Solutions of the most important questions for 10 Class CBSE SA-1 and SA-2
In this post, we have brought for you the most important questions of your NCERT book, as it is clear from the most recent years’ SA-1 and SA-2 exam question papers that the greater number of the questions have been asked , are from NCERT book, After you would have experienced the exercises of NCERT book, repeat these most important questions more than once for getting additional benefits. All questions are fathomed by a subject expert, questions are planned according to the CBSE standards with the goal of perfecting you so you need not be confounded. Design of questions is chosen in an ordered way that will help you in your study plan for SA-1 and SA-2 CBSE board.
Here we have introduced just those questions wherein the majority of the students face a few sorts of troubles, and irrespective of whether you are a genius, normal or week in maths it will support everyone, our principal target is to give you an idea that how might you accomplish the phenomenal performances that you merit. Toward the finish of the post please reveal to us your concern and how did you like these solutions of 10 th class maths solution of most important questions for SA1/SA-2 CBSE board exam.
Solutions of the most important questions for 10 Class CBSE SA-1 and SA-2
Q1. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
According to Euclid’s division lemma, any positive integer is written in the form of bq +r where 0 ≤ r < b.
Let b = 3, so possible values of r = 0, 1, 2 (as 0 ≤ r<3)
We get corresponding values of positive integers by placing b=3 and values of r one by one in the expression bq + r.
(3q)² = 3(3q²) = 3m (where m = 3q², another positive integer)
(3q +1)² = 9q² +6q +1 = 3(3q² +2q) +1 = 3m +1(where m = 3q²+2q another positive integer)
(3q +2)² = 9q² +12q +4 = 9q² +12q +3+ 1= 3(3q² +4q +1) +1 = 3m +1(where m = 3q² +4q +1, is another positive integer)
Hence it has been proved that squire of any positive integers either of the form 3m or 3m +1.
Q2. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
According to Euclid’s division lemma, a positive integer is written in the form bq + r where 0 ≤ r < b.
Let b = 9, then 0 ≤ r < 9, so possible values of r will be →0,1,2,3,4,5,6,7,8
Correspoding values of positive intgers are →9q,9q +1, 9q +2, 9q +3, 9q +4, 9q +5, 9q +6, 9q +7, 9q +8
Cubing each term we have
(9q)³ = 9(81 q³) = 9m (where m = 9q³ is another positive integer)
(9q + 1)³ = (9q)³ + 3(9q)² + 3(9q) + 1 = 9(81q³ + 27q² + 3q) + 1 = 9m + 1(where m =81q³ + 27q² + 3q is another positive integer)
(9q +2)³ = (9q)³ +3(9q)²×2 +3×2²×9q +2³= 9(81q³ +54q +12q) +8 = 9m+8 (where m = 81q³ +54q +12q, is another positive integer)
Therefore it is evident from above that cubes of any positive integer is written in the form of 9m or 9m +1 or 9m +8.
Q3.Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.
If α and β are the two zeroes of a quadratic polynomial then the quadratic polynomial is expressed as follows.
x² –(α +β)x + αβ
we are given α +β =–3, αβ =2
x² –(–3)x + 2
Q4.Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).
If two of the zeroes of the given polynomial are √(5/3) and – √(5/3) then [x- √(5/3)] and[x +√(5/3)] will be its two of the factors.
The product of both zeroes = [x- √(5/3)] [x +√(5/3)] = x² – √(5/3)
Therefore the product of both factor will be the factor of the given polynomial so dividing the polynomial by
Factorizing the quotient 3x²+6x + 3
3x² + 3x +3x +3
3x(x +1) + 3(x +1)
(x +1)(3x +1)
x = –1, x = –1/3
Two other zeroes of the given polynomial are –1 and –1/3
Q.5-On dividing x³–3x²+x +2 by a polynomial g(x), the quotient and remainder were (x-2) and (-2x+4) respectively. Find g(x).
p(x) = x³–3x²+x +2, q(x) = x-2, r(x) = -2x+4
p(x) = g(x). q(x) +r(x)
g(x) = [p(x) -r(x)]/q(x)
g(x) = [(x³–3x²+x +2) -( -2x+4)/(x-2)
= (x³ -3x² -3x -2)/(x -2)
Therefore g(x) = x³ –x +1
Q6. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.
