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Integer Diff

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by pemdas » Mon Dec 26, 2011 2:24 am
find x-y, first
x>3 is x-3>0 and
y<10 is y-10<0 or -y+10>0 add them up
x-y>-7

then
x<6 is x-6<0 and
y>6 is y-6>0 or -y+6<0 add them up
x-y<0

so -7<x-y<0, since difference doesn't imply subtract y from x, but rather only difference, we have difference as 6 between 0 and 7.

d
karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?

3
4
5
6
7
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by karthikpandian19 » Mon Dec 26, 2011 3:33 am
Oh....that's a new way of doing...thank you for that
pemdas wrote:find x-y, first
x>3 is x-3>0 and
y<10 is y-10<0 or -y+10>0 add them up
x-y>-7

then
x<6 is x-6<0 and
y>6 is y-6>0 or -y+6<0 add them up
x-y<0

so -7<x-y<0, since difference doesn't imply subtract y from x, but rather only difference, we have difference as 6 between 0 and 7.

d
karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?

3
4
5
6
7
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by arpan20690 » Mon Dec 26, 2011 3:38 am
This question does not need to played with inequalities.
get the maximum difference between x and y. Y is greater than x.
Take x as less as possible. and y as large as possible.
let take x = 3.01 and y = 9.99
so, y-x = 6.98
We can not touch 7 as this will violate the given condition. So, maximum Possible
Integer Value is 6.
Introduction of BOX ([]) function would make this even more
simpler.
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by LalaB » Mon Dec 26, 2011 4:47 am
karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?

3
4
5
6
7
not sure, that it is a good idea, but still if I have no time, I will think this way-

3<x<6<y<10, then what is the greatest possible positive integer

x can be between 3.01 and 5.99 (approx)

y can be between 6.01 and 9.99 (approx)


so ,the greatest positive INTEGER difference of y and x is 9.99-3.99=6
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by karthikpandian19 » Mon Dec 26, 2011 5:00 am
@pemdas

After seeing the explanations, this problem cannot be solved with Inequalities.

OA is 6
LalaB wrote:
karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?

3
4
5
6
7
not sure, that it is a good idea, but still if I have no time, I will think this way-

3<x<6<y<10, then what is the greatest possible positive integer

x can be between 3.01 and 5.99 (approx)

y can be between 6.01 and 9.99 (approx)


so ,the greatest positive INTEGER difference of y and x is 9.99-3.99=6
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by pemdas » Mon Dec 26, 2011 7:56 am
2<a<3<b<4

this is similar question from Kaplan which asks to solve for (1/3)ab

care to solve this question by assessing min/max values for a and b?
I don't mean inequalities are must here, but they help clear the way
karthikpandian19 wrote:@pemdas

After seeing the explanations, this problem cannot be solved with Inequalities.

OA is 6
LalaB wrote:
karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?

3
4
5
6
7
not sure, that it is a good idea, but still if I have no time, I will think this way-

3<x<6<y<10, then what is the greatest possible positive integer

x can be between 3.01 and 5.99 (approx)

y can be between 6.01 and 9.99 (approx)


so ,the greatest positive INTEGER difference of y and x is 9.99-3.99=6
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by LalaB » Mon Dec 26, 2011 8:08 am
@karthikpandian19
pemdas shows the conceptual way of problem solving

as I have mentioned my way is not perfect (in some cases it is even bad). my method is laid more on common sense, than on solid math knowledge.

pemdas wrote:2<a<3<b<4

this is similar question from Kaplan which asks to solve for (1/3)ab

care to solve this question by assessing min/max values for a and b?
I don't mean inequalities are must here, but they help clear the way
2<a<3
3<b<4
6<ab<12

2<1/3ab<4

using a calculator and my previous way of problem solving,we can find out whether the solution above is right.


@pemdas, is my answ ok?
Last edited by LalaB on Mon Dec 26, 2011 8:41 am, edited 1 time in total.
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by pemdas » Mon Dec 26, 2011 8:41 am
in general an answer found is correct, but the way two inequalities were multiplied was aberrant. Instead, i would separate the signs and put >0