Commerce Exam > Commerce Notes > Mathematics (Maths) Class 11 > Sample Solution Paper 5 - Math, Class 11
Sample Solution Paper 5 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
Page 2
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
Page 3
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
CBSE XI | Mathematics
Sample Paper – 5 Solution
1 1 x
9! 10! 11!
1 1 x
9! 10 9! 11 10 9!
1 1 x 1
1
9! 10 11 10 9!
11 x
10 11 10
x 121
??
??
? ? ?
? ? ? ?
??
? ? ? ?
?
? ? ? ?
?
?
?
10.
Given that the first term is 1.
Also given that each term is the sum of all the terms which follow it.
Let 1, r, r
2
,… be an infinite G.P., where r is the common ratio.
Sum of terms of an infinite G.P.,
a
S
1r
?
?
Here, a = r
Thus,
r
S
1r
?
?
From the given statement of the problem, we have,
?
?
? ? ?
? ? ?
??
??
r
1
1r
1 r r
rr1
2r 1
1
r
2
Thus the required G.P. is:
1 1 1 1
1, , , , ,...
2 4 8 16
11.
27
,k, are in G.P.
72
??
2
2
27
k1
72
k1
k1
27
Whenk 1;G.P.: ,1,
72
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
? ? ?
Page 4
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
CBSE XI | Mathematics
Sample Paper – 5 Solution
1 1 x
9! 10! 11!
1 1 x
9! 10 9! 11 10 9!
1 1 x 1
1
9! 10 11 10 9!
11 x
10 11 10
x 121
??
??
? ? ?
? ? ? ?
??
? ? ? ?
?
? ? ? ?
?
?
?
10.
Given that the first term is 1.
Also given that each term is the sum of all the terms which follow it.
Let 1, r, r
2
,… be an infinite G.P., where r is the common ratio.
Sum of terms of an infinite G.P.,
a
S
1r
?
?
Here, a = r
Thus,
r
S
1r
?
?
From the given statement of the problem, we have,
?
?
? ? ?
? ? ?
??
??
r
1
1r
1 r r
rr1
2r 1
1
r
2
Thus the required G.P. is:
1 1 1 1
1, , , , ,...
2 4 8 16
11.
27
,k, are in G.P.
72
??
2
2
27
k1
72
k1
k1
27
Whenk 1;G.P.: ,1,
72
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
? ? ?
CBSE XI | Mathematics
Sample Paper – 5 Solution
17
r
2
2
7
27
Whenk 1;GP : , 1,
72
17
r
2
2
7
? ? ?
?
? ? ? ? ?
?
??
?
OR
3, 6, 12,…
a = 3, r = 2 and n = 7
S7 =
77
r 1 2 1
a 3 3 127 381
r 1 2 1
? ? ? ? ??
? ? ? ?
? ? ? ?
??
? ? ? ?
12. an =
(1 2 3 ....n) n(n 1)
n 2n
? ? ? ?
?
Sn =
n
n
a ?
n
i1
2
1
(n 1)
2
1 n(n 1) n
2 2 2
(n n) n
42
n(n 3)
4
?
? ??
?
??
?
??
?
?
SECTION – C
13.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
22
22
22
1 i 1 i
1+i 1 i
1 i 1+i 1 i 1 i
1 i 2i 1 i 2i
1 i 1 i
1 1 2i 1 1 2i 2i 2i
1 i 1 i
? ? ?
?
??
? ? ?
? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
??
Page 5
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
CBSE XI | Mathematics
Sample Paper – 5 Solution
1 1 x
9! 10! 11!
1 1 x
9! 10 9! 11 10 9!
1 1 x 1
1
9! 10 11 10 9!
11 x
10 11 10
x 121
??
??
? ? ?
? ? ? ?
??
? ? ? ?
?
? ? ? ?
?
?
?
10.
Given that the first term is 1.
Also given that each term is the sum of all the terms which follow it.
Let 1, r, r
2
,… be an infinite G.P., where r is the common ratio.
