Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

 

 ncert cbse chapter 10 straight lines miscellaneous exercise

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3. 

let m be the slope of the required line

using formula slope = [(- coefficient of x) / (coefficient of y)]

Slope of  the given line [ x – 2y = 3 ] 

is m2 = [(-1)/(-2)] = [1/2]

using formula for angle between two lines

tanu = | [m1 - m2] / [ 1 + m1*m2]| 

given angle is 45 degrees

tan(45 degrees) = | [m - (1/2)] / [ 1 + m*(1/2)]| 

1  = | [2m-1] / [2+m] |

because of the absolute value sign

(+1) = [2m-1] / [2+m]  OR  (-1) = [2m-1] / [2+m] 

solving

2+m = 2m -1  or  -2-m = 2m-1

m=3  or m=(-1/3)

Required line passes through (3, 2) 

using point slope form

equation is

[y-2] =[3][x-3]  or [y-2] =[(-1/3)][x-3]

y-2 =3x-9   or 3y-6 =(- x)+3

3x-y-7=0  or x+3y-9=0

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the
line to be a plane mirror.

solution 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

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A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

 ncert cbse chapter 10 straight lines miscellaneous exercise

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Since A lies on the x axis, the y coordinate of A is 0

Let A =(k,0) 

let P=(1,2)

Q =(5,3)

both PA and AQ make equal angles with the normal line (perpendicular to the x axis) at A due to  the law of reflection.

If AQ makes an angle u with the positive direction of the x axis, then PA makes an angle of (180-u) with the positive direction of the x axis.

using the definition of slope

 

for PA, P=(1,2), A =(k,0)

tan(u) =[0-2] / [k-1] --------------------(1)

for QA , Q =(5,3) , A =(k,0)

tan[180-u]  = [0-3] / [k-5]

but tan[180-u]  = {-tan (u)}

we get 

{-tan (u)} = [0-3] / [k-5] ------------(2)

using eqn (1)  in  (2)

-{[0-2] / [k-1] } = [0-3] / [k-5]

2(k-5) = (-3)(k-1)

2k-10 = -3k+3

5k =13

k =[13/5]

therefore

A=( (13/5)  , 0 )

-----------------------------------------------------------------------------------------------------------------

21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

 

first line is

[ 9x + 6y – 7 = 0 ] dividing  by 3 to make the coefficients in the two given equations the same

3x +2y -(7/3) =0 -------------(1)

second line is  

 3x + 2y + 6 =0 --------------(2)

 

A line parallet to these lines will be of the form

3x+2y +k =0 ----------------(3) 

distance between (1) and (3)

d1 = | k +(7/3) | / sqrt[(3^2) +(2^2)]

distance between (2) and (3)

d2 = | k -6 | / sqrt[(3^2) +(2^2)]

 

for equidistant line d1 = d2

 equating d1 and d2 and cancelling off the denominator

| k +(7/3) | = | k -6 | 

k +(7/3) =  k -6   OR  k +(7/3) = -(k -6 )

(7/3) =(-6) which is not possible

OR 

k +(7/3) = -(k -6 )

2k =6 -(7/3) 

2k = (11/3)

k = (11/6)

substitute in (3)

3x+2y +(11/6) =0

 

18 x  + 12y +11 =0

 

 

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means.

A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

We have to find the point of intersection of the lines

2x – 3y + 4 = 0

3x + 4y – 5 = 0  

[ 2x - 3y = (-4) ] x 3

[ 3x + 4y = 5 ] x 2

6x -9y = (-12)

6x +8y =10

subtracting

-17y =-22

 

y=(22/17)

substitute in 

3x + 4y = 5

3x = 5 -{88/17}=[-3/17]

x =(-1/17)

 

The person is standing at [ (-1/17) , (22/17) ] 

The shortest path is perpendicular to 6x – 7y + 8 = 0

Perpendicular line is 7x+6y+k=0

This passes through [ (-1/17) , (22/17) ] if

7 (-1/17) +6(22/17) +k =0

k = (-125/17)

Perpendicular line is 7x+6y+k=0

changes to  7x+6y+(-125/17)=0 

or 119x+102y-125=0 

119x+102y=125 is the required path.

