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9 months ago

Stanford University Study Abroad

Which MBA specializations are popular at Stanford?

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Nishtha Bhatnagar

Contributor-Level 9

As per a ranking body, Stanford is popular for various programs like Aerospace Engineering, Biomedical, Chemical, Civil, Computer, Materials, Cybersecurity, Data Analytics, Software Engineering, and more. Some popular programs with the US News subject rankings, are given below:

Popular programs	US News Rankings 2026
Computer Science	#2
Artificial Intelligence	#3
Cybersecurity	#5
Data Analytics/ Science	#2
Programming Language	#3
Software Engineering	#3

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9 months ago

Ahmedabad Institute of Business Management Business Management Studies

Who can enroll for course admission at Ahmedabad Institute of Business Management?

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Parul Shukla

Contributor-Level 10

Candidates seeking admission to the Ahmedabad Institute of Business Management, must fulfil the college's eligibility criteria. For the BBA course, applicants must have completed Class 12 or higher, and for the MBA course, they must have completed graduation or higher. The selection process for all courses is based on institute-level entrance exam/Gujarat Common Admission Services.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

If $A$ is the arithmetic mean and $G_{1}, G_{2}$ be two geometric means between any two numbers, then prove that

$2 A = \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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9 months ago

Jyothy Institute of Commerce and Management Institute Of Commerce

What is the eligibility criteria for BBA at Jyothy Institute of Commerce and Management, Bangalore?

0 Follower 2 Views

Anushka Bidhi

Contributor-Level 10

Jyothy Institute of Commerce and Management offers an opportunity for interested students to study BBA. Admission to this programme is merit-based. It is also important to fulfill the application process. The table here outlines the eligibility criteria:

Course	Eligibility Criteria
BBA	Students should have completed their 2nd PUC from a recognised university or class 12 from a valid board.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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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9 months ago

Stanford University Study Abroad

How can an Indian get into the Stanford MBA?

0 Follower 8 Views

Esha Tiwari

Contributor-Level 9

Indian students can get into the university by having a UG degree, releted work experience, and a GPA of 3.5 on 4 point scale. Students have to submit LOR, essay, SOP, transcripts, resume, and English language requirements.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

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

0 Follower 13 Views

Payal Gupta

Contributor-Level 10

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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9 months ago

Study Abroad South Korea Cost of living

Can I study in South Korea for FREE?

0 Follower 19 Views

Akash Gaur

Contributor-Level 10

YES it is now possible to study in South Korea for no tuition fee as part of Seoul Tech Scholarship (STS). The scholarship is an initiative by the South Korea's Seoul Metropolitan Government (SMG), that allows PG students enrolled in STEM based education the chance to pursue their education, without paying anything, along with some other additional benefits.

If you want to know how to apply, then check out: Seoul Tech Scholarship 2025 Eligibility

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

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.]

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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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9 months ago

Study Abroad South Korea Cost of living

Can I study in South Korea for free?

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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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9 months ago

Indian Institute for Aeronautical Engineering & Information Technology (IIAEIT) Colleges in Pune

What topics can candidates expect to be covered in BTech course at IIAEIT?

0 Follower 1 View

Sejal Baveja

Contributor-Level 10

BTech is offered in three different specialisations at Indian Institute for Aeronautical Engineering and Information Technology. These are Aerospace Engineering, Aeronautical Engineering, Avionics Engineering. All these are offered in full-time mode. As the course progresses, students study core aerospace subjects such as Aerodynamics, Propulsion Systems, Aircraft Structures, and Flight Mechanics. Modules on Avionics, Control Systems, and Computational Fluid Dynamics are also included.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

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

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

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]

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

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

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f w e f i x 3 a t u n i t p l a c e, t h e n t h e t o t a l p o s s i b l e o u t c o m e s = 3! \\ I f w e f i x 4, 5 a n d 6 a t u n i t p l a c e, t h i s i s e a c h c a s e, t o t a l p o s s i b l e n u m b e r s a r e 3! \\ Required s u m o f u n i t d i g i t s o f a l l s u c h n u m b e r s i s \\ = 3 \times 3! + 4 \times 3! + 5 \times 3! + 6 \times 3! = (3 + 4 + 5 + 6) \times 3! \\ = 18 \times 3! = 18 \times 3 \times 2 \times 1 = 108 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

If the seventh terms from the beginning and the end in the expansion of are equal, then $n$ equals _________________ .

