RD SHARMA SOLUTION CHAPTER- 25 Parabola I CLASS 11TH MATHEMATICS-EDUGROW

Chapter 25 Parabola Exercise Ex. 25.1

Question 1(i)

Solution 1(i)

Question 1(iii)

Solution 1(iii)

Question 1(ii)

Solution 1(ii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

L a t u s space R e c t u m space equals L e n g t h space o f space p e r p e n d i c u l a r space f r o m space f o c u s space left parenthesis 2 comma 3 right parenthesis space o n space d i r e c t r i x space x minus 4 y plus 3 equals 0
equals 2 open vertical bar fraction numerator 2 minus 12 plus 3 over denominator square root of 1 plus 16 end root end fraction close vertical bar
equals 2 open vertical bar fraction numerator minus 7 over denominator square root of 17 end fraction close vertical bar
equals fraction numerator 14 over denominator square root of 17 end fraction

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

      Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 4(vi)

Solution 4(vi)

Question 4(vii)

Solution 4(vii)

Question 4(viii)

Solution 4(viii)

Question 4(ix)

Solution 4(ix)

Question 5

Solution 5

Question 6

Find the area of the triangle formed by the lines joining the vertex of the parabola x squared space equals space 12 y  to the ends of its latus- rectumSolution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Find the equation of the lines joining the vertex of the parabola y2 = 6x to the point on it which have abscissa 24Solution 13

Question 14

Find the coordinates of points on the parabola y2 = 8x whose focal distance is 4Solution 14

In given parabola

a=2

Given focal distance=a+x=4, so x=2

So points are (2, 4) and (2, -4)Question 15

Find the length  of the lines segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle theta to the axisSolution 15

Question 16

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.Solution 16

Question 17

If the line y = mx + 1 is tangent to the parabola y2 = 4x, then find the value of m.Solution 17

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RD SHARMA SOLUTION CHAPTER- 24 The Circle ICLASS 11TH MATHEMATICS-EDUGROW

Chapter 24 The Circle Exercise Ex. 24.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 7(iv)

Solution 7(iv)

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

If the lines 2x-3y = 5 and 3x-4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.Solution 14

Question 15

Solution 15

Question 16

Find the equation of the circle having (1, -2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x +5y = 18.Solution 16

Question 17

If the lines 3x-4y+4 = 0 and 6x-8y-7 = 0 are tangents to a circle, then find the radius of the circle.Solution 17

Question 18

Solution 18

Question 19

The circle x2+y2-2x-2y+1 = 0 is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in new-position.Solution 19

Question 20

Solution 20

Question 21

Solution 21

Chapter 23 The Circle Exercise Ex. 24.2

Question 14

If a circle passes through the point (0, 0), (a, 0), (0, b), then find the coordinates of its centre.Solution 14

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

A space c i r c l e space p a s sin g space t h r o u g h space P left parenthesis 3 comma minus 2 right parenthesis space a n d space Q left parenthesis minus 2 comma 0 right parenthesis space a n d space h a v i n g space i t s space c e n t r e space o n space 2 x minus y equals 3.
L e t space t h e space e q u a t i o n space o f space t h e space c i r c l e space b e space x squared plus y squared plus 2 g x plus 2 f y plus c equals 0.
S i n c e space t h e space c i r c l e space p a s s e s space t h r o u g h space left parenthesis 3 comma minus 2 right parenthesis space a n d A l s o space space left parenthesis minus 2 comma 0 right parenthesis space t h e r e f o r e
9 plus 4 plus 6 g minus 4 f plus c equals 0....... left parenthesis i right parenthesis
4 plus 0 minus 4 g plus 0 plus c equals 0........ left parenthesis i i right parenthesis
A l s o space t h e space c e n t r e space o f space t h e space c i r c l e space l i e s space o n space 2 x minus y equals 3
minus 2 g plus f equals 3......... left parenthesis i i i right parenthesis
S o l v i n g space e q u a t i o n s space left parenthesis i right parenthesis comma left parenthesis i i right parenthesis space a n d space left parenthesis i i i right parenthesis comma space w e space g e t
g equals 3 over 2 comma space f equals space 6 space a n d space c equals 2
T h e r e f o r e space t h e space e q u a t i o n space o f space t h e space c i r c l e space i s
x squared plus y squared plus 3 x plus 12 y plus 2 equals 0

