एक समान लंबाई के दो मोमबत्ती हैं जिनको जलने पर क्रमसा: 5 घंटे or 4 घंटे में समाप्त हो जाती है माना प्रत्येक मोमबत्ती नियमित गति से जलती है मोमबत्ती के जलने के कितने समय बाद पहली और दूसरी मोमबत्ती का अनुपात 3:2 रह जाएंगे ?
There are two candles of the same length, which are finished in the sequence: 5 hours or 4 hours after burning: Each candle burns at regular speed. How many times after the candle burn, the ratio of first and second candles will remain at 3: 2?
☆ your options are ☆
(I) 3 hours
(II) 3:45
(III) 20/6
(Iv) Non of the above
Answers
I hope my answer is correct
The answer is d) none of the above
Because if we are doing with a) 3 hours then the answer is coming 5:4
If we are doing with b) 3:45 then the answer again is coming 5:4
And if we are doing with c) 20/6 then then the answer is coming 12:5
Therefore the answer is none of these
Thanks
Method - 1:
Let the candles be A and B.
(i) Given that Candle A takes 5 hours to burn, part of the candle burn in 1 hour = (1/5).
(ii) Given that Candle B takes 4 hours to burn, part of the candle burn in 1 hour = (1/4).
Let us assume that After 'x' hours the ratio of A and B will remain at 3:2
Therefore, the answer is 20/7 - Option(D)
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Method - 2:
Given that two candles are of same length.
Let their length (or) height be 'h'.
(i)
Given that first candle takes 5 hours to burn completely.
So, the first candle shortens by h/5 per hour.
(ii)
Given that second candle takes 4 hours to burn completely.
So, the second candle shortens h/4 per hour.
Now,
Let 'x' be time taken when ratio of first & second candles will remain 3:2.
So, in x hours:
⇒ Part of first candle burns = hx/5.
⇒ Part of second candle burns = hx/4.
According to the given statement.
⇒ (h - hx/5)/(h - hx/4) = 3/2
⇒ h(1 - x/5)/h(1 - x/4) = 3/2
⇒ (1 - x/5)/(1 - x/4) = 3/2
⇒ 2 - 2x/5 = 3 - 3x/4
⇒ -8x = -15x + 20
⇒ 7x = 20.
⇒ x = 20/7.
Therefore, the answer is 20/7 - Option (D)
Hope this helps!