Latest CSC Aptitude Question SOLUTION: If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day, how much coal will be required for 8 engines, each running 13hours a day, it being given that 3 eng

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If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours
day, how much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?

Asked In CSC (11 years ago)

Unsolved Read Solution (6)

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lucky lovely  (11 years ago)

26metric tonnes.
Sol: We shall compare each item with the quantity of coal.
Less engines, less coal consumed (direct)
More working hours, more coal consumed (direct)
If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.
If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.
Less rate of consumption, less coal consumed (direct).
Therefore, number of engines 9:8 :: 24:x
Working hours 8:13 :: 24:x
Rate of consumption 1/3:1/4 :: 24:x.
9*8*1/3*x= 8*13*1/4*24 or x= 26.
Therefore, required consumption of coal 26 metric tonnes.

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sasi  (11 years ago)

3 engines of former type for one hour consumes 3*24/(9*8)=1 ton
i.e..4 engines of latter type consumes 1 ton for one hour
hence 8 engines consumes 2 tons for one hour
for 13 hours it is 13*2=26

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Subham Mohapatra  (9 years ago)

26

9engines consumes 24 metric ton for 8hrs
9 engines for 1hr=3
1 engine for 1hr=0.33

3former engine = later 4
1engine consumes for 1hr=0.33
3 engine former/4 engine later in 1hr = 0.33*3=0.99
8 engine in 13hrs = 0.99*2*13=26

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chandu  (10 years ago)

three engines of former type for one hour consumes 3*24/(9*8)=1 ton
i.e.four engines of latter type consumes 1 ton for one hour
hence 8 engines consumes 2 tons for one hour
for 13 hours it is 13*2=twenty six.is that okay friends?

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Komal Jain  (9 years ago)

consumption of one engine for 1 hr=x
for 8 hrs it would be 8x
for 9 engines 9*8*x=24 (given)
hence x=1/3
for latter type consumption of one engine for 1 hr=y
nw it is given 4*y=3*x hence y=3x/4
for 13 hrs consumption =13y
for 8 engines= 8*13*y = 8*13*3*x/4
put x=1/3 we get 26 metric tonnes

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NAGENDRA KUMAR  (8 years ago)

9 engines - 8 hr =24 ton coal
1 engines -1 hr =24/(9*8)=1/3
we have to find:-
8 engine -13 hr=?
given that:-
3 engine for former(of 8 working hr engin)=4 *engine for latter(of 13 working hr engine)
4* engine for latter in 1 hr=3* 1/3=>1
so,
8 engine -13 hr - 1 day= 2*13=26
Ans
26 ton

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