Mixture & Percentage problem

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GMAT12: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Apr 25, 2009 5:05 am; Thanked: 1 times

Mixture & Percentage problem

by GMAT12 » Mon May 25, 2009 7:00 am

1) A glass of wine contains three parts of wine and one part of water. How much of the mixture must be withdrawn and water substituted so that the resulting mixture will be half wine and half water?

a. 1/2
b. 1/3
c. 1/4
d. 1/5
e.None

2) A business man prints two types of prices on his goods,one for cash payments and other for those who pay after one month. If the rate of interest is 4 percent per annum, what is the ratio of the two prices?

a. 200:291
b.392.394
c.3000:3010
d.4000:4567
e.None

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Source: Beat The GMAT — Problem Solving | Mon May 25, 2009 7:00 am

avanishjoshi: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Fri Jun 20, 2008 2:06 am; Thanked: 2 times

by avanishjoshi » Mon May 25, 2009 7:18 am

I have solved the first one using the alligation.I donno how to solve the second one.

Wine solution contains 3 wine and 1 water.
We are adding water to make 1/2 wine.

3(parts of wine in first solutn) 0(parts of wine in 2nd soln ie water)

1/2

0- 1/2 1/2-3

0 minus 1/2 : 1/2 -3 gives the ratio of the weights

0-1/2 : 1/2: 3 = [spoiler]1:5[/spoiler] - which is the answer.

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GMAT12: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Apr 25, 2009 5:05 am; Thanked: 1 times

by GMAT12 » Mon May 25, 2009 8:45 am

i didn't understand the explanation

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mikeCoolBoy: Master | Next Rank: 500 Posts; Posts: 119; Joined: Tue Jan 20, 2009 10:16 pm; Thanked: 9 times; GMAT Score:730

by mikeCoolBoy » Mon May 25, 2009 9:26 am

For the first problem I get B 1/3

we're told that we have a mixture that is 3/4 alcohol and 1/4 water then let's assume that we have 12 liters of mixture and therefore we have 9 liters of alcohol and 3 liters of water and we want to end up with 6 liters of alcohol and 6 liters of water.

This means that we need to remove 3 liters of alcohol. Every time you remove a liter from the mixture you're removing 3/4 liters of alcohol so
if X are the liters that we need to remove from the mixture.

3 = 3/4X then X = 4 which is 1/3 of 12.

if you want to set the equations, it would be something like this. We have to equal the fraction of alcohol to the fraction of water.

we initially have 3/4 of alcohol and we remove 3/4 of the proportion X
so the final amount of alcohol would be 3/4 - 3/4X.

we initially have 1/4 of water and we add X but at the same time we remove 1/4X so if fact we add 3/4X

setting the equations

3/4 - 3/4X = 1/4 + 3/4X --> 2/4 = 6/4X ---> X = 1/3

For the second problem I get C.

Let's say that the product costs $100.
The price for the people who pay in cash is $100
and the price for the people who pay in one month is $100 + 100*0.04/12 = 100 + 1/3 = 301/3

the ratio then is 100 / 301/3 --> 300/301 ---> 3000/3010.