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# A company’s four cars running 10 hrs a day consume 1200 lts of fuel in

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A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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09 Sep 2015, 02:13
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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

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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09 Sep 2015, 04:19
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Bunuel wrote:
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

Kudos for a correct solution.

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

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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09 Sep 2015, 05:18
1
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Bunuel wrote:
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

Kudos for a correct solution.

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.

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A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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09 Sep 2015, 11:06
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Bunuel wrote:
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

Kudos for a correct solution.

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

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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09 Sep 2015, 14:45
1
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Bunuel wrote:
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

Kudos for a correct solution.

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]

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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10 Sep 2015, 09:59
1
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Bunuel wrote:
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

Kudos for a correct solution.

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

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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10 Sep 2015, 20:39
GMATinsight wrote:
Bunuel wrote:
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

Kudos for a correct solution.

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

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

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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12 Sep 2015, 00:09
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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 . :-)
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A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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12 Sep 2015, 06:43
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Mo2men wrote:
GMATinsight wrote:
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

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

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

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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13 Sep 2015, 08:39
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Bunuel wrote:
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

Kudos for a correct solution.

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.

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A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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23 Mar 2016, 21:54
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
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

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A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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24 Mar 2016, 00:12
(4 c)(10 d)(10 h)=400 car hours
1200 liters/400 car hours=3 liters per car per hour
(4c)(6d)(12h)(3 lch)=864 liters
(5 c)(6 d)(12 h)(3 lch)(80%)=864 liters
(864)(2)=1728 liters

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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23 Apr 2017, 15:03
4 cars -> 1 hour for consumption will cost ( 1200 : 10 ) : 10 = 12 fuel
-> 12 hour / day for 6 day will cost 12 * 12 * 6 = 864
=> 5 cars (864 / 4 )* 5
=> 5 cars with less 20% fuel consumption 864 * (5/4) * (4/5) = 864

Total = 864 + 864

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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05 May 2017, 10:15
4 cars....running 10 hrs per day...for 10 days....total lit consumed=1200
1 car consumed in 1 hr each day=1200/(4*10*10)=3 lit
rented car consumed in 1 hr each day=3*80%=2.4 lit
9 car...running 12 hrs per day...for 6 days...lit consume=[4(owned)*3*12*6]+[5(rented)*2.4*12*6]=1728 lit

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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14 Aug 2017, 10:03
my solution
X - fuel consumption rate
4*x*10*10 = 1200 solve for x = 3 per mile
In next 6 days for those 4 cars
4*3*6*12=12*72=864
So for the new 5 cars fuel consumption rate is 20% of 3, which makes 12/5
Then 5*12/5*6*12=12*72=864
Total 864 + 864 = 1728

Actually when you get total fuel consumption for the first group of cars there is no need to calculate for the other one, as fuel consumption rate decreases for 20% but the number of car increases by the same 20% from 4 to 5. Just double 864

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in [#permalink]

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18 Aug 2017, 08:57
Bunuel wrote:
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

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.

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Re: A company’s four cars running 10 hrs a day consume 1200 lts of fuel in   [#permalink] 18 Aug 2017, 08:57
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