Page 49 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

Exemplar Solution Class 11th

A box contains 2 white balls, 3 black balls, and 4 red balls. The number of ways three balls can be drawn from the box if at least one black ball is to be included in the draw is ______.

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 2 w h i t e, 3 b l a c k a n d 4 r e d b a l l s \\ I t i s g i v e n t h a t a t l e a s t 1 b l a c k b a l l i s t o b e i n c l u d e d . \\ ∴ Required n u m b e r o f w a y s = {}_{3}{C_{1}} \times {}_{6}{C_{2}} + {}_{3}{C_{2}} \times {}_{6}{C_{1}} + {}_{3}{C_{3}} \\ = 3 \times 15 + 3 \times 6 + 1 = 45 + 18 + 1 = 64 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 64 . \end{array}$

Upvote

Exemplar Solution Class 11th

A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee?
[Hint: At least 3 men and 2 women: The number of ways = ¹⁰C₃ × ¹⁰C₃ + ¹⁰C₄ × ⁷C₂. For 2 particular women to be always there: the number of ways = ¹⁰C₄ × ¹⁰C₃ × ⁵C₁. The total number of committees when two particular women are never together = Total – together.]

14 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 10 m e n a n d 7 w o m e n o u t o f w h i c h a c o m m i t t e e o f 6 i s t o b e f o r m e d w h i c h c o n t a i n \\ a t l e a s t 3 m e n a n d 2 w o m e n \\ T h e r e f o r e, N u m b e r o f w a y s = {}_{10}{C_{3}} \times {}_{7}{C_{3}} + {}_{10}{C_{4}} \times {}_{7}{C_{2}} \\ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} + \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1} \\ = 120 \times 35 + 210 \times 21 = 4200 + 4410 = 8610 \\ I f 2 p a r t i c u l a r w o m e n t o b e a l w a y s p r e s e n t, t h e n t h e n u m b e r o f w a y s = {}_{10}{C_{4}} \times {}_{5}{C_{0}} + {}_{10}{C_{3}} \times {}_{5}{C_{1}} \\ = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times 1 + \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times 5 = 210 + 120 \times 5 \\ = 210 + 600 = 810 \\ ∴ T o t a l n u m b e r o f c o m m i t t e e = 8610 - 810 = 7800 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 7800 . \end{array}$

Upvote

Exemplar Solution Class 11th

The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two signs ‘–’ occur together is ______.

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e f o l l o w i n g m a y b e t h e a r r a n g e m e n t o f (-) a n d (+) \\ (-) (+) (-) (+) (-) (+) (-) (+) (-) (+) (-) (+) (-) \\ T h e r e f o r e,' +' s i g n c a n b e a r r a n g e d o n l y i s 1 w a y b e c a u s e a l l a r e i d e n t i c a l . \\ a n d 4 (-) s i g n s c a n b e a r r a n g e d a t 7 p l a c e s i n {}_{7}{C_{4}} w a y s \\ ∴ T o t a l n u m b e r o f w a y s = {}_{7}{C_{4}} \times 1 = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} \times 1 = 35 w a y s \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 35 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is $216 c m^{3}$ and the total surface area $252 c m^{2}$ is . The length of the longest edge is:

(a) 12 

(b) 6

(c) 18

(d) 3

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 an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e l e n g t h, b r e a d t h a n d h e i g h t o f a rectangular b l o c k b e \frac{a}{r}, a a n d a r . \\ Since t h e y a r e i n G . P \\ ∴ V o l u m e = l \times b \times h \\ 216 = \frac{a}{r} \times a \times a r \\ \Rightarrow a^{3} = 216 \Rightarrow a = 6 \\ N o w t o t a l s u r f a c e a r e a = 2 [l b + b h + h l] \\ 252 = 2 [\frac{a}{r} . a + a . a r + \frac{a}{r} . a r] \\ \Rightarrow 252 = 2 [\frac{a^{2}}{r} + a^{2} r + a^{2}] \Rightarrow 252 = 2 a^{2} [\frac{1}{r} + r + 1] \\ \Rightarrow 252 = 2 {(6)}^{2} [\frac{1 + r^{2} + r}{r}] \Rightarrow 252 = 72 [\frac{1 + r^{2} + r}{r}] \\ \Rightarrow \frac{252}{72} = [\frac{1 + r^{2} + r}{r}] \Rightarrow \frac{7}{2} = \frac{1 + r^{2} + r}{r} \\ \Rightarrow 2 + 2 r + 2 r^{2} = 7 r \Rightarrow 2 r^{2} - 5 r + 2 = 0 \\ \Rightarrow 2 r^{2} - 4 r - r + 2 = 0 \Rightarrow 2 r (r - 2) - 1 (r - 2) = 0 \\ \Rightarrow (r - 2) (2 r - 1) = 0 \Rightarrow r - 2 = 0 a n d 2 r - 1 = 0 \\ \Rightarrow r = 2, \frac{1}{2} \\ T h e r e f o r e, t h e t h r e e e d g e a r e : \\ I f r = 2 t h e n e d g e s a r e 3, 6, 12 \\ I f r = \frac{1}{2} t h e n e d g e s a r e 12, 6, 3 \\ S o, t h e l e n g t h o f t h e l o n g e s t e d g e = 12 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Upvote

Exemplar Solution Class 11th

In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams participating in the championship is ______.

