This thread is dedicated to finding the best Method/Approach to a question. Almost everyone can get almost every question on the GMAT right, given enough time to solve it. What makes this test difficult is that one has to solve questions in increasing difficulty in a time frame that is constantly reducing.
A few suggestions:
1. Do not get jumpy about the answers, the purpose is method, and how we can reduce the time taken.
2. Do not post bits like IMO C and end it at that. Do explain how you arrived at C, and how that is the fastest method for you.
3. Do READ before you post, so as to not repeat already discussed methods.
Cheers.
'Though this be madness yet there is method in it.'
A certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one-gallon or half- gallon cans, what is the least amount of paint, in gallons, that must be purchased in order to measure out the portions needed for the mixture?
(A) 2
(B) 2 and a 1/2
(C) 3
(D) 3 and a 1/2
(E) 4
OA : B
Now, this is how I (wrongly) solved this question. I figured that in all 5+3 = 8 parts of paint are needed, and the smallest part can be 1/2 a gallon, thus the answer must be 8 x 1/2 = 4, and marked E.
Do suggest your own appraoches to this question. Also state what difficulty bin do you think this belongs to, for example, in my opinion this is a 600-700 bin question.
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(A) 2
(B) 2 and a 1/2
(C) 3
(D) 3 and a 1/2
(E) 4
OA : B
Now, this is how I (wrongly) solved this question. I figured that in all 5+3 = 8 parts of paint are needed, and the smallest part can be 1/2 a gallon, thus the answer must be 8 x 1/2 = 4, and marked E.
Do suggest your own appraoches to this question. Also state what difficulty bin do you think this belongs to, for example, in my opinion this is a 600-700 bin question.
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for 2 gallons of the gray paint, we need:
5/8*2=5/4=1 and 1/4 gallons of black paint
=>1 gallon can and 1 half gallon can of black paint....1
3/8*4=3/4 gallons of white paint
=>1 gallon can of white paint.......2
adding 1 and 2 we need 2 and half gallons of paint (white and black)
hence B
5/8*2=5/4=1 and 1/4 gallons of black paint
=>1 gallon can and 1 half gallon can of black paint....1
3/8*4=3/4 gallons of white paint
=>1 gallon can of white paint.......2
adding 1 and 2 we need 2 and half gallons of paint (white and black)
hence B
I started similar to the way you did: 3x + 5x = 2, then 8x=2
x= 1/4
So you substitute in the original equation and you get 3/4 + 5/4 = 2
Then you can interpret that you will need a little less than one gallon of white paint (3/4),but as you can only buy one gallon and half gallons you need 1 gallon and you will need a little more than one gallon of black paint (5/4) but as you can only buy one gallon and half gallons you need 1 gallon and a half. So there you have 2 1/2 as the answer
x= 1/4
So you substitute in the original equation and you get 3/4 + 5/4 = 2
Then you can interpret that you will need a little less than one gallon of white paint (3/4),but as you can only buy one gallon and half gallons you need 1 gallon and you will need a little more than one gallon of black paint (5/4) but as you can only buy one gallon and half gallons you need 1 gallon and a half. So there you have 2 1/2 as the answer
Source: MGMAT
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
A) 5
B) 5(x-y)
C) 20x
D) 20y
E) 35x
Please let's just focus on the method.
OA will follow.
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
A) 5
B) 5(x-y)
C) 20x
D) 20y
E) 35x
Please let's just focus on the method.
OA will follow.
- ssmiles08
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I took a jab at it and went for C. Little unsure but both 35x and 20y must be divisible by the GCF.
A) 5 is a factor of both 35x and 20y
B) 35x/[5(x-y)] = 7x/(x-y) ---> can be an integer depending on what integers x and y we pick.
20y/[[5(x-y)] = 4y/(x-y) ---> can also be an integer depending on what integers x and y we pick.
C) 35x/20x = 7/4 --->sticks out like a sore thumb b/c it is not an integer.
D) 35x/20y = 7x/4y ---> can be an integer depending on what integers x and y we pick.
20x/20y = x/y --->can be an integer depending on what integers x and y we pick.
E) 35x/35x is divisble
20y/35x = 4y/7x ---> can be an integer depending on what integers x and y we pick.
A) 5 is a factor of both 35x and 20y
B) 35x/[5(x-y)] = 7x/(x-y) ---> can be an integer depending on what integers x and y we pick.
20y/[[5(x-y)] = 4y/(x-y) ---> can also be an integer depending on what integers x and y we pick.
C) 35x/20x = 7/4 --->sticks out like a sore thumb b/c it is not an integer.
D) 35x/20y = 7x/4y ---> can be an integer depending on what integers x and y we pick.
20x/20y = x/y --->can be an integer depending on what integers x and y we pick.
E) 35x/35x is divisble
20y/35x = 4y/7x ---> can be an integer depending on what integers x and y we pick.
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IMO Caxat wrote:Source: MGMAT
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
A) 5
B) 5(x-y)
C) 20x
D) 20y
E) 35x
Please let's just focus on the method.
OA will follow.
when we find GCF, we use the common prime factors:
for GCF of 8 and 18
8=2^3
18=2*3^2
common prime factor is 2 and lowest power is 1. so GCF is 2
now
35x=5*7*x
20y=2^2*5*y
A) 5 is possible because it is a prime factor
B) Possible depending upon x and y
C)20x, now for this we need a 2^2 term in the 35x which is not there.
