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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

The two successive terms in the expansion of ${(1 + x)}^{24}$ whose coefficients are in the ratio 1:4 are:

(a) 3^rd and 4^th

(b) 4^th and 5^th

(c) 5^th and 6^th

(d) 6^th and 7^th

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t r^{t h} a n d {(r + 1)}^{t h} b e t w o s u c c e s s i v e t e r m s i n t h e expansion {(1 + x)}^{24} \\ ∴ T_{r + 1} = {}_{24}{C_{r}} . x^{r} \\ a n d T_{r + 2} = T_{r + 1 + 1} = {}_{24}{C_{r + 1}} . x^{r + 1} \\ A s p e r q u e s t i o n, w e h a v e \frac{{}_{24}{C_{r}}}{{}_{24}{C_{r + 1}}} = \frac{1}{4} \\ \Rightarrow \frac{\frac{24!}{r! (24 - r)!}}{\frac{24!}{(r + 1)! (24 - r - 1)!}} = \frac{1}{4} \\ \Rightarrow \frac{24!}{r! (24 - r)!} \times \frac{(r + 1)! (24 - r - 1)!}{24!} = \frac{1}{4} \\ \Rightarrow \frac{(r + 1) . r! (24 - r - 1)!}{r! (24 - r) (24 - r - 1)!} = \frac{1}{4} \\ \Rightarrow \frac{r + 1}{24 - r} = \frac{1}{4} \Rightarrow 4 r + 4 = 24 - r \\ \Rightarrow 5 r = 20 \Rightarrow r = 4 \\ ∴ T_{4 + 1} = T_{5} a n d T_{4 + 2} = T_{6} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

IIT Dhanbad (ISM) - Indian Institute of Technology (ISM)

Can I get admission to IIT Dhanbad with a rank of 10,000?

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Akanksha Arora

Contributor-Level 6

IIT Dhanbad JEE Advanced cutoff 2025 has been released for the third round. The cutoff was released for BTch and Integrated BTech courses. The cutoff ranks for the General AI category were between the range of 4014 and 16143. As per the cutoff range, it is possible to get admission at 10,000 range. At 10,000 rank, student can get admission to a number of specialisations such as B.Tech. in Environmental Engineering, B.Tech. in Petroleum Engineering, B.Tech. in Chemical Engineering etc. However, if the student wants to get admission to the top branches such as CSE and Electrical, then admission at 10,000 rank is difficult.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Given the integers $r > 1, n > 2$ , and the coefficients of the $(3 r)$ th and $(r + 2)$ nd terms in the binomial expansion of ${(1 + x)}^{2 n}$ are equal, then:

