Exemplar Solution

If 2515 is divided by 13, the remainder is _________________ .

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t 2 5^{15} = {(26 - 1)}^{15} \\ = {}_{15}{C_{0}} {(26)}^{15} {(- 1)}^{0} + {}_{15}{C_{1}} {(26)}^{14} {(- 1)}^{1} + {}_{15}{C_{2}} {(26)}^{13} {(- 1)}^{2} + \dots + {}_{15}{C_{15}} {(- 1)}^{15} \\ = 2 6^{15} - 15 {(26)}^{14} + \dots - 1 - 13 + 13 \\ = 2 6^{15} - 15 {(26)}^{14} + \dots - 13 + 12 \\ = 13 λ + 12 \\ ∴ T h e r e m a i n d e r = 12 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 12 . \end{array}$

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New answer posted•a year ago

Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is:

(a) 11

(b) 12

(c) 13

(d) 14

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e t o t a l n u m b e r o f p e r s o n s i n a r o o m b e n since, t w o p e r s o n s m a k e 1 h a n d s h a k e . \\ ∴ T h e n u m b e r o f h a n d s h a k e s = {}_{n}{C_{2}} \\ S o, {}_{n}{C_{2}} = 66 \\ \Rightarrow \frac{n!}{2! (n - 2)!} = 66 \Rightarrow \frac{n (n - 1) (n - 2)!}{2 \times 1 \times (n - 2)!} = 66 \\ \Rightarrow \frac{n (n - 1)}{2} = 66 \Rightarrow n^{2} - n = 132 \\ \Rightarrow n^{2} - n - 132 = 0 \Rightarrow n^{2} - 12 n + 11 n - 132 = 0 \\ \Rightarrow n (n - 12) + 11 (n - 12) = 0 \Rightarrow (n - 12) (n + 11) = 0 \\ \Rightarrow n - 12 = 0, n + 11 = 0 \Rightarrow n = 12, n = - 11 \\ \Rightarrow n = 12 (? n \neq - 11) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

The position of the term independent of $x$ in the expansion of $\sqrt {\frac{x}{3} + \frac{3}{{2 x}^{2}}}^{10}$ is _________________ .

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expansion i s {(\sqrt[]{\frac{x}{3}} + \frac{3}{2 x^{2}})}^{10} \\ T_{r + 1} = {}_{10}{C_{r}} {(\sqrt[]{\frac{x}{3}})}^{10 - r} {(\frac{3}{2 x^{2}})}^{r} = {}_{10}{C_{r}} {(\frac{x}{3})}^{\frac{10 - r}{2}} {(\frac{3}{2})}^{r} . \frac{1}{x^{2 r}} \\ = {}_{10}{C_{r}} {(\frac{1}{3})}^{\frac{10 - r}{2}} . x^{\frac{10 - r}{2}} {(\frac{3}{2})}^{r} . \frac{1}{x^{2 r}} \\ = {}_{10}{C_{r}} {(\frac{1}{3})}^{\frac{10 - r}{2}} . x^{\frac{10 - r}{2} - 2 r} {(\frac{3}{2})}^{r} \\ = {}_{10}{C_{r}} {(\frac{1}{3})}^{\frac{10 - r}{2}} . x^{\frac{10 - r - 4 r}{2}} {(\frac{3}{2})}^{r} \\ F o r i n d e p e n d e n t o f x, w e g e t \\ \frac{10 - r - 4 r}{2} = 0 \Rightarrow 10 - 5 r = 0 \Rightarrow r = 2 \\ S o, t h e p o s i t i o n o f t h e t e r m i n d e p e n d e n t o f x i s 3 r d t e r m . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s T h i r d t e r m . \end{array}$

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New answer posted•a year ago

The ratio of the coefficients of $x^{p}$ and $x^{q}$ in the expansion of ${(1 + x)}^{p + q}$ is _________________ .

[Hint: : ^p^{+ q}C_p = ^p^{+ q}C_q

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expansion i s {(1 + x)}^{p + q} \\ T_{r + 1} = {}_{p + q}{C_{r}} x^{r} \\ P u t r = p \\ = {}_{p + q}{C_{p}} x^{p} \\ ∴ t h e c o e f f i c i e n t s o f x^{p} = {}_{p + q}{C_{p}} \\ S i m i l a r l y, c o e f f i c i e n t s o f x^{q} = {}_{p + q}{C_{q}} \\ {}_{p + q}{C_{p}} = \frac{(p + q)!}{p! (p + q - p)!} = \frac{(p + q)!}{p! q!} \\ a n d {}_{p + q}{C_{q}} = \frac{(p + q)!}{q! (p + q - q)!} = \frac{(p + q)!}{p! q!} \\ S o, t h e r a t i o i s 1 : 1 . \end{array}$

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New answer posted•a year ago

A five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4, and 5 without repetitions. The total number of ways this can be done is:

(a) 216

(b) 600

(c) 240

(d) 3125

[Hint: 5-digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since the sum of digits in these cases is divisible by 3.]