D)20y-possible
E)35x -possible
The powers of two are bloody impolite!!
"Thanks scoobydooby and cfarrera.
I made the mistake of assuming that because one can buy only 1 or half a gallon of paint, one can't even add a quantity less than that to the mixture.
"
Dude I made the same dang assumption too. I thought, copol so they have 2 1/2 gallons but how can they mix that to get a 3:5 proportion? I remember that I solved puzzles in the past which involved mixing solutions exactly according to the proportion or so. But this problem was much easier. They didn't care how the solution will be mixed. They only cared for the minimum amount of paint needed!!!. Dude what can we do to get rid of such assumptions??
I made the mistake of assuming that because one can buy only 1 or half a gallon of paint, one can't even add a quantity less than that to the mixture.
"
Dude I made the same dang assumption too. I thought, copol so they have 2 1/2 gallons but how can they mix that to get a 3:5 proportion? I remember that I solved puzzles in the past which involved mixing solutions exactly according to the proportion or so. But this problem was much easier. They didn't care how the solution will be mixed. They only cared for the minimum amount of paint needed!!!. Dude what can we do to get rid of such assumptions??
200 or 800. It don't matter no more.
Source MGMAT
x, y, x + y, x – 4y, xy, 2y
For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be
A) (x+y)/2
B) y+3
C) y
D) (3y)/2
E) (x/3) + y
OA is B.
But it's the method of getting the answer in under 100 seconds that I can't figure out.
x, y, x + y, x – 4y, xy, 2y
For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be
A) (x+y)/2
B) y+3
C) y
D) (3y)/2
E) (x/3) + y
OA is B.
But it's the method of getting the answer in under 100 seconds that I can't figure out.
- ssmiles08
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I dont know what method you used, but I used the following method and it took under 2 minutes.axat wrote:Source MGMAT
x, y, x + y, x – 4y, xy, 2y
For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be
A) (x+y)/2
B) y+3
C) y
D) (3y)/2
E) (x/3) + y
OA is B.
But it's the method of getting the answer in under 100 seconds that I can't figure out.
figure out x: if you write it out in addition form, you can see that all the y's cancel out except for xy
so you get 3x +xy = 6(y+3)
x = 6.
Since we know y >6 we can just gaze at the following numbers and see that x – 4y is the smallest b/c it is a negative number.
so the order in ascending form is:
x – 4y, x, y, x+y, 2y, xy
you can tell x+y < 2y b/c it is 6 + y < 2y = y > 6
the median is just (2y + x)/2
= y + 6/2 = y + 3
- Vemuri
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Well, I am confused & appreciate some clarification. The question stem says that x and y are positive integers. They are not necessarily prime numbers.tohellandback wrote: IMO C
when we find GCF, we use the common prime factors:
for GCF of 8 and 18
8=2^3
18=2*3^2
common prime factor is 2 and lowest power is 1. so GCF is 2
now
35x=5*7*x
20y=2^2*5*y
A) 5 is possible because it is a prime factor Agreed
B) Possible depending upon x and y Can you explain? Are we assuming that (x-y) will be a number that will be 35x as well as in 20y?
C)20x, now for this we need a 2^2 term in the 35x which is not there. (or) x is not in 20y. Is that right?
D)20y-possible I am stumped. How is this possible? 20y should also be in 35x to be a common factor
E)35x -possible I am stumped. How is this possible? 35x should also be in 20y to be a common factor
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Vemuri wrote:Well, I am confused & appreciate some clarification. The question stem says that x and y are positive integers. They are not necessarily prime numbers.tohellandback wrote: IMO C
when we find GCF, we use the common prime factors:
for GCF of 8 and 18
8=2^3
18=2*3^2
common prime factor is 2 and lowest power is 1. so GCF is 2
now
35x=5*7*x
20y=2^2*5*y
A) 5 is possible because it is a prime factor Agreed
B) Possible depending upon x and y Can you explain? Are we assuming that (x-y) will be a number that will be 35x as well as in 20y?
vemuri,
don't get stumped
C)20x, now for this we need a 2^2 term in the 35x which is not there. (or) x is not in 20y. Is that right?
D)20y-possible I am stumped. How is this possible? 20y should also be in 35x to be a common factor
E)35x -possible I am stumped. How is this possible? 35x should also be in 20y to be a common factor
A) agreed as you said
B) take the example x=21,y=14
35x=5*7*3*7
20y=4*5*2*7
GCF is 5*7 (7 is 21-14, thats what I meant when i said "depends upon the values of x and y")
C) Hold on I will give you a "big explanation"
D) possible because it is possible that between two numbers one of them can be the GCF.
I will give you an example, you figure out the rest.
y=3
=12
well, let me explain
since 20y contains 2^2, 5 and y. it can be the GCF if 35x contains all of these. now 35x already has 5 so, any value of x that contains 2^2 and as factors will make it possible
E) same explanation as above
20y already contains 5. so any value of y that has x and 7 as factors will make it possible
ex: x=3
y=21
by now I think you have understood why C is right, but still
35x=5*7*x
20x would have been possible only if there was a 2^2 in the factorization.
now you can argue that what if x has that 2^2 so precious to me
let x=4
now 20x is 2^2*5*2^2
but you original number 35x doesn't have two (2^2)s. It has only one 2^2 and that is from x. so 20x cannot divide it. Now you see why i mentioned there should be the prime numbers in the factorization.
Hope I am clear. Thanks
The powers of two are bloody impolite!!