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Exemplar Solution Class 11th

A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f s t u d e n t s i n e a c h c l a s s = 20 \\ W e h a v e t o s e l e c t a t l e a s t 5 s t u d e n t s f r o m e a c h c l a s s . \\ W e h a v e t h e f o l l o w i n g c a s e s . \\ (i) 5 s t u d e n t s f r o m X I c l a s s a n d 6 s t u d e n t s f r o m X I I c l a s s \\ (i i) 6 s t u d e n t s f r o m X I c l a s s a n d 5 s t u d e n t s f r o m X I I c l a s s \\ S o, n u m b e r o f w a y s o f s e l e c t i o n o f a t e a m o f 11 p l a y e r s \\ = {}_{20}{C_{5}} \times {}_{20}{C_{6}} + {}_{20}{C_{6}} \times {}_{20}{C_{5}} = 2 [{}_{20}{C_{5}} \times {}_{20}{C_{6}}] \\ H e n c e, t h e r e q u i r e d w a y s o f s e l e c t i o n = 2 [{}_{20}{C_{5}} \times {}_{20}{C_{6}}] \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

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?

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A s p e r t h e g i v e n information w e h a v e t h e f o l l o w i n g d i a g r a m \\ S t a r t i n g point = S \\ Distance t r a v e l l e d t o b r i n g t h e f i r s t p o t a t o = 24 + 24 = 48 m \\ Distance t r a v e l l e d t o b r i n g t h e second p o t a t o = 2 (24 + 4) = 56 m \\ Distance t r a v e l l e d t o b r i n g t h e t h i r d p o t a t o = 2 (24 + 4 + 4) = 64 m \\ T h e r f o r e, t h e s e r i e s w i l l b e = 48, 56, 64, \dots \\ w h i c h a n ? ? A . P i n w h i c h a = 48, d = 56 - 48 = 8 \\ W e h a v e t o f i n d t h e t o t a l distance t o b r i n g a l l t h e p o t a t o e s b a c k, s o, n = 20 \\ ∴ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{20} = \frac{20}{2} [2 \times 48 + (20 - 1) 8] = 10 [96 + 152] \\ = 10 \times 248 = 2480 m \\ H e n c e, t h e r e q u i r e d distance = 2480 m \end{array}$

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Exemplar Solution Class 11th

If $x^{p}$ occurs in the expansion of ${(\frac{x^{2} + 1}{x})}^{2 n},$ prove that its coefficient is $\frac{4 n - p}{3} \frac{2 n + p}{3} .$

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(x^{2} + \frac{1}{x})}^{2 n} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{2 n}{C_{r}} {(x^{2})}^{2 n - r} {(\frac{1}{x})}^{r} = {}_{2 n}{C_{r}} {(x)}^{4 n - 2 r} . \frac{1}{x^{r}} \\ = {}_{2 n}{C_{r}} {(x)}^{4 n - 2 r - r} = {}_{2 n}{C_{r}} {(x)}^{4 n - 3 r} \\ I f x^{p} o c c u r s i n {(x^{2} + \frac{1}{x})}^{2 n} \\ t h e n 4 n - 3 r = p \Rightarrow 3 r = 4 n - p \\ \Rightarrow r = \frac{4 n - p}{3} \\ ∴ C o e f f i c i e n t o f x^{p} = {}_{2 n}{C_{r}} = {}_{2 n}{C_{\frac{4 n - p}{3}}} \\ = \frac{(2 n)!}{(\frac{4 n - p}{3})! (2 n - \frac{4 n - p}{3})!} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{6 n - 4 n + p}{3})!} \\ = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!} \\ H e n c e, c o e f f i c i e n t o f x^{p} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!} \end{array}$

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Exemplar Solution Class 11th

In how many ways can a football team of 11 players be selected from 16 players? How many of them will:

(i) Include 2 particular players?

(ii) Exclude 2 particular players?