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 8

Solution 8

Question 9

Solution 9

I f space a comma space b comma space c space a r e space i n space A P comma space t h e n space b equals fraction numerator a plus c over denominator 2 end fraction
F o r space a equals 1 comma b equals space 4 comma c equals space 7 comma space fraction numerator 1 plus 7 over denominator 2 end fraction equals 4 equals b comma space t h e r e f o r e space 1 comma space 4 comma space 7 space a r e space i n space A P.
T h e space c e n t r e s space o f space t h e space t h r e e space c i r c l e s space l i e space i n space A P.

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 7(iv)

Find the equation of the circle which circumscribes the triangle formed by the lines.

iv. y = x + 2, 3y = 4x and 2y = 3x.Solution 7(iv)

Question 15

Find the equation of the circle which passes through the point (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.Solution 15

Chapter 24 The Circle Exercise Ex. 24.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

F i n d space t h e space e q u a t i o n space o f space a space c i r c l e space c i r c u m s c r i b i n g space t h e space r e c tan g l e space w h o s e space s i d e s space a r e space x minus 3 y equals 4 comma
3 x plus y equals 22 comma space x minus 3 y equals 14 space a n d space 3 x plus y equals 62.

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 10

T h e space l i n e space 2 x minus y plus 6 equals 0 space m e e t s space t h e space c i r c l e space x squared plus y squared minus 2 y minus 9 equals 0 space a t space A space a n d space B. space F i n d space t h e space e q u a t i o n
o f space t h e space c i r c l e space o n space A B space a s space d i a m e t e r.

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 9

ABCD is a square whose side is a; taking AB and AD as axes, prove that the equation of the circle circumscribing the square is x2 + y2 – a (x + y) = 0Solution 9

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RD SHARMA SOLUTION CHAPTER- 23 The Straight Lines I CLASS 11TH MATHEMATICS-EDUGROW

Chapter 23 The Straight Lines Exercise Ex. 23.1

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Chapter 23 The Straight Lines Exercise Ex. 23.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Chapter 23 The Straight Lines Exercise Ex. 23.3

Question 1

Solution 1

Question 2

Solution 2

y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
S i n c e space t h e space l i n e space c u t s space t h e space x minus a x i s space a t space open parentheses minus 3 comma 0 close parentheses space w i t h space s l o p e space minus 2 comma space w e space h a v e comma
y minus 0 equals minus 2 open parentheses x plus 3 close parentheses
rightwards double arrow y equals minus 2 x minus 6
rightwards double arrow 2 x plus y plus 6 equals 0

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 23 The Straight Lines Exercise Ex. 23.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

T h e space l i n e space p a s s e s space t h r o u g h space t h e space p o i n t space open parentheses 2 comma 0 close parentheses.
A l s o space i t s space i n c l i n a t i o n space t o space y minus a x i s space i s space 135 degree.
T h a t space i s comma space t h e space i n c l i n a t i o n space o f space t h e space g i v e n space l i n e space w i t h space t h e space x minus a x i s space i s space 180 degree minus 135 degree.
T h a t space i s comma space t h e space s l o p e space o f space t h e space g i v e n space l i n e space i s space 45 degree
T h e space e q u a t i o n space o f space t h e space l i n e space h a v i n g space s l o p e space apostrophe m apostrophe space a n d space p a s sin g space t h r o u g h space t h e
p o i n t space open parentheses x subscript 1 comma y subscript 1 close parentheses space i s space y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
T h e r e f o r e comma space t h e space r e q u i r e d space e q u a t i o n space i s
y minus 0 equals tan 45 degree open parentheses x minus 2 close parentheses
rightwards double arrow y equals 1 cross times open parentheses x minus 2 close parentheses
rightwards double arrow y equals x minus 2
rightwards double arrow x minus y minus 2 equals 0

