BdMO National Junior 2008/2
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- Posts:1007
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Faria sets of her bike to Nazia's house.At exactly the same time,Nazia sets off to Faria's house along the same straight road in her car.A while later,they pass each other (neither spotting the other) and shortly after ,Nazia arrives at Faria's house and find that she is not there.Nazia waits for $22$ minutes and then heads back along the same road,arriving at her own place at exactly the same time as Faria.Faria traveled at the same speed the whole time whereas Nazia traveled $4$ times as fast as Faria on the way of Faria's house and $5$ times as fast on the way back .How many time did it take Faria to reach Nazia's house?
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2008/2
Answer:$\fbox {40}$
Solution:
Let,
Faria took $t$ minute to reach Nazia's house.
Distance from Faria's house to Nazia's house is $x$
For Faria, \[x=vt\Rightarrow t=\dfrac xv\]
For Nazia,Total distance \[x+x=2x\]
For first time \[x=4vt_1\Rightarrow t_1=\dfrac {x}{4v}\]
For second time \[x=5vt_2\Rightarrow t_2=\dfrac {x}{5v}\]
According the question,
\[t_1+t_2+22=t\]
\[\Rightarrow \dfrac{x}{4v}+\dfrac {x}{5v}+22=\dfrac {x}{v} \]
\[\Rightarrow \dfrac{x}{v}-\dfrac{x}{4v}-\dfrac {x}{5v}=22 \]
\[\Rightarrow \dfrac{11x}{20v}=22 \]
\[\Rightarrow \dfrac{x}{v}=40 \]
\[\Rightarrow t=40 \]
Solution:
Let,
Faria took $t$ minute to reach Nazia's house.
Distance from Faria's house to Nazia's house is $x$
For Faria, \[x=vt\Rightarrow t=\dfrac xv\]
For Nazia,Total distance \[x+x=2x\]
For first time \[x=4vt_1\Rightarrow t_1=\dfrac {x}{4v}\]
For second time \[x=5vt_2\Rightarrow t_2=\dfrac {x}{5v}\]
According the question,
\[t_1+t_2+22=t\]
\[\Rightarrow \dfrac{x}{4v}+\dfrac {x}{5v}+22=\dfrac {x}{v} \]
\[\Rightarrow \dfrac{x}{v}-\dfrac{x}{4v}-\dfrac {x}{5v}=22 \]
\[\Rightarrow \dfrac{11x}{20v}=22 \]
\[\Rightarrow \dfrac{x}{v}=40 \]
\[\Rightarrow t=40 \]