Problem Solving

Folks, here is the next one...

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

I think one can use options but that's taking lot of time for me......I hope someone here will come with an explanation....sureshbala, any clue to solve this..

sureshbala wrote:Folks, here is the next one...

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

Answer is (B).

According to the question, i have derived the below equation: (Assume the capacity of the vessel is x gallons)

(x-8-(x-8)/x*8)/(8-8/x*8+8)=9/40

After the above equation is simplified, it becomes: 40x^2-784*x+3136=0

x=14 or 5.6

As x must be great than 8, 5.6 is discarded. So x=14.

B
i got 14 and 5.6.

remember doing similar sort of puzzle long time ago....forgot but i guess there is some short way of doing this.
bala will definitely be knowing that if its possible.plz share.

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

To solve problems of these kind, we have a formula,

{(P-Q)/P}^n = resultant ratio, Where 'P' is the original liquid and Q is the amount taken over and replaced....'n' is the number of times the action performed.

therefore, in this problem, P is unknown, Q is 8 gallons, and the resultant ratio is 9/(40+9), that is 9/49, and n is 2

{p-8/p}^2 = 9/49, solve for P and you get P = 14.

{(P-8)/P}^2

sureshbala wrote:Folks, here is the next one...

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

Folks, the simple concept here is "The percentage (or the fraction) of the water that you loose will always be constant". I mean to say that in the first transaction if you have lost x% of the available water, in the second transaction also you will loose the same x% of the available water .

Let us look at an example.

Let's say we have 100 gallons of water initially and each time we are replacing 20 gallons of water with milk. (i.e we are replacing 20%)

After the first transaction, the amount of water will be 80 and milk will be 20.

Now from this solution if we take out 20 gallons, these 20 gallons will contain water and milk in the ratio 80:20 i.e 4:1. So the amount of water lost = 4/5(20) = 16 litres, which is 20% of the water available after the first transaction.

So try to remember this concept and I am sure the above question can be answered quickly.

Let's say that the fraction of the water that we are left with after the first transaction = x/y.

So given that x/y * x/y = 9/49

Hence x/y = 3/7.

So we are being left with 3/7 of the available water after every transaction, which means that we have taken out 4/7 of the water initially. This is given as 8 litres.

Hence 4/7 (v) = 8 litres. So V = 14 litres

sureshbala wrote:

sureshbala wrote:Folks, here is the next one...

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

Folks, the simple concept here is "The percentage (or the fraction) of the water that you loose will always be constant". I mean to say that in the first transaction if you have lost x% of the available water, in the second transaction also you will loose the same x% of the available water .

Let us look at an example.

Let's say we have 100 gallons of water initially and each time we are replacing 20 gallons of water with milk. (i.e we are replacing 20%)

After the first transaction, the amount of water will be 80 and milk will be 20.

Now from this solution if we take out 20 gallons, these 20 gallons will contain water and milk in the ratio 80:20 i.e 4:1. So the amount of water lost = 4/5(20) = 16 litres, which is 20% of the water available after the first transaction.

So try to remember this concept and I am sure the above question can be answered quickly.

Let's say that the fraction of the water that we are left with after the first transaction = x/y.

So given that x/y * x/y = 9/49

Hence x/y = 3/7.

So we are being left with 3/7 of the available water after every transaction, which means that we have taken out 4/7 of the water initially. This is given as 8 litres.

Hence 4/7 (v) = 8 litres. So V = 14 litres

yes..now this is clear to me and i can imagine how a formula is derived for the same...thanq sureshbala

Here is the next one....

Find the total number of 4 digit numbers which contain 36 in them and are divisible by 36.

A. 16
B. 15
C. 14
D. 13
E. None of these

sureshbala wrote:Here is the next one....

Find the total number of 4 digit numbers which contain 36 in them and are divisible by 36.

A. 16
B. 15
C. 14
D. 13
E. None of these

14.
Therefore C.
The Possibilities :

1. ab36

2. a36b

3. 36ab


36 = 4*9

(a,b) could be

(2,7)
(4,5)
(6,3)
(8,1)
(0,9)

using these values we get the following number of values for

1. ab36 = 9 (0936 = three digit number. Rest are four digit)

2. a36b = 3 (6b should be divisible by 4, therefore 0,4 and 8 satisfy )

3. 36ab = 2 ( since 3636 is already covered in 1 therefore only two new values 3600 and 3672)

For a number to be divisible by 36 it should be divisible by both 4 and 9
A number is said to be divisible by 9 if the sumof digits is mulitple of 9
and a number is said to be divisible by 4 if last two digits are divisible by 4

lets try out different cases

Case 1: the number is ab36
for any value of ab the number is divisible by 4 and for it to be divisibl by 9 a+b should be 9, (9,0)(8,1)(7,2)(6,3)(5,4)(4,5)(3,6)(2,7)(1,8) total 9 numbers

