| View previous topic :: View next topic |
| Author |
Message |
Quant.Database
Joined: 04 Sep 2009 Posts: 67
|
|
| Back to top |
|
 |
therani.arch
Joined: 19 Jul 2010 Posts: 65
|
Posted: Thu Jul 22, 2010 3:04 pm Post subject: |
|
|
1. 1/25
2. Something <8
3. C
4. bk/a, a-k
5. 4
6.
7. A (If that is 'pie' not 'n')
8. B
9 Col A=30, Col B=? |
|
|
| Back to top |
|
 |
devineni
Joined: 21 Jul 2010 Posts: 5
|
Posted: Thu Jul 22, 2010 5:19 pm Post subject: 4rth ans |
|
|
hii brother
can u explain how u got x= bk/a for the 4rth question _________________ Devineni |
|
|
| Back to top |
|
 |
devineni
Joined: 21 Jul 2010 Posts: 5
|
Posted: Thu Jul 22, 2010 5:27 pm Post subject: |
|
|
7th question ans will be C
if n= constant or pie
1/3 * n * (2r)^2 * 2h = 8/3*n* r*h = A
8*(cylinder A volume) =B
both the values are equal so answer is C _________________ Devineni |
|
|
| Back to top |
|
 |
devineni
Joined: 21 Jul 2010 Posts: 5
|
Posted: Thu Jul 22, 2010 5:31 pm Post subject: |
|
|
can u explain the 8 question _________________ Devineni |
|
|
| Back to top |
|
 |
therani.arch
Joined: 19 Jul 2010 Posts: 65
|
Posted: Fri Jul 23, 2010 3:49 am Post subject: |
|
|
Hi devineni,
For the forth one, i equated the angle of both the triangle, so that makes it
k/x = (a-k)/(b-x)
Solving this you can find the answer.
P.S: 'x' is the value of x-coordinate.
For the Seventh one,
The volume of cylinder is (pie*r^2*h), for cone is 1/3*(pie*r^2*h)
For the 8th one,
No one side of the triangle is greater than sum of 2 other sides, Hence the side AC is <9.
That makes it lesser than other 2 angles of triangle ABC.
which means <60
Cheers |
|
|
| Back to top |
|
 |
rangu.naveenkumar
Joined: 18 Jul 2010 Posts: 2 Location: Hyderabad
|
Posted: Fri Jul 23, 2010 10:59 am Post subject: |
|
|
| please explain me 6th problem.... |
|
|
| Back to top |
|
 |
nehareddy4u
Joined: 23 Jul 2010 Posts: 66
|
Posted: Fri Jul 23, 2010 3:26 pm Post subject: |
|
|
1. 1/25
2. 6.29 (approx) or (<7)
3. C
4. any one explain it.. i got only the 'y' point
5. 4
6. 8 pi
7. D
8. A |
|
|
| Back to top |
|
 |
therani.arch
Joined: 19 Jul 2010 Posts: 65
|
Posted: Fri Jul 23, 2010 3:55 pm Post subject: |
|
|
Hi nehareddy4u,
Could you explain the 6th one.
Thanks |
|
|
| Back to top |
|
 |
prabudhakshin
Joined: 21 Jul 2010 Posts: 2
|
Posted: Fri Jul 23, 2010 6:43 pm Post subject: |
|
|
| therani.arch wrote: | Hi nehareddy4u,
Could you explain the 6th one.
Thanks |
The idea is u got to find the length of the arc and subtract from the perimeter of the circle.
1. Draw a line connecting the centres of the circle (call it as BASE line)
2. From one of the centre, draw two lines(length = radius = 3) to each of the intersections (where circles intersect) - call it as LINE
3. From one of the intersection point, draw perpendicular line to the BASE line - it will divide the BASE into two equal halves (1.5 each)
4. Now, cos theta = 1.5/3 = 1/2 => theta = 60 degree
5. So LINE makes 60 deg with the BASE
6. So the length of the arc = ((2 * pi * 3)/360) * 120 = 2 pi
7. Length of the other arc = 2 pi
8. Circum = (2 * pi * 3) + (2 * pi * 3) - (2 *pi) - (2* pi) |
|
|
| Back to top |
|
 |
therani.arch
Joined: 19 Jul 2010 Posts: 65
|
Posted: Fri Jul 23, 2010 7:12 pm Post subject: |
|
|
Got it.
Thanks a lot |
|
|
| Back to top |
|
 |
sunil24sunil
Joined: 17 Jul 2010 Posts: 1
|
Posted: Mon Jul 26, 2010 11:30 am Post subject: 6th questionnnnn answer |
|
|
| therani.arch wrote: | Got it.
Thanks a lot |
6*pi + 6*pi - 3/2*pi-3/2*pi = 9*pi
pakkkkkkkkkkkaaaaaaaaa solution i think any doubts post messageeee |
|
|
| Back to top |
|
 |
nandana
Joined: 18 Jul 2010 Posts: 6
|
Posted: Mon Jul 26, 2010 3:13 pm Post subject: |
|
|
| can anyone can exlain 1st que |
|
|
| Back to top |
|
 |
nandana
Joined: 18 Jul 2010 Posts: 6
|
Posted: Mon Jul 26, 2010 3:36 pm Post subject: |
|
|
| can u xplain 6que |
|
|
| Back to top |
|
 |
nehareddy4u
Joined: 23 Jul 2010 Posts: 66
|
Posted: Thu Jul 29, 2010 4:53 pm Post subject: |
|
|
| nandana wrote: | | can u xplain 6que |
2 circles with radius 3..
so the distance between the two centers will aslo be 3...
draw a line..
next.. draw a line from the center to the upper intersection of the circles.. which length is 3 (i.e.. since radius).. similarly with the 2nd circle also..
now it forms a equilateral triangle.. v know angle is 60'.. so find the length of the arc with angel as 60' and radius as 3.. u will get (pi).. there are 4 pi..
so total circumference - 4 pi.. ==> 8pi..
total circumference of 2 circles = 2(pi)(3) + 2(pi)(3) ==> 12 pi.. |
|
|
| Back to top |
|
 |
|