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Please can some one help me to solve these problems

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Please can some one help me to solve these problems

by asamanta » Thu Jun 12, 2008 11:09 am
P1

The ratio of Income of A:B is 4:7 and the ratio of thier savings are 6:11. If A saves 1/3 of his income then how much does B save ?

Answer is 19/33 but I cant get there ? Does anyone know how to solve this ?

P2


A vessel contains 5 parts of milk and 1 part water. how much of the mixture is to be removed and replaced by water to make the resulting mixture 1/2 water and 1/2 milk ?

Answer is 2 2/5 i think...can some one please solve both these problems
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Re: Please can some one help me to solve these problems

by durgesh79 » Thu Jun 12, 2008 8:40 pm
asamanta wrote:P1

The ratio of Income of A:B is 4:7 and the ratio of thier savings are 6:11. If A saves 1/3 of his income then how much does B save ?

Answer is 19/33 but I cant get there ? Does anyone know how to solve this ?

P2


A vessel contains 5 parts of milk and 1 part water. how much of the mixture is to be removed and replaced by water to make the resulting mixture 1/2 water and 1/2 milk ?

Answer is 2 2/5 i think...can some one please solve both these problems
P1
lets assume the income of A is 4x and B is 7x
and savings of A is 6y and B is 11y

now as per question 6y = 1/3 (4x)
y/x = 4/18

we have to find out "how much B save ?" not possible.
however it is possible to find out "how much B saves as a ratio of his income ?"

11y / 7x = 22/63 ... not matching with the answer ---> something wrong with the question. Please recheck.

P2.

currently we have.

5 parts milk : 1 part water, so total we have 6 units in the vessel

finally we'll have equal qty of milk and water, --> 3 parts milk and 3 part water

so for milk we have to go from 5 to 3, means remove 2 units of milk.
for removing 2 units of milk we'll have to remove 2/5 units of water also, becuase water and milk are in a mixture, we cant actully remove only pure milk.

So mixture to be removed, 2 + 2/5. Answer
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by asamanta » Thu Jun 12, 2008 10:31 pm
Yes you are right on first one but i think the second one needs a better explanation. Not sure...
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by Ian Stewart » Fri Jun 13, 2008 3:46 am
The explanation durgesh gives for Qn 2 is correct- I like his solution, because it's faster than doing the question algebraically. But we can do the question algebraically, in any case. We have 5 parts milk, 1 part water. Say we will remove T parts of mixture, then add T parts of water. When we remove T parts of mixture, (5/6)*T is milk, and (1/6)*T is water, because the mixture is in a 5:1 ratio. We then add back T parts of pure water. Thus,

____OLD --> NEW
milk: 5 --> 5 - 5T/6
water: 1 --> 1 - T/6 + T

We know that the milk and water are equal in the new mixture:

5 - 5T/6 = 1 - T/6 + T
4 = 10T/6
24 = 10T
2.4 = T
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by nadib002 » Thu Apr 28, 2011 4:15 pm
Hello Ian,

For the 2nd question ( mixture of milk and water), I used the alligation method. I took the ratio of water and solved for it. I got 2/5 ( the amount of water that needs to be removed). But i do-not seem to understand why we add 2 for the milk.

Could you please help
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