A companys four cars running 10 hrs a day consume 1200 lts of fuel in

CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)

i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

New car consumes 20% less fuel (work)
i.e. work of new car = 80% work of Old car
i.e. New car = 0.8* Old car
i.e. 5 New cars = 0.8*5 Old car = 4 Old card
i.e. 4 Old and 5 New cars = 4 old + 4 old cars = 8 Old cars
Time = 12 hours for 6 days = 6*12

i.e. \(\frac{[4 * (10*10)]}{1200} = \frac{[8 * (6*12)]}{W_2}\)

i.e. \(W_2 = 1728\)Lt

Answer: option D

A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?

(A) 1200 lt
(B) 1555 lt
(C) 1664 lt
(D) 1728 lt
(E) 4800 lt

Lets keep it simple without time and work usual approach .

Given that
4 cars running 10 hrs a day consume 1200 lts. of fuel in 10 days.
1 car consumption per hour per day = 1200 /4 *10 *10 = 3 litre


Now question say new car consume 20% less fuel than the company’s four cars = 80/100 of 3 = 2.4 ( 20 percent less than 3)

Hence we calculate total consumption for next 6 days, the company will need to run 5 new cars for 12 hrs = 2.4 *12 *6*5 =864
similarly = old 4 car consumption for next 6 days for 12 hrs = 3*6*12*4 = 864


hence total is = 864 +864 = 1728 lt

Ans is D .

I too dont like time and work ...so i try to solve in a logical manner wherever possible . :-)
_________________

In the next 6 days, the same 4 cars will consume= 4/4*1200*12/10*6/10=864 litres
And, in the next 6 days, 5 cars consuming 20% less fuel will consume=1200*5/4*12/10*6/10*80/100=864 litres
Therefore, total fuel consumed=1728 litres
Answer D

Solution: Let the amount of fuel consumed by company car and extra car be x and y lts/car/hour.
y = 0.8x

Amount of fuel consumed by 4 company cars running for 10 days for 10 hours = 4*10*10*x = 1200 ==> x= 3 and thus y = 2.4.
The no. of hours traveled by each car for next 6 days = 6*12 = 72.

Amount of fuel consumed by 4 company cars and 5 extra cars in next 6 days = 4*72*3 + 5*72*2.4 = 72(24) = 1728 lts.

Option D.

4[cars] ---- 10[hrs] ----1200[lts]----10[days]
split the 9 cars into 4 and 5 cars and calculate the consumption
4[cars] ---12[hrs] ---- x [lts] ---6[days]

x= 1200*(4/4)*(12/10)*(6/10) = 864[lts] for 4 cars

for 5 cars
4[cars] ---- 10[hrs] ----1200[lts]----10[days]
5[cars] ---- 12[hrs] ---- y[lts]----6[days]

y = 1080[lts]
as 5 cars consume 20% less fuel, fule consumption of 5 cars = 1080*0.8 = 864[lts]

total = 864 + 864 = 1728[lts]

let x be the fuel consumed by one car in one hour
4 cars run for 10 days at 10 hrs a day. So they will run 100 hrs in total.
So, fuel consumed per hr by each car will be
100 * 4x = 1200
or 4x=12
or x= 3

now 5 extra cars are hired which consume 20 % less fuel. let the consumption of fuel per hour of one new car be y
then
y=\(\frac{4}{5}\) * x
or y=\(\frac{4}{5}\)*3
Consumption of 5 new cars will be
\(\frac{4}{5}\) * 3 *5 = 12

So previous 4 cars consume same fuel as 5 new cars per hour i.e 12 lts.
Now these 9 cars run for 6 days at 12hr/day. So they run for total 72 hours.
thus total fuel consumption will be
12*72+12*72 = 1728

Answer:-D

Hi GMATinsight,

Can you elaborate more about the rule you use above? What is m/c power? when to use this rule? How is it relate to the usual rule work= rate * time? Is the m/c power is the same as rate?

looking forward to your great explanation

Thanks

The basic Understanding of the Rule is here

1) MORE work requires MORE man Power/Machine Power AND LESS work requires LESS man Power/Machine Power
i.e. Manpower/Machine Power is DIRECTLY proportional to the amount of work done (to be done)

Let, M is Man Power/Machine Power
W is the amount of work done (to be done)
then,
\(M = K_1*W\)
where, \(K_1\) is a constant


2) MORE Time requires LESS man Power/Machine Power AND LESS time requires MORE man Power/Machine Power
i.e. Manpower/Machine Power is INVERSELY proportional to the Time in which work is to be done