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t a n u m b e r i s d i v i s i b l e b y 3 w h e n t h e s u m o f i t s d i g i t s i s d i v i s i b l e b y 3 . \\ I f w e t a k e t h e d i g i t s 0, 1, 2, 4, 5, t h e n t h e s u m o f t h e d i g i t s = 0 + 1 + 2 + 4 + 5 = 12 w h i c h i s \\ d i v i s i b l e b y 3 . \\ S o, t h e 5 d i g i t n u m b e r s using t h e d i g i t s 0, 1, 2, 4 a n d 5 \\ \underset{4}{T T H} \underset{4}{T H} \underset{3}{H} \underset{2}{T} \underset{1}{O} \\ = 4 \times 4 \times 3 \times 2 \times 1 = 96 \\ a n d i f w e t a k e t h e d i g i t s 1, 2, 3, 4, 5, t h e n t h e i r s u m \\ = 1 + 2 + 3 + 4 + 5 = 15 w h i c h i s d i v i s i b l e b y 3 . \\ S o, t h e 5 d i g i t n u m b e r s c a n b e f o r m e d using t h e d i g i t s 1, 2, 3, 4, 5 i s 5! w a y s \\ = 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ T o t a l n u m b e r o f w a y s = 96 + 120 = 216 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

The middle term in the expansion of ${(a^{3} + b a)}^{28}$ is _________________ .

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} N u m b e r o f t e r m s i n t h e expansion {(a^{3} + b a)}^{28} = 28 + 1 = 29 (o d d) \\ ∴ M i d d l e t e r m = \frac{29 + 1}{2} = 1 5^{t h} t e r m \\ ∴ T_{15} = T_{14 + 1} = {}_{28}{C_{14}} {(a^{3})}^{28 - 14} . {(b a)}^{14} = {}_{28}{C_{14}} {(a)}^{42} . b^{14} . a^{14} \\ = {}_{28}{C_{14}} a^{56} . b^{14} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {}_{28}{C_{14}} a^{56} . b^{14} . \end{array}$

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New answer posted•a year ago

The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to:

(a) 60

(b) 120

(c) 7200

(d) 720

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t t o t a l n u m b e r s o f v o w e l s = 4 \\ a n d t o t a l n u m b e r s o f c o n s o n a n t s = 5 \\ T o t a l n u m b e r o f w o r d s f o r m e d b y 2 v o w e l s a n d 3 c o n s o n a n t s \\ = {}_{4}{C_{2}} \times {}_{5}{C_{3}} = \frac{4!}{2! 2!} \times \frac{5!}{3! 2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{5 \times 4 \times 3!}{3! \times 2} \\ = 6 \times 10 = 60 \\ N o w p e r m u t a t i o n o f 2 v o w e l s a n d 3 c o n s o n a n t s = 5! \\ = 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ S o, t h e t o t a l n u m b e r o f w o r d s = 60 \times 120 = 7200 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

The coefficient of $a^{- 6} b^{4}$ in the expansion of ${(\frac{1}{a} + \frac{2 b}{3})}^{10}$ is _________________ .

**[Hint: T₅ = ¹⁰C₄ 1/a ^b -2b/3 ⁴ = 1120/27 a^-6b⁴]**

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expansion i s {(\frac{1}{a} - \frac{2 b}{3})}^{10} \\ F r o m a^{- 6} b^{4}, w e c a n t a k e r = 4 \\ ∴ T_{5} = T_{4 + 1} = {}_{10}{C_{4}} {(\frac{1}{a})}^{10 - 4} . {(- \frac{2 b}{3})}^{4} = {}_{10}{C_{4}} {(\frac{1}{a})}^{6} . {(- \frac{2}{3})}^{4} . b^{4} \\ = \frac{10.9.8.7}{4.3.2.1} \times \frac{16}{81} . a^{- 6} b^{4} = 210 \times \frac{16}{81} . a^{- 6} b^{4} \\ = \frac{1120}{27} . a^{- 6} b^{4} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{1120}{27} . \end{array}$

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New answer posted•a year ago

The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, and 6 taken all at a time is:

(a) 432

(b) 108

(c) 36

(d) 18

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by