# Official Quant Thread for CAT 2011 [Part 8]Quantitative Report

Please continue here with all the quant queries and their discussions.
:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...
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1)
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
19+39+59............
25*(38+980)
1018*25=20360+5090=25450..sorry bhai no pen paper so yahi calculation kar di..
1.) The number is of the form 20K+19
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 ??
The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.
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 !!!
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yup answer is 70 but i didnt get u pls explain me again. :-(
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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Wilson bhaia ne kaha hai ki (p-2)!/p gives remainder 1 where p is a prime number
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 .
For these questions.. Follow this methodology
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...
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69!mod71=1
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
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theek hua yaar..ab bas sab bhagwaan bharose hai...
Bhai dekho:
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...
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sorry for spam..par mantri ji papaer kaisa hua??
varun bhai kya akele akele bethe ho???
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ya it is 7
For no green balls = 1 way
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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