Question

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[Maths]
Ch: Simultaneous Linear Equations

Q) On Diwali eve, two candles, one of which is 3 cm longer than the other, are lighted. The longer one is lighted at 5.30 p.m. and the shorter at 7 p.m. At 9.30 p.m. they both
are of the same length. The longer one burns out at 11.30 p.m. and the shorter one at 11 p.m. How long was each candle originally?

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Answer 1

Topic

Linear Equations

To Find

Length of both candles

Solution

Considerations

Let two candles be 'A' and 'B'.

Let length of candle 'A' = x cm

Let length of candle 'B' = ( x + 3 ) cm

Let disappearing rate of candle A be 'v' cm / h.

Let disappearing rate of candle B be v' cm / h.

Solving

Longer one it means candle 'B' was lighted at 5:30 pm.

Length of candle 'B' at 9:30 pm ( after 4 hours )

Original length - Disappeared Length

( x + 3 ) - 4v'

( Disappeared Length = Rate of Disappearing × Time )

The shorter on means candle A was lighted at 7 pm.

Length of candle 'A' at 9:30 pm ( after 2.5 hours)

Original Length - Disappeared length

x - 2.5v

It is given that length of both candles A and B are equal at 9:30 pm

So,

x + 3 - 4v' = x - 2.5v

3 - 4v = -2.5v

3 = 4v' - 2.5v . . . . . . . equation 1

The longer one means candle B burns out at 11:30 pm ( after 6 hours )

Burn out mean whole candle is disappeared or we can say disappeared length is equal to original length of candle.

Original Length = Disappeared length

x + 3 = 6v' . . . . . equation 2

The shorter one means candle A burns out at 11:00 pm ( after 4 hours )

Original Length = Disappeared length

x = 4v . . . . . . . equation 3

Put value of x from equation 3 into equation 2

x + 3 = 6v'

4v + 3 = 6v'

6v' - 4v = 3 . . . . . . equation 4

Now, solve equation 1 and 4

( 4v' - 2.5v = 3 ) × ( 4 )

( 6v' - 4v = 3 ) × ( -2.5 )

Equations changes to,

16v' - 10v = 12

- 15v' + 10v = -7.5

Add both equation

16v' - 15v' - 10v + 10v = 12 - 7.5

v' = 4.5

So, rate of disappearing of candle B is 4.5 cm / hour.

Put value of v' in any above written equation to get value of v.

6v' - 4v = 3

6(4.5) - 4v = 3

27 - 4v = 3

4v = 24

v = 6

So, rate of disappearing of candle A is 6 cm / hour.

Calculating Length of Candles

Length of Candle A

x = 4v

x = 4(6)

x = 24 cm

Length of Candle B

x + 3 = 6v'

x + 3 = 6(4.5)

x + 3 = 27 cm

Answer

So, length of candles are 24 cm and 27 cm.

Answer 2

Answer:

$\huge{\bf{\underline{\red{Solution:}}}}$

Length of Shorten Candle = x

Length of longer Candle = x + 3

Rate of Burning = ${\dfrac{\textbf{x}}{{\textbf{4}}}} \: \: \:7 \: \: \: {\dfrac{{\textbf{x +3}}}{{\textbf{6 }}}}$cm

After 9:30pm:-

Their Heights are:- $2.5 \times \frac{x}{4} = \frac{5x}{8} \: and \: 4 \times \frac{(x + 3)}{6} = \frac{2x + 6}{3}$

$\Rightarrow$ According to Question

$x - \Large\frac{5x}{8} = (x + 3) - \frac{(2x + 6)}{3}$

$\Rightarrow$${\Large\frac{3x}{8} = \frac{x + 3}{3}}$

$\Rightarrow$ 9x = 8x + 24

$\:\:\:\:\:\:\:\:\:\:\:\:$x = 24

Other

=x + 3

=24 + 3

= 27

Hence, The length of candles are 24cm and 27cm