Class 10 Exam > Class 10 Notes > RD Sharma Solutions for Class 10 Mathematics > Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6)
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics PDF Download
Page No 5.25
Ques.31. Which term of the arithmetic progression 8, 14, 20, 26, ... will be 72 more than its 41st term.
Ans. In the given problem, let us first find the 41st term of the given A.P.
A.P. is 8, 14, 20, 26 …
Here,
First-term (a) = 8
Common difference of the A.P. (d) = 14 - 8 = 6
Now, as we know,
an = a + (n - 1)d
So, for 41st term (n = 41),
a41 = 8 + (41 - 1)(6)
= 8 + 40(6)
= 8 + 240
= 248
Let us take the term which is 72 more than the 41st term as an. So,
an = 72 + a41
= 72 + 248
= 320
Also, an = a + (n - 1)d
320 = 8 + (n - 1)6
320 = 8 + 6n - 6
320 = 2 + 6n
320 - 2 = 6n
Further simplifying, we get,
318 = 6n
n = 318/6
n = 53
Therefore, the 53rd term of the given A.P. is 72 more than the 41st term.
Ans. In the given problem, let us first find the 41st term of the given A.P.
A.P. is 8, 14, 20, 26 …
Here,
First-term (a) = 8
Common difference of the A.P. (d) = 14 - 8 = 6
Now, as we know,
an = a + (n - 1)d
So, for 41st term (n = 41),
a41 = 8 + (41 - 1)(6)
= 8 + 40(6)
= 8 + 240
= 248
Let us take the term which is 72 more than the 41st term as an. So,
an = 72 + a41
= 72 + 248
= 320
Also, an = a + (n - 1)d
320 = 8 + (n - 1)6
320 = 8 + 6n - 6
320 = 2 + 6n
320 - 2 = 6n
Further simplifying, we get,
318 = 6n
n = 318/6
n = 53
Therefore, the 53rd term of the given A.P. is 72 more than the 41st term.
Page No 5.26
Ques.32. Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its 36th term.
Ans. In the given problem, let us first find the 36st term of the given A.P.
A.P. is 9, 12, 15, 18 …
Here,
First term (a) = 9
Common difference of the A.P. (d) = 12 - 9 = 3
Now, as we know,
an = a + (n - 1)d
So, for 36th term (n = 36),
a36 = 9 + (36 - 1)(3)
= 9 + 35(3)
= 9 + 105
= 114
Let us take the term which is 39 more than the 36th term as an. So,
an = 39 + a36
= 39 + 114
= 153
Also, an = a + (n - 1)d
153 = 9 + (n - 1)3
153 = 9 + 3n + 3
153 = 6 + 3n
153 - 6 = 3n
Further simplifying, we get,
147 = 3n
n = 147/3
n = 49
Therefore, the 49th term of the given A.P. is 39 more than the 36th term.
Ques.33. Find the 8th term from the end of the A.P. 7, 10, 13, ..., 184.
Ans. In the given problem, we need to find the 8th term from the end for the given A.P.
We have the A.P as 7, 10, 13 …184
Here, to find the 8th term from the end let us first find the total number of terms.
Let us take the total number of terms as n.
So,
First term (a) = 7
Last term (an) = 184
Common difference (d) = 10 - 7 = 3
Now, as we know,
an = a + (n - 1)d
So, for the last term,
184 = 7 + (n - 1)3
184 = 7 + 3n - 3
184 = 4 + 3n
184 - 4 = 3n
Further simplifying,
180 = 3n
n = 180/3
n = 60
So, the 8th term from the end means the 53rd term from the beginning.
So, for the 53rd term (n = 53)
a53 = 7 + (53 - 1)3
= 7 + (52)3
= 7 + 156
= 163
Therefore, the 8th term from the end of the given A.P. is 163.
Ques.34. Find the 10th term from the end of the A.P. 8, 10, 12, ..., 126.
Ans. In the given problem, we need to find the 10th term from the end for the given A.P.
We have the A.P as 8, 10, 12 …126
Here, to find the 10th term from the end let us first find the total number of terms. Let us take the total number of terms as n.
