
cos(A-B)= v3/2 = cos 30, so A-B= 30……………(ii)
From (i) & (ii) ?A = 60°
And ?B = 30°
OR
cos? - sin ?
cos?+sin ?
=
1-v3
1+v3
Dividing the numerator and denominator of LHS by cos?, we get
1 - tan ?
1+tan ?
=
1-v3
1+v3
Which on simplification (or comparison) gives tan? = v3
Or ?= 60°
½
½
½
½
½
½
½
½
SECTION - C
26
Let us assume 5 + 2v3 is rational, then it must be in the form of p/q where p and
q are co-prime integers and q ? 0
i.e 5 + 2v3 = p/q
So v3 =
?? -5?? 2?? ……………………(i)
Since p, q, 5 and 2 are integers and q ? 0, HS of equation (i) is rational. But
LHS of (i) is v3 which is irrational. This is not possible.
This contradiction has arisen due to our wrong assumption that 5 + 2v3 is
rational. So, 5 + 2v3 is irrational.
1
½
½
½
½
27 Let a and ß be the zeros of the polynomial 2x
2
-5x -3
Then a + ß = 5/2
And aß = -3/2.
Let 2a and 2ß be the zeros x
2
+ px +q
Then 2a + 2ß = -p
2(a + ß) = -p
2 x 5/2 =-p
So p = -5
And 2a x 2ß = q
4 aß = q
So q = 4 x-3/2
= -6
½
½
½
½
½
½
Page 4
1
SAMPLE QUESTION PAPER
MARKING SCHEME
SUBJECT: MATHEMATICS- STANDARD
CLASS X
SECTION - A
1 (c) 35
1
2 (b) x
2
–(p+1)x +p=0
1
3 (b) 2/3
1
4 (d) 2
1
5 (c) (2,-1)
1
6 (d) 2:3
1
7 (b) tan 30°
1
8 (b) 2
1
9
(c) x=
????
?? +??
1
10 (c) 8cm
1
11 (d) 3v3cm
1
12 (d) 9p cm
2
1
13 (c) 96 cm
2
1
14 (b) 12
1
15 (d) 7000
1
16 (b) 25
1
17 (c) 11/36
1
18 (a) 1/3
1
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct
explanation of assertion (A)
1
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
2
SECTION – B
21 Adding the two equations and dividing by 10, we get : x+y = 10
Subtracting the two equations and dividing by -2, we get : x-y =1
Solving these two new equations, we get, x = 11/2
y = 9/2
½
½
½
½
22 In ?ABC,
?1 = ?2
? AB = BD ………………………(i)
Given,
AD/AE = AC/BD
Using equation (i), we get
AD/AE = AC/AB ……………….(ii)
In ?BAE and ?CAD, by equation (ii),
AC/AB = AD/AE
?A= ?A (common)
? ?BAE ~ ?CAD [By SAS similarity criterion]
½
½
½
½
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent)
?AOB = 105° (By angle sum property of a triangle)
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the
angle subtended by the arc at the centre)
½
½
1
24
We know that, in 60 minutes, the tip of minute hand moves 360°
In 1 minute, it will move =360°/60 = 6°
? From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?
of 210° and radius of 6 cm
=
210
360
x p x 6
2
=
7
12
x
22
7
x 6 x 6
=66cm
2
OR
Let the measure of ?A, ?B, ?C and ?D be ?1, ?2, ?3 and ?4 respectively
Required area = Area of sector with centre A + Area of sector with centre B
+ Area of sector with centre C + Area of sector with centre D
½
½
½
½
½
3
=
?? 1
360
x p x 7
2
+
?? 2
360
x p x 7
2
+
?? 3
360
x p x 7
2
+
?? 4
360
x p x 7
2
=
(?? 1 + ?? 2 + ?? 3 + ?? 4)
360
x p x 7
2
=
(?????? )
360
x
????
7
x 7x 7 ( By angle sum property of a triangle)
= 154 cm
2
½
½
½
25
sin(A+B) =1 = sin 90, so A+B = 90……………….(i)
cos(A-B)= v3/2 = cos 30, so A-B= 30……………(ii)
From (i) & (ii) ?A = 60°
And ?B = 30°
OR
cos? - sin ?
cos?+sin ?