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t t h e t o t a l n u m b e r o f p l a y e r s = 16 \\ W e h a v e t o s e l e c t 11 p l a y e r s o u t o f 16 p l a y e r s . \\ (i) I f 2 p l a y e r s a r e i n c l u d e d, t h e n n u m b e r o f w a y s o f s e l e c t i o n = {}_{16 - 2}{C_{11 - 2}} = {}_{14}{C_{9}} \\ (i i) I f 2 p l a y e r s a r e e x c l u d e d, t h e n n u m b e r o f w a y s o f s e l e c t i o n = {}_{16 - 2}{C_{11}} = {}_{14}{C_{11}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s o f s e l e c t i o n a r e \\ (i) {}_{14}{C_{9}} (i i) {}_{14}{C_{11}} \end{array}$

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Exemplar Solution Class 11th

In the expansion of ${(x + a)}^{n}$ if the sum of odd terms is denoted by $O$ and the sum of even terms by $E$ , prove that:

$O^{2} - E^{2} = {(x^{2} - a^{2})}^{n} .$

$4 O E = {(x + a)}^{2 n} - {(x - a)}^{2 n} .$

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} (i) T h e g i v e n expression i s {(x + a)}^{n} \\ {(x + a)}^{n} = {}_{n}{C_{0}} x^{n} a^{0} + {}_{n}{C_{1}} x^{n - 1} a + {}_{n}{C_{2}} x^{n - 2} a^{2} + {}_{n}{C_{3}} x^{n - 3} a^{3} + \dots + {}_{n}{C_{n}} a^{n} \\ S u m o f o d d t e r m s, \\ O = {}_{n}{C_{0}} x^{n} + {}_{n}{C_{2}} x^{n - 2} a^{2} + {}_{n}{C_{4}} x^{n - 4} a^{4} + \dots \\ a n d t h e s u m o f e v e n t e r m s, \\ E = {}_{n}{C_{1}} x^{n - 1} a + {}_{n}{C_{3}} x^{n - 3} a^{3} + {}_{n}{C_{5}} x^{n - 5} a^{5} + \dots \\ N o w {(x + a)}^{n} = O + E \dots (i) \\ S i m i l a r l y, {(x - a)}^{n} = O - E \dots (i i) \\ M u l t i p l y i n g e q n . (i) a n d e q n . (i i), w e g e t \\ {(x + a)}^{n} {(x - a)}^{n} = (O + E) (O - E) \\ {(x^{2} - a^{2})}^{n} = O^{2} - E^{2} \\ H e n c e, O^{2} - E^{2} = {(x^{2} - a^{2})}^{n} \\ (i i) 4 O E = {(O + E)}^{2} - {(O - E)}^{2} \\ = {[{(x + a)}^{n}]}^{2} - {[{(x - a)}^{n}]}^{2} \\ = {[x + a]}^{2 n} - {[x - a]}^{2 n} \\ H e n c e, 4 O E = {(x + a)}^{2 n} - {(x - a)}^{2 n} \end{array}$

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Exemplar Solution Class 11th

Find $n$ in the binomial expansion $,$ if the ratio of the 7th term from the beginning to the 7th term from the end is $\frac{1}{6} .$