(a) $n = 2 r$

(b) $n = 3 r$

(c) $n = 2 r + 1$

(d) None of these

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t r > 1 a n d n > 2 \\ t h e n T_{3 r} = T_{3 r - 1 + 1} = {}_{2 n}{C_{3 r - 1}} . x^{3 r - 1} \\ a n d T_{r + 2} = T_{r + 1 + 1} = {}_{2 n}{C_{r + 1}} . x^{r + 1} \\ A s p e r q u e s t i o n, w e h a v e \\ {}_{2 n}{C_{3 r - 1}} = {}_{2 n}{C_{r + 1}} \\ \Rightarrow 3 r - 1 + r + 1 = 2 n [? {}_{n}{C_{p}} = {}_{n}{C_{q}} \Rightarrow n = p + q] \\ \Rightarrow 4 r = 2 n \\ \Rightarrow n = 2 r \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ are in A.P., where $a_{i} > 0$ for all $i$ , show that $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{1}{a_{1}} - \frac{1}{a_{n}} .$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{n} a r e i n A . P . \\ ∴ C o m m o n d i f f e r e n c e d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = a_{n} - a_{n - 1} \\ I f a_{2} - a_{1} = d t h e n \sqrt[]{a_{2}^{2}} - \sqrt[]{a_{1}^{2}} = d \\ \Rightarrow (\sqrt[]{a_{2}} - \sqrt[]{a_{1}}) (\sqrt[]{a_{2}} + \sqrt[]{a_{1}}) = d [? a^{2} - b^{2} = (a + b) (a - b)] \\ \Rightarrow \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} = \frac{\sqrt[]{a_{2}} - \sqrt[]{a_{1}}}{d} \\ S i m i l a r l y, \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} = \frac{\sqrt[]{a_{3}} - \sqrt[]{a_{2}}}{d} \\ \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} = \frac{\sqrt[]{a_{4}} - \sqrt[]{a_{3}}}{d} \\ \dots \dots \dots \\ \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{\sqrt[]{a_{n}} - \sqrt[]{a_{n - 1}}}{d} \\ A d d i n g t h e a b o v e t e r m s, w e g e t \\ \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} + \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} + \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} + \dots + \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{1}{d} [\begin{array}{l} \sqrt[]{a_{2}} - \sqrt[]{a_{1}} + \sqrt[]{a_{3}} - \sqrt[]{a_{2}} + \\ \sqrt[]{a_{4}} - \sqrt[]{a_{3}} + \dots + \sqrt[]{a_{n}} - \sqrt[]{a_{n - 1}} \end{array}] \\ \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} + \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} + \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} + \dots + \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{1}{d} [\sqrt[]{a_{n}} - \sqrt[]{a_{1}}] \dots (i) \\ N o w a_{n} = a_{1} + (n - 1) d \\ \Rightarrow a_{n} - a_{1} = (n - 1) d \\ \Rightarrow \sqrt[]{a_{n}^{2}} - \sqrt[]{a_{1}^{2}} = (n - 1) d \\ \Rightarrow (\sqrt[]{a_{n}} + \sqrt[]{a_{1}}) (\sqrt[]{a_{n}} - \sqrt[]{a_{1}}) = (n - 1) d \\ \Rightarrow \sqrt[]{a_{n}} - \sqrt[]{a_{1}} = \frac{(n - 1) d}{\sqrt[]{a_{n}} + \sqrt[]{a_{1}}} \\ \Rightarrow \frac{\sqrt[]{a_{n}} - \sqrt[]{a_{1}}}{d} = \frac{(n - 1)}{\sqrt[]{a_{n}} + \sqrt[]{a_{1}}} \dots (i i) \\ F r o m e q n . (i) a n d e q n . (i i) w e g e t \\ \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} + \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} + \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} + \dots + \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{(n - 1)}{\sqrt[]{a_{n}} + \sqrt[]{a_{1}}} \\ H e n c e p r o v e d . \end{array}$

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9 months ago

What are the eligibility requirements to enroll at courses offered by Adani Skill Development?

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saurya snehal

Contributor-Level 10

Candidates wishing to secure a seat at any programme offered by Adani Skills Development

Class 12 from a recognised board.
The minimum age requirement to be eligible for the course is 13.

Candidates who meet the above mentioned points are eligible for admissions to the courses offered by Adani Skill Development.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

The total number of terms in the expansion of ${(x + a)}^{100} + {(x - a)}^{100}$ after simplification is:

(a) 50

(b) 202

(c) 51

(d) None of these

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} N u m b e r o f t e r m s i n t h e expansion o f {(x + a)}^{100} = 101 \\ N u m b e r o f t e r m s i n t h e expansion o f {(x - a)}^{100} = 101 \\ N o w 50 t e r m s o f expansion w i l l c a n c e l o u t w i t h n e g a t i v e 50 t e r m s o f {(x - a)}^{100} \\ S o, t h e r e m a i n i n g 51 t e r m s o f f i r s t expansion w i l l b e a d d e d t o 51 t e r m s o f o t h e r . \\ T h e r e f o r e, t h e n u m b e r o f t e r m s = 51 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) No girls.

(ii) At least one boy and one girl.