So,
First-term (a) = 8
Last term (an) = 126
Common difference (d) = 10 - 8 = 2
Now, as we know,
an = a + (n - 1)d
So, for the last term,
126 = 8 + (n - 1)2
126 = 8 + 2n - 2
126 = 6 + 2n
126 - 6 = 2n
Further simplifying,
120 = 2n
n = 120/2
n = 60
So, the 10th term from the end means the 51st term from the beginning.
So, for the 51st term (n = 51)
a51 = 8 + (51 - 1)2
= 8 + (50)2
= 8 + 100
= 108
Therefore, the 10th term from the end of the given A.P. is 108.
Ques.35. The sum 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P.
Ans. In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 44. We have to find the A.P
We can write this as,
a4 + a8 = 24 ...(1)
a6 + a10 = 44 ...(2)
We need to find the A.P
For the given A.P., let us take the first term as a and the common difference as d
As we know,
an = a + (n - 1)d
For 4th term (n = 4),
a4 = a + (4 - 1)d
= a + 3d
For 8th term (n = 8),
a8 = a + (8 - 1)d
= a + 7d
So, on substituting the above values in (1), we get,
(a + 3d)+(a + 7d) = 24
2a + 10d = 24 ...(3)
Also, for 6th term (n = 6),
a6 = a + (6 - 1)d
= a + 5d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
So, on substituting the above values in (2), we get,
(a + 5d) + (a + 9d) = 44
2a + 14d = 44 ...(4)
Next we simplify (3) and (4). On subtracting (3) from (4), we get,
(2a + 14d)-(2a + 10d) = 44 - 24
2a + 14d - 2a - 10d = 20
4d = 20
d = 20/4
d = 5
Further, using the value of d in equation (3), we get,
a + 10(5) = 24
2a + 50 = 24
2a = 24 - 50
2a = -26
a = -13
So here, a = -13 and d = 5
Therefore, the A.P. is -13, -8, -3, 2, ... .
Ques.36. Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?
Ans. In the given problem, let us first find the 21st term of the given A.P.
A.P. is 3, 15, 27, 39 …
Here,
First-term (a) = 3
Common difference of the A.P. (d) = 15 - 3 = 12
Now, as we know,
an = a + (n - 1)d
So, for 21st term (n = 21),
a21 = 3 + (21 - 1)(12)
= 3 + 20(12)
= 3 + 240
= 243
Let us take the term which is 120 more than the 21st term as an. So,
an = 120 + a21
= 120 + 243
= 363
Also, an = a + (n - 1)d
363 = 3 +(n - 1)12
363 = 3 + 12n - 12
363 = 9 + 12n
363 + 9 =12n
Further simplifying, we get,
372 = 12n
n = 370/12
n = 31
Therefore, the 31st term of the given A.P. is 120 more than the 21st term.
Ques.37. The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43. find the nth term.
Ans. Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a17 = 5 + 2a8
⇒ a + (17 − 1)d = 5 + 2(a + (8 − 1)d)
⇒ a + 16d = 5 + 2a + 14d
⇒ 16d − 14d = 5 + 2a − a
⇒ 2d = 5 + a
⇒ a = 2d − 5 .... (1)
Also, a11 = 43
⇒ a + (11 − 1)d = 43
⇒ a + 10d = 43 ....(2)
On substituting the values of (1) in (2), we get
2d − 5 + 10d = 43
⇒ 12d = 5 + 43
⇒ 12d = 48
⇒ d = 4
⇒ a = 2 × 4 − 5 [From (1)]
⇒ a = 3
∴ an = a + (n − 1)d
= 3 + (n − 1)4
= 3 + 4n − 4
= 4n − 1
Thus, the nth term of the given A.P. is 4n − 1.
Ques.38. Find the number of all three-digit natural numbers which are divisible by 9.
Ans. First three-digit number that is divisible by 9 is 108.
Next number is 108 + 9 = 117.
And the last three-digit number that is divisible by 9 is 999.
Thus, the progression will be 108, 117, .... , 999.
All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.
We know that, nth term = an = a + (n − 1)d
According to the question,
999 = 108 + (n − 1)9
⇒ 108 + 9n − 9 = 999
⇒ 99 + 9n = 999
⇒ 9n = 999 − 99
⇒ 9n = 900
⇒ n = 100
Thus, the number of all three digit natural numbers which are divisible by 9 is 100.
