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### IIT are ranked 4'th in top 100 engineering universities (a survey)

THES (Times Higher Education Supplement) came with an exclusive ranking of universities worldwide. IIT made it to the list at a very impressive position. In the category, "International comparisons: Top 100 Engineering and IT Universities"
http://www.thes.co.uk/statistics/international_comparisons/2004/top_100_england.aspx ,
IIT are at 4'th position. The three ahead of IIT are:
3. Stanford (Normalized score = 150.71)
2. MIT (Normalized score = 191.15)
1. UCB (Normalized score = 200.00)
Also remarkable is the fact that IIT scroed 149.34 compared to Stanford's 150.71.

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$