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There are two candles of equal lengths and of different
thickness. The thicker one lasts of six hours. The thinner
2 hours less than the thicker one. Ramesh lights the two
candles at the same time. When he went to bed he saw the
thicker one is twice the length of the thinner one. How
long ago did Ramesh light the two candles .
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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.
| Is This Answer Correct ? | 151 Yes | 28 No |
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
| Is This Answer Correct ? | 64 Yes | 22 No |
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
| Is This Answer Correct ? | 34 Yes | 13 No |
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.
| Is This Answer Correct ? | 30 Yes | 9 No |
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.
| Is This Answer Correct ? | 23 Yes | 7 No |
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..
| Is This Answer Correct ? | 19 Yes | 8 No |
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
| Is This Answer Correct ? | 8 Yes | 3 No |
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.
| Is This Answer Correct ? | 5 Yes | 2 No |
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