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Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink] New post 19 Apr 2013, 12:08
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

You guys are right I edited the answer, thanks!
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Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink] New post 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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Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink] New post 19 Apr 2013, 12:45
Expert's post
RyanCJW wrote:
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.


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%.

Thanks!
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Re: There are two buses [#permalink] New post 19 Apr 2013, 12:53
Expert's post
Zarrolou wrote:
F=first bus
S=second bud

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.

Hope this helps!
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Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink] New post 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 man sold a strawberry juice can and pineapple juice can fo [#permalink] New post 19 Apr 2013, 12:55
Nice one!

S+n%S is the transaltion of sold for n% profit and S is the original cost

(S+0.1S)+(P+0.25P)=760
(S+0.1S)+(P+0.25P)=767.5

Solve those and find S=300 P=350
E
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Re: .A train covered a certain distance at uniform speed [#permalink] New post 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.

D=ST -----------> 1

Situation 2

D = (S+6)(T-4) -----------> 2

Situation 3

D=(S-6)(T+6) -----------> 3

Solving 1 & 2
6T-4S-24 = 0

Solving 1 & 3
6S-6T-36=0

solve last 2 equations to get C 720.
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Re: Albert invested a some amount at rate of 3% per quaterly sim [#permalink] New post 19 Apr 2013, 14:20
Few confusions in the question.

3% per quaterly simple interest

does this mean, Simple Interest @ 3% per annum compounded quarterly? or 3% per quarter ?

same goes for the other interest amount
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Re: After covering a distance of 30 miles with a uniform speed a [#permalink] New post 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

ANSWER D
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Re: After covering a distance of 30 miles with a uniform speed a [#permalink] New post 19 Apr 2013, 15:01
Assuming, the train would have reached on time without defect. How is the train reaching 9 mins earlier with reduced speed? :?

Assuming, the train was not reaching on time anyways.
Here is my solution. :roll:

0 - 30 miles: speed of train is same (in both scenarios) & hence time taken to cover is same

30 - 48 miles: scenario 1, train is running at reduced speed & hence more time. Scenario 2, train is running at original speed.

So, there is time difference in covering this distance.

beyond 48 miles: trains are running at same speed. So, no time difference in covering any distance.

Conclusion: 54 mins difference is coming only during 18 mile stretch.

As per my understanding, answer could be any distance greater than 48 miles. :lol:
I will wait for others to comment.

You can calculate the original speed of the train. :P

(18/x) - (18/(4x/5)) = 54/60

you can solve for x. which is the original speed of the train.
:P
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Re: After covering a distance of 30 miles with a uniform speed a [#permalink] New post 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 ???? :-D :-D :-D
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Re: A child on a controlled diet is fed 300 ounce of a mixture [#permalink] New post 19 Apr 2013, 16:30
Just a different approach with a shortcut method.


10% 15% Participating percentages of the component


38/3 (38/300)x 100 Resulting percentage of the component


15-(38/3) 10-38/3 Ration of a and b.


Thus the mixtures are used in the ratio 7:8

which will give 7/15 x 300 of a which is 140
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Re: There are two buses [#permalink] New post 19 Apr 2013, 16:33
First Equation:-
F-10 = S+10


Second Equation
S+40 = 2S-40

which gives S=80 and F=100
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Re: the charges for a 5 day trip by tourisim bus for one full an [#permalink] New post 19 Apr 2013, 16:45
F - Fare Charges
B- Boarding Charges

(1+3/4)F + 2B = 1440
(2+3/4)F + 3B = 2220

Solving the two equations we will get F as 480 and B as 300
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Re: the charges for a 5 day trip by tourisim bus for one full an [#permalink] New post 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
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Re: Albert invested a some amount at rate of 3% per quaterly sim [#permalink] New post 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

B = 4400

Not sure of the Ans choices.
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Re: Albert invested a some amount at rate of 3% per quaterly sim   [#permalink] 19 Apr 2013, 19:46
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