## Problem Of The Week 34

**Rakesh has the habit of always pouring his tea from the cup into the saucer before drinking it. He fills both the cup and the saucer to only 90% of their capacity (subject to the availability of tea). He also does not drink any tea which is below the 15% mark in the saucer. If he has to pour the tea from the cup into the saucer at least three times before emptying the cup (each time drinking from the saucer till it reaches the minimum level), then what is the maximum possible ratio of the volume of the cup to that of the saucer respectively? ****(1) 3 : 1 (2) 9 : 4 (3) 5 : 2 (4) 8 : 3 (5) None of these.**

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0.9 Vc = 3*(0.75)Vs

Vc/Vs=5/2

tyroSeptember 28, 2008 at 3:03 am

It would actually be option (4)8/3

The first time the tea is poured from the cup to the saucer it would be

0.9 Vs and subsequently 0.75Vs twice.

Hence

0.9 Vc = 0.9Vs + 2 (0.75)Vs

Vc/Vs = 8/3

tyroSeptember 28, 2008 at 5:17 am

The second one is correct tyro!

outtimedSeptember 28, 2008 at 2:57 pm

Isn’t 8/3 the minimum ratio? When the saucer is too small relative to cup,more number of refilling is required to empty the cup. When the saucer is as large as possible relative to cup, minimum no. of transfer is required ie 3 which is assumed in this solution. So when saucer size is as large as possible wrt cup, the ratio should be minimum,right? Am I missing something?

riyaziqSeptember 29, 2008 at 8:14 am

my answer is 8/3

i assumed the volumes of cup and saucer as X & Y respectively

now since the saucer was initially empty and cup can be filled to the max of 0.9X

the eqn i get is 0.9X-0.9Y-0.75Y-0.75Y=0(FOR MAX RATIO)

thus i got X/Y =8/3

if u find any error then please let me know

wolverineOctober 1, 2008 at 3:56 am

yupp

8/3 is absolutely right !

outtimedOctober 1, 2008 at 11:18 am

@outtimed

i think our ideologies synchronize……!!!

wolverineOctober 1, 2008 at 5:21 pm