Maths-

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#### The circle with centre at and radius 2 is

#### The correct answer is:

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### Related Questions to study

Maths-

#### Statement-I : If

Statement-II :If

Which of the above statements is true

#### Statement-I : If

Statement-II :If

Which of the above statements is true

Maths-General

Maths-

#### If x, y are rational number such that Then

#### If x, y are rational number such that Then

Maths-General

maths-

#### A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

Number of groups having 4 boys and 1 girl

= (

and number of groups having 3 boys and 2 girls

= (

Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

We are given 4g

= (

^{4}C_{4}) (^{g}C_{1}) = gand number of groups having 3 boys and 2 girls

= (

^{4}C_{3}) (^{g}C_{2}) = 2g(g – 1)Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

^{2}– 3gWe are given 4g

^{2}– 3g = 85 g = 5.#### A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

maths-General

Number of groups having 4 boys and 1 girl

= (

and number of groups having 3 boys and 2 girls

= (

Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

We are given 4g

= (

^{4}C_{4}) (^{g}C_{1}) = gand number of groups having 3 boys and 2 girls

= (

^{4}C_{3}) (^{g}C_{2}) = 2g(g – 1)Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

^{2}– 3gWe are given 4g

^{2}– 3g = 85 g = 5.maths-

#### There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

Let the number of yellow balls be x, that of black be 2x and that of green be y. Then

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

#### There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

maths-General

Let the number of yellow balls be x, that of black be 2x and that of green be y. Then

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

maths-

#### If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a^{3} + b^{3} + c^{3} must be divisible by -

We have (a

= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

6 | (a

^{3}– a) + (b^{3}– b) + (c^{3}– c)= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

^{3}– a) + (b^{3}– b) + (c^{3}– c)}6 | (a

^{3}+ b^{3}+ c^{3}) as 6|(a + b + c).#### If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a^{3} + b^{3} + c^{3} must be divisible by -

maths-General

We have (a

= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

6 | (a

^{3}– a) + (b^{3}– b) + (c^{3}– c)= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

^{3}– a) + (b^{3}– b) + (c^{3}– c)}6 | (a

^{3}+ b^{3}+ c^{3}) as 6|(a + b + c).maths-

#### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

For 0 r 66, 0 <

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

#### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

maths-General

For 0 r 66, 0 <

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

maths-

#### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

Let t = z + 1

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

= 91 – 36 – 28 + 3 = 30.

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

^{25}in [(x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (x^{12}+ x^{13}+…..+ x^{18}) (1 + x + x^{2}+ …..)]= Coefficient of x

^{12}in (1 + x + x^{2}+ x^{3}+ x^{4}) (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}+ x^{6}) (1 + x + x^{2}+ …..)= Coefficient of x

^{12}in (1 – x)^{–1}= Coefficient of x

^{12}in (1 – x^{5}) (1 – x^{6}) (1 – x)^{–3}= Coefficient of x

^{12}in (1 – x^{5}– x^{6}+ x^{11}) (1 – x)^{–3}=

^{12+3–1}C_{3–1}–^{7+3–1}C_{3–1}–^{6+3–1}C_{3–1}+^{1+3–1}C_{3–1}=

^{14}C_{2}–^{9}C_{2}–^{8}C_{2}+^{3}C_{2}= 91 – 36 – 28 + 3 = 30.

#### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

maths-General

Let t = z + 1

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

= 91 – 36 – 28 + 3 = 30.

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

^{25}in [(x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (x^{12}+ x^{13}+…..+ x^{18}) (1 + x + x^{2}+ …..)]= Coefficient of x

^{12}in (1 + x + x^{2}+ x^{3}+ x^{4}) (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}+ x^{6}) (1 + x + x^{2}+ …..)= Coefficient of x

^{12}in (1 – x)^{–1}= Coefficient of x

^{12}in (1 – x^{5}) (1 – x^{6}) (1 – x)^{–3}= Coefficient of x

^{12}in (1 – x^{5}– x^{6}+ x^{11}) (1 – x)^{–3}=

^{12+3–1}C_{3–1}–^{7+3–1}C_{3–1}–^{6+3–1}C_{3–1}+^{1+3–1}C_{3–1}=

^{14}C_{2}–^{9}C_{2}–^{8}C_{2}+^{3}C_{2}= 91 – 36 – 28 + 3 = 30.

maths-

#### Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

Required number of ways D(r)

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

#### Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

maths-General

Required number of ways D(r)

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

=

= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)

= 10 + 20 + 45 + 44

= 119.