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Chapter 23 The Straight Lines Exercise Ex. 23.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Solution 3

Question 4

Solution 4

T h e space r e c tan g l e space A B C D space w i l l space h a v e space d i a g o n a l s space A C space a n d space B D
A C space p a s s e s space t h r o u g h space A open parentheses a comma b close parentheses space a n d space C open parentheses a apostrophe comma b apostrophe close parentheses. space
T h u s space e q u a t i o n space o f space A C space i s :
fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction
rightwards double arrow fraction numerator y minus b over denominator b apostrophe minus b end fraction equals fraction numerator x minus a over denominator a apostrophe minus a end fraction
rightwards double arrow open parentheses y minus b close parentheses open parentheses a apostrophe minus a close parentheses equals open parentheses x minus a close parentheses open parentheses b apostrophe minus b close parentheses
rightwards double arrow y open parentheses a apostrophe minus a close parentheses minus a apostrophe b plus a b equals x open parentheses b apostrophe minus b close parentheses minus a b apostrophe plus a b
rightwards double arrow y open parentheses a apostrophe minus a close parentheses equals x open parentheses b apostrophe minus b close parentheses minus a b apostrophe plus a apostrophe b
rightwards double arrow y open parentheses a apostrophe minus a close parentheses minus x open parentheses b apostrophe minus b close parentheses equals a apostrophe b minus a b apostrophe

B D space p a s s e s space t h r o u g h space B open parentheses a apostrophe comma b close parentheses space a n d space D open parentheses a comma b apostrophe close parentheses. space
T h u s space e q u a t i o n space o f space B D space i s :
fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction
rightwards double arrow fraction numerator y minus b over denominator b apostrophe minus b end fraction equals fraction numerator x minus a apostrophe over denominator a minus a apostrophe end fraction
rightwards double arrow open parentheses y minus b close parentheses open parentheses a minus a apostrophe close parentheses equals open parentheses x minus a apostrophe close parentheses open parentheses b apostrophe minus b close parentheses
rightwards double arrow minus y open parentheses a apostrophe minus a close parentheses minus a b plus a apostrophe b equals x open parentheses b apostrophe minus b close parentheses minus a apostrophe b apostrophe plus a apostrophe b
rightwards double arrow a apostrophe b apostrophe minus a b equals x open parentheses b apostrophe minus b close parentheses plus y open parentheses a apostrophe minus a close parentheses
rightwards double arrow x open parentheses b apostrophe minus b close parentheses plus y open parentheses a apostrophe minus a close parentheses equals a apostrophe b apostrophe minus a b

Question 5

Find the equation of the side BC of the triangle ABC whose vertices are A (-1, -2), B (0, 1) and (2, 0) respectively. Also, find the equation of the median through (-1, -2).Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

L equals 4 over 1875 C plus 124.942 minus 4 cross times 20 over 1875
rightwards double arrow L equals 4 over 1875 C plus 124.899

Question 12

Solution 12

Question 13

Solution 13

rightwards double arrow y minus 3 equals 1 third open parentheses x minus 4 close parentheses
rightwards double arrow 3 open parentheses y minus 3 close parentheses equals x minus 4
rightwards double arrow x minus 3 y plus 9 minus 4 equals 0
rightwards double arrow x minus 3 y plus 5 equals 0

Question 14

Solution 14

Question 15

Find the equations of the diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1.Solution 15

Chapter 23 The Straight Lines Exercise Ex. 23.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

3Question 4

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by +8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.Solution 4

Question 5

Solution 5

Question 6

Find the equation of the line which passing through the point (-4, 3) and the portion of the line intercepted  between the axes is divided internally in the ratio 5:3 by this pointSolution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

