@shattereddream A 6-cm long cigarette will burn itself up in 15 min if no puff is taken. For every puff, it burns three times as fast as during the duration of the puff. if the cigarette burns itself in 13 min, how many puffs has the smoker taken if his average puff lasted 3 sec ?
@venomizer said:A 6-cm long cigarette will burn itself up in 15 min if no puff is taken. For every puff, it burns three times as fast as during the duration of the puff. if the cigarette burns itself in 13 min, how many puffs has the smoker taken if his average puff lasted 3 sec ?
20 ?
@pratskool said:x^2 - y^2 = 1215^2--> (x-y)(x+y) = 1215^2x-y 1215all factors of 1215^2 less than 1215...so all factors of 1215 except 1215 which is 11...and factors of 1215^2 which are not of 1215 but less than it...5^2, 5^2*3^1,5^2*3^2, 5^2*3^3,3^6...so 16
i have a doubt bro. take dis example
12^2 = 144 .... 2 of its factors can be 16,9
144 = 16x9 (one is less than 12 and the other greater than 12
now if i say from this that
x + y = 16
x - y = 9
i get x = 25/2 ... a fractional value of x
dont u think the way u calculated , such values have been counted in it.. ?
@venomizer said:@shattereddreamA 6-cm long cigarette will burn itself up in 15 min if no puff is taken. For every puff, it burns three times as fast as during the duration of the puff. if the cigarette burns itself in 13 min, how many puffs has the smoker taken if his average puff lasted 3 sec ?
5 puffs?
@venomizer said:howdont have sol
number of puffs be n
normal rate = 6/15x60 = 1/150 cm/s
total time of puffs = 3n seconds
remaining time = 13x60 - 3n = 780 - 3n
since total length is 6cm
(780 - 3n)/150 + 3n/50 = 6
n = 20
@techgeek2050 said:i have a doubt bro. take dis example12^2 = 144 .... 2 of its factors can be 16,9144 = 16x9 (one is less than 12 and the other greater than 12now if i say from this that x + y = 16x - y = 9i get x = 25/2 ... a fractional value of xdont u think the way u calculated , such values have been counted in it.. ?
all factors are odd in this case... x + y is odd x -y is odd... so adding both would give even and hence...
@techgeek2050 said:number of puffs be nnormal rate = 6/15x60 = 1/150 cm/stotal time of puffs = 3n secondsremaining time = 13x60 - 3n = 780 - 3nsince total length is 6cm(780 - 3n)/150 + 3n/50 = 6n = 20
i also got 20.. this is a question someone posted in prep section... OA is given as 15 
@techgeek2050 said:i have a doubt bro. take dis example12^2 = 144 .... 2 of its factors can be 16,9144 = 16x9 (one is less than 12 and the other greater than 12now if i say from this that x + y = 16x - y = 9i get x = 25/2 ... a fractional value of xdont u think the way u calculated , such values have been counted in it.. ?
1215=3^5*5=x^2-y^2=(x+y)(x-y)
there are 6 cases
1215*1
3^5*5
3^4*15
3^3*45
3^2*135
3^1*405
bt shrtest side can be either base or prepndicular so total 12 cases
there are 6 cases
1215*1
3^5*5
3^4*15
3^3*45
3^2*135
3^1*405
bt shrtest side can be either base or prepndicular so total 12 cases
15min for 6cm
1sec for 1/150cm
in 1 puff or 3sec he smokes thrice the normal rate.
in normal 3sec it would have been 1/50. since it is given thrice as fast so 3/50.
in 5 puffs, he;ll take 15 sec. and the cig will burn 0.9 cm
length left = 5.1
time left = 12min and 45sec or 720+45=765sec
we know 1sec for 1/150.
so 765sec for 765/150... which comes out the length left 5.1
hence 5 puffs.
please correct me if i m wrong. :)
@ChirpiBird said:15min for 6cm1sec for 1/150cmin 1 puff or 3sec he smokes thrice the normal rate.in normal 3sec it would have been 1/50. since it is given thrice as fast so 3/50.in 5 puffs, he;ll take 15 sec. and the cig will burn 0.9 cmlength left = 5.1time left = 12min and 45sec or 720+45=765secwe know 1sec for 1/150.so 765sec for 765/150... which comes out the length left 5.1 hence 5 puffs.please correct me if i m wrong.
it is given the rate becomes thrice. u have multiplied by 3 twice. 
@ishu1991 said:1215=3^5*5=x^2-y^2=(x+y)(x-y)there are 6 cases 1215*13^5*53^4*153^3*453^2*1353^1*405bt shrtest side can be either base or prepndicular so total 12 cases
Factor would be 3^10*5^2 due to square
@iimSekar said:Difference is 2 min i.e. 60 seconds.Speed is 3 times. 60/3 = 20 Puffs
bhai 2min = 120 seconds
@venomizer said:@shattereddreamA 6-cm long cigarette will burn itself up in 15 min if no puff is taken. For every puff, it burns three times as fast as during the duration of the puff. if the cigarette burns itself in 13 min, how many puffs has the smoker taken if his average puff lasted 3 sec ?
Sorry didnt got the notification
cigarette is burning at = 4 mm/min
cigarette burning during puff= 3 x 4=12 mm/min
let t be total time duration for puff
then 12t + 4(13-t) = 60
t=1 min
so no of puffs=20.
cigarette is burning at = 4 mm/min
cigarette burning during puff= 3 x 4=12 mm/min
let t be total time duration for puff
then 12t + 4(13-t) = 60
t=1 min
so no of puffs=20.
@shattereddream said:Sorry didnt got the notification cigarette is burning at = 4 mm/mincigarette burning during puff= 3 x 4=12 mm/minlet t be total time duration for puffthen 12t + 4(13-t) = 60 t=1 min so no of puffs=20.
sorry kis kiye bhai..
@shattereddream said:Sorry didnt got the notification cigarette is burning at = 4 mm/mincigarette burning during puff= 3 x 4=12 mm/minlet t be total time duration for puffthen 12t + 4(13-t) = 60 t=1 min so no of puffs=20.
i am nt getting it please explain it in more detail
@ishu1991 said:i am nt getting it please explain it in more detail
dekho
length = 60 mm. to burning rate 4mm/ minute hua
and during puff three times se burn hota hai ie 3 x 4 = 12 mm/min
let t be total time duration for puff
now frame an equation
length = 60 mm. to burning rate 4mm/ minute hua
and during puff three times se burn hota hai ie 3 x 4 = 12 mm/min
let t be total time duration for puff
now frame an equation
@shattereddream said:dekho length = 60 mm. to burning rate 4mm/ minute hua and during puff three times se burn hota hai ie 3 x 4 = 12 mm/minlet t be total time duration for puffnow frame an equation
sryyyy i m nt geetn equation hw dumb i m 