If zeroes of a polynomial are given by α, β and γ then the polynomial is written as follows
x³ – (α + β + γ)x² + (αβ + βγ + γα )x – αβγ
α = 2, β=–7, γ = –14
x³ –(2 –7–14)x²+ (2 × –7 –7×–14 –14×2)x + 2×–7×–14
x³ +19x² + 56x +196
Q7.If the zeroes of the polynomial x3 – 3x2 + x + 1 are a–b, a, a+b, find a and b.
α + β + γ=a –b + a + a + b = 3a
αβ + βγ + γα = (a –b)a + a(a +b) + (a –b)(a +b) =a² +ab +a²+ab +a²–b²= 3a²– b² +2ab
αβγ = (a–b) a (a+b) = a³–ab²
So, the polynomial will be as follows
x³– 3ax² + (3a²– b² +2ab)x – a³+ ab²
Comparing it with the given polynomial x3 – 3x2 + x + 1
x³– 3ax² + (3a²– b² +2ab)x – a³–ab² = x3 – 3x2 + x + 1
– 3a = –3
a = 1
– a³+ab² = 1
–(1)³ + 1 × b² = 1
b² = 2
b = ±√2
Therefore a = 1, b= ±√2
Q8. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2±√3, find other zeroes.
Two zeroes of the polynomial are 2 +√3 and 2 –√3, therefore two of the factors are [x -(2 +√3)] and [x -(2 –√3)]
Hence, one of the factor of the polynomial will be = [x –(2 +√3)][x –( 2 –√3)] =[(x–2 )–√3][(x–2) +√3] = (x –2)² –(√3)² = x² +4 –4x –3 = x² –4x +1
Therefore, dividing the polynomial by x² –4x +1
x²– 2x – 35
x² – 7x + 5x – 35
x(x – 7) + 5(x – 7)
(x – 7)( x + 5)
Therefore its other two zeroes will be x =7 and x = – 5
Q9. If the polynomial X4– 6x3 + 16x2– 25x + 10 is divided by another polynomial x² –2x +k the remainder comes out to be x +a find k and a.
Comparing it with the reminder given to us
(2k – 9)x + (10 – 8k + k²) = x + a
2k – 9 = 1
k = 5
a =10 – 8k + k² =10 –8 ×5 + 5² = 10–40 + 25 = –5
Therfore value of k = 5 and a = –5
Q10. Solve the following equations
2√x + 3√y = 0
3√x + 2√y = 0
The given pair of the equation is
2√x + 3√y = 0 …..(i)
3√x + 2√y = 0 …..(ii)
Multiplying eq.(i) by 2 and eq.(ii) by 3 and then subtracting (i) from (ii)
Substituting the value of y =0 in equation (1) we get x = 0
So, equations will become
2a + 3b = 2 → (1)
4a – 9b = –1→(2)
For making the coefficient of ‘a’ equal in both equations, multiplying (1) by 2 and subtracting it from (2)
As we have supposed
Therefore value of x =9 and y = 4
Q12. Solve the following equations
Substituting 1/x = a and 1/y = b in the equations (1) and (2) we get following equations (3) and (4)
4a + 3y = 14…….(2), 3a – 4y = 23……..(3)
Multiplying equation (2) by 3 and equation (3) by 4 , we get equation (5) and (6), subtracting equation (6) from (5)
Substituting y = –2, in equation (2)
4a + 3 × –2 = 14
a = 5
As we have supposed
Therefore x = 1/5 and y = –2
Q13.Solve the following equations
Substituting 1/(x +y) and 1/(x –y) by a and b we get the equations (3) and (4)
10a + 2b = 4………(3) 15a – 5b = –2…….(4)
Multiplying equation (3) by 5 and equation (4) by 2 we get equation (5) and (6), adding (6) to equation (5)
Putting the value of ‘a’ in the equation (3)
we get b = 1
As we have supposed 1/(x +y) = a and 1/(x –y) = b
x +y = 5……….(7)
x – y =1……….(8)
Adding both of the equation (7) and (8) we get
x =3 and y= 2
Q14. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solution. Let the speed of current = x km/h and the speed of boat in still water is y km/h.
