@raopradeep said:bhaiyo plz explain
f(x)=x-3, g(x)=x-2. hence h(x)=x-3+x-2=2x-5
now 2x-5
x
x^4y^2=1024
find least value of x+y is
6
12
9
5
@KaranGarcia said:@hatemongerYa I could decipher that...I took 1st round sum of ranks as 65 then 2nd round 33 and so on...Seems tedious!....what if upsets happen??Couldnt solve the questions !!..Please do try the questions and lemme know..
5. c
6. a
7. b.
8. .... am i right?
@raopradeep said:x^4y^2=1024find least value of x+y is 61295
AM >= GM
(x/4) + (x/4) + (x/4 )+ (x/4) + (y/2) + (y/2)/6 >= rt ((2^10)/(2^8)(2^2))^1/6
x + y = 6
An university has to select an examiner from a list of 60 persons. 25 are women, 35 men, 15 know French, 45 do not, 20 of them are teachers and 40 are not. What is the probability that the University selects a French knowing woman teacher?
1. 2/25
2. 5/144
3. 3/50
4. 6/125
5. 4/25
1. 2/25
2. 5/144
3. 3/50
4. 6/125
5. 4/25
@catahead said:We need to subtract the number of multiples of 2 ie 200 So we are left with 400-200=200 Now we need to remove the multiples of 3 which are odd, so 400/6=66So we are left with 200-66=134Now we need to remove multiples of odd multiples of 5 which are 395/10=39So we are left with 134-39=95But we also subtracted multiples of 15 which are multiples of 3 and hence have already been subtracted Number of multiples of 15 = 26So we are left with 95+26 = 121Now we need to remove the odd multiples of 7 which are 400/14=28But in these 28, we have multiples of 21,35 have already been removed , so we need to find them But again the LCM(3,5,7)=105 are been considered more than once400/21=19400/35=11400/105=3So 121 - 28 + 19 + 11 - 3 = 120Whats the OA?
Hi,
In Second step when you are dividing it by 6. It contains numbers multiple of 3 as well as multiples of 2. Which are already removed in step 1.
I am not sure whats the exact answer.
@ishu1991 said:An university has to select an examiner from a list of 60 persons. 25 are women, 35 men, 15 know French, 45 do not, 20 of them are teachers and 40 are not. What is the probability that the University selects a French knowing woman teacher?1. 2/252. 5/1443. 3/504. 6/1255. 4/25
1/4 * 1/3 *5/12 =5/144?
@ishu1991 said:An university has to select an examiner from a list of 60 persons. 25 are women, 35 men, 15 know French, 45 do not, 20 of them are teachers and 40 are not. What is the probability that the University selects a French knowing woman teacher?1. 2/252. 5/1443. 3/504. 6/1255. 4/25
25/60 * 15/60 * 20/60
= 5/12 * 1/4 * 1/3
=5/144
@ananyboss said:Hi,In Second step when you are dividing it by 6. It contains numbers multiple of 3 as well as multiples of 2. Which are already removed in step 1.I am not sure whats the exact answer.
No , I am not considering even numbers.
I need 3,9,15,21,...
So divide by 6(common diference), i.e 400/6= 66 terms shud be removed
I need 3,9,15,21,...
So divide by 6(common diference), i.e 400/6= 66 terms shud be removed
@naveenkrs said:In triangle ABC, D, E, and F are the trisection points of AB, BC, and CA nearer A,B,C, respectively. Let BF and AE meet at J. Let CD and AE meet at K and CD and BF meet at L. Find 1) BJ : JF :: 3:42) AJ : JE :: 6:13) DK : KL : LC :: 1:3:34) EJ : JK : KA :: 1:3:35) FL : LJ : JB :: 1:3:3
@ishu1991 said:An university has to select an examiner from a list of 60 persons. 25 are women, 35 men, 15 know French, 45 do not, 20 of them are teachers and 40 are not. What is the probability that the University selects a French knowing woman teacher?1. 2/252. 5/1443. 3/504. 6/1255. 4/25
CBD unless the intersection data is provided......