=
1-v3
1+v3
Dividing the numerator and denominator of LHS by cos?, we get
1 - tan ?
1+tan ?
=
1-v3
1+v3
Which on simplification (or comparison) gives tan? = v3
Or ?= 60°
½
½
½
½
½
½
½
½
SECTION - C
26
Let us assume 5 + 2v3 is rational, then it must be in the form of p/q where p and
q are co-prime integers and q ? 0
i.e 5 + 2v3 = p/q
So v3 =
?? -5?? 2?? ……………………(i)
Since p, q, 5 and 2 are integers and q ? 0, HS of equation (i) is rational. But
LHS of (i) is v3 which is irrational. This is not possible.
This contradiction has arisen due to our wrong assumption that 5 + 2v3 is
rational. So, 5 + 2v3 is irrational.
1
½
½
½
½
27 Let a and ß be the zeros of the polynomial 2x
2
-5x -3
Then a + ß = 5/2
And aß = -3/2.
Let 2a and 2ß be the zeros x
2
+ px +q
Then 2a + 2ß = -p
2(a + ß) = -p
2 x 5/2 =-p
So p = -5
And 2a x 2ß = q
4 aß = q
So q = 4 x-3/2
= -6
½
½
½
½
½
½
4
28
Let the actual speed of the train be x km/hr and let the actual time taken be y hours.
Distance covered is xy km
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e.,
when speed is (x+6)km/hr, time of journey is (y-4) hours.
? Distance covered =(x+6)(y-4)
?xy=(x+6)(y-4)
?-4x+6y-24=0
?-2x+3y-12=0 …………………………….(i)
Similarly xy=(x-6)(y+6)
?6x-6y-36=0
?x-y-6=0 ………………………………………(ii)
Solving (i) and (ii) we get x=30 and y=24
Putting the values of x and y in equation (i), we obtain
Distance =(30×24)km =720km.
Hence, the length of the journey is 720km.
½
½
½
1
½
OR
Let the number of chocolates in lot A be x
And let the number of chocolates in lot B be y
? total number of chocolates =x+y
Price of 1 chocolate = ? 2/3 , so for x chocolates =
?? ?? x
and price of y chocolates at the rate of ? 1 per chocolate =y.
? by the given condition
?? ?? x +y=400
?2x+3y=1200 ..............(i)
Similarly x+
?? ?? y = 460
?5x+4y=2300 ........ (ii)
Solving (i) and (ii) we get
x=300 and y=200
?x+y=300+200=500
So, Anuj had 500 chocolates.
½
½
½
1
½
29 LHS : sin
3
?/ cos
3
? + cos
3
?/ sin
3
?
1+ sin
2
?/cos
2
? 1+ cos
2
?/ sin
2
?
½
Page 5
1
SAMPLE QUESTION PAPER
MARKING SCHEME
SUBJECT: MATHEMATICS- STANDARD
CLASS X
SECTION - A
1 (c) 35
1
2 (b) x
2
–(p+1)x +p=0
1
3 (b) 2/3
1
4 (d) 2
1
5 (c) (2,-1)
1
6 (d) 2:3
1
7 (b) tan 30°
1
8 (b) 2
1
9
(c) x=
????
?? +??
1
10 (c) 8cm
1
11 (d) 3v3cm
1
12 (d) 9p cm
2
1
13 (c) 96 cm
2
1
14 (b) 12
1
15 (d) 7000
1
16 (b) 25
1
17 (c) 11/36
1
18 (a) 1/3
1
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct
explanation of assertion (A)
1
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
2
SECTION – B
21 Adding the two equations and dividing by 10, we get : x+y = 10
Subtracting the two equations and dividing by -2, we get : x-y =1
Solving these two new equations, we get, x = 11/2
y = 9/2
½
½
½
½
22 In ?ABC,
?1 = ?2
? AB = BD ………………………(i)
Given,
AD/AE = AC/BD
Using equation (i), we get
AD/AE = AC/AB ……………….(ii)
In ?BAE and ?CAD, by equation (ii),
AC/AB = AD/AE
?A= ?A (common)
? ?BAE ~ ?CAD [By SAS similarity criterion]
½
½
½
½
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent)
?AOB = 105° (By angle sum property of a triangle)
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the
angle subtended by the arc at the centre)
½
½
1
24
We know that, in 60 minutes, the tip of minute hand moves 360°
In 1 minute, it will move =360°/60 = 6°
? From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?
