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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?

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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?

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

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09 Sep 2015, 05:18

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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?

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.

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?

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
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Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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09 Sep 2015, 14:45

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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?

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

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10 Sep 2015, 09:59

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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?

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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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?

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

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?

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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?

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 . :-)
_________________

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?

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

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

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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?

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

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 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

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 answer: D

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

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?

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
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