20 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(+ \frac{1}{})}^{n} \\ = {(2^{1 / 3} + \frac{1}{3^{1 / 3}})}^{n} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ T_{7} = T_{6 + 1} = {}_{n}{C_{6}} {(2^{\frac{1}{3}})}^{n - 6} {(\frac{1}{3^{\frac{1}{3}}})}^{6} \\ = {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . (\frac{1}{3^{2}}) = {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2} \\ 7 t h t e r m f r o m t h e e n d = {(n - 7 + 2)}^{t h} t e r m f r o m t h e b e g i n n i n g \\ = {(n - 5)}^{t h} t e r m f r o m t h e b e g i n n i n g \\ S o, T_{n - 6 + 1} = {}_{n}{C_{n - 6}} {(2^{\frac{1}{3}})}^{n - n + 6} {(\frac{1}{3^{\frac{1}{3}}})}^{n - 6} \\ = {}_{n}{C_{n - 6}} {(2)}^{2} . (\frac{1}{3^{\frac{n - 6}{3}}}) = {}_{n}{C_{n - 6}} {(2)}^{2} . {(3)}^{\frac{6 - n}{3}} \\ A c c o r d i n g t o t h e q u e s t i o n, w e g e t \\ \frac{{}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2}}{{}_{n}{C_{n - 6}} {(2)}^{2} . {(3)}^{\frac{6 - n}{3}}} = \frac{1}{6} \\ \Rightarrow \frac{{}_{n}{C_{n - 6}} {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2}}{{}_{n}{C_{n - 6}} {(2)}^{2} . {(3)}^{\frac{6 - n}{3}}} = \frac{1}{6} \Rightarrow {(2)}^{\frac{n - 6}{3} - 2} . {(3)}^{- 2 - \frac{6 - n}{3}} = \frac{1}{6} \\ \Rightarrow {(2)}^{\frac{n - 6 - 6}{3}} . {(3)}^{\frac{- 6 - 6 + n}{3}} = \frac{1}{6} \Rightarrow {(2)}^{\frac{n - 12}{3}} . {(3)}^{\frac{n - 12}{3}} = {(6)}^{- 1} \\ \Rightarrow {(6)}^{\frac{n - 12}{3}} = {(6)}^{- 1} \Rightarrow \frac{n - 12}{3} = - 1 \\ \Rightarrow n - 12 = - 3 \Rightarrow n = 12 - 3 = 9 \\ H e n c e, t h e r e q u i r e d v a l u e o f n i s 9 . \end{array}$

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Exemplar Solution Class 11th

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:

(i) They can be of any colour.

(ii) Two must be white and two red.

(iii) They must all be of the same colour.

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f m a r b l e s = 6 w h i t e + 5 r e d = 11 m a r b l e s \\ (i) Since, w e h a v e t o d r a w 4 m a r b l e s o f a n y c o l o u r f r o m t h e 11 m a r b l e s \\ ∴ Required n u m b e r o f w a y s = {}_{11}{C_{4}} \\ (i i) I f 2 m u s t b e w h i t e a n d 2 m u s t b e r e d, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{2}} \times {}_{5}{C_{2}} \\ (i i i) I f a l l t h e 4 m a r b l e s a r e o f t h e s a m e c o l o u r, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{4}} + {}_{5}{C_{4}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) {}_{11}{C_{4}} (i i) {}_{6}{C_{2}} \times {}_{5}{C_{2}} (i i i) {}_{6}{C_{4}} + {}_{5}{C_{4}} \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

12 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e s i d e o f t h e f i r s t e q u i l a t e r a l Δ A B C = 20 c m \\ B y j o i n i n g t h e m i d points o f t h e s i d e s o f t h i s t r i a n g l e, w e g e t t h e second e q u i l a t e r a l t r i a n g l e \\ w h i c h e a c h s i d e = \frac{20}{2} = 10 c m [\begin{array}{l} ? T h e l i n e j o i n i n g t h e m i d -points o f t w o s i d e s o f a t r i a n g l e \\ i s \frac{1}{2} a n d p a r a l l e l t o t h e t h i r d s i d e o f t h e t r i a n g l e \end{array}] \\ S i m i l a r l y, e a c h s i d e o f t h e t h i r d e q u i l a t e r a l t r i a n g l e = \frac{10}{2} = 5 c m \\ ∴ P e r i m e t e r o f f i r s t t r i a n g l e = 20 \times 3 = 60 c m \\ P e r i m e t e r o f second t r i a n g l e = 10 \times 3 = 30 c m \\ a n d t h e p e r i m e t e r o f t h i r d t r i a n g l e = 5 \times 3 = 15 c m \\ T h e r e f o r e, t h e s e r i e s w i l l b e 60, 30, 15, \dots \\ w h i c h i s G . P . i n w h i c h a = 60, a n d r = \frac{30}{60} = \frac{1}{2} \\ N o w, w e h a v e t o f i n d t h e p e r i m e t e r o f s i x t h i n s c r i b e d e q u i l a t e r a l t r i a n g l e \\ ∴ a_{6} = a r^{6 - 1} \\ = 60 \times {(\frac{1}{2})}^{5} = 60 \times \frac{1}{32} = \frac{15}{8} c m \\ H e n c e, t h e r e q u i r e d p e r i m e t e r = \frac{15}{8} c m \end{array}$