of 210° and radius of 6 cm
=
210
360
x p x 6
2
=
7
12
x
22
7
x 6 x 6
=66cm
2
OR
Let the measure of ?A, ?B, ?C and ?D be ?1, ?2, ?3 and ?4 respectively
Required area = Area of sector with centre A + Area of sector with centre B
+ Area of sector with centre C + Area of sector with centre D
½
½
½
½
½
3
=
?? 1
360
x p x 7
2
+
?? 2
360
x p x 7
2
+
?? 3
360
x p x 7
2
+
?? 4
360
x p x 7
2
=
(?? 1 + ?? 2 + ?? 3 + ?? 4)
360
x p x 7
2
=
(?????? )
360
x
????
7
x 7x 7 ( By angle sum property of a triangle)
= 154 cm
2
½
½
½
25
sin(A+B) =1 = sin 90, so A+B = 90……………….(i)
cos(A-B)= v3/2 = cos 30, so A-B= 30……………(ii)
From (i) & (ii) ?A = 60°
And ?B = 30°
OR
cos? - sin ?
cos?+sin ?
=
1-v3
1+v3
Dividing the numerator and denominator of LHS by cos?, we get
1 - tan ?
1+tan ?
=
1-v3
1+v3
Which on simplification (or comparison) gives tan? = v3
Or ?= 60°
½
½
½
½
½
½
½
½
SECTION - C
26
Let us assume 5 + 2v3 is rational, then it must be in the form of p/q where p and
q are co-prime integers and q ? 0
i.e 5 + 2v3 = p/q
So v3 =
?? -5?? 2?? ……………………(i)
Since p, q, 5 and 2 are integers and q ? 0, HS of equation (i) is rational. But
LHS of (i) is v3 which is irrational. This is not possible.
This contradiction has arisen due to our wrong assumption that 5 + 2v3 is
rational. So, 5 + 2v3 is irrational.
1
½
½
½
½
27 Let a and ß be the zeros of the polynomial 2x
2
-5x -3
Then a + ß = 5/2
And aß = -3/2.
Let 2a and 2ß be the zeros x
2
+ px +q
Then 2a + 2ß = -p
2(a + ß) = -p
2 x 5/2 =-p
So p = -5
And 2a x 2ß = q
4 aß = q
So q = 4 x-3/2
= -6
½
½
½
½
½
½
4
28
Let the actual speed of the train be x km/hr and let the actual time taken be y hours.
Distance covered is xy km
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e.,
when speed is (x+6)km/hr, time of journey is (y-4) hours.
? Distance covered =(x+6)(y-4)
?xy=(x+6)(y-4)
?-4x+6y-24=0
?-2x+3y-12=0 …………………………….(i)
Similarly xy=(x-6)(y+6)
?6x-6y-36=0
?x-y-6=0 ………………………………………(ii)
Solving (i) and (ii) we get x=30 and y=24
Putting the values of x and y in equation (i), we obtain
Distance =(30×24)km =720km.
Hence, the length of the journey is 720km.
½
½
½
1
½
OR
Let the number of chocolates in lot A be x
And let the number of chocolates in lot B be y
? total number of chocolates =x+y
Price of 1 chocolate = ? 2/3 , so for x chocolates =
?? ?? x
and price of y chocolates at the rate of ? 1 per chocolate =y.
? by the given condition
?? ?? x +y=400
?2x+3y=1200 ..............(i)
Similarly x+
?? ?? y = 460
?5x+4y=2300 ........ (ii)
Solving (i) and (ii) we get
x=300 and y=200
?x+y=300+200=500
So, Anuj had 500 chocolates.
½
½
½
1
½
29 LHS : sin
3
?/ cos
3
? + cos
3
?/ sin
3
?
1+ sin
2
?/cos
2
? 1+ cos
2
?/ sin
2
?
