The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the near wheel?

20 ft

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25 ft

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750 ft

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900 ft

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Solution

The correct option is D 900 ft
$C_{1} = 30 f t, C_{2} = 36 f e e t$
Let $C_{2}$ makes $n$ number of revolutions, then $C_{1}$ makes $(n + 5)$ revolutions.
Distance traveled by both the wheel is same, therefore
$\Rightarrow n C_{2} = (n + 5) C_{2} \Rightarrow 36 n = (n + 5) 30 \Rightarrow 36 n - 30 n = 150$
$\Rightarrow n = \frac{150}{6} = 25$
$∴$ Distance covered $= 25 \times 36 = 900$ ft.

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The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 36 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the near wheel?

The correct option is D 900 ftC1=30ft,C2=36feetLet C2 makes n number of revolutions, then C1 makes (n+5) revolutions.Distance traveled by both the wheel is same, therefore ⇒nC2=(n+5)C2⇒36n=(n+5)30⇒36n−30n=150⇒n=1506=25∴ Distance covered =25×36=900ft.