@Logrhythm @Logrhythm said:15??options se karlo bhai ye wala toh... btw joyjit bhai...V ke fan ho aap
yup truely deeply and madly a fan of V as well as eVey
@Logrhythm @ChirpiBird said:none of these? 1/30 + 2/30+ 3/30 + ... 299/301/30 (1+2+3+... 299)1/30*((299* 300)/2)= 299*5= 1495??
even i did this way but answer given is 400
@ishu1991 said:110) What is the sum of all positive rationals p/30 (in lowest terms) which are less than 10?(a) 360 (b) 400 (c) 450 (d) none of the foregoing
this is nothing but sum of all numbers less than and co-prime to 300/30...
(150*300*1/2*2/3*4/5)/30 = 12000/30 = 400...
@ishu1991 said:110) What is the sum of all positive rationals p/30 (in lowest terms) which are less than 10?(a) 360 (b) 400 (c) 450 (d) none of the foregoing
1/30 + 2/30 + .........299/30
1/30(1 + 2 + .......299)
1/30(299*300/2)
1/30(299*150)
299*5 = 1495??
so option d ?
@vbhvgupta said:colonel major and general started a work for rs 816. colonel and major did 8/17 of the total work , while major and general did 12/17 of work. what is the amount of the least efficient person?
kya baat aaj major aur general ko ley aaye ho tum yahan?? 
@saurav.kgp said:@DeAdLy -post the solution plz.....m not able to solve it..
correct hai kya mera answer?? 
@DeAdLy said:kya baat aaj major aur general ko ley aaye ho tum yahan??
are han yar...time n work shuru kia hai aaj se.
@saurabhlumarrai said:can u please elaborate this relationship-990*1/2*2/3*4/5...i mean to say why multiplied only with 1/2*2/3*4/5.can it be something else for different digit, like the same question changes to 2000 then what will be the multiplicative factor...??
here we have to eliminate factors of 2,3,5
in case of 2000 if we have to eliminate the same
then (1980*1/2*2/3*4/5) + 6
plus 6 is for the 6 nos which are not divisible by 2,3, or 5 (1981,1987,1991,1993,1997 and 1999)
in case of 2000 if we have to eliminate the same
then (1980*1/2*2/3*4/5) + 6
plus 6 is for the 6 nos which are not divisible by 2,3, or 5 (1981,1987,1991,1993,1997 and 1999)
@vbhvgupta said:colonel major and general started a work for rs 816. colonel and major did 8/17 of the total work , while major and general did 12/17 of work. what is the amount of the least efficient person?
say total work = 17 units
=>C+M+G = 17; C+M =8; M+G = 12
=> G = 9; C=5 and M=3
=> M gets (3/17)*816 = 3*48 = 144.
ATDH.
@vbhvgupta said:are han yar...time n work shuru kia hai aaj se.
aat hun main thodi der main join karne...saath milkar aaj saare time ka work pura kar denge
major ko 144 dekar nikal to kaam se
@Logrhythm said:this is nothing but sum of all numbers less than and co-prime to 300/30...(150*300*1/2*2/3*4/5)/30 = 12000/30 = 400...
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If positive integers a, b, c are such that the quadratic equation ax^2 - bx + c = 0 has two distinct real roots in the open interval (0, 1), find the minimum value of a.(a) 2 (b) 3 (c) 5 (d) 9
... plz post the solution bhai..
If a, b, c, d are real numbers with a^2 + b^2 + c^2 + d^2 = 100, then what is the maximum value of 2a + 3b + 6c + 24d?(a) 240 (b) 250 (c) 300 (d) none of the foregoing
@ishu1991 said:If a, b, c, d are real numbers with a^2 + b^2 + c^2 + d^2 = 100, then what is the maximum value of 2a + 3b + 6c + 24d?(a) 240 (b) 250 (c) 300 (d) none of the foregoing
240 ?? the expression can attain max. value for d having the max. value, which can be 10..
@pratskool said:240 ?? the expression can attain max. value for d having the max. value, which can be 10..
no OA is 250
@abhi0988 said:@ishu1991by cauchy schwartz inequality(a^2+b^2)(c^2+d^2)>=(ac+bd)^2so 250
can you plz elaborate
@joyjitpal said:can you plz elaborate
Cauchy-Schwarz's inequality: (a1^2 + a2^2 + a3^3 +...)*(b1^2 + b2^2 + b3^3 +...) >= (a1*b1 + a2*b2 + a3*b3 + ...)^2
(a^2 + b^2 + c^2 + d^2)*(2^2 + 3^2 + 6^2 + 24^2) >= (2a + 3b + 6c + 24d)^2
100*625 >= (2a + 3b + 6c + 24d)^2
250
(a^2 + b^2 + c^2 + d^2)*(2^2 + 3^2 + 6^2 + 24^2) >= (2a + 3b + 6c + 24d)^2
100*625 >= (2a + 3b + 6c + 24d)^2
250
@ishu1991 said:If a, b, c, d are real numbers with a^2 + b^2 + c^2 + d^2 = 100, then what is the maximum value of 2a + 3b + 6c + 24d?(a) 240 (b) 250 (c) 300 (d) none of the foregoing
cauchy inequality use karlo...
(a^2+b^2+c^2+d^2+....)(x^2+y^2+z^2+....) >= (ax+bx+cz+...)^2
=> (a^2+b^2+c^2+d^2)(2^2+3^2+6^2+24^2) >= (2a+3b+6c+24d)^2
(100)(625) >= (2a+3b+6c+24d)^2
(2a+3b+6c+24d) =