(iii) At least three girls.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} W e h a v e 4 g i r l s a n d 7 b o y s a n d a t e a m o f 5 m e m b e r s i s t o b e s e l e c t e d . \\ (i) I f n o g i r l i s s e l e c t e d, t h e n a l l t h e 5 m e m b e r s a r e t o b e s e l e c t e d \\ o u t o f 7 b o y s i . e . {}_{7}{C_{5}} = \frac{7!}{5! 2!} = \frac{7 \times 6.5!}{5! \times 2} = 21 w a y s \\ (i i) W h e n a t l e a s t o n e b o y a n d o n e g i r l a r e t o b e s e l e c t e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{1}} \times {}_{7}{C_{4}} + {}_{4}{C_{2}} \times {}_{7}{C_{3}} + {}_{4}{C_{3}} \times {}_{7}{C_{2}} + {}_{4}{C_{4}} \times {}_{7}{C_{1}} \\ = 4 \times \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} + \frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} + 4 \times \frac{7 \times 6}{2 \times 1} + 1 \times 7 \\ = 4 \times 35 + 6 \times 35 + 4 \times 21 + 7 = 140 + 210 + 84 + 7 = 441 w a y s \\ (i i i) W h e n a t l e a s t 3 g i r l s a r e i n c l u d e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{3}} \times {}_{7}{C_{2}} + {}_{4}{C_{4}} \times {}_{7}{C_{1}} \\ = 4 \times \frac{7 \times 6}{2 \times 1} + 1 \times 7 = 84 + 7 = 91 w a y s \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) 21 w a y s (i i) 441 w a y s (i i i) 91 w a y s \end{array}$

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9 months ago

btech

What specialisations are offered for BTech programme at KAU?

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Rachit Kumar

Contributor-Level 10

Kerala Agricultural University offers three specialisations in the BTech course that spread across fields of agriculture and allied sciences. At the time of application, students must choose their choice of specialisation. Mentioned below are the specialisations offered by the university:

BTech in Food Technology
BTech in Agricultural Engineering
BTech in Biotechnology

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Find the term independent of $x$ in the expansion of ${(1 + x + 2 x^{3})}^{}$ 3/2 x² – 1/3x ⁹.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n expression i s (1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ L e t u s c o n s i d e r {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ T_{r + 1} = {}_{9}{C_{r}} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r} = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(x)}^{18 - 2 r} {(- \frac{1}{3})}^{r} . \frac{1}{x^{r}} \\ = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(x)}^{18 - 2 r - r} {(- \frac{1}{3})}^{r} = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{18 - 3 r} \\ S o, t h e g e n e r a l t e r m i n t h e expansion o f \\ (1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{18 - 3 r} + {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{19 - 3 r} + 2 . {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{21 - 3 r} \\ F o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ P u t 18 - 3 r = 0, 19 - 3 r = 0 a n d 21 - 3 r = 0 w e g e t \\ r = 6, r = \frac{19}{3} a n d r = 7 \\ T h e p o s s i b l e v a l u e o f r a r e 6 a n d 7 (? r \neq \frac{19}{3}) \\ ∴ T h e t e r m i n d e p e n d e n t o f x i s \\ = {}_{9}{C_{6}} {(\frac{3}{2})}^{9 - 6} {(- \frac{1}{3})}^{6} + . {}_{9}{C_{7}} {(\frac{3}{2})}^{9 - 7} {(- \frac{1}{3})}^{7} \\ = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!} . \frac{3^{3}}{2^{3}} . \frac{1}{3^{6}} - 2 . \frac{9 \times 8 \times 7!}{2 \times 1 \times 7!} . \frac{3^{2}}{2^{2}} . \frac{1}{3^{7}} \\ = \frac{84}{8} . \frac{1}{3^{3}} - \frac{36}{4} . \frac{2}{3^{5}} = \frac{7}{18} - \frac{2}{27} = \frac{21 - 4}{54} = \frac{17}{54} \\ H e n c e, t h e r e q u i r e d t e r m i s \frac{17}{54} \end{array}$

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9 months ago

What is the selection criteria for the courses at Adani Skill Development?