Ques.39. The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a19 = 3a6
⇒ a + (19 − 1)d = 3(a + (6 − 1)d)
⇒ a + 18d = 3a + 15d
⇒ 18d − 15d = 3a − a
⇒ 3d = 2a
⇒ a = (3/2)d .... (1)
Also, a9 = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19 ....(2)
On substituting the values of (1) in (2), we get
(3/2)d + 8d = 19
⇒ 3d + 16d = 19 × 2
⇒ 19d = 38
⇒ d = 2
⇒ a = (3/2)×2 [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .... .
Ques.40. The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a9 = 6a2
⇒ a + (9 − 1)d = 6(a + (2 − 1)d)
⇒ a + 8d = 6a + 6d
⇒ 8d − 6d = 6a − a
⇒ 2d = 5a
⇒ a = (2/5)d .... (1)
Also, a5 = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22 ....(2)
On substituting the values of (1) in (2), we get
(2/5)d + 4d = 22
⇒ 2d + 20d = 22 × 5
⇒ 22d = 110
⇒ d = 5
⇒ a = (2/5)×5 [From (1)]
⇒ a = 2
Thus, the A.P. is 2, 7, 12, 17, .... .
Ans. In the given problem, let us first find the 36st term of the given A.P.
A.P. is 9, 12, 15, 18 …
Here,
First term (a) = 9
Common difference of the A.P. (d) = 12 - 9 = 3
Now, as we know,
an = a + (n - 1)d
So, for 36th term (n = 36),
a36 = 9 + (36 - 1)(3)
= 9 + 35(3)
= 9 + 105
= 114
Let us take the term which is 39 more than the 36th term as an. So,
an = 39 + a36
= 39 + 114
= 153
Also, an = a + (n - 1)d
153 = 9 + (n - 1)3
153 = 9 + 3n + 3
153 = 6 + 3n
153 - 6 = 3n
Further simplifying, we get,
147 = 3n
n = 147/3
n = 49
Therefore, the 49th term of the given A.P. is 39 more than the 36th term.
Ques.33. Find the 8th term from the end of the A.P. 7, 10, 13, ..., 184.
Ans. In the given problem, we need to find the 8th term from the end for the given A.P.
We have the A.P as 7, 10, 13 …184
Here, to find the 8th term from the end let us first find the total number of terms.
Let us take the total number of terms as n.
So,
First term (a) = 7
Last term (an) = 184
Common difference (d) = 10 - 7 = 3
Now, as we know,
an = a + (n - 1)d
So, for the last term,
184 = 7 + (n - 1)3
184 = 7 + 3n - 3
184 = 4 + 3n
184 - 4 = 3n
Further simplifying,
180 = 3n
n = 180/3
n = 60
So, the 8th term from the end means the 53rd term from the beginning.
So, for the 53rd term (n = 53)
a53 = 7 + (53 - 1)3
= 7 + (52)3
= 7 + 156
= 163
Therefore, the 8th term from the end of the given A.P. is 163.
Ques.34. Find the 10th term from the end of the A.P. 8, 10, 12, ..., 126.
Ans. In the given problem, we need to find the 10th term from the end for the given A.P.
We have the A.P as 8, 10, 12 …126
Here, to find the 10th term from the end let us first find the total number of terms. Let us take the total number of terms as n.
So,
First-term (a) = 8
Last term (an) = 126
Common difference (d) = 10 - 8 = 2
Now, as we know,
an = a + (n - 1)d
So, for the last term,
126 = 8 + (n - 1)2
126 = 8 + 2n - 2
126 = 6 + 2n
126 - 6 = 2n
Further simplifying,
120 = 2n
n = 120/2
n = 60
So, the 10th term from the end means the 51st term from the beginning.
So, for the 51st term (n = 51)
a51 = 8 + (51 - 1)2
= 8 + (50)2
= 8 + 100
= 108
Therefore, the 10th term from the end of the given A.P. is 108.
Ques.35. The sum 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the A.P.
