Aptitude Discussion

Q. |
The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel? |

✖ A. |
950 ft |

✖ B. |
1450 ft |

✔ C. |
1200 ft |

✖ D. |
800 ft |

**Solution:**

Option(**C**) is correct

Let the total distance travelled by the cart be $x$ ft

Then,

\(\dfrac{x}{40}-\dfrac{x}{48}=5\)

\(\Rightarrow \dfrac{6x-5x}{240}=5\)

⇒ $x$ = **1200 ft**

**Akash Bansal**

*()
*

**Vejayanantham TR**

*()
*

Front = 40

Back = 48

So, 48-40 = 8

8 * 5(rotation) makes =40 ie one rotation of front.

So, for 5 times the rotation of front more than rear ,

48*25 = 1200

for every revolution of rear wheel, front wheel rotates 1.2 revolutions.

Therefore, front wheel goes 0.2 revolutions more than rear wheel.

-> for 5 more revolutions,

number of revolutions of rear wheel= (1/0.2)x5 = 25 revolutions

distance travelled by cart= 25 x 48= 1200 Ft.