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New answer posted

a year ago

Class 11th Exemplar Solution

Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the points.
[Hint: Number of straight lines = ¹⁸C₂ – ⁵C2 + 1.]

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f points = 12 \\ O u t o f 18 n u m b e r s, 5 a r e c o l l i n e a r a n d w e g e t a s t r a i g h t l i n e b y j o i n i n g a n y t w o points. \\ ∴ T o t a l n u m b e r o f s t r a i g h t l i n e f o r m e d b y j o i n i n g 2 points o u t o f 18 points = {}_{18}{C_{2}} \\ N u m b e r o f s t r a i g h t l i n e s f o r m e d b y j o i n i n g 2 points o u t o f 5 points = {}_{5}{C_{2}} \\ B u t 5 points a r e c o l l i n e a r a n d w e g e t o n l y o n e l i n e w h e n t h e y a r e j o i n e d p a i r w i s e . \\ S o, t h e r e q u i r e d n u m b e r o f s t r a i g h t l i n e s a r e \\ = {}_{18}{C_{2}} - {}_{5}{C_{2}} + 1 = \frac{18.17}{2.1} - \frac{5.4}{2.1} + 1 = 153 - 10 + 1 = 144 \\ H e n c e, t o t a l n u m b e r o f s t r a i g h t l i n e s = 144 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f q u e s t i o n s = 12 \\ N u m b e r o f q u e s t i o n s i n e a c h g r o u p = 6 \\ 7 q u e s t i o n s a r e t o b e a t t e m p t e d b u t n o t m o r e t h a n 5 q u e s t i o n s f r o m e i t h e r g r o u p \\ ∴ T o t a l n u m b e r o f w a y s = {}_{6}{C_{5}} * {}_{6}{C_{2}} + {}_{6}{C_{4}} * {}_{6}{C_{3}} + {}_{6}{C_{3}} * {}_{6}{C_{4}} + {}_{6}{C_{2}} * {}_{6}{C_{5}} \\ = 2 [{}_{6}{C_{5}} * {}_{6}{C_{2}} + {}_{6}{C_{4}} * {}_{6}{C_{3}}] = 2 [6 * \frac{6.5}{2.1} + \frac{6.5.4.3}{4.3.2.1} * \frac{6.5.4}{3.2.1}] \\ = 2 [6 * 15 + 15 * 20] = 2 [90 + 300] = 2 * 390 = 780 \\ H e n c e, t h e t o t a l n u m b e r o f w a y s = 780 \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary, then what is the rank of the word RACHIT?
[Hint: In each case number of words beginning with A, C, H, I is 5!]

0 Follower 22 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T h e a l p h a b e t i c a l o r d e r o f R A C H I T i s A, C, H, I, R a n d T \\ N u m b e r o f w o r d s b e g i n n i n g w i t h A = 5! \\ N u m b e r o f w o r d s b e g i n n i n g w i t h C = 5! \\ N u m b e r o f w o r d s b e g i n n i n g w i t h H = 5! \\ N u m b e r o f w o r d s b e g i n n i n g w i t h I = 5! \\ a n d N u m b e r o f w o r d s b e g i n n i n g w i t h R i . e . R A C H I T = 1 \\ ∴ T h e r a n k o f t h e w o r d' R A C H I T' i n t h e d i c t i o n a r y \\ = 5! + 5! + 5! + 5! + 1 = 4 * 5! + 1 \\ = 4 * 5.4.3.2.1 + 1 = 4 * 120 + 1 = 480 + 1 = 481 \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.

[Hint: 2 women occupy the chair, from 1 to 4 in ⁴P₂ ways and 3 men occupy the remaining chairs in ⁶P₃ ways.]