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e n u m b e r o f p a r t i c i p a t i n g t e a m s b e n . \\ G i v e n t h a t e v e r y t w o t e a m s p l a y e d o n e m a t c h w i t h e a c h o t h e r . \\ ∴ T o t a l n u m b e r o f m a t c h e s p l a y e d = {}_{n}{C_{2}} \\ S o {}_{n}{C_{2}} = 153 \\ \Rightarrow \frac{n (n - 1)}{2} = 153 \Rightarrow n^{2} - n = 306 \\ \Rightarrow n^{2} - n - 306 = 0 \Rightarrow n^{2} - 18 n + 17 n - 306 = 0 \\ \Rightarrow n (n - 18) + 17 (n - 18) = 0 \Rightarrow (n - 18) (n + 17) = 0 \\ \Rightarrow n - 18 = 0 a n d n + 17 = 0 \Rightarrow n = 18, n \neq - 17 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 18 . \end{array}$

Upvote

Exemplar Solution Class 11th

The number of six-digit numbers, all digits of which are odd, is ______.

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} O u t o f t h e d i g i t s 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 t h e o d d d i g i t s a r e 1, 3, 5, 7, 9 . \\ T h e r f o r e, n u m b e r o f 6 d i g i t s n u m b e r s = {(5)}^{6} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {(5)}^{6} . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If $t_{n}$ denotes the $n$ -th term of the series $2 + 3 + 6 + 11 + 18 + \dots$ , then $t_{50}$ is:

(a) $492 - 1$ 

(b) $492$ 

(c) $502 + 1$ 

(d) $492 + 2$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t S_{n} = 2 + 3 + 6 + 11 + 18 + \dots + t_{50} \\ Using m e t h o d o f d i f f e r e n c e, w e g e t \\ S_{n} = 2 + 3 + 6 + 11 + 18 + \dots + t_{50} \dots (i) \\ a n d S_{n} = 0 + 2 + 3 + 6 + 11 + \dots + t_{49} + t_{50} \dots (i i) \\ S u b t r a c t i n g e q n . (i i) f r o m (i), w e g e t \\ 0 = 2 + 1 + 3 + 5 + 7 + \dots - t_{50} t e r m s \\ \Rightarrow t_{50} = 2 + (1 + 3 + 5 + 7 + \dots u p t o 49 t e r m s) \\ \Rightarrow t_{50} = 2 + \frac{49}{2} [2 \times 1 + (49 - 1) 2] = 2 + \frac{49}{2} [2 + 96] \\ = 2 + \frac{49}{2} \times 98 = 2 + 49 \times 49 = 4 9^{2} + 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution Class 11th

Three balls are drawn from a bag containing 5 red, 4 white, and 3 black balls. The number of ways in which this can be done if at least 2 are red is ______.

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 5 r e d, 4 w h i t e a n d 3 b l a c k b a l l s o u t o f w h i c h a t l e a s t 2 r e d b a l l s a r e t o b e d r a w n . \\ ∴ N u m b e r o f w a y s = {}_{5}{C_{2}} \times {}_{7}{C_{1}} + {}_{5}{C_{3}} \\ = 10 \times 7 + 10 = 70 + 10 = 80 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 80 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Let $S_{n}$ denote the sum of the cubes of the first $n$ natural numbers and $s_{n}$ denote the sum of the first $n$ natural numbers. Then


$\frac{1}{S_{n}} \sum_{r = 1} r^{3} s_{r} =$

(a) $\frac{n (n + 1) (n + 2)}{6}$

(b) $\frac{n (n + 1)}{2}$

(c) $\frac{2 n^{3} + 3 n^{2} + n}{6}$

(d) None of these

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 an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \sum_{i = 1}^{n} \frac{S_{r}}{s_{r}} = \frac{S_{1}}{s_{1}} + \frac{S_{2}}{s_{2}} + \frac{S_{3}}{s_{3}} + \dots + \frac{S_{n}}{s_{n}} \\ L e t T_{n} b e t h e n t h t e r m o f t h e a b o v e s e r i e s \\ ∴ T_{n} = \frac{S_{n}}{s_{n}} = \frac{{[\frac{n (n + 1)}{2}]}^{2}}{\frac{n (n + 1)}{2}} = \frac{n (n + 1)}{2} = \frac{n^{2} + n}{2} \\ N o w s u m o f t h e g i v e n s e r i e s \\ \sum_{}^{} T_{n} = \frac{1}{2} \sum_{}^{} [n^{2} + n] = \frac{1}{2} [\sum_{}^{} n^{2} + \sum_{}^{} n] \\ = \frac{1}{2} [\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}] \\ = \frac{1}{2} . \frac{n (n + 1)}{2} [\frac{2 n + 1}{3} + 1] = \frac{n (n + 1)}{4} [\frac{2 n + 1 + 3}{3}] \\ = \frac{n (n + 1)}{4} . \frac{(2 n + 4)}{3} = \frac{n (n + 1) (n + 2)}{6} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

The minimum value of $4^{x} + 4^{1 - x}, x \in ℝ$ , is:

(a) 2

(b) 4

(c) 1

(d) 0

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t A M \geq G M \\ ∴ \frac{4^{x} + 4^{1 - x}}{2} \geq \sqrt[]{4^{x} . 4^{1 - x}} \\ \Rightarrow 4^{x} + 4^{1 - x} \geq 2 \sqrt[]{4^{x + 1 - x}} \\ \Rightarrow 4^{x} + 4^{1 - x} \geq 2.2 = \Rightarrow 4^{x} + 4^{1 - x} \geq 4 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers

A box contains 2 white balls, 3 black balls, and 4 red balls. The number of ways three balls can be drawn from the box if at least one black ball is to be included in the draw is ______.

The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two signs ‘–’ occur together is ______.

In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams participating in the championship is ______.

The number of six-digit numbers, all digits of which are odd, is ______.

Three balls are drawn from a bag containing 5 red, 4 white, and 3 black balls. The number of ways in which this can be done if at least 2 are red is ______.

The minimum value of $4^{x} + 4^{1 - x}, x \in ℝ$ , is:

(a) 2

(b) 4

(c) 1

(d) 0