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Exemplar Solution Class 11th

18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

19 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 18 m i c e w e r e p l a c e d e q u a l l y i n t w o experimental g r o u p s a n d \\ o n e c o n t r o l g r o u p i . e . 3 g r o u p s \\ ∴ T h e r e q u i r e d n u m b e r o f a r r a n g e m e n t s = \frac{T o t a l a r r a n g e m e n t s}{E q u a l l y l i k e l y a r r a n g e m e n t s} \\ = \frac{18!}{6! 6! 6!} = \frac{18!}{{(6!)}^{3}} \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t s = \frac{18!}{{(6!)}^{3}} \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

We know the sum of the interior angles of a triangle is . 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.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} Since, t h e s u m o f a l l interior a n g l e s o f a p o l y g o n o f n s i d e s = (2 n - 4) \times 9 0^{0} \\ ∴ S u m o f interior a n g l e s o f a p o l y g o n o f 3 s i d e s = (2 \times 3 - 4) \times 9 0^{0} = 18 0^{0} \\ S u m o f interior a n g l e s o f a p o l y g o n o f 4 s i d e s = (2 \times 4 - 4) \times 9 0^{0} = 36 0^{0} \\ S i m i l a r l y, t h e s u m o f interior a n g l e s o f a p o l y g o n o f s i d e s 5, 6, 7, \dots a r e 54 0^{0}, 72 0^{0}, 90 0^{0}, \dots \\ T h e r e f o r e, t h e s e r i e s w i l l b e 18 0^{0}, 36 0^{0}, 54 0^{0}, 72 0^{0}, 90 0^{0}, \dots w h i c h i s A . P . \\ H e r e a = 18 0^{0}, d = 18 0^{0} \\ W e h a v e t o f i n d t h e s u m o f a l l interior a n g l e s o f a p o l y g o n o f 21 s i d e s i . e ., 19 t h t e r m \\ a_{n} = a + (n - 1) d \\ a_{19} = 18 0^{0} + (19 - 1) 18 0^{0} = 18 0^{0} + 18 \times 18 0^{0} \\ = 18 0^{0} + 324 0^{0} = 342 0^{0} \\ H e n c e, t h e r e q u i r e d s u m o f interior a n g l e s = 342 0^{0} . \end{array}$

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A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?

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?

If $x^{p}$ occurs in the expansion of ${(\frac{x^{2} + 1}{x})}^{2 n},$ prove that its coefficient is $\frac{4 n - p}{3} \frac{2 n + p}{3} .$

In how many ways can a football team of 11 players be selected from 16 players? How many of them will:

(i) Include 2 particular players?

(ii) Exclude 2 particular players?

In the expansion of ${(x + a)}^{n}$ if the sum of odd terms is denoted by $O$ and the sum of even terms by $E$ , prove that:

$O^{2} - E^{2} = {(x^{2} - a^{2})}^{n} .$

$4 O E = {(x + a)}^{2 n} - {(x - a)}^{2 n} .$

Find $n$ in the binomial expansion $,$ if the ratio of the 7th term from the beginning to the 7th term from the end is $\frac{1}{6} .$

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:

(i) They can be of any colour.

(ii) Two must be white and two red.

(iii) They must all be of the same colour.

A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

We know the sum of the interior angles of a triangle is . 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.