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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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Full syllabus notes, lecture & questions for Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solution (2022-23) | Mathematics (Maths) Class 10 - Class 10 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 10 | Best notes, free PDF download
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Additional Information about Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solution (2022-23) for Class 10 Preparation
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The content of Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solution (2022-23) is prepared as per the latest Class 10 syllabus.
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