P o i n t space left parenthesis h comma k right parenthesis space d i v i d e s space t h e space l i n e space s e g m e n t space i n space t h e space r a t i o space 1 : 2
T h u s comma space u sin g space s e c t i o n space p o i n t space f o r m u l a comma space w e space h a v e
h equals fraction numerator 2 cross times a plus 1 cross times 0 over denominator 1 plus 2 end fraction
a n d
k equals fraction numerator 2 cross times 0 plus 1 cross times b over denominator 1 plus 2 end fraction
T h e r e f o r e comma space w e space h a v e comma
h equals fraction numerator 2 a over denominator 3 end fraction space a n d space k equals b over 3
rightwards double arrow a equals fraction numerator 3 h over denominator 2 end fraction a n d space b equals 3 k
T h u s comma space t h e space c o r r e s p o n d i n g space p o i n t s space o f space A space a n d space B space a r e space open parentheses fraction numerator 3 h over denominator 2 end fraction comma 0 close parentheses space a n d space open parentheses 0 comma 3 k close parentheses
T h u s comma space t h e space e q u a t i o n space o f space t h e space l i n e space j o i n i n g space t h e space p o i n t s space A space a n d space B space i s
fraction numerator y minus 3 k over denominator 3 k minus 0 end fraction equals fraction numerator x minus 0 over denominator 0 minus fraction numerator 3 h over denominator 2 end fraction end fraction
rightwards double arrow minus fraction numerator 3 h over denominator 2 end fraction open parentheses y minus 3 k close parentheses equals x cross times 3 k
rightwards double arrow minus 3 h y plus 9 h k equals 6 k x
rightwards double arrow 2 k x plus h y equals 3 k h

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Chapter 23 The Straight Lines Exercise Ex. 23.7

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 300.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 The Straight Lines Exercise Ex. 23.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 23 The Straight Lines Exercise Ex. 23.9

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 23 The Straight Lines Exercise Ex. 23.10

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

 Question 10

Solution 10

Question 11

Solution 11

 Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

 Question 15

Solution 15

Question 16

Find the equations of the line passing through the intersection of the lines 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.Solution 16

Question 17

Find the equation of the line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.Solution 17

Chapter 23 The Straight Lines Exercise Ex. 23.11

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 23 The Straight Lines Exercise Ex. 23.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Find the equation of the straight line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0.Solution 27

Chapter 23 The Straight Lines Exercise Ex. 23.13

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 The Straight Lines Exercise Ex. 23.14

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Chapter 23 The Straight Lines Exercise Ex. 23.15

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Chapter 23 The Straight Lines Exercise Ex. 23.16

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Find the ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0.Solution 6

Chapter 23 The Straight Lines Exercise Ex. 23.17

Question 1

Deduce the condition for these lines to form a rhombus.Solution 1

Question 2

Solution 2

Question 3

Solution 3

Chapter 23 The Straight Lines Exercise Ex. 23.18

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Consider the following figure:

T h e space e q u a t i o n space o f space A B space i s
y minus 2 equals 1 fifth open parentheses x minus 1 close parentheses
rightwards double arrow 5 y minus 10 equals x minus 1
rightwards double arrow x minus 5 y plus 9 equals 0
A n d space t h e space e q u a t i o n space o f space B C space i s
y minus 8 equals minus 5 open parentheses x minus 5 close parentheses
rightwards double arrow y minus 8 equals minus 5 x plus 25
rightwards double arrow 5 x plus y minus 33 equals 0

Chapter 23 The Straight Lines Exercise Ex. 23.19

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

rightwards double arrow fraction numerator 6 x minus 21 y plus 33 plus 7 x plus 21 y minus 56 over denominator 3 end fraction equals 0
rightwards double arrow 6 x minus 21 y plus 33 plus 7 x plus 21 y minus 56 equals 0
rightwards double arrow 13 x minus 23 equals 0
rightwards double arrow 13 x equals 23