maths-

#### In how many ways can we get a sum of at most 17 by throwing six distinct dice -

x

When 1 x

Let x

x

Clearly x

= Coefficient of x

(1 + x + x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6} 17When 1 x

_{i}6, i = 1, 2, 3, …..6Let x

_{7}be a variable such thatx

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}+ x_{7}= 17Clearly x

_{7}0 Required number of ways= Coefficient of x

^{17}in (x^{1}+ x^{2}+ ….. + x^{6})^{6}(1 + x + x

^{2}+ …..)= Coefficient of x

^{11}in= Coefficient of x

^{11}in (1–^{ 6}C_{1}x^{6}+^{6}C_{2}x^{12}……) (1 – x)^{–7}= Coefficient of x

^{11 }in (1 – x)^{–7}–^{6}C_{1}× coefficient of x^{5}in (1 – x)^{–7}=

^{11+7–1}C_{7–1}–^{6}C_{1}×^{7+5–1}C_{7–1}=

^{17}C_{6}– 6 ×^{11}C_{6}= 9604.#### In how many ways can we get a sum of at most 17 by throwing six distinct dice -

maths-General

x

When 1 x

Let x

x

Clearly x

= Coefficient of x

(1 + x + x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6} 17When 1 x

_{i}6, i = 1, 2, 3, …..6Let x

_{7}be a variable such thatx

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}+ x_{7}= 17Clearly x

_{7}0 Required number of ways= Coefficient of x

^{17}in (x^{1}+ x^{2}+ ….. + x^{6})^{6}(1 + x + x

^{2}+ …..)= Coefficient of x

^{11}in= Coefficient of x

^{11}in (1–^{ 6}C_{1}x^{6}+^{6}C_{2}x^{12}……) (1 – x)^{–7}= Coefficient of x

^{11 }in (1 – x)^{–7}–^{6}C_{1}× coefficient of x^{5}in (1 – x)^{–7}=

^{11+7–1}C_{7–1}–^{6}C_{1}×^{7+5–1}C_{7–1}=

^{17}C_{6}– 6 ×^{11}C_{6}= 9604.maths-

#### The number of non negative integral solutions of equation 3x + y + z = 24

3x + y + z = 24, x 0, y 0, z 0

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

=

= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

^{24–3k+2–1}C_{2–1}=

^{25–3k}C_{1}= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

#### The number of non negative integral solutions of equation 3x + y + z = 24

maths-General

3x + y + z = 24, x 0, y 0, z 0

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

=

= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

Let x = k y + z = 24 – 3k …(1)

24 – 3k 0 k 8

0 k 8

For fixed value of k the number of solutions of (1) is

^{24–3k+2–1}C_{2–1}=

^{25–3k}C_{1}= 25 – 3k

Hence number of solutions

= 25 × 9 – = 225 – 108 = 117.

maths-

#### Sum of divisors of 2^{5 }·3^{7 }·5^{3 }· 7^{2} is –

Any divisor of 2

Hence the sum of the divisors

= (1 + 2 + …… + 2

=

=.

^{5}· 3^{7}· 5^{3}· 7^{2}is of the type of 2*3*^{l}^{m}5^{n}7^{p}, where 0*l*5, 0 m 7, 0 n 3 and 0 p 2Hence the sum of the divisors

= (1 + 2 + …… + 2

^{5}) (1 + 3 + ……. + 3^{7}) (1 + 5 + 5^{2}+ 5^{3}) (1 + 7 + 7^{2})=

=.

#### Sum of divisors of 2^{5 }·3^{7 }·5^{3 }· 7^{2} is –

maths-General

Any divisor of 2

Hence the sum of the divisors

= (1 + 2 + …… + 2

=

=.

^{5}· 3^{7}· 5^{3}· 7^{2}is of the type of 2*3*^{l}^{m}5^{n}7^{p}, where 0*l*5, 0 m 7, 0 n 3 and 0 p 2Hence the sum of the divisors

= (1 + 2 + …… + 2

^{5}) (1 + 3 + ……. + 3^{7}) (1 + 5 + 5^{2}+ 5^{3}) (1 + 7 + 7^{2})=

=.

maths-

#### The length of the perpendicular from the pole to the straight line is

#### The length of the perpendicular from the pole to the straight line is

maths-General

maths-

#### The condition for the lines and to be perpendicular is

#### The condition for the lines and to be perpendicular is

maths-General

Maths-

#### If f : R →R; f(x) = sin x + x, then the value of (f^{-1} (x)) dx, is equal to

#### If f : R →R; f(x) = sin x + x, then the value of (f^{-1} (x)) dx, is equal to

Maths-General

maths-

#### The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is

#### The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is

maths-General