two water taps together can fill a tank in in 75 upon 8 hours. the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. find the time in which each tap can separately fill the tank.
Answers
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Answer:
Answer:
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Tap of smaller diameter can fill the tank in 25 hours separately.
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Explanation−
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Let tap of smaller diameter can fill the tank in x hours.
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So, tap of larger diameter can fill the tank in (x - 10) hours ( as mention in the question )
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In 1 hour, tap of smaller diameter can fill tank in \dfrac{1}{x}
x
1
hours.
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Similarly, In 1 hour, tap of larger diameter can fill tank in
x−10
1
hours.
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It is given that, both taps can fill the tank in
8
75
hours. So, in hour, both taps can fill tank in
75
8
hours.
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x
1
+
x−10
1
=
75
8
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★ Taking LCM,
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x(x−10)
x−10+x
=
75
8
x
2
−10x
2x−10
=
75
8
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★ By cross multiplying,
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150x−750=8x
2
−80
2
−230x+750=0
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★ Taking 2 as common
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: \implies⟹ \rm{4x^2-115x+375=0}4x
2
−115x+375=0
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★ Solving the Quadratic equation, by splitting middle term.
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: \implies⟹ \rm{4x^2-100x-15x+375=0}4x
2
−100x−15x+375=0
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We get,
x=
4
15
and25
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4
15
,
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[/tex]:\implies⟹ \rm{x-10\:=\:\dfrac{15}{4}\:-\:10}x−10=[/tex]
4
15
−10
x−10=
4
−25
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Since time can't be negative, so we reject this value.
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\therefore∴
x=25
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Therefore, tap of smaller diameter can fill the tank in 25 hours separately.