½
5
= sin
3
?/ cos
3
? + cos
3
?/ sin
3
?
(cos
2
? + sin
2
?)/cos
2
? (sin
2
? + cos
2
?)/ sin
2
?
= sin
3
? + cos
3
?
cos? sin?
= sin
4
? + cos
4
?
cos?sin?
= (sin
2
? + cos
2
?)
2
– 2 sin
2
?cos
2
?
cos?sin?
= 1 - 2 sin
2
?cos
2
?
cos?sin?
= 1 - 2 sin
2
?cos
2
?
cos?sin? cos?sin?
= sec?cosec? – 2sin?cos?
= RHS
½
½
½
½
½
30
Let ABCD be the rhombus circumscribing the circle
with centre O, such that AB, BC, CD and DA touch
the circle at points P, Q, R and S respectively.
We know that the tangents drawn to a circle from an
exterior point are equal in length.
? AP = AS………….(1)
BP = BQ……………(2)
CR = CQ …………...(3)
DR = DS……………(4).
Adding (1), (2), (3) and (4) we get
AP+BP+CR+DR = AS+BQ+CQ+DS
(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
? AB+CD=AD+BC-----------(5)
Since AB=DC and AD=BC (opposite sides of parallelogram ABCD)
putting in (5) we get, 2AB=2AD
or AB = AD.
? AB=BC=DC=AD
Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a
rhombus
1
1
½
½
OR
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FAQs on Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Standard) - 1 - CBSE Sample Papers For Class 10
| 1. What is the format of the CBSE Sample Question Paper for Class 10 Mathematics (Standard) for the academic year 2022-23? |  |
Ans. The CBSE Sample Question Paper for Class 10 Mathematics (Standard) for the academic year 2022-23 follows the pattern and format prescribed by the Central Board of Secondary Education (CBSE). It consists of various sections, including objective-type questions, short answer type questions, and long answer type questions. The paper is designed to assess students' understanding of the concepts and their ability to solve mathematical problems effectively.
| 2. How can the CBSE Sample Question Paper for Class 10 Mathematics (Standard) help students in their exam preparation? | |
Ans. The CBSE Sample Question Paper for Class 10 Mathematics (Standard) serves as a valuable resource for students' exam preparation. It provides them with an overview of the question paper pattern, marking scheme, and the type of questions that can be expected in the actual exam. By practicing these sample papers, students can become familiar with the exam format, improve their time management skills, and identify their strengths and weaknesses in different topics of mathematics.
| 3. Are the solutions provided for the CBSE Sample Question Paper for Class 10 Mathematics (Standard) useful for self-assessment? | |
Ans. Yes, the solutions provided for the CBSE Sample Question Paper for Class 10 Mathematics (Standard) are indeed useful for self-assessment. These solutions are prepared by subject matter experts and provide step-by-step explanations for solving each question. By referring to these solutions, students can compare their answers, identify any mistakes or gaps in their understanding, and rectify them. It helps in evaluating their performance and enhancing their problem-solving skills.
| 4. Can the CBSE Sample Question Paper for Class 10 Mathematics (Standard) be used as a reference for understanding the difficulty level of the actual exam? | |
Ans. Yes, the CBSE Sample Question Paper for Class 10 Mathematics (Standard) can be used as a reference for understanding the difficulty level of the actual exam. The sample papers are designed to replicate the level of difficulty and complexity that students can expect in the final examination. By attempting these sample papers, students can gauge their preparedness and get an idea of the type of questions they may encounter in the real exam.
| 5. Is it advisable to solve the CBSE Sample Question Paper for Class 10 Mathematics (Standard) under exam conditions? | |
Ans. Yes, it is highly advisable to solve the CBSE Sample Question Paper for Class 10 Mathematics (Standard) under exam conditions. By attempting the sample paper within the specified time limit and following the exam rules, students can simulate the actual exam environment. It helps them develop their speed, accuracy, and time management skills, which are crucial for performing well in the final examination. Additionally, solving the sample paper under exam conditions boosts students' confidence and reduces exam-related anxiety.
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Document Description: Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Standard) - 1 for Class 10 2024 is part of CBSE Sample Papers For Class 10 preparation.
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