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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 (Standard): CBSE (Official) Marking Scheme (2022-23) - CBSE Sample Papers For Class 10
| 1. What is the marking scheme for CBSE Class 10 Mathematics (Standard) exam in 2022-23? |  |
Ans. The marking scheme for CBSE Class 10 Mathematics (Standard) exam in 2022-23 is the official guidelines provided by the CBSE board. It outlines the distribution of marks across different sections or topics of the exam, as well as the criteria for awarding marks to the students' answers. This scheme helps teachers and examiners evaluate the students' performance accurately and consistently.
| 2. How are marks distributed in the CBSE Class 10 Mathematics (Standard) exam? | |
Ans. The marks in the CBSE Class 10 Mathematics (Standard) exam are distributed across different sections or topics according to the marking scheme. For example, the scheme may specify that a certain percentage of marks will be allocated to algebra, a certain percentage to geometry, and so on. The distribution of marks may vary from year to year, so it is important for students to refer to the latest marking scheme for accurate information.
| 3. What is the importance of following the CBSE Class 10 Mathematics (Standard) marking scheme? | |
Ans. Following the CBSE Class 10 Mathematics (Standard) marking scheme is crucial for both students and teachers. It ensures that the evaluation process is fair and consistent across different schools and examiners. By adhering to the marking scheme, students know what is expected of them and can prepare accordingly. Teachers can also use the scheme as a guideline to assess the students' performance accurately and provide appropriate feedback.
| 4. How can students use the CBSE Class 10 Mathematics (Standard) marking scheme to improve their exam preparation? | |
Ans. Students can use the CBSE Class 10 Mathematics (Standard) marking scheme to improve their exam preparation in several ways. Firstly, they can identify the weightage given to different sections or topics and allocate their study time accordingly. They can focus more on topics that carry more marks and allocate sufficient time for practice. Additionally, students can refer to the marking scheme while solving sample papers or previous year's question papers to understand the pattern of marks distribution and the type of questions that are likely to be asked.
| 5. Where can students find the CBSE Class 10 Mathematics (Standard) marking scheme for the 2022-23 exam? | |
Ans. The CBSE Class 10 Mathematics (Standard) marking scheme for the 2022-23 exam can be found on the official website of the CBSE board. The board usually releases the marking scheme along with the sample question papers or the official exam notification. Students can download the marking scheme from the website and refer to it while preparing for the exam. It is important to ensure that the marking scheme being referred to is the latest one released by the CBSE board.
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Document Description: Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme (2022-23) for Class 10 2024 is part of CBSE Sample Papers For Class 10 preparation.
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