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10th Maths 4.2

NCERT Class 10th solution of Exercise 4.1 NCERT Class 10th solution of Exercise 4.3 NCERT Class 10th solution of Exercise 4.4 NCERT Class 10th Maths Projects Exercise 4.2 Q1. Find the roots of the following quadratic equation by factorisation : i) x2-3x-10=0, ii) 2x2+x-6=0, iii) 2x2+7x+52=0, iv) 2x2-x+18=0, v) 100x2-20x+1=0. Sol. : i) x2-3x-10=0 x2-5x+2x-10=0 x(x-5)+2(x-5)=0 (x-5)(x+2)=0 x-5=0 x=5 x+2=0 x=-2 Answer : x=5,x=-2. ii) 2x2+x-6=0 2x2+4x-3x-6=0 2x(x+2)-3(x+2)=0 (x+2)(2x-3)=0 x+2=0 x=-2, 2x-3=0 x=32 Answer : x=-2,x=32. iii) 2x2+7x+52=0 2x2+5x+2x+52=0 x(2x+5)+2(2x+5)=0 (2x+5)(x+2)=0 2x+5=0 x=-52 x+2=0 x=-2 Answer : x=-52,x=-2. iv) 2x2-x+18=0 16x2-8x+1=0 16x2-4x-4x+1=0 4x(4x-1)-1(4x-1)=0 (4x-1)(4x-1)=0 (4x-1)2=0 4x-1=0 x=14 Answer : each x=14. v) 100x2-20x+1=0 100x2-10x-10x+1=0 `10x(10x-1)-1(10x-1)=0...

10th Maths 4.1

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NCERT Class 10th solution of Exercise 4.2 NCERT Class 10th solution of Exercise 4.3 NCERT Class 10th solution of Exercise 4.4 NCERT Class 10th Maths Projects Exercise 4.1 Q1. Check whether the following are quadratic equations : i) (x+1)2=2(x-3) ii) x2-2x=(-2)(3-x) iii) (x-2)(x+1)=(x-1)(x+3) iv) (x-3)(2x+1)=x(x+5) v) (2x-1)(x-3)=(x+5)(x-1) vi) x2+3x+1=(x-2)2 vii) (x+2)3=2x(x2-1) viii) x3-4x2-x+1=(x-2)3 Sol. : i) (x+1)2=2(x-3) x2+2x+1=2x-6 x2+2x+1-2x+6=0 x2+7=0 Answer : Yes. ii) x2-2x=(-2)(3-x) x2-2x=-6+2x x2-2x+6-2x=0 x2-4x+6=0 Answer : Yes. iii) (x-2)(x+1)=(x-1)(x+3) x(x+1)-2(x+1)=x(x+3)-1(x+3) x2+x-2x-2=x2+3x-x-3 x2-x-2=x2+2x-3 x2-x-2-x2-2x+3=0 -3x+1=0 3x-1=0 Answer : No. iv) (x-3)(2x+1)=x(x+5) x(2x+1)-3(2x+1)=x2+5x 2x2+x-6x-3-x2-5x=0 x2-10x-3=0 Answer : Yes. v) (2x-1)(x-3)=(x+5)(x-1) 2x(x-3)-1(x-3)=x(x-1)+5(x-1) 2x2-6x-x+3=x2-x+5x-5 2x2-7x+3=x2+4x-5 2x2-7x+3-x2-4x+5=0 x2-11x+8=0 Answe...

10th Maths 3.6

NCERT Class 10th solution of Exercise 3.1 NCERT Class 10th solution of Exercise 3.2 NCERT Class 10th solution of Exercise 3.3 NCERT Class 10th solution of Exercise 3.4 NCERT Class 10th solution of Exercise 3.5 NCERT Class 10th solution of Exercise 4.1 NCERT Class 10th Maths Projects Exercise 3.6 Q1. Solve the following pairs of equations by reducing them to a pair of linear equations : i) 12x+13y=2;13x+12y=136 ii) 2x+3y=2;4x-9y=-1 iii) 4x+3y=14;3x-4y=23 iv) 5x-1+1y-2=2;6x-1-3y-2=1 v) 7x-2yxy=5;8x+7yxy=15 vi) 6x+3y=6xy;2x+4y=5xy vii) 10x+y+2x-y=4;15x+y-5x-y=-2 viii) 13x+y+13x-y=34;12(3x+y)-12(3x-y)=-18 Sol. : i) 12x+13y=2___________(1) 13x+12y=136_________(2) Let 1x=p  and 1y=q then p2+q3=2 3p+2q=12________________(3) p3+q2=136 2p+3q=13________________(4) from equation (3) q=12-3p2 Put in equation (4) 2p+3(12-3p2)=13 `4p+36-9p=2...

