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by rupsk » Fri Sep 02, 2011 8:25 am
A certain amount of solution contains 18% alcohol. 8 litres of the solution is taken out and replaced with water. The resultant solution contains 15% alcohol. Find the volume(in litres) of the solution

Possible Answers
Selected Possible Answer

A. 40


B. 50


C. 48


D. 44


E. None of these

I had taken below procedure but my answer is no where near to the solution.
let x be the solution then current alcohol is 18%x
so 18%-15% of x is 8lit
3% of x = 8
solve it and get x=800/3
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by knight247 » Fri Sep 02, 2011 8:38 am
Let x be the original quantity of the solution

Since it is 18% alcohol, the alcohol content in it will be:
18x/100

Now, out of the original x litres, 8 litres is taken out. So what remains is (x-8) litres
Now, the alcohol % of x-8 litres is stil 18%. So the actual alcohol content will be

18(x-8)/100....(1)

Now 8 litre of only water are added back to the solution so x-8+8= x which is what remains. This makes the solution 15% alcohol.

So, the alcohol content is 15x/100....(2)

However, the content of alcohol has still remained the same as before the 8 litres of water were added. It is only the proportion of the alcohol in the entire solution that has changed.

Equating 1 and 2 we get 18(x-8)/100=15x/100
18(x-8)=15x
18x-18*8=15x
3x=18*8
x=6*8=48 Hence [spoiler]D[/spoiler]
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by rupsk » Fri Sep 02, 2011 8:41 am
thanks it looks like i m missing on simple problems. need some brain wash
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by rupsk » Fri Sep 02, 2011 9:58 am
one more

A shopkeeper increases the price of an article by X% and then decreases it by X%. As a result the price of the article is reduced by $180. After one more such change the price is further reduced by $153 Find the original price of the article.

Possible Answers
Selected Possible Answer

A. $1200


B. $1100


C. $1400


D. $1000


E. $1500
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by rupsk » Fri Sep 02, 2011 10:05 am
one more

A milk vendor sells milk at Cost Price but still gains 20%. Find the ratio of milk and water in every gallon that he sells.

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Selected Possible Answer

A. 4:1


B. 5:1


C. 5:2


D. 6:1


E. None of these
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by knight247 » Fri Sep 02, 2011 10:51 am
This problem tends to become a little complicated when using variables. So I'd prefer solving such a problem by plugging in numbers

Let the vendor originally have purchased 100Gallons of milk for $100 i.e. at the rate of $1/Gallon

His total CP=$100 and he makes a 20% profit on his CP. So his SP=CP+20(CP)/100=100+20(100)/100=$120

So, his CP=$100 and his SP=$120

But, out of this $120, $100 is milk so the remaining $20 has to be the contribution of the water. At the rate of $1/Gallon there are 20 Gallons of water.

Ratio of milk to water is 100:20 i.e. 5:1 Hence B
Last edited by knight247 on Fri Sep 02, 2011 11:03 am, edited 2 times in total.
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by rupsk » Fri Sep 02, 2011 10:54 am
thax knight i tried with variable but i got lost.
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by hdk2104 » Thu May 17, 2012 3:06 am
rupsk wrote:one more

A shopkeeper increases the price of an article by X% and then decreases it by X%. As a result the price of the article is reduced by $180. After one more such change the price is further reduced by $153 Find the original price of the article.

Possible Answers
Selected Possible Answer

A. $1200


B. $1100


C. $1400


D. $1000


E. $1500
If the price is increased by x% and then decreased by x% there would be a decrease in the price and let the decrease by y%.

Let P be the original price of the article.

Given that y%(P) = 180..........(1)

Now the current price is P - 180 and this price is further reduced by y%

Given that y%(P - 180) = 153......(2)

From (1) and (2), we get y = 15

Thus we have 15%(P) = 180

Hence P = 1200.
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