link to previous thread :-

http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html

:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...

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link to previous thread :-

http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html

:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...

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terms common are 19,39,59,..419

a=19 and d=20

sum of 50 terms= 19*50+*20=50(19+490)=50*509=25450

2)its 6!/2=360

25*(38+980)

1018*25=20360+5090=25450..sorry bhai no pen paper so yahi calculation kar di..

Sum = 10*(n-1)*(n) + 19n ====> (n-1)(n) because n varies from 0 to 1

= 10*49*50 + 19*50

= 25450

2.) Is it 360 ??

25450

24550

50900

49100

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope al...

sry yaar typo tha..i mean isse ganda typo nahi ho sakta kabhi:splat:

70 ???

pranaam sabhiko !!!

pranaam sabhiko !!!

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69*68*67*66*65*64*61!mod71=1

-2*-3*-4*-5*-6*-7*69!*mod71=1

-1*69!mod71=1

actually i didnt get the bold ones earlier but after reading varun bhai post samaj a gaya

thanxx to u n @varun bhai

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so 69! = 1 Mod(71)

69*68*67*66*65*64*63! = 1 Mod(71)

-2*-3*-4*-5*-6*-7 * 63! = 1 Mod(71)

120*42*63! = 1 Mod(71)

49*42*63! = 1 Mod(71)

70*63! = 1 Mod(71)

-1*63! = 1 Mod(71)

so 63! = -1 Mod(71) ...

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Varun bhai would u pls explain this in detail as i dint get .

Given Equations are :

3M + 4G + 5W = 750

6M + 9G + 10W = 1580

Multiply first and second by A and B respectively

We'll get:

M(3A-6B) + G(4A-9B) + W(5A-10B) = 750*A - 1580*B

Required to find:

6M + 4G + 10W

Compare the co-effi...

3

69*68*67*66*65*64*63!mod71=1

-2*-3*-4*-5*-6*-7*63!*mod71=1

-1*63!mod71=1

63!mod71=-1=70..hope no mistakes

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Only for positive real values and so 7 it is..correcto.

find remainder when 63! divided by 71

3m+4g+5w=750

6m+9g+10w=1580

Solving these guava=80

6m+10w=860

so 6m+4g+10w=1180

6m+9g+10w=1580

Solving these guava=80

6m+10w=860

so 6m+4g+10w=1180

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theek hua yaar..ab bas sab bhagwaan bharose hai...

Case 1: When no green ball is there

That is possible in 1 way

Case 2: 1 Green ball

It can occupy any 6 places so 6 ways

Case 3: 2 Green balls

_G_G_

Three gaps a,b,c

a+b+c = 4, where b>0

so a+b+c = 3 => C(5,2) = 10 ways

Case 4: 3 Green balls

Green balls...

2

sorry for spam..par mantri ji papaer kaisa hua??

varun bhai kya akele akele bethe ho???

varun bhai kya akele akele bethe ho???

1

ya it is 7

For 1 green ball = 6 ways

For 2 green balls = 4+3+2+1 = 10 ways

For 3 green balls = 4 ways

Total = 21 ways

bhai ye explain kar do i always have problem in permutation/combination

geetting 1180..2*ii-i se aa jayega..

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