mnotageek

 Profile Answers by mnotageek 

• Jul 19th, 2007

 

ok let us suppose 100 kg of fuel is being wasted.. if i use the A then 30 kg will be saved n 70 kg will be wasted. now if i bring in B it will save 40% of 70 n (70*0.6)=42 kg will be wasted. Now if i incorporate C then it will save 10%of 42 i.e..,(42*0.9) 37.8 kg will be wasted... so 100-37.8=62.2 kg will be saved.. so ans is 62.2%

ramesh

• Sep 22nd, 2007

 

Answer will be 62.2 because,
= (70/100)*(60/100)*(90/100)
= 0.378
=37.8 in%
=>100-37.8
=62.2

a.karthickeie

 Profile Answers by a.karthickeie 

• Jul 21st, 2008

 

combination of three mechanism is
30+40+10=80/3
=26.666=27

sudipsinha123

 Profile Answers by sudipsinha123 

• Jul 13th, 2009

 

The Fuel economy efficiency for A,B,C will be 1/3,1/4 and 1/10 respectively

Hence the combined Fuel Economy efficiency will be (1/3+1/4+1/10)=41/60

in % form it will be (41/60)x100=68% approx.

Siddhubias

 Profile Answers by Siddhubias 

• Apr 21st, 2010

 

Let each mechanism solution is 100 litre.

In "A" mechanism, wastage is (100-30)/100 = 70 %
In "B" mechanism, wastage is (100-40)/100 = 60 %
In "C" mechanism, wastage is (100-10)/100 = 90 %

If all are combined,total solution = 300 litres
                   Total wastage  = 220 litres
                   Total used fuel = 80 litres

So, wastage in 100 litres = 80/3 litres
               % wastage = 26.6666.... %
               = 27 % (approx)

Mudit

• Jan 24th, 2017

 

How can you apply efficiency on wasted fuel?

Anirudh Dayma

• Jul 11th, 2017

 

assume total =100 then 30% = 70, 40% reduction = 3/5 and 10% reduction = 9/10. Therefore total reduction = 70X(3/5)X(9/10) = 37.8 therefore ans = 100 - 37.8 = 62.2