The circumference of front wheel of a car is $10 d m$ and that of the hind wheel in $16 d m$ , How much revolution more will the first wheel make then the second in covering a distance of $96 k m$ ?

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Solution

Revolutions of first wheel $= \frac{T o t a l D i s t a n c e}{C i r c u m f e r e n c e} = \frac{96 \times 1000 \times 10}{2 \times π \times 10} = 15278$
Revolutions of second wheel $= \frac{96 \times 1000 \times 10}{2 \times π \times 16} = 9549$
$\Rightarrow 15278 - 9549$
$\Rightarrow 5729$ Revolutions more than second wheel.

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The circumference of front wheel of a car is 10 dm and that of the hind wheel in 16 dm, How much revolution more will the first wheel make then the second in covering a distance of 96 km?

Revolutions of first wheel =TotalDistanceCircumference=96×1000×102×π×10=15278Revolutions of second wheel =96×1000×102×π×16=9549⇒15278−9549⇒5729 Revolutions more than second wheel.

The circumference of front wheel of a car is $10 d m$ and that of the hind wheel in $16 d m$ , How much revolution more will the first wheel make then the second in covering a distance of $96 k m$ ?