Ans. In the given problem, the sum of 4th and 8th term is 24 and the sum of 6th and 10th term is 44. We have to find the A.P
We can write this as,
a4 + a8 = 24 ...(1)
a6 + a10 = 44 ...(2)
We need to find the A.P
For the given A.P., let us take the first term as a and the common difference as d
As we know,
an = a + (n - 1)d
For 4th term (n = 4),
a4 = a + (4 - 1)d
= a + 3d
For 8th term (n = 8),
a8 = a + (8 - 1)d
= a + 7d
So, on substituting the above values in (1), we get,
(a + 3d)+(a + 7d) = 24
2a + 10d = 24 ...(3)
Also, for 6th term (n = 6),
a6 = a + (6 - 1)d
= a + 5d
For 10th term (n = 10),
a10 = a + (10 - 1)d
= a + 9d
So, on substituting the above values in (2), we get,
(a + 5d) + (a + 9d) = 44
2a + 14d = 44 ...(4)
Next we simplify (3) and (4). On subtracting (3) from (4), we get,
(2a + 14d)-(2a + 10d) = 44 - 24
2a + 14d - 2a - 10d = 20
4d = 20
d = 20/4
d = 5
Further, using the value of d in equation (3), we get,
a + 10(5) = 24
2a + 50 = 24
2a = 24 - 50
2a = -26
a = -13
So here, a = -13 and d = 5
Therefore, the A.P. is -13, -8, -3, 2, ... .
Ques.36. Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?
Ans. In the given problem, let us first find the 21st term of the given A.P.
A.P. is 3, 15, 27, 39 …
Here,
First-term (a) = 3
Common difference of the A.P. (d) = 15 - 3 = 12
Now, as we know,
an = a + (n - 1)d
So, for 21st term (n = 21),
a21 = 3 + (21 - 1)(12)
= 3 + 20(12)
= 3 + 240
= 243
Let us take the term which is 120 more than the 21st term as an. So,
an = 120 + a21
= 120 + 243
= 363
Also, an = a + (n - 1)d
363 = 3 +(n - 1)12
363 = 3 + 12n - 12
363 = 9 + 12n
363 + 9 =12n
Further simplifying, we get,
372 = 12n
n = 370/12
n = 31
Therefore, the 31st term of the given A.P. is 120 more than the 21st term.
Ques.37. The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43. find the nth term.
Ans. Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a17 = 5 + 2a8
⇒ a + (17 − 1)d = 5 + 2(a + (8 − 1)d)
⇒ a + 16d = 5 + 2a + 14d
⇒ 16d − 14d = 5 + 2a − a
⇒ 2d = 5 + a
⇒ a = 2d − 5 .... (1)
Also, a11 = 43
⇒ a + (11 − 1)d = 43
⇒ a + 10d = 43 ....(2)
On substituting the values of (1) in (2), we get
2d − 5 + 10d = 43
⇒ 12d = 5 + 43
⇒ 12d = 48
⇒ d = 4
⇒ a = 2 × 4 − 5 [From (1)]
⇒ a = 3
∴ an = a + (n − 1)d
= 3 + (n − 1)4
= 3 + 4n − 4
= 4n − 1
Thus, the nth term of the given A.P. is 4n − 1.
Ques.38. Find the number of all three-digit natural numbers which are divisible by 9.
Ans. First three-digit number that is divisible by 9 is 108.
Next number is 108 + 9 = 117.
And the last three-digit number that is divisible by 9 is 999.
Thus, the progression will be 108, 117, .... , 999.
All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.
We know that, nth term = an = a + (n − 1)d
According to the question,
999 = 108 + (n − 1)9
⇒ 108 + 9n − 9 = 999
⇒ 99 + 9n = 999
⇒ 9n = 999 − 99
⇒ 9n = 900
⇒ n = 100
Thus, the number of all three digit natural numbers which are divisible by 9 is 100.
Ques.39. The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a19 = 3a6
⇒ a + (19 − 1)d = 3(a + (6 − 1)d)
⇒ a + 18d = 3a + 15d
⇒ 18d − 15d = 3a − a
⇒ 3d = 2a
⇒ a = (3/2)d .... (1)
Also, a9 = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19 ....(2)
On substituting the values of (1) in (2), we get
(3/2)d + 8d = 19
⇒ 3d + 16d = 19 × 2
⇒ 19d = 38
⇒ d = 2
⇒ a = (3/2)×2 [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .... .