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saurya snehal

Contributor-Level 10

Candidates are evaluated over their performance in Class 12 and Class 10. The selection criteria for the courses at Adani Skill Development is based on merit.Candidates are required to pay the course fee if they receive the final admission offer from the institute. The application form is evaluated of all the candidates to ensure their competency for the courses on offer.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

In a cricket tournament, 16 school teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t t h e p r i z e a m o u n t g o t b y f i r s t p l a c e t e a m b e a \\ Since, t h e p r i z e m o n e y i n c r e a s e s b y t h e s a m e a m o u n t f o r s u c c e s s i v e f i n i s h i n g p l a c e s, \\ T h e r e f o r e, t h e s e r i e s w i l l b e A . P . \\ ∴ a_{n} = 275, n = 16 a n d S_{16} = 8000 \\ a_{n} = a + (n - 1) d \\ 275 = a + (16 - 1) (- d) [? C o m m o n d i f f e r e n c e d i s (-) a s t h e s e r i e s i s decreasing] \\ \Rightarrow 275 = a - 15 d \dots (i) \\ N o w S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{16} = \frac{16}{2} [2 a + 15 (- d)] \\ \Rightarrow 8000 = 8 [2 a - 15 d] = 2 a - 15 d = 1000 \dots (i i) \\ S o l v i n g e q . (i) a n d e q . (i i) w e g e t \\ a = 725 a n d d = 30 \\ H e n c e, t h e r e q u i r e d a w a r d r e c e i v e d b y f i r s t p l a c e t e r m = 725 . \end{array}$

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9 months ago

Elphinstone College B.Sc.

I scored less than 50% in my Class 12 board examination. Can I get admission into Elphinstone College BSc courses?

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Saurabh Khanduri

Contributor-Level 10

Yes, students with a minimum aggregate of 45% in their Class 12 board examination can get admissions into Elphinstone College BSc courses. However, some BSc course specialisations at Elphinstone College may require a minimum 50% aggregate in Class 12 exams.

We suggest checking the eligibility criteria for each BSc course at Elphinstone College.

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9 months ago

Ahmedabad Institute of Business Management Class 12th

Does Ahmedabad Institute of Business Management accept Class 12th marks?

0 Follower 3 Views

Shikha Arora

Contributor-Level 10

Candidates seeking admission to the BBA and MBA programme can enroll for admission with Class 12 and graduation marks. Candidates must complete Class 12 to enroll for BBA course. Similarly, for MBA course, candidates are selected based on graduation scores. The college offers only one UG course, i.e., BBA. AIBM Ahmedabad admissions are entrance-based/Gujarat Common Admission Services.

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9 months ago

Bihar BCECE

I have got 92 rank in my BCECE exam. Will I be able to get admission in MIT in Muzaffarpur?

0 Follower 2 Views

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

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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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

KAU - Kerala Agricultural University B.Tech

How can I get admission get admission to BTech at Kerala Agricultural University?

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Piyush Jain

Contributor-Level 10

The KAU BTech admission process can be divided into five broad steps. Interested students can follow the below-mentioned steps to get admission to their preferred BTech course:

1. Eligibility check: Any candidate who has passed Class 12 from the Science stream with a minimum of 50% aggregate is eligible to apply.

2. Application process: Those who are eligible must fill out the application form available on the official website, with all the requisite documents and scores in the accepted entrance exam.

3. Merit list check: After the application process, students must stay updated with the merit list. Shortlisted students move on

...more

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

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?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

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?

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Payal Gupta

Contributor-Level 10

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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9 months ago

DE-CODE Pattern DE-CODE Eligibility

Can I edit the DE-CODE application form after submission?

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Sayeba Naushad

Contributor-Level 10

Yes, a correction window is usually provided for a limited period after submission of DE-CODE Application form. You can log in to your portal and make necessary edits within that timeframe.

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