[Hint: $T_{7} = T_{n - 7 + 2}$ , = ⁿC⁶ (2 /3)^n-6 (1/3 ^{1/3 6}) = ⁿC^{n – 6} (2 ^1/3 ⁶ (1/3 ^1/3) ⁿ^{- 6} solve for $n .$ ]

= (2 ^1/3)ⁿ^{– 12} = 1/3 ^{1/3 n -12} = only problem when n – 12 = 0 = n =12].

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n expansion i s {(+ \frac{1}{})}^{n} \\ ∴ T_{7} = T_{6 + 1} = {}_{n}{C_{6}} {(2^{1 / 3})}^{n - 6} . \frac{1}{{(3^{1 / 3})}^{6}} = {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} {(- 1)}^{r} . \frac{1}{{(3)}^{2}} \\ N o w t h e T_{7} f r o m t h e e n d = T_{7} f r o m t h e b e g i n n i n g i n {(\frac{1}{} +)}^{n} . \\ ∴ T_{7} = T_{6 + 1} = {}_{n}{C_{6}} {(\frac{1}{3^{1 / 3}})}^{n - 6} . {(2^{1 / 3})}^{6} \\ A s p e r t h e q u e s t i o n s, w e g e t \\ {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . (\frac{1}{3^{2}}) = {}_{n}{C_{6}} . \frac{1}{3^{\frac{n - 6}{3}}} . {(2)}^{2} \\ \Rightarrow {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2} = {(3)}^{- (\frac{n - 6}{3})} . {(2)}^{2} \\ \Rightarrow {(2)}^{\frac{n - 6}{3} - 2} . {(3)}^{- 2 + \frac{n - 6}{3}} = 1 \\ \Rightarrow {(2)}^{\frac{n - 12}{3}} . {(3)}^{\frac{n - 12}{3}} = 1 \\ \Rightarrow {(6)}^{\frac{n - 12}{3}} = {(6)}^{0} \\ \Rightarrow \frac{n - 12}{3} = 0 \Rightarrow n = 12 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 12 . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

Find the $r$ -th term of an A.P. sum of whose first $n$ terms is $2 n + 3 n^{2}$ .
[Hint: $a_{n} = S_{n} - S_{n - 1}$ ]

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t S_{n} = 2 n + 3 n^{2} \\ \Rightarrow S_{1} = 2 \times 1 + 3 {(1)}^{2} = 5 \\ \Rightarrow S_{2} = 2 \times 2 + 3 \times 4 = 16 \\ \Rightarrow S_{3} = 2 \times 3 + 3 \times 9 = 33 \\ \dots \dots \dots \\ ∴ S_{1} = a_{1} = 5 \\ S_{2} - S_{1} = a_{2} = 16 - 5 = 11 \\ ∴ d = a_{2} - a_{1} = 11 - 5 = 6 \\ N o w T_{r} = a_{1} + (r - 1) d \\ = 5 + (r - 1) 6 = 5 + 6 r - 6 = 6 r - 1 \\ H e n c e, t h e r e q u i r e d r t h t e r m i s 6 r - 1 \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

The number of different four-digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is:

(a) 120

(b) 96

(c) 24

(d) 100

0 Follower 8 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F o u r - d i g i t n u m b e r s a r e t o b e f o r m e d f r o m t h e d i g i t s 2, 3, 4, 7 w i t h o u t r e p e t i t i o n \\ S o, t h e r e q u i r e d 4 - d i g i t n u m b e r s = {}_{4}{P_{4}} = 4! = 4 \times 3 \times 2 \times 1 = 24 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

Business Management Studies BBA

How to apply in offline mode for BBA at CMJ University?

0 Follower 1 View

Manori Sahni

Contributor-Level 9

Candidates who meet the eligibility criteria can apply for CMJ University BBA course. The application forms are accepted in both online and offline modes. Mentioned below are the steps to be followed for the offline application submission:

Go through the prospectus and then select the course of your choice.
Fill up the admission form carefully.
Attach copies of all qualification certificates and marksheets.
Send the completed application form along with four stamp-size photographs and demand draft of the first instalment fees to the address mentioned below:

CMJ University
Modrina Mansion
Laitumkhrah
Shillong-793 003
Meghalaya

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