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Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink]

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19 Apr 2013, 12:38

Zarrolou wrote:

\(A+B=300\) Total food \(0.1A+0.15B=38\) Only proteins

\(B=300-A\) \(0.1A+0.15(300-A)=38\) solve this

\(A=140\) B

Isn't the answer B here? Sorry if I missed something but that's the answer I got and it looks to be the answer you got, Zarrolou - but the buttons at top reflect C being correct.
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\(A+B=300\) Total food \(0.1A+0.15B=38\) Only proteins

\(B=300-A\) \(0.1A+0.15(300-A)=38\) solve this

\(A=140\) B

Isn't the answer B here? Sorry if I missed something but that's the answer I got and it looks to be the answer you got, Zarrolou - but the buttons at top reflect C being correct.

Totally B, if it's not reflecting that in the buttons then they will surely be adjusted.

Very easy to double check, which is always a good idea before selecting "next". 140x10% = 14 oz from food a. That leaves 38-14 = 24 oz from food b in 160 oz of food, exactly 15%.

\(F-10=S+10\) \(2*(S-20)=F+20\) \(2S-40=F+20\) \(2S-F=60\) do some annoying math

\(F=100, S=80\)

A

Based on the fact that you can move passangers and get a 2/1 split, this means the total number of passengers must be divisible by 3. You can also eyeball the numbers and realize 300 or 600 will be too big. This means it should be 180. Plug it in logically and you must have 80 and 100 to satisfy equation 1, and the same split works for equation 2.

Algebra always works but you can always use concepts if algebra isn't your strong suit.

Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink]

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19 Apr 2013, 12:54

abhishek1990p wrote:

A child on a controlled diet is fed 300 ounce of a mixture of two foods a and b food a contains 10% protein and food b contains 15% protein if child diet contains exactly 38 ounce of protein daily. How many ounce of food a are in the mixture?

a.100 b.140 c.150 d.160 e.200

Oh, just to show how I got it (a little different from above)

A is 10% protein and B is 15%.

I just ran through the answers quickly.

A - If 100 oz of food A, thats 10 oz of protein. Which leaves 200 oz of food B and 30 oz protein, too much. B - If 140 oz of food A, thats 14 oz of protein. 160 oz of food B is 16+8 (10%+5%) and we're at 38!

It was relatively easy to run through these so you could probably start from anywhere and work your way around but I solved this in less than a minute this way.
_________________

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Re: .A train covered a certain distance at uniform speed [#permalink]

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19 Apr 2013, 13:50

abhishek1990p wrote:

11.A train covered a certain distance at uniform speed if train would have been 6miles/hour faster it would have taken 4 hrs less than the scheduled time and if the train were slower by 6 miles/hour it would have taken 6 hrs more than the scheduled time find the distance in miles of journey a.840 b.780 c.720 d.680 e.630

C

3 situations are give. Distance is same in all 3.

Situation 1 Let 'S', 'T' & 'D' are the standard Speed, time & distance of the train.

Re: After covering a distance of 30 miles with a uniform speed a [#permalink]

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19 Apr 2013, 14:51

I can bet this is not a real GMAT question, as it is absolutely impossible to solve this in less than 3 or 4 minutes.

The solution:

Speed x Time = Distance

We have two equations here to find the total distace L:

1) \(\frac{1}{5}x*\frac{3}{4}=d_1\) , where \(d_1\) is the distance required to finish the trip once the defect appears (arriving 45 minutes late condition).

2) \(\frac{1}{5}x*\frac{36}{60}=d_2\) , where \(d_2\) is the distance required to finish the trip once the defect appears in the second condition or hint (arriving 45-9=36 minutes late condition).

Therefore, we will have two equations:

1) \(30+\frac{3}{20}x=L\)

1) \(48+\frac{6}{50}x=L\)

Solving for x, we have that x=600, and finally L=120 miles

Re: After covering a distance of 30 miles with a uniform speed a [#permalink]

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19 Apr 2013, 16:06

Let x be the distance covered in the second part of the journey.

From the first case we will get (x/v) + 3/4 = x/(4v/5) where v is the original speed of the train. 4v/5 is the reduced speed. We can get x = 3v from this equation.

From the second case we will get (x-18)/v + (45 min - 9 min) = 5(x-18)/4v .Put x = 3v You will get x value as 90 miles. SO the total distance is 120 miles.

This is a very lengthy question.Does anyone have a short version of the solution ????
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Re: the charges for a 5 day trip by tourisim bus for one full an [#permalink]

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19 Apr 2013, 19:36

abhishek1990p wrote:

6.the charges for a 5 day trip by tourisim bus for one full and half ticket is 1440 iccluding of boarding charges which are same for a full as well as a half ticket the charges for the same trip for 2 full and one half ticket the charges are 2220the fare for half ticket is 75% of the full ticket find the difference betwn the fare and boarding charges for one full ticket a.480 b.360 c.300 d.240 e.180

Ok. So variables are Fare and Boarding. Charges for full is 4x and half is 3x. Now,

(4x+3x) + 2 Boarding = 1440 7x+2B= 1440 Similary, (4x*2+3x) + 3B = 2220 11x+3B = 2220 Solve for X and B, we need to find 4X - B = 180
_________________

Re: Albert invested a some amount at rate of 3% per quaterly sim [#permalink]

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19 Apr 2013, 19:46

abhishek1990p wrote:

Albert invested a some amount at rate of 3% per quaterly simple interest and other amount at the rate of 7.5% half yearly simple interest. he received 1500$ as interest for a year. if he had interchanged the amount invested he would have received 78$ more as interest for the year. Find the smallest investment

a.7062.5$ b.5762.5$ c.7562.5$ d.5002.0$ e.7002.0$

Ok. It is simple interest. Quarterly 3% is equal to 12 % yearly. Similarly, 7.5% is 15% yearly. Now, let amounts be A and B. Then. 12% A + 15% B = 1500 12% B + 15% A = 1578

Solving for A and B. (A - B)/100 = 78/3 = 26

Now put A = 2600 + B in eq 1 12/100(2600+B) + 15/100 (B) = 1500

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