 

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

 

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

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show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer

 ncert cbse chapter 8 binomial theorem exercise 8.1

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer 

(x+1)^(n+1) = [x^(n+1)] +C[(n+1),1][x^(n)]+...+C[(n+1),(n-1)][x^2] +C[(n+1),n][x] + 1

 

C[(n+1),n] =C[(n+1),1] = (n+1)

C[(n+1),(n-1)]= C[(n+1),2] = (n+1)(n)/2

(x+1)^(n+1) = [x^(n+1)] +(n+1)[x^(n)]+...+[(n+1)(n)/2][x^2]+(n+1)[x] + 1

put x=8

 (8+1)^(n+1) = [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]+(n+1)[8] + 1

concentrating on the last 2 terms only

9^(n+1) =  { [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]} + [8n+8]+1

9^(n+1) =  { [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]} + 8n+9

re-arranging

 9^(n+1) - 8n - 9 ={ [8^(n+1)] +(n+1)[8^(n)]+...+[(n+1)(n)/2][8^2]}

note that C[(n+1),(n-1)]= C[(n+1),2] = (n+1)(n)/2 is an integer

so after taking factor of (8^2) = 64 out,

9^(n+1) - 8n - 9 = 64*[8^(n-1)+ ...+[(n+1)(n)/2] ] =divisible by 64

8. evaluate (101)^4

using expansion with 

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

x=100

y=1

(100+1)^4 = (100^4)+4(100^3)(1)+6(100^2)(1^2)+4(100^1)(1^3)+(1^4)

(101)^4 =100000000+4000000+60000+400+1

(101)^4 =104060401

 

chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

 

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution 

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

solution

 

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution  

 

8.find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [(fourth root of 2) + {1/(fourth root of 3)}]^n is (sqrt6):1 

solution

 

exercise 8.2

 

 5. find the 4th term in the expansion of (x-2y)^12

solution

7. Find the middle terms in the expansion of [3 - ((x^3) / 6)]^7

solution

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

solution

 

 10.The coefficients of the (r-1)th, rth, (r+1)th  terms in the expansion of [(x+1)^n] is in the ratio 1:3:5. Find n and r.

solution 

 

exercise 8.1

8. evaluate (101)^4

solution

13.show that 9^(n+1) - 8n -9 is divisible by 64 whenever n is a positive integer 

solution

 

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find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

 ncert cbse chapter 8 binomial theorem miscellaneous exercise    

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

using expansion with 

C(4,1) =4=C(4,3)

C(4,2)=6

(x+y)^4 = (x^4)+4(x^3)(y)+6(x^2)(y^2)+4(x)(y^3)+(y^4)

(x-y)^4 = (x^4)-4(x^3)(y)+6(x^2)(y^2)-4(x)(y^3)+(y^4)

adding the two expansions 

(x+y)^4 +(x-y)^4= 2[(x^4) +6(x^2)(y^2) +(y^4) ]

use x =(a^2) y = sqrt{(a^2)-1}

[(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4

 = 2[{(a^2)^4}+6{(a^2)^2}{[ sqrt{(a^2)-1}]^2} +{[ sqrt{(a^2)-1}]^4}]

 =2[ (a^8) +6(a^4)((a^2)-1) +{(a^2)-1}^2 ]

=2 [ (a^8) +6(a^6) -6(a^4) +{(a^4) -2(a^2)+1} ]

=2[ (a^8) +6(a^6) -5(a^4)-2(a^2)+1  ]

=2(a^8) +12(a^6) -10(a^4)- 4(a^2)+2

chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

 

5.evaluate { (sqrt(3) + sqrt(2))^6 } - { (sqrt(3) - sqrt(2))^6 }

solution 

 6.find the value of [(a^2)+sqrt{(a^2)-1}]^4 + [(a^2)-sqrt{(a^2)-1}]^4 

solution

 

7.find an approximate value of (0.99^5) using the first three terms of its expansion

solution  

exercise 8.2

Q8) Find the  middle terms in the expansion of [(x/3)+9y)]^10

solution  

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

 

 

find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

 ncert cbse chapter 8 binomial theorem miscellaneous exercise 

 3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

using binomial expansion

 (1+2x)^6 = 1 +C(6,1)(2x)+C(6,2)(2x)^2+C(6,3)(2x)^3 +C(6,4)(2x)^4 

+C(6,5)(2x)^5+C(6,6)(2x)^6

 C(6,1) = 6 =C(6,5)

C(6,2)=15=C(6,4)

C(6,3)=20

C(6,6)=1

(1+2x)^6 =

 1 +6(2x) +15(4(x^2))+20(8(x^3))+15(16(x^4))+6(32(x^5))+1(32(x^6))

=1+12x+60(x^2)+160(x^3)+240(x^4)+192(x^5)+32(x^6)

 

(1-x)^7=

1-C(7,1)(x)+C(7,2)(x^2)-C(7,3)(x^3)+C(7,4)(x^4)

-C(7,5)(x^5)+C(7,6)(x^6)-C(7,7)(x^7) 

C(7,1)=7=C(7,6) 

C(7,2)=21=C(7,5) 

C(7,3)=35=C(7,4)  

C(7,7)=1

 (1-x)^7=1 -7x+21(x^2)-35(x^3)+35(x^4)-21(x^5)+7(x^6)-1(x^7)

 

now multiplying the two expansions and concentrate only on the terms containing

(x^5)

