@abhijit90 said:@saurav205 bcom in single sitting exams..........is it valid for cat
any undergrad degree is valid...
and this is the wrong thread to post your query...Kindly post your queries here :
@saurav205 said:bhai logic??
bhai koi logic nahi pata, mere pass kuch shortcuts hai usme ye bhi ek hai...
rest are:
|x| + |y| + |z| = a has total number of integral solutions = 4*a^2 + 2.
|x| + |y| = p has a total number of integral solutions = 4*p.
|X| + |Y| = a; then 2*a^2 is the area enclosed between these lines.
|x-a| + |y-b| = k has 4k integral solutions.
|x| + |y| = p has a total number of integral solutions = 4*p.
|X| + |Y| = a; then 2*a^2 is the area enclosed between these lines.
|x-a| + |y-b| = k has 4k integral solutions.
PS - Pls dnt blame me if these fail in some cases...coz ye meine derive karke nahi dekhe hai..
@Logrhythm said:x=0,y=0,1,2,3,...,25 -> 26 points..x=1,y=0,1,2,3,...,24 -> 25 points..x=2,y=0,1,2,3,...,24 -> 25 points..x=3,y=0,1,2,3,...,23 -> 24 points.....x=47,y=1 -> 2 pointsx=48,y=1 -> 2 pointsx=49,y=0 -> 1 pointx=50,y=0 -> 1 pointtotal = 26 + 2(1+2+3+...+25) = 676 points??
OA = 676
f(x/(2x+3)) =3x, find f(1)
f(x/(2x+3)) =3x, find f(1)
edited my post .. what a typo 
Pavan, Mohan and Sohan were three friends. Pavan bought a cornetto, which is a large ice-cream in the shape of a perfect cone. He first had the top 70(10/27) % of it (by volume) and then gave it to Mohan. Mohan then had the top 70(10/27) % (again by volume) of the remaining part and gave the rest to Sohan. The level of the ice-cream then was 4 cm from the bottom, i.e., the vertex. Find the height of the initial cone (in cm).
8.1
9.9
9
7.2
8.1
9.9
9
7.2
Two circles of equal radii are drawn, without any overlap, in a semicircle of diameter 4 cm. What is the radius (in cm) of the largest possible circles that the semicircle can accommodate in such manner?
0.414
0.828
0.172
0.586
0.414
0.828
0.172
0.586
@swapnil4ever2u said:edited my post .. what a typoPavan, Mohan and Sohan were three friends. Pavan bought a cornetto, which is a large ice-cream in the shape of a perfect cone. He first had the top 70(10/27) % of it (by volume) and then gave it to Mohan. Mohan then had the top 70(10/27) % (again by volume) of the remaining part and gave the rest to Sohan. The level of the ice-cream then was 4 cm from the bottom, i.e., the vertex. Find the height of the initial cone (in cm).8.19.997.2
9...They left 2/3rd and 2/3rd of height = 4/9 of height.
regards
scrabbler
regards
scrabbler
@swapnil4ever2u said:Two circles of equal radii are drawn, without any overlap, in a semicircle of diameter 4 cm. What is the radius (in cm) of the largest possible circles that the semicircle can accommodate in such manner?0.4140.8280.1720.586
0.828...doubled the fig to get 4 circles in a circle 😃 Then easy.
regards
scrabbler
regards
scrabbler
@swapnil4ever2u said:Two circles of equal radii are drawn, without any overlap, in a semicircle of diameter 4 cm. What is the radius (in cm) of the largest possible circles that the semicircle can accommodate in such manner?0.4140.8280.1720.586
.828??
@scrabbler said:0.828...doubled the fig to get 4 circles in a circle Then easy.regardsscrabbler
but I didnt maximise it or something...
drew the figure...
used simple geometry..
I think I got lucky with this one...how did you maximise it??
@saurav205 said:but I didnt maximise it or something...drew the figure...used simple geometry..I think I got lucky with this one...how did you maximise it??
Not maximise...just reflected the fig around the dia of semicircle. Then it is 4 circles in one big circle....easy-peasy :)
Chal me off to bed now...kal office :(
regards
scrabbler
Chal me off to bed now...kal office :(
regards
scrabbler
@scrabbler said:Not maximise...just reflected the fig around the dia of semicircle. Then it is 4 circles in one big circle....easy-peasy Chal me off to bed now...kal office regardsscrabbler
yea yea...
same here...:)
@swapnil4ever2u said:@scrabbler how r they left with 2/3 of height.. dint get this part..
use geometry bhai...similar triangles...
and equate the areas as per the given question
cone 1 - cone 2...etc kar ke...
@swapnil4ever2u said:@scrabbler how r they left with 2/3 of height.. dint get this part..
Fraction is 1900/27% = 19/27...so remaining is 8/27 of vol => 2/3 of height.
regards
scrabbler
regards
scrabbler
@swapnil4ever2u said:Two circles of equal radii are drawn, without any overlap, in a semicircle of diameter 4 cm. What is the radius (in cm) of the largest possible circles that the semicircle can accommodate in such manner?0.4140.8280.1720.586
Sorry for the pathetic drawing skills, but kaam chala lo...