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

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RD SHARMA SOLUTION CHAPTER- 22 Brief Review of Cartesian System of Rectangular Coordinates I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Exercise Ex. 22.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Exercise Ex. 22.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Exercise Ex. 22.3

Question 1

Solution 1


Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 8

Solution 8

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RD SHARMA SOLUTION CHAPTER- 21 Some Special Series I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 21 Some Special Series Exercise Ex. 21.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 9

Solution 9

Chapter 21 Some Special Series Exercise Ex. 21.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

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RD SHARMA SOLUTION CHAPTER- 20 Geometric Progressions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 20 Geometric Progressions Exercise Ex. 20.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Chapter 20 Geometric Progressions Exercise Ex. 20.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 20 Geometric Progressions Exercise Ex. 20.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Find the sum of the geom etric series:

begin mathsize 11px style 3 over 5 plus 4 over 5 squared plus 3 over 5 cubed plus 4 over 5 to the power of 4 plus.... space to space 2 straight n space terms semicolon end style

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

1, -a, a2, – a3 , ….. to n terms (a ≠ 1)Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

A person has 2 parents, 4 grandparents, 8 great grand parents,  and so on. Find the number his ancestors during the ten generations preceding his own.Solution 18

Question 19

(n – 1) Sn = 1n + 2n + 3n + ….+ nnSolution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Chapter 20 Geometric Progressions Exercise Ex. 20.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 20 Geometric Progressions Exercise Ex. 20.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(I)

Solution 8(I)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 8(v)

Solution 8(v)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 10(iii)

Solution 10(iii)

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 11(iv)

Solution 11(iv)

Question 12

Solution 12

Question 13

Solution 13

Question 14

If the 4th, 10th, and 16th terms of a G.P. are x, y, and z respectively. Prove that x, y, z are in G.P.Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

If pth, qth, and rth terms of an A.P. and G.P. are both a, b, and c respectively, show that ab-c bc-a ca-b = 1.Solution 23

Chapter 20 Geometric Progressions Exercise Ex. 20.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

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RD SHARMA SOLUTION CHAPTER-19 Arithmetic Progressions I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 19 Arithmetic Progressions Exercise Ex. 19.1

Question 1

Solution 1

Question 2

Solution 2

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Solution 7

Question 8

Solution 8

Question 3

Find the first four terns of the sequence defined by a1 = 3 and, an = 3an– 1 + 2, for all n > 1Solution 3

Chapter 19 Arithmetic Progressions Exercise Ex. 19.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Chapter 19 Arithmetic Progressions Exercise Ex. 19.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Arithmetic Progressions Exercise Ex. 19.4

Question 1

Solution 1

( vii ) 

            Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 15

Find the rth term of an A.P., the sum of whose first n terms is 3n2 + 2n.Solution 15

Chapter 19 Arithmetic Progressions Exercise Ex. 19.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Show that x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are consecutive terms of an A.P., if x, y and z are in A.P.Solution 7

Chapter 19 Arithmetic Progressions Exercise Ex. 19.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.Solution 9

Chapter 19 Arithmetic Progressions Exercise Ex. 19.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?Solution 12

Question 13

We know that the sum of the interior angles of a triangle is 180o. Show that the sums of the interior angles of polygons with 3, 4, 5, 6,…. sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.Solution 13

Question 14

In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?Solution 14

Question 15(i)

A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.

i. Find his salary for the tenth month.Solution 15(i)

Question 15(ii)

A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.