Case2 : the number is a36b
for the number to be divisible by 4 b can take values 0, 4,8. a can take values such that a+b = 9 (9,0)(5,4)(1,8) total 3 numbers

case 3 : the number is 36ab
the numbers are (9,0)(7,2)(0,0) total 3 numbers

total number of numbers from all the cases 9+3+3 = 15

The Original answer could be B

NareshCS wrote:For a number to be divisible by 36 it should be divisible by both 4 and 9
A number is said to be divisible by 9 if the sumof digits is mulitple of 9
and a number is said to be divisible by 4 if last two digits are divisible by 4

lets try out different cases

Case 1: the number is ab36
for any value of ab the number is divisible by 4 and for it to be divisibl by 9 a+b should be 9, (9,0)(8,1)(7,2)(6,3)(5,4)(4,5)(3,6)(2,7)(1,8) total 9 numbers

Case2 : the number is a36b
for the number to be divisible by 4 b can take values 0, 4,8. a can take values such that a+b = 9 (9,0)(5,4)(1,8) total 3 numbers
Thanks!


case 3 : the number is 36ab
the numbers are (9,0)(7,2)(0,0) total 3 numbers

total number of numbers from all the cases 9+3+3 = 15

The Original answer could be B

neither 3690 nor 3609 satisfies the condition!

sureshbala wrote:Folks, here is the next apple.....



In the above figure, PB/PC = 1/2 LCBA = 45° while LAPC = 60°. Find LACB.

Lets consider <A = 3x
take triangle ABC
so,
3x + 45 + <C = 180
thus,
3x +<C =135.........(1)

Now take triangle, ACP
Since the line AP divides the side BC in the ratio 2:1,so the line AP also divides <CAB in the ratio of 2:1 with <CAP being 2x and the other being x

so we have another equation as

2x + <C +60 = 180
2X +<C =120............(2)

Solving (1) & (2)

We get x=15
3x= <CAB =45
Thus <A +<B+<C =180
<A=<B=45
SO <CAB =90

Please check out the mistake

sureshbala wrote:Here is the next one....

Find the total number of 4 digit numbers which contain 36 in them and are divisible by 36.

A. 16
B. 15
C. 14
D. 13
E. None of these

A number is divisible by 36, if it is divisible by both 4 and 9.

Divisibility Rule of 4: The number formed by the last digits of the number must be divisible by 4.

Dvisibililty Rule of 9: Sum of the digits of the number must be divisible by 9.

Case 1: Number is of the form 36xy.

It is clear that if the number xy is divisible by 36, then the number 36xy will be divisible by 36.

so xy could be 00, 36 and 72.

Hence 3 numbers are possible in this case

Case 2: Number is of the form x36y.

Now if we take care of the value of y, such that 6y is divisible by 4, we can place the value for x such that the number is dvisible by 9 as well.

So y can take 0, 4 and 8 for which x will take 9, 5 and 1 respectively.

Hence 3 numbers in this case as well

Case 3: Number is of the form xy36.

Since this number is divisible by 4, all we have to see is that this number is divisible by 9. Also, since 3+6 =9, the sum of x and y must be such that it is divisible by 9.

So for x+ y = 9, we have (9,0) (8,1)........(1,8). i.e. a total of 9 cases

Also x+y = 18, we have (9,9)

Hence there are 10 numbers in this case.

So totally 3+3+10 = 16 possibilities. But of this number 3636 is counted in the first case as well as the last case.

Hence total number of possibilities = 15.

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

This question is typically a CAT(Common Admission Test) Problem.

Although I am yet to appear for GMAT, but I have done quiet a lot of Prob Solving as well as Quant(as it is commonly refered in CAT preparation in India)

And I know the similarity and differences between the GMAT and CAT.

To my knowledge this question is irrelevant from point of view of GMAT.

Experts please help to clarify.

gmat740 wrote:

From a vessel containing pure water 8 gallons of water is taken out and replaced by pure milk. Again 8 gallons of the resultant mixture is taken out and was substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity(in gallons) of the vessel.

A. 20
B. 14
C. 33
D. 21
E. 30

This question is typically a CAT(Common Admission Test) Problem.

Although I am yet to appear for GMAT, but I have done quiet a lot of Prob Solving as well as Quant(as it is commonly refered in CAT preparation in India)

And I know the similarity and differences between the GMAT and CAT.

To my knowledge this question is irrelevant from point of view of GMAT.

Experts please help to clarify.

Dear gmat740,

I think this is not much different from GMAT standard. One of my friend who took GMAT recently told me that he found some hard questions from mixtures. I think there is nothing wrong to learn the next level. But again as you said it depends upon one's own requriments. Anyway, i think there are one or two questions in this thread which are definitely tough and may not match with GMAT standard but I think we need not worry much about that and solve the questions in which we are interested in.

I hope Sureshbala will definitely consider this and bring in the neccessary
changes to make this thread more favorable to everyone on BTG