Two candles of equal height but different thickness are lighted the first ones of in 6 hours and the second and 8 hours how long after lighting the candle with the first candle behalf of the height of the second one
Answers
Given: Two candles of equal height but different thickness are lighted the first ones of in 6 hours and the second and 8 hours.
To find: How long after lighting the candle with the first candle behalf of the height of the second one?
Solution:
- Now we have given two candles with equal height. So let the height be L.
- Now we have given that the candles are lighted.
- The first one burns in 6 hours which means:
in one hour it will burn = L/6
- The second one burns in 8 hours which means:
in one hour it will burn = L/8
- Now let the time taken to burn the first candle at half the height of second one be t.
- So in t time the candles burnt will be Lt/6 and Lt/8
- So according to the question, we get:
L - Lt/8 = 2(L - Lt/6)
L - Lt/8 = 2L - Lt/3
- Cancelling L from both sides, we get:
1 - t/8 = 2 - t/3
t/3 - t/8 = 1
5t/24 = 1
t = 24/5
Answer:
So in 24/5 hours, the the first candle will be half of the height of the second one.
Answer:
TIME TAKEN= 4.8 HOURS
Step-by-step explanation:
LET THE HEIGHT OF THE CANDLE BE H CM
TIME TAKE BY FIRST CANDLE TO BURN COMPLETELY=6 HOURS
TIME TAKEN BY SECOND CANDLE TO BURN COMPLETELY=9 HOURS
HEIGHT OF THE FIRST CANDLE AFTER BURNING FOR 1 HOUR= H/6
HEIGHT OF THE SECOND CANDLE AFTER BURNING FOR 1 HOUR=H/8
LET THE REQUIRED TIME BE T
HEIGHT OF FIRST CANDLE AFTER BURNING FOR T HOURS=TH/6
REMAINING HEIGHT OF FIRST CANDLE=(H-TH/6)
HEIGHT OF THE SECOND CANDLE AFTER BURNING FOR T HOURS=TH/8
REMAING HEIGHT OF SECOND CANDLE=(H-TH/8)
ACCORDING TO QUESTION
HEIGHT OF FIRST CANDLE = 1/2 OF THE SECOND CANDLE AFTER T HOURS
(H-TH/6)=1/2(H-TH/8)
H(1-T/6)=1/2H(1-T/8)
1-T/6=1/2(1-T/8)
1-T/6=1/2-T/16
1-1/2=-T/16+T/6
1/2=5T/48
T=1*48/2*5
T=48/10
T=4.8 HOURS