Ascent Education Home
• CAT 2008 for IIMs
CAT - Intensive Contact Class @ Chennai
» CAT - Weekend Math & DI Crash Course
» CAT - Correspondence
• India
• Abroad
» Mock CAT Series
» CAT - Math & DI Boot Camp
» Math Shortcut Workshops
• TANCET 2008 @ Chennai
• TANCET 2008 @ Kovai
• XAT 2008
• GRE-GMAT
• US B Schools
• Faculty List
• Question A Day

Mensuration - Quant/Math - CAT 2008

  1. Algebra
  2. Progressions
  3. Averages
  4. Clocks and Calendars
  5. Data Sufficiency
  6. English Grammar
  7. Function
  8. Geometry
  9. Interest
  10. Mensuration
  11. Mixtures & Alligations
  12. Number System
  13. Percentages
  14. Permutation & Combination
  15. Pipes & Cisterns And Work & Time
  16. Probability
  17. Profit & Loss
  18. Races
  19. Ratio, Proportion
  20. Speed, Time & Distance
  21. Trigonometry
  22. Miscellaneous
  23. General Knowledge
Question 4 the day: September 30, 2003

The question for the day is from the topic of Mensuration.
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 travelled by the cart, when the front wheel has done five more revolutions than the rear wheel?

(1) 20 ft (2) 25 ft
(3) 750 ft (4) 900 ft
Correct Answer - (4)

Solution:


The circumference of the front wheel is 30 ft and that of the rear wheel is 36 feet.

Let the rear wheel make n revolutions. At this time, the front wheel should have made n+5 revolutions.

As both the wheels would have covered the same distance, n*36 = (n+5)*30

36n = 30n + 150

6n = 150

n = 25.

Distance covered = 25*36 = 900 ft.




Page top        

 
  © 2002 - 07 ASCENT Education all rights reserved.
Website Maintained by Chrysalis Technologies - Medical Transcription