@Anivesh90 said:May 21?
nopes..please check again
@soumitrabengeri said:The price of Darjeeling tea (in Rs/Kg) is 100+0.1n on the nth day of 2007 (n = 1,2,.....100) and then remains constant. On the other hand the price of Ooty tea (in Rs/Kg) is 89+0.15n on the nth day of 2007(n = 1,2,.....365)On which date in 2007 will their prices be equal?1) May 212) April 113) May 204) June 30
may20
110=89+.15n
=>n=21/.15=140=31+28+31+30+20=>20may
110=89+.15n
=>n=21/.15=140=31+28+31+30+20=>20may
@staaalinnn said:Its an AIMCAT question I suppose..
oops..I forgot to give the answer at that moment..it should be May 20 I guess !!!
@soumitrabengeri said:
can you please share your approach?
From the options i figured out n>100
So dar tea price=Rs 110
110=89+.15n
n=140 and hence may 20..
@soumitrabengeri said:The price of Darjeeling tea (in Rs/Kg) is 100+0.1n on the nth day of 2007 (n = 1,2,.....100) and then remains constant. On the other hand the price of Ooty tea (in Rs/Kg) is 89+0.15n on the nth day of 2007(n = 1,2,.....365)On which date in 2007 will their prices be equal?1) May 212) April 113) May 204) June 30
All dates after 100 days. So, D tea= 110. Now, For O tea to be 110, n should be equal to 140. J31+F28+M31+A30+M20= 140. Ans: May 20
Minimum possible perimeter of a triangle with integer sides in which one of the angles is 120 degrees.
1) 12
2) 15
3) 10
4) 14
1) 12
2) 15
3) 10
4) 14
@krum said:Minimum possible perimeter of a triangle with integer sides in which one of the angles is 120 degrees.1) 12 2) 15 3) 10 4) 14
getting answer as 15..but this is a very crude method to follow
applying the cosine rule
a^2+b^2+ab = c^2
Trial and error...3,5,7
If there is any other way, please tag me in the solution
@sauravd2001 said:@soumitrabengeri bhai hats off awesome one.........pulled something out of box........congo
Thank you but i feel that there must be a better way to do it than this..the trial and error method won't work in the actual exam..
@soumitrabengeri said:Thank you but i feel that there must be a better way to do it than this..the trial and error method won't work in the actual exam..
@sauravd2001 said:@soumitrabengeri bhai hats off awesome one.........pulled something out of box........congo
2abcos120=a^2+b^2-c^2
=>c^2=a^2+b^2+ab
=>a^2+b^2+ab is a perfect square
without loss of generality
case 1: b=1
=>a^2+a+1 ; not a perfect square
case 2: b=2
=>a^2+2a+4=>(a+2)^2
=>c=a+2
but for a,b,c to be sides of a triangle c
=> (a+2)
case 3: b=3
=>a^2+3a+9
taking a=5=>25+15+9=49=>c=7
so sides are 5,3,7
@soumitrabengeri , @sauravd2001 , @bullseyes
@krum said:Minimum possible perimeter of a triangle with integer sides in which one of the angles is 120 degrees.1) 12 2) 15 3) 10 4) 14
15.
for a triangle to have integer sides wd one angle as 120, following condition of sides must exist.
a= m(sq) + m.n + n(sq)
b= 2.m.n + n(sq)
c= m(sq)-n(sq)
add all three
a+b+c = (2m+n)(m+n)
for minimum perimeter. m=2,n=1
hence 15
PS : i learnt this concept a while back.
@bullseyes i used the same concept like blind man ......i took m =1 and n=2......and didnot see that c will become negative............. :/.......but good one u puled it out good
what will be the remainder when 2(13!) + 14! is divided by 13! - 1?
A.16
B.14
C.12
D.10
Please help with this question and also explain the concept behind finding a remainder in case of factorials.