In downstream the speed of boat will be = distance/time= 20/2 = 10 km/h
According to first condition
x + y = 10………..(1)
According to the second condition, the speed of the boat in upstream = distance/time = 4/2 =2 km/h
x – y = 2………(2)
Adding both of the equation (1) and (2)
2x = 12
Placing the value of x = 6 in (1), we have
y = 4
Therefore speed of boat in still water =4 km/h and speed of current = 6 km/h
Q15. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Let the time taken by 1 woman alone to finish the work = x days and time taken by 1 man alone = y days
x days are the time taken by a woman alone in doing → 1 piece of embroidery work
In 1 day she will do = th part of that work
In 1 day 2 women will do = th part
Y days are the time taken by man alone in doing → 1 piece of embroidery work
In 1 day he will do = th part of that work
In 1 day 5 men will do = th part
It is given to us that 2 women and 5 men finish the work = in 4 day
∴In 1 day they will do = th part
According to first condition the equation will become as following
Similarly, according to the second condition, the equation will be
So, we get the following equations
Multiplying equations (3) by 3 and equation (4) by 2 and then subtracting (3) from (4)
Putting b= 1/36 in equation (3)
putting this value of b = 1/36 in b = 1/y, we get y = 36
Placing the value of a in a = 1/x, we get x = 18
Therefore a woman will take 36 days and a man alone will take 18 days to complete the embroidery work
You can download pdf of most important questions for SA-1/SA-2 from the following link that will definitely help you in achieving your target of excellent marks.
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions of class 9 maths
|Chapter 1- Number System||Chapter 9-Areas of parallelogram and triangles|
|Chapter 2-Polynomial||Chapter 10-Circles|
|Chapter 3- Coordinate Geometry||Chapter 11-Construction|
|Chapter 4- Linear equations in two variables||Chapter 12-Heron’s Formula|
|Chapter 5- Introduction to Euclid’s Geometry||Chapter 13-Surface Areas and Volumes|
|Chapter 6-Lines and Angles||Chapter 14-Statistics|
|Chapter 7-Triangles||Chapter 15-Probability|
|Chapter 8- Quadrilateral|
NCERT Solutions of class 9 science
NCERT Solutions of class 10 maths
NCERT solutions of class 10 science
Solutions of class 10 last years Science question papers
NCERT solutions of class 11 maths
|Chapter 1-Sets||Chapter 9-Sequences and Series|
|Chapter 2- Relations and functions||Chapter 10- Straight Lines|
|Chapter 3- Trigonometry||Chapter 11-Conic Sections|
|Chapter 4-Principle of mathematical induction||Chapter 12-Introduction to three Dimensional Geometry|
|Chapter 5-Complex numbers||Chapter 13- Limits and Derivatives|
|Chapter 6- Linear Inequalities||Chapter 14-Mathematical Reasoning|
|Chapter 7- Permutations and Combinations||Chapter 15- Statistics|
|Chapter 8- Binomial Theorem||Chapter 16- Probability|
NCERT solutions of class 12 maths
|Chapter 1-Relations and Functions||Chapter 9-Differential Equations|
|Chapter 2-Inverse Trigonometric Functions||Chapter 10-Vector Algebra|
|Chapter 3-Matrices||Chapter 11 – Three Dimensional Geometry|
|Chapter 4-Determinants||Chapter 12-Linear Programming|
|Chapter 5- Continuity and Differentiability||Chapter 13-Probability|
|Chapter 6- Application of Derivation||CBSE Class 12- Question paper of maths 2021 with solutions|
|Chapter 7- Integrals|
|Chapter 8-Application of Integrals|
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