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} W e h a v e 2 w o m e n a n d 3 m e n \\ F i r s t w o m e n c h o o s e t h e c h a i r s a m o n g s t t h e c h a i r s 1 t o 4 i . e . \\ T o t a l n u m b e r o f c h a i r s = 4 \\ S o, t h e n u m b e r o f a r r a n g e m e n t s = {}_{4}{P_{2}} w a y s \\ N o w 3 m e n c h o o s e f r o m t h e r e m a i n i n g 6 c h a i r s \\ S o, t h e n u m b e r o f a r r a n g e m e n t s = {}_{6}{P_{3}} w a y s \\ ∴ T o t a l n u m b e r o f a r r a n g e m e n t s = {}_{4}{P_{2}} * {}_{6}{P_{3}} \\ = \frac{4!}{(4 - 2)!} * \frac{6!}{(6 - 3)!} = \frac{4!}{2!} * \frac{6!}{3!} \\ = \frac{4.3.2!}{2!} * \frac{6.5.4.3!}{3!} = 12 * 120 = 1440 \\ H e n c e, t h e t o t a l n u m b e r o f p o s s i b l e a r r a n g e m e n t s a r e 1440 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Choose the correct option for $x$ :

(a) $x \in (- \infty, - 2)$

(b) $x \in (- \infty, - 2]$

(c) $x \in (- 2, \infty]$

(d) $x \in [- 2, \infty)$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h r e p r e s e n t s a l l r e a l v a l u e s o f x l e s s t h a n a n d e q u a l t o - 2 . \\ S o, x \in (- \infty, - 2] \\ H e r e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Choose the correct option for $x$ :

(a) $x \in (- \infty, \frac{7}{2})$

(b) $x \in (- \infty, \frac{7}{2}]$

(c) $x \in [\frac{7}{2}, \infty)$

(d) $x \in (\frac{7}{2}, \infty)$

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h r e p r e s e n t s a l l t h e v a l u e s o f x l e s s t h a n \frac{7}{2} o n t h e r e a l n u m b e r l i n e . \\ S o, x \in (- \infty, \frac{7}{2}) \\ H e r e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Choose the correct option for $x$ :

(a) $x \in (\frac{9}{2}, \infty)$

(b) $x \in [\frac{9}{2}, \infty)$

(c) $x \in (- \infty, \frac{9}{2}]$

(d) $x \in (- \infty, \frac{9}{2})$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h h a s a l l r e a l v a l u e s o f x g r e a t e r t h a n a n d e q u a l t o \frac{9}{2} . \\ S o, x \geq \frac{9}{2} \Rightarrow x \in [\frac{9}{2}, \infty) \\ H e r e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Choose the correct option for $x$ :

(a) $x \in (- \infty, 5)$

(b) $x \in (- \infty, 5]$

(c) $x \in [5, \infty)$

(d) $x \in (5, \infty)$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h r e p r e s e n t s a l l v a l u e s o f x g r e a t e r t h a n 5 i n c l u d i n g 5 o n t h e r e a l n u m b e r l i n e \\ S o, x \in (5, \infty) \\ H e r e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The inequality representing the following graph is:

Fig. 6.7

(a) $? x ? < 5$

(b) $? x ? \leq 5$

(c) $? x ? > 5$

(d) $? x ? \geq 5$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h r e p r e s e n t s \\ x > - 5 a n d x < 5 \\ C o m b i n i n g t h e t w o i n e q u a l i t i e s \\ | x | < 5 \\ H e r e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $x + 2 \leq 9$ , then

(a) $x \in (- 7, 11)$

(b) $x \in [- 11, 7]$

(c) $x \in (- \infty, - 7) \cup (11, \infty)$

(d) $x \in (- \infty, - 7) \cup [11, \infty)$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t | x + 2 | \leq 9 \\ \Rightarrow - 9 \leq x + 2 \leq 9 [? | x | \leq a \Rightarrow - a \leq x \leq a] \\ \Rightarrow - 9 - 2 \leq x \leq 9 - 2 \\ \Rightarrow - 11 \leq x \leq 7 \\ \Rightarrow x \in [- 11, 7] \\ H e r e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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