min will be 0 and max will be 15/60=1/4
Sum of two roots of equation 4x^4-24x^3+31x^2+6x-8 = 0 is 0. Then sum of reciprocals of the roots of reciprocals of above equation is:
1. 6
2. 4
3. 3/4
4. -3/4
5. 1/4
min will be 0 and max will be 15/60=1/4
Sum of two roots of equation 4x^4-24x^3+31x^2+6x-8 = 0 is 0. Then sum of reciprocals of the roots of reciprocals of above equation is:
1. 6
2. 4
3. 3/4
4. -3/4
5. 1/4
@bs0409 said:
Sum of two roots of equation 4x^4-24x^3+31x^2+6x-8 = 0 is 0. Then sum of reciprocals of the roots of reciprocals of above equation is:1. 62. 43. 3/44. -3/45. 1/4
Let say that two roots are a and -a
==> 4x^4-24x^3+31x^2+6x-8 = (x^2-a^2)(b*x^2+c*x+d)
or, 4x^4-24x^3+31x^2+6x-8 = b*x^4+c*x^3+(d-a^2*b)*x^2-a^2*c*x-a^2*d
by comparing the coefficients, b=4, c=-24, -a^2*-24 = 6 => a=1/2 or -1/2
-a^2*d=-8=> d= 8*4=32
==> b*x^2+c*x+d = 4*x^2-24*x+32=0 or x= 4, 2
==> sum of the reciprocals of the roots = 2-2+1/4+1/2 = 3/4
@bs0409 said:CBD unless the intersection data is provided......min will be 0 and max will be 15/60=1/4Sum of two roots of equation 4x^4-24x^3+31x^2+6x-8 = 0 is 0. Then sum of reciprocals of the roots of reciprocals of above equation is:1. 62. 43. 3/44. -3/45. 1/4
Is it 3/4??
sum(pqr) = -6/4 = -3/2
prod(pqrs) = -8/4 = -2
so, 1/p+1/q+1/r+1/s = sum(pqr)/prod(pqrs) = 3/4
@vijay_chandola said:Let say that two roots are a and -a==> 4x^4-24x^3+31x^2+6x-8 = (x^2-a^2)(b*x^2+c*x+d)or, 4x^4-24x^3+31x^2+6x-8 = b*x^4+c*x^3+(d-a^2*b)*x^2-a^2*c*x-a^2*dby comparing the coefficients, b=4, c=-24, -a^2*-24 = 6 => a=1/2 or -1/2-a^2*d=-8=> d= 8*4=32==> b*x^2+c*x+d = 4*x^2-24*x+32=0 or x= 4, 2==> sum of the reciprocals of the roots = 2-2+1/4+1/2 = 3/4
@Logrhythm said:Is it 3/4?? sum(pqr) = -6/4 = -3/2prod(pqrs) = -8/4 = -2so, 1/p+1/q+1/r+1/s = sum(pqr)/prod(pqrs) = 3/4
Correct.............
If the minimum value of 4(x^2) + px - 3 is realized for x=a, where a>0, which of the following is necessarily true?
a) 8a+pb) 6a+p>0
c) 11a+2p>0
d) 15a+2p
@bs0409 said:Correct.............If the minimum value of 4(x^2) + px - 3 is realized for x=a, where a>0, which of the following is necessarily true?a) 8a+p0c) 11a+2p>0d) 15a+2p
d/dx -> 8x+p=0
=> x = -p/8 = a
hence, option a and c = 0 (not true)
option b -> 6a+p = 6a+(-8a) = -2a is not > 0
option d -> 15a+2(-8a) = -a
hene, option d holds true...
Guys can someone tell me where do they study VA at (forum link) on PG?? I searched but found no thread.....VA bi toh padhni hai :)