Ques.40. The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
Ans. Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a9 = 6a2
⇒ a + (9 − 1)d = 6(a + (2 − 1)d)
⇒ a + 8d = 6a + 6d
⇒ 8d − 6d = 6a − a
⇒ 2d = 5a
⇒ a = (2/5)d .... (1)
Also, a5 = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22 ....(2)
On substituting the values of (1) in (2), we get
(2/5)d + 4d = 22
⇒ 2d + 20d = 22 × 5
⇒ 22d = 110
⇒ d = 5
⇒ a = (2/5)×5 [From (1)]
⇒ a = 2
Thus, the A.P. is 2, 7, 12, 17, .... .
The document Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics is a part of the Class 10 Course RD Sharma Solutions for Class 10 Mathematics.
All you need of Class 10 at this link: Class 10
FAQs on Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) - RD Sharma Solutions for Class 10 Mathematics
| 1. How do you solve quadratic equations using the quadratic formula? |  |
Ans. To solve a quadratic equation using the quadratic formula, first identify the coefficients of the equation (a, b, and c). Then, substitute these values into the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. Calculate the discriminant (b² - 4ac) to determine the nature of the roots (real, imaginary, or equal). Finally, solve for 'x' by using the formula and simplifying the expression.
| 2. What are the different methods to solve quadratic equations? | |
Ans. There are several methods to solve quadratic equations, including factoring, completing the square, using the quadratic formula, graphing, and using the method of substitution. Each method has its advantages depending on the complexity of the equation and the ease of application.
| 3. How can you determine the number of solutions of a quadratic equation? | |
Ans. The number of solutions of a quadratic equation can be determined by calculating the discriminant using the formula b² - 4ac. If the discriminant is greater than zero, the equation has two distinct real solutions. If the discriminant is equal to zero, the equation has one real solution. If the discriminant is less than zero, the equation has two complex conjugate solutions.
| 4. Can a quadratic equation have no real solutions? | |
Ans. Yes, a quadratic equation can have no real solutions if the discriminant (b² - 4ac) is negative. In this case, the quadratic equation will have two complex conjugate solutions which are not real numbers.
| 5. How do you graph a quadratic equation to find its solutions? | |
Ans. To graph a quadratic equation and find its solutions, plot the equation as a parabola on a graph. The x-intercepts of the parabola represent the solutions of the quadratic equation. By analyzing the graph, you can determine whether the equation has real or imaginary solutions based on the x-intercepts.
Related Exams
About this Document
|
4.77/5 Rating |
|
Nov 24, 2024 Last updated |
Document Description: Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) for Class 10 2024 is part of RD Sharma Solutions for Class 10 Mathematics preparation.
The notes and questions for Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) have been prepared according to the Class 10 exam syllabus. Information about Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) covers topics
like and Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6).
Introduction of Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) in English is available as part of our RD Sharma Solutions for Class 10 Mathematics
for Class 10 & Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) in Hindi for RD Sharma Solutions for Class 10 Mathematics course.
Download more important topics related with notes, lectures and mock test series for Class 10
Exam by signing up for free. Class 10: Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Description
Full syllabus notes, lecture & questions for Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics - Class 10 | Plus excerises question with solution to help you revise complete syllabus for RD Sharma Solutions for Class 10 Mathematics | Best notes, free PDF download
Information about Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6)
In this doc you can find the meaning of Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) defined & explained in the simplest way possible. Besides explaining types of
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) theory, EduRev gives you an ample number of questions to practice Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) tests, examples and also practice Class 10
tests
|
Explore Courses for Class 10 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
shortcuts and tricks
,Free
,mock tests for examination
,Extra Questions
,Chapter 5 - Quadratic Equations
,video lectures
,Chapter 5 - Quadratic Equations
,Sample Paper
,Important questions
,Objective type Questions
,study material
,Chapter 5 - Quadratic Equations
,practice quizzes
,Summary
,Viva Questions
,MCQs
,ppt
,Semester Notes
,RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
,past year papers
,RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
,Exam
,RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
,Previous Year Questions with Solutions
;
Additional Information about Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) for Class 10 Preparation
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Free PDF Download
The Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) is an invaluable resource that delves deep into the core of the Class 10 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 Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) now and kickstart your journey towards success in the Class 10 exam.
Importance of Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6)
The importance of Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) cannot be overstated, especially for Class 10 aspirants.
This document holds the key to success in the Class 10 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.
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Notes
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6).
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, Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Notes on EduRev are your ultimate resource for success.
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Class 10 Questions
The "Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 10 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 Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) on the App
Students of Class 10 can study Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6),
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 Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-6) is prepared as per the latest Class 10 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