Let, M is Man Power/Machine Power
T is the Time in which the work is to be done
then,
\(M = \frac{K_2}{T}\)
where, \(K_2\) is a constant

Combining the two relationships gives us

\(M = K_1*K_2*\frac{W}{T}\)

i.e. \(\frac{M*T}{W} = Constant\)

i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

VERITAS PREP OFFICIAL SOLUTION:

First let’s try to figure out what is meant by ‘consume 20% less fuel than the company’s cars’. It means that if company’s each car consumes 1 lt per hour, the hired cars consume only 4/5 lt per hour. So renting 5 more cars is equivalent to renting 4 cars which are same as the company’s cars. Hence, the total number of cars that will be run for the next 6 days is 8 company-equivalent cars.

4 cars running 10 hrs for 10 days consume 1200 lt of fuel

8 cars running 12 hrs for 6 days consume x lt of fuel

\(x = 1200*(\frac{8}{4)}*(\frac{12}{10})*(\frac{6}{10}) = 1728\) lt

We multiply by 8/4 because more cars implies more fuel so we multiply by a number greater than 1.

We multiply by 12/10 because more hours implies more fuel so we multiply by a number greater than 1.

We multiply by 6/10 because fewer days implies less fuel so we multiply by a number smaller than 1.

Answer (D)
_________________

Lets first calculate the fuel consumed by each of the four cars per hr.

Condition 1
1. Total fuel consumed 1200ltrs in 10 days
2. Total cars=4. fuel consumed by each car= 1200/4= 300ltrs for total of 10 days.
3. Fuel consumed by each car per day= total fuel consumed in total no of days/No of days= 300/10=30ltrs
4. Fuel consumed by each car per hr=fuel consumed per day/No of of Hrs the car is run=30/10=3ltrs per hr

Condition 2
1. Additional cars=5
2. From point 4 the fuel consumed by each car for 6 days running for 12 hrs per day will be No of cars of days X no of hrs X fuel consumed per day=5x6x12x3=1080 Ltrs
3. The new cars are 20% fuel efficient. The fuel we have calculated in step 2 above is based on fuel efficiency of earlier cars. So new cars will consume 20% less fuel which means
20% of 1080 lts=216. Subtract this from 1080 ltrs=864 ltrs
4. Total fuel consumed= fuel consumed by earlier 4 cars+ fuel consumed by new additional 5 cars running for 12 hrs daily for 6 days.
= 4(No of cars)X6(No of days)x12(no of hrs)x3(fuel consumed per hr)+ 864(calculated above point 3)= 864+864=1728ltrs

We are given that the four cars consume 1200 L of fuel working 10 hours a day for 10 days, or, in total, 100 hours; therefore, 1200/100 = 12 L of fuel was consumed by four cars per hour. Thus, each of the existing cars consume 12/4 = 3 L of fuel per hour.

In the next 6 days, since the existing cars will be working for a total of 6 x 12 = 72 hours, they will consume 72 x 3 x 4 = 864 L of fuel.

Since each of the 5 newly bought cars consumes 20% less fuel than the existing cars do, each of the newly bought cars consumes (0.8)(3) = 2.4 L of fuel per hour. In the next 6 days, working for a total of 72 hours, the newly bought cars will consume another 72 x 2.4 x 5 = 864 L of fuel.

In total, 864 + 864 = 1728 L of fuel will be consumed in the next 6 days.

Answer: D
_________________

Consider giving Kudos if the solution below helps :D

Posted from my mobile device

Hi All,

We're told that a company’s 4 cars running 10 hrs a day consume 1200 liters of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. We're asked for the number of liters of fuel that will be consumed in the next 6 days. This question is a variation on a "Work" question and comes down to taking the proper notes and doing the necessary Arithmetic.

To start, we need to determine the amount of fuel that 1 car will use in 1 hour.

We know that 4 cars will run 10 hours a day for 10 days. That's (4)(10)(10) = 400 hours of car-running time
During that time, 1200 liters of fuel will be spent. Thus, we know that 1200/400 = 3 liters of fuel will be used by 1 car in 1 hour.

We're told that 5 additional cars will be rented, but those additional 5 cars consume 20% LESS fuel than the company's 4 cars. Each car will run 12 hours a day for the next 6 days. We can break the calculation down into 2 pieces:

(4 cars)(3 liters/hour)(12 hours/day)(6 days) = 864 liters of fuel
(5 cars)(2.4 liters/hour)(12 hours/day)(6 days) = 864 liters of fuel
Total = 864 + 864 = 1728 liters of fuel

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________