On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, close relative, comes up for air on average once every five minutes. The number of times a bottle-dolphin would come up for air in a 24 hour period is approximately what percent greater than number of times a beluga whale would come up for air in that same period?

Question

On average, the bottle-nosed dolphin comes up for air once every two minutes; the beluga whale, close relative, comes up for air on average once every five minutes. The number of times a bottle-dolphin would come up for air in a 24 hour period is approximately what percent greater than number of times a beluga whale would come up for air in that same period?

Shohan

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2025-01-04 04:42

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Shohan · Accepted Answer · 2024-12-14T13:31:11+06:00

$T h e f i r s t p u m p c a n d r a i n i n 3 h o u r s \frac{1}{2} o f t h e p o n d$

$∴ T h e f i r s t p u m p c a n d r a i n i n 1 h o u r s = \frac{1}{2 \times 3} = \frac{1}{6} o f t h e p o n d$

$B o t h p u m p s c a n d r a i n i n 1 \frac{1}{2} o r \frac{3}{2} h o u r s = 1 - \frac{1}{2} = \frac{1}{2} o f t h e p o n d .$

$∴ B o t h p u m p s c a n d r a i n i n 1 h o u r s = \frac{1}{2} \times \frac{1}{3} = \frac{1}{3} o f t h e p o n d .$

$∴ W o r k i n g a l o n e t h e s e c o n d p u m p c a n d r a i n i n 1 h o u r = \frac{1}{3} - \frac{1}{6} = \frac{2 - 1}{6} = \frac{1}{6} p a r t o f t h e p o n d$

$∴ T h e s e c o n d p u m p s c a n e m p t y \frac{1}{6} o f t h e p o n d i n 1 h o u r$

$∴ T h e s e c o n d p u m p c a n e m p t y 1 o r f u l l p o n d n = \frac{6 \times 1}{1} = 6 h o u r s .$