Answer / mathsskumar_sarma2000

let x be lenghth of candles.

thicker one---- in 1 hr 1/6th will go.
thinner one--- in 1 hr 1/4th will go.

let he lit for n hrs.

so in n hrs thicker one goes n/6.
and thinner one goes n/4.

so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.

Answer / yogesh wagh

Hrs Remaining Candle (Thicker) Remaining
Candle (Thiner)
Lighted 1 1
1 0.833333333 0.75
2 0.666666667 0.5
3 0.5 0.25 Answer
4 0.333333333 0
5 0.166666667
6 0


Its clear from above table that the Ramesh light the two
candles Before 3 Hrs

Answer / kunal patil-pune

as per data,
thicker one can consume 1/6 in an hour
smaller can consume 1/4 in an hour.
consider after an hour,
thicker one consume 1/6 & remains 5/6.
shorter one consume 1/4 & remains 3/4.
AFTER 2 HOURS,
thicker one consums 2/6 & remains 4/6.
shorter one consumes 2/4 & remns 2/4.
AFTER 3 HOURS,
thicker one consumes 3/6& remains 3/6.
shorter one consumes 3/4 & remains 1/4.
THIS CLEARLY MEANS THAT,

THICKER IS 50% OFORIGIONAL LENGTH.
&
THINNER ONE IS 25% OF ORIGIONAL LENGTH.
u can comment me on - kunal1985engg@yahoo.co.in

Answer / subhakar prabhu

The equations for the lengths of the candles (in terms of time)

Thick length = 1 - t/6
Thin length = 1 - t/4
At time T, (1 - T/6) = 2*(1-T/4)
This gives T = 3.

Answer / pappu kumar yadav

Let the length of both the candles be"l".By the qstn in 6
hrs thicker candle consumes "l" length.Hence in "t" time it
consumes"(l/6)t".(where "t" s d duration during whch bth d
candles glow n chkd).Similarly during"t" time thinner
candle consumes"(l/4)t".This implies that after "t" time
length of thicker candle remaining =l-(lt/6).Length of
thinner candle remaining=l-(lt/4).Now by the question we
have (1/2)*[l-(lt/6)]=l-(lt/4). on solving this we get t=3
Hours. Hence bth d candles were lit 3 hours before.

Answer / abhijnan dasgupta

let their original lengths be 1 each...let no. of hrs
elapsed since he lighted the candles be x....now since they
are consumed in 6 and 4 hrs respectively their rate of
decrease=1/6 per hr and 1/4 per hr respectively...so we need
(1-x/6)/(1-x/4)=2/1...which gives...x=3 hrs..

Answer / ashish rawat

let the length of candle is – L
let ramesh light candles x hours ago
thinner one lasts = 4h
length burnt in one hour = L/4
length burnt in x hour = xL/4
remaining length = L-(xL/4) = (4-x)L/4

thicker one lasts = 6h
length burnt in one hour = L/6
length burnt in x hour = xL/6
remaining length = L-(xL/6) = (6-x)L/6

now according to conditions after x hours , thicker candle
length = 2*thinner candle length

(6-x)L/6 = 2* (4-x)L/4
(6-x) = 3*(4-x)
2x = 12-6 = 6

x = 3 hours

Answer / sunil

let us assume the total units is lcm(6, 4)= 12 units;
candle 1 lasts 6 hours => in 1 hour the candle 1 lasts 2 units,
caldle 2 lasts 6-2 = 4 hours => in 1 hour the candle 2 lasts
3 units.
there fore remaining units for candle 1 after t hours is 12-2t
remaining units for candle 2 after t hours is 12-3t

there fore (12 - 2t) = 2(12 - 3t) [from the question]
=> 12 - 2t = 24 - 6t
=> 6t - 2t = 24 - 12
=> 4t = 12
=> t = 3 hours.
3 hours ago ramesh light the two candles.

Answer / gunaramesh

3 hrs

Answer / mujeeb

3 hours

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