What is his total earnings during the first year?Solution 15(ii)

Question 16

A man saved Rs. 66000 in 20 years. In each succeeding year after the first year he saved Rs. 200 more then what he saved in the previous year. How much did he save in the first year?Solution 16

Question 17

In a cricket team tournament 16 teams participated. A sum of Rs. 8000 is to be awarded among themselves as prize money. If the last place team is awarded Rs. 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?Solution 17

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RD SHARMA SOLUTION CHAPTER- 18 Binomial Theorem I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 18 Binomial Theorem Exercise Ex. 18.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Show that 24n + 4 – 15n – 16, where n Î N is divisible by 225.Solution 12

Chapter 18 Binomial Theorem Exercise Ex. 18.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

Question 9(vi)

Solution 9(vi)

Question 9(vii)

Solution 9(vii)

Question 9(viii)

Find the coefficient of x in the expansion of

 (1 – 3x + 7x2) (1 – x)16.Solution 9(viii)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

Solution 13(i)

Question 13(ii)

Solution 13(ii)

Question 13(iii)

Solution 13(iii)

Question 13(iv)

Solution 13(iv)

Question 14(i)

Solution 14(i)

Question 14(ii)

Solution 14(ii)

Question 14(iii)

Solution 14(iii)

Question 14(iv)

Solution 14(iv)

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Question 15(v)

Solution 15(v)

Question 15(vi)

Solution 15(vi)

Question 15(vii)

Solution 15(vii)

Question 15(viii)

Find the middle term (s) in expansion of:

Solution 15(viii)

Question 15(ix)

Find the middle term (s) in expansion of:

Solution 15(ix)

Question 15(x)

Find the middle term (s) in expansion of:

Solution 15(x)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 16(vii)

Solution 16(vii)

Question 16(viii)

Solution 16(viii)

Question 16(ix)

Solution 16(ix)

Question 16(x)

Solution 16(x)

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

If the seventh term from the beginning and end in the binomial expansion of   are equal, find n.Solution 39

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RD SHARMA SOLUTION CHAPTER-17 Combinations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 17 Combinations Exercise Ex. 17.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

If n + 2C8 : n – 2P4 = 57 : 16, find n.Solution 10

Question 11

Solution 11

Question 12

If nC2nC5 and nC6 are in A.P., then find nSolution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

rightwards double arrow 42504

Question 20 (i)

Solution 20 (i)

Question 20(ii)

Solution 20(ii)

Question 20(iii)

Solution 20(iii)

Question 20(iv)

Solution 20(iv)

Chapter 17 Combinations Exercise Ex. 17.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Chapter 17 Combinations Exercise Ex. 17.3

Question 1

Solution 1

Question 3

Solution 3

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 2

Solution 2

Question 4

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.Solution 4

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RD SHARMA SOLUTION CHAPTER- 16 Permutations I CLASS 11TH MATHEMATICS-EDUGROWN

Chapter 16 Permutations Exercise Ex. 16.1

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 12

Solution 12

Chapter 16 Permutations Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

How many three-digit numbers are there whit no digit repeated?Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19(i)

Solution 19(i)

Question 19(ii)

Solution 19(ii)

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?Solution 24Any number less than 1000 may be any of a number from one-digit number, two-digit number and three-digit number.

One-digit odd number:

3 possible ways are there. These numbers are 3 or 5 or 7.

Two-digit odd number:

Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.

So, there are 3  2 = 6 such 2-digit numbers.

Three-digit odd number:

Ignore the presence of zero at ones place for some instance.

Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways.

So, there are a total of 3  3  2 = 18 numbers of 3-digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero.

To obtain the even 3-digit numbers, ones place can be filled up in 1 way (only 0 to be filled), hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tens place in 2 ways (using remaining 2 digits after filling up hundreds place).

So, there are a total of 1  3  2 = 6 even 3-digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed)

So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 –  6 = 12.

Therefore, odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

How many for digit natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, and 4, if the digits can repeat?Solution 33

The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.

Consider four digit natural numbers whose digit at thousandths place is 1.

Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Number of four digit natural numbers whose digit at thousandths place is 1 = 4  4  4 = 64

Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4  4  4  = 64

Now, consider four digit natural numbers whose digit at thousandths place is 4:

Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.

Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4  4 + 4  4 + 4 + 1 = 37

Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.Question 34

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7, and 9 when on digit is repeated? How many of them are divisible by 10?Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

In how many ways can 5 different balls be distributed among three boxes?Solution 46

Question 47(i)

Solution 47(i)

Question 47(ii)

Solution 47(ii)

Question 47(iii)

Solution 47(iii)

Question 48

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.Solution 48

Each lamps has two possibilities either it can be switched on or off.

There are 10 lamps in the hall.

So the total numbers of possibilities are 210.

To illuminate the hall we require at least one lamp is to be switched on.

There is one possibility when all the lamps are switched off. If all the bulbs are switched off then hall will not be illuminated.

So the number of ways in which the hall can be illuminated is 210-1.

Chapter 16 Permutations Exercise Ex. 16.3

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(i)

Solution 1(i)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.Solution 32

Chapter 16 Permutations Exercise Ex. 16.4

Question 1

Solution 1

Question 2

Solution 2

T h e r e space a r e space 7 space l e t t e r s space i n space t h e space w o r d space apostrophe space S T R A N G E apostrophe comma space i n c l u d i n g space 2 space v o w e l s space left parenthesis A comma E right parenthesis space a n d space 5 space c o n s o n a n t s space left parenthesis S comma T comma R comma N comma G right parenthesis.
left parenthesis i right parenthesis space C o n s i d e r i n g space 2 space v o w e l s space a s space o n e space l e t t e r comma space w e space h a v e space 6 space l e t t e r s space w h i c h space c a n space b e space a r r a n g e d space i n space to the power of 6 p subscript 6 equals 6 factorial space w a y s
A comma E space c a n space b e space p u t space t o g e t h e r space i n space 2 factorial space w a y s.

H e n c e comma space r e q u i r e d space space n u m b e r space o f space w o r d s space
equals space 6 factorial cross times 2 factorial
equals space 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1 cross times 2
equals 720 cross times 2 space
equals 1440.

left parenthesis i i right parenthesis space T h e space t o t a l space n u m b e r space o f space w o r d s space f o r m e d space b y space u sin g space a l l space t h e space l e t t e r s space o f space t h e space w o r d s space apostrophe S T R A N G E apostrophe
i s space to the power of 7 p subscript 7 equals 7 factorial
equals 7 cross times 6 cross times 5 cross times 4 cross times 3 cross times 2 cross times 1
equals 5040.
S o comma space t h e space t o t a l space n u m b e r space o f space w o r d s space i n space w h i c h space v o w e l s space a r e space n e v e r space t o g e t h e r
equals T o t a l space n u m b e r space o f space w o r d s space minus space n u m b e r space o f space w o r d s space i n space w h i c h space v o w e l s space a r e space a l w a y s space space t o g e t h e r
equals 5040 minus 1440
equals 3600

left parenthesis i i i right parenthesis space T h e r e space a r e space 7 space l e t t e r s space i n space t h e space w o r d space apostrophe S T R A N G E apostrophe. space o u t space o f space t h e s e space l e t t e r s space apostrophe A apostrophe space a n d space apostrophe E apostrophe space a r e space t h e space v o w e l s.
T h e r e space a r e space 4 space o d d space p l a c e s space i n space t h e space w o r d space apostrophe S T R A N G E apostrophe. space T h e space t w o space v o w e l s space c a n space b e space a r r a n g e d space i n space to the power of 4 p subscript 2 space w a y s.
T h e space r e m a i n i n g space 5 space c o n s o n a n t s space c a n space b e space a r r a n g e d space a m o n g space t h e m s e l v e s space i n space to the power of 5 p subscript 5 space w a y s.
T h e space t o t a l space n u m b e r space o f space a r r a n g e m e n t s
equals to the power of 4 p subscript 2 cross times to the power of 5 p subscript 5
equals fraction numerator 4 factorial over denominator 2 factorial end fraction cross times 5 factorial
equals 1440

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 16 Permutations Exercise Ex. 16.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

How many permulations of the letters of the world ‘MADHUBANI’ do not begain with M but end with I?Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

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