10th Maths 3.5

NCERT Class 10th solution of Exercise 3.1 NCERT Class 10th solution of Exercise 3.2 NCERT Class 10th solution of Exercise 3.3 NCERT Class 10th solution of Exercise 3.4 NCERT Class 10th solution of Exercise 3.6 NCERT Class 10th Maths Projects Exercise 3.5 Q1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method. (i) x-3y-3=0;3x-9y-2=0 (ii) 2x+y=5;3x+2y=8 (iii) 3x-5y=20;6x-10y=40 (iv) x-3y-7=0;3x-3y-15=0 Sol. : i) x-3y-3=0_______(1) 3x-9y-2=0______(2) a1a2=13,b1b2=-3-9=13, and c1c2=-3-2=32 a1a2=b1b2c1c2 Answer : No solution. ii) 2x+y=5________(1) 3x+2y=8_______(2) a1a2=23,b1b2=12, and c1c2=-5-8=58 a1a2b1b2c1c2 By Cross Multiplication Method : xb1c2-b2c1=yc1a2-c2a1=1a1b2-a2b1 x(1)(-8)-(2)(-5)=y(-5)(3)-(-8)(2)=1(2)(2)-(3)(1) `x/...

10th Maths 3.4

NCERT Class 10th solution of Exercise 3.1 NCERT Class 10th solution of Exercise 3.2 NCERT Class 10th solution of Exercise 3.3 NCERT Class 10th solution of Exercise 3.5 NCERT Class 10th solution of Exercise 3.6 NCERT Class 10th Maths Projects Exercise 3.4 Q1. Solve the following pair of linear equations by the elimination method and the substitution method : i) x+y=5 and 2x-3y=4 ii) 3x+4y=10 and 2x-2y=2 iii) 3x-5y-4=0 and 9x=2y+7 iv) x2+2y3=-1 and x-y3=3 Sol. : i) By Elimination Method : x+y=5___________(1) 2x-3y=4__________(2) Multiply equation (1) by 3 and equation (2) by 1, 3x+3y=15________(3) 2x-3y=4__________(4) By Equation (3) + (4) 3x+3y+2x-3y=15+4 5x=19 x=195 Put in equation (1) 195+y=5 y=5-195 y=25-195 y=65 By Substitution Method : from equation (1) y=5-x__________(5) Substituting this value in equation (2) 2x-3(5-x)=4 2x-15+3x=4 5x=4+15 5x=19 x=195 put in equation (5) y=5-195 y=25-195 y=65 Answer : `...

10th Maths 3.3

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Exercise 3.3 NCERT Class 10th solution of Exercise 3.1 NCERT Class 10th solution of Exercise 3.2 NCERT Class 10th solution of Exercise 3.4 NCERT Class 10th solution of Exercise 3.5 NCERT Class 10th solution of Exercise 3.6 NCERT Class 10th Maths Projects Exercise 3.3 Q1. Solve the following pair of linear equations by the substitution method : i) x+y=14;x-y=4 ii) s-t=3;s3+t2=6 iii) 3x-y=3;9x-3y=9 iv) 0.2x+0.3y=1.3;0.4x+0.5y=2.3 v) 2x+3y=0;3x-8y=0 vi) 3x2-5y3=-2;x3+y2=136 Sol. : i) x+y=14________(1) x-y=4 _________(2) from equation (2) x=y+4_________(3) Substituting this value in equation (1) y+4+y=14 2y=14-4 2y=10 y=102 y=5 Put in equation (3) x=14-5 x=9 Answer : x=9,y=5. ii)  s-t=3___________(1) s3+t2=6_______(2) from equation (1) s=t+3__________(3) Substituting this value in equation (2) t+33+t2=6 2(t+3)+3t6=6 2t+6+3t=36 5t=36-6 5t=30 t=305 t=6 Put in equation (3) s=6+3 s=9 An...

10th Maths 3.2

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NCERT Class 10th solution of Exercise 3.1 NCERT Class 10th solution of Exercise 3.3 NCERT Class 10th solution of Exercise 3.4 NCERT Class 10th solution of Exercise 3.5 NCERT Class 10th solution of Exercise 3.6 NCERT Class 10th Maths Projects Exercise 3.2 Q1. From the pair of linear equations in the following problems, and find their solutions graphically. i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that one pen. Sol. : i) Let the number of boys in Mathematical qize are x, and the number of girls in Mathematical qize are y According to question x+y=10______(1) and y=x+4_______(2) from equation (1) when x=4 4+y=10 y=10-4 y=6 when x=6 6+y=10 y=10-6 y=4 from equation (2) when x=0 y=0+4 y=4...