1* {-21(x^5)}+{+12x}*{+35(x^4}+{+60(x^2)}*{-35(x^3)}+{+160(x^3)}{+21(x^2)}+{+240(x^4)}{-7x}+{+192(x^5)}{1}

 

coefficient of  (x^5) is

1*(-21)+(12)*(35)+(60)(-35)+(160)(21)+(240)(-7) +192*1

=(-21)+420-2100+3360-1680+192

=171

 

chapter 8 binomial theorem miscellaneous exercise

 1.Find a , b and n in the expansion of (a+b)^n if the first three terms in the expansion are 729, 7290, 30375

solution

 

2.  Find a if the coefficients of (x^2)  & (x^3) in the expansion of {(3+ax)^9} are equal 

solution

  3.find the coefficient of {x^5} in the expansion of{(1+2x)^6}{(1-x)^7}

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

 

The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

ncert cbse chapter 7 permutations miscellaneous exercise

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

Because there are 5 vowels,

the 2 vowels can be selected in C(5,2) = [(5*4)/(1*2)] = 10 ways

Because there are 21 consonants

the 2 consonants can be selected in C(21,2) = [(21*20)/(1*2)] = 210 ways

After selecting the 4 required letters now we can arrange them in

(4!) = 4*3*2*1 = 24 ways

Required number of words = 10*210*24 =50400

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

let the selections be denoted by (x,y) where x denotes the questions from part I

and y the questions from part II

atleast 3 means 3 or more

(3,5) or (4,4) or (5,3) are the different possibilities with their respective number of

possible selections being

C(5,3) * C(7,5) = [(5*4)/(1*2)]*[(7*6)/(1*2)]=210 ways

C(5,4) * C(7,4) = [5]*[(7*6*5)/(1*2*3)]=175 ways

C(5,5) * C(7,3) = [1]*[(7*6*5)/(1*2*3)]=35 ways

Total number of ways = 210+175+35 =420 ways

Here note that C[n,r] =C[n,(n-r)]

gives

C[5,3]=C[5,2]

C[7,5]=C[7,2]

C[5.4]=C[5,1]=5

C[7,4]=C[7,3]

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

solution

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

ncert cbse chapter7permutations miscellaneous exercise

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

first arrange the 11 letters in alphabetical order

AA   2 A's

E

II  2 I's

M

NN 2 N's

O

T

X

Therefore before the first word starting with E, words starting with the letter

A will be arranged in alphabetical order in the dictionary.

fix one A in the first slot

remaining 10 letters contain 2 I's and 2 N's [ one of the 2 A's is used for the first slot ]

So number of words starting with the letter A

= [(10!)/ {(2!)*(2!)}] =907200

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

The one's place should contain 0 for the number to be divisible by 10

Fix the digit 0 in the one's place Therefore the one's place can be filled in only one way.

The other 5 slots can be filled in 5! = 120 ways

Answer = 1*120 = 120

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

ncert cbse 11th maths excercise 7.3 chapter 7

ncert cbse 11th maths  excercise 7..3 chapter 7

exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

There are 12 letters in the word PERMUTATIONS

i)words start with P and end with S

If the word start with P and end with S

first slot and 12th slot can be filled in 1 way each.

After removing P and S,

For the remaining 10 slots there are 10 items

with  2 T's,

So number of words starting with P and end with S

= 1 *[(10!)/(2!)]*1 = 1814400

ii)vowels are all together

vowels in the given word are E,U,A,I,O

If the vowels are together, think of the 5 vowels as a single block.

Inside the block the vowels can be arranged in (5!) ways

Now this block of vowels with the remaining 7 consonants

make up 7+1=8 pieces with 2 T's

which can be arranged in [(8!)/(2!)] ways

Number of words in which the vowels are together = (5!)*[(8!)/(2!)]

=120*20160=2419200

iii)there are always 4 letters between P and S

When there are 4 letters between P and S

one particular arrangement will be

P x x x x S # # # #

in this example first and fifth slots can be filled in 1 way each

and the remaining 10 slots can be filled with 10 remaining letters

containing 2 T's in = 1 *[(10!)/(2!)]*1 = 1814400 ways

Now possible re-arrangements  are

1        P x x x x S # # # ###

2        #P x x x x S # # # ##

3         ##P x x x x S # # ##

4        ###P x x x x S # ##

5        ####P x x x x S##

6        #####P x x x x S#

7        ######P x x x x S

Now the position of P and S can be interchanged

8       S x x x x P # # # ###

9        #S x x x x P # # # ##

10         ##S x x x x P # # ##

11        ###S x x x x P # ##

12        ####S x x x x P##

13        #####S x x x x P#

14        ######Sx x x x P

A total of 14 arrangements

number of words where

there are always 4 letters between P and S

=1814400*14

=25401600

 

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

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