If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how much coal (in m

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21 May 2015 How much coal will be needed

If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how much coal (in metric tonnes) will be needed for 8 engines, each running 10 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type?

OPtion

1) 26.68
2) 25/8
3) 6.48
4) 77/9
5) 25/3
6) 77/3
7) 11.58
8) 27
9) 18
10)8

Solution

More engines, more coal (Direct)
More hours a day, more coal (Direct)
More rate of consumption, more coal (Direct)
Engines ------------ 5:8
Hrs/day ------------ 9:10
Rae of consumption - (1/3):(1/4)
6:x
Therefore
x = (8*10*(1/4)*6*(1/5)*(1/9)*3) = 8 metric tonnes.
Option 10)

Correct Option: 10 Best Solution (18)

ABHINAV PUSHPAM   9 years AGO

[(5*9/6)*4]=[(8*10/X)*3]
on solving x=8
answer is 8

Like? Yes (10) | No (1)  

Aditi Tenguria   9 years AGO

LET THE REQUIRED CONSUMPTION OF COAL BE "X"UNIT
NO.OF ENGINES=5:8
WORKING HOURS=9:10::6:X
RATE OF CONSUMPTION=1/3:1/4
....
5*9*1/3*X =8*10*6*1/4
=>X=8 METRIC TONNES

Like? Yes (9) | No   

Shiva Teja   9 years AGO

Coal consumed by one engine of type 1
=6/(5*9)
=2/15 metric tonnes per hour
3 engines of type 1 consume as much as 4 engines of type 2
Therefore,
Coal consumed by one engine of type 2=(3/4)*(2/15)
=1/10 MT/hour

Coal consumed by 8 engines (of type 2),running 10 hours a day
=(1/10)*8*10
=8 metric tonnes.

Like? Yes (8) | No   

Sundar Kumar   9 years AGO

Engines 5 : 8
hours / day 9 : 10
Rate of consumption 1/3:1/4

Therefore x= 8*10*(1/4)*6*(1/5)*(1/9)*3
x=8 metric tonnes

Answer: Option: 10) 8

Like? Yes (3) | No   

Karthick Raja   9 years AGO

RATIO

ENGINE 5:8
HOURS 9:10 (PER DAYS)
3 engines of the former type consume as much as 4 engines of latter type (1/3):(1/4)

THESE THREE RATIOS ARE EQUAL TO 6:x

x=(8*10*(1/4)*6*(1/5)*(1/9)*3)
X=8 METRIC TONNES

Like? Yes (2) | No   

Dileepkumar Palli   9 years AGO

5 engines consumes 6 metric tonnes of coal in 9 hours a day.
so 1 engine consumes of coal in 1 hour of a day is 6/(5*9).
and 3 former engines is equal to 4 latter engines.
so 8 latter engines= 6 former engines.
so 8 latter engines used coal in 10 hours a day= 6*(1 former engine consumes coal in 1 hour a day)*10 hours.
=6*(6/(5*9))*10.
=8.
so ans is 8

Like? Yes (2) | No   

SUSHANT KUMAR   9 years AGO

engine=5:8
hours/day=9:10
rate of consumption=1/3:1/4
coal=6:x
so x=8

Like? Yes (2) | No (1)  

Mohit   9 years AGO

after solving the answer is found to be this 8 metric tonne

Like? Yes (1) | No   

vivek   9 years AGO

let 3 engines of former type consumes 1 unit in 1 hr.
then, 4 engines of former type consumes 1 unit in 1 hr.
let the required consumption of coal be x unit.
so according to chain rule,
5*9*4*x= 8*10*3*6
or, x= 2*2*2=8 metric tones

Like? Yes (2) | No (1)  

padam   9 years AGO

engine 5:8
hours/day 9:10
rate of consumption 1/3:1/4
tonnes 6:x
x=8*10*1/4*6*1/5*1/9*3=8 answer option 10

Like? Yes (1) | No   

Ashwin Dinesh   9 years AGO

6*(8/5)*(10/9)*(3/4) = 8

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sompon temtheerakij   9 years AGO

8 engines , 10 hr/day consume = 8x10x6/5x9 = 10 2/3 tons
=10 2/3 x 3/4 = 8 tons

Like? Yes (1) | No   

karan   9 years AGO

3 engines of former type consume : 3.6 metric tonne (6/5=1.2*3=3.6)
4engines of latter type=3engines of former type=3.6 metric tonne
7engine consume=3.6+3.6=7.2 metric tonne
1engine consume=1.3 metric tonne
8engine consume=7.2+1.38.55=77/9 metric tonne

Like? Yes (1) | No   

siddharth prakash singh   9 years AGO

using chain rule.
more engine consume more coal (direct relation)
more hour of running consume more coal (direct relation)
more rate more consuming of coal (direct relation)
so, 5:8,9:10,1/3:1/4::6:x
giving x=8

Like? Yes (1) | No   

Divyabala R   9 years AGO

Its given that 3 engines of former equals the 4 engines of latter in coal intake so we need to find the solution using former i.e,
from the given 5E[9H] = 6MT of coal;

so we should find 6E[10H] = ? which gives the answer for 8E[10h] of latter .

Consider 5E[9H] = 6 MT of coal then,

1E[1H] = 6 / (5* 9) = 2/15

6E[1h] = [2/15] * 6 =12/15

6E[10H] = [12/15] * 10 = 8

Thats it then we can pass this to 8E[10H] of the latter .

So the answer is 8.

Like? Yes (2) | No (1)  

sivaramakrishna   9 years AGO

5f*9=6==>f=6/45, 3f=4s==>s=3f/4=3*6/45*4
substitute s in the equation
8s*10=?==>8

Like? Yes (1) | No   

YUVRAJ BALINGE   9 years AGO

Consumtion is directly proportional to no. of engine and time
of hour it works.
Consumption of former type engine=2/15 metric tonne/hour/engine
Consumption of later=8

Like? Yes (1) | No   

aman aditya   9 years AGO

First type of machine work 9*5=45 hr
energy consumed per hr=6/45
second type of machine work 10*8=80hr
as 3 of first machine take as much as energy of 4 machine of second type.
so,
(2*3*20)/(45*4)=8

Like? Yes (1) | No (1)