Sum of terms of an infinite G.P.,
a
S
1r
?
?
Here, a = r
Thus,
r
S
1r
?
?
From the given statement of the problem, we have,
?
?
? ? ?
? ? ?
??
??
r
1
1r
1 r r
rr1
2r 1
1
r
2
Thus the required G.P. is:
1 1 1 1
1, , , , ,...
2 4 8 16
11.
27
,k, are in G.P.
72
??
2
2
27
k1
72
k1
k1
27
Whenk 1;G.P.: ,1,
72
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
? ? ?
CBSE XI | Mathematics
Sample Paper – 5 Solution
17
r
2
2
7
27
Whenk 1;GP : , 1,
72
17
r
2
2
7
? ? ?
?
? ? ? ? ?
?
??
?
OR
3, 6, 12,…
a = 3, r = 2 and n = 7
S7 =
77
r 1 2 1
a 3 3 127 381
r 1 2 1
? ? ? ? ??
? ? ? ?
? ? ? ?
??
? ? ? ?
12. an =
(1 2 3 ....n) n(n 1)
n 2n
? ? ? ?
?
Sn =
n
n
a ?
n
i1
2
1
(n 1)
2
1 n(n 1) n
2 2 2
(n n) n
42
n(n 3)
4
?
? ??
?
??
?
??
?
?
SECTION – C
13.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
22
22
22
1 i 1 i
1+i 1 i
1 i 1+i 1 i 1 i
1 i 2i 1 i 2i
1 i 1 i
1 1 2i 1 1 2i 2i 2i
1 i 1 i
? ? ?
?
??
? ? ?
? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
??
CBSE XI | Mathematics
Sample Paper – 5 Solution
? ? ? ?
22
2i 2i 4i
2i
1 ( 1) 2
1+i 1 i
2i 0 2 4 2
1 i 1+i
?
? ? ?
??
?
? ? ? ? ? ?
?
OR
1 2 1 2
Consider the given equation:
(a + ib) (c + id) (e + if) (g + ih) = A + iB
Let us take modulus on both sides ,
(a + ib) (c + id) (e + if) (g + ih) = A + iB
We know, z z z z
(a + ib) (c + id) (e + if) (g
?
?
? ? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2 2 2 2 2 2 2
22
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
+ ih) = A + iB
(a + ib) (c + id) (e + if) (g + ih) = A + iB
a b . c d . e f . g h A B
a b . c d . e f . g h A B
a b . c d . e f . g h A B
Thus, the value of A B a b . c d . e f .
?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ?
22
gh ?
14. Consider the given quadratic equation:
2
2
2
9x 12 x + 20 = 0
20
3x 4x 0
3
b b 4ac
x
2a
?
? ? ? ?
? ? ?
?
? ? ? ? ? ? ?
? ? ?
? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ?
2
20
( 4) ( 4) 4 3
4 16 4 20
3
x
2.3 6
4 16 80 4 64 4 8 1 2 4i
6 6 6 3
24
xi
33
Read More
|
75 videos|238 docs|91 tests
|
FAQs on Sample Solution Paper 5 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What is the Pythagorean theorem? |  |
Ans. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. It can be expressed as a^2 + b^2 = c^2, where a and b are the lengths of the two legs and c is the length of the hypotenuse.
| 2. How is the Pythagorean theorem useful in real life? | |
Ans. The Pythagorean theorem is widely used in various fields such as construction, architecture, engineering, and navigation. It helps in calculating distances, finding the length of diagonals, determining the height of buildings, and solving problems related to angles and distances in real-life situations.
| 3. Can the Pythagorean theorem be applied to non-right-angled triangles? | |
Ans. No, the Pythagorean theorem is only applicable to right-angled triangles. In non-right-angled triangles, other trigonometric functions such as sine, cosine, and tangent are used to find the relationships between the sides and angles.
| 4. How can I use the Pythagorean theorem to find the length of the hypotenuse? | |
Ans. To find the length of the hypotenuse using the Pythagorean theorem, you need to know the lengths of the other two sides (legs) of the right-angled triangle. Square the lengths of the legs, add them together, and then take the square root of the sum. The result will be the length of the hypotenuse.
| 5. Are there any practical applications of the Pythagorean theorem? | |
Ans. Yes, the Pythagorean theorem has numerous practical applications. It is used in construction to ensure that buildings are structurally sound and to calculate the lengths of diagonals. It is also used in navigation to determine distances and angles. Additionally, it is used in various fields of science and engineering to solve problems involving right-angled triangles.
About this Document
|
4.94/5 Rating |
|
Dec 18, 2024 Last updated |
Document Description: Sample Solution Paper 5 - Math, Class 11 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation.
The notes and questions for Sample Solution Paper 5 - Math, Class 11 have been prepared according to the Commerce exam syllabus. Information about Sample Solution Paper 5 - Math, Class 11 covers topics
like and Sample Solution Paper 5 - Math, Class 11 Example, for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Sample Solution Paper 5 - Math, Class 11.
Introduction of Sample Solution Paper 5 - Math, Class 11 in English is available as part of our Mathematics (Maths) Class 11
for Commerce & Sample Solution Paper 5 - Math, Class 11 in Hindi for Mathematics (Maths) Class 11 course.
Download more important topics related with notes, lectures and mock test series for Commerce
Exam by signing up for free. Commerce: Sample Solution Paper 5 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce
Description
Full syllabus notes, lecture & questions for Sample Solution Paper 5 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce - Commerce | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 11 | Best notes, free PDF download
Information about Sample Solution Paper 5 - Math, Class 11
In this doc you can find the meaning of Sample Solution Paper 5 - Math, Class 11 defined & explained in the simplest way possible. Besides explaining types of
Sample Solution Paper 5 - Math, Class 11 theory, EduRev gives you an ample number of questions to practice Sample Solution Paper 5 - Math, Class 11 tests, examples and also practice Commerce
tests
|
Explore Courses for Commerce exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Sample Solution Paper 5 - Math
,Sample Solution Paper 5 - Math
,Sample Solution Paper 5 - Math
,mock tests for examination
,Previous Year Questions with Solutions
,past year papers
,ppt
,study material
,Summary
,Free
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,Objective type Questions
,shortcuts and tricks
,Extra Questions
,MCQs
,Semester Notes
,Important questions
,Viva Questions
,Class 11 | Mathematics (Maths) Class 11 - Commerce
,Exam
,video lectures
,Sample Paper
,practice quizzes
,Class 11 | Mathematics (Maths) Class 11 - Commerce
;
Additional Information about Sample Solution Paper 5 - Math, Class 11 for Commerce Preparation
Sample Solution Paper 5 - Math, Class 11 Free PDF Download
The Sample Solution Paper 5 - Math, Class 11 is an invaluable resource that delves deep into the core of the Commerce exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Sample Solution Paper 5 - Math, Class 11 now and kickstart your journey towards success in the Commerce exam.
Importance of Sample Solution Paper 5 - Math, Class 11
The importance of Sample Solution Paper 5 - Math, Class 11 cannot be overstated, especially for Commerce aspirants.
This document holds the key to success in the Commerce exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Sample Solution Paper 5 - Math, Class 11 Notes
Sample Solution Paper 5 - Math, Class 11 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Sample Solution Paper 5 - Math, Class 11.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Sample Solution Paper 5 - Math, Class 11 Notes on EduRev are your ultimate resource for success.
Sample Solution Paper 5 - Math, Class 11 Commerce Questions
The "Sample Solution Paper 5 - Math, Class 11 Commerce Questions" guide is a valuable resource for all aspiring students preparing for the
Commerce exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Sample Solution Paper 5 - Math, Class 11 on the App
Students of Commerce can study Sample Solution Paper 5 - Math, Class 11 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Sample Solution Paper 5 - Math, Class 11,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Sample Solution Paper 5 - Math, Class 11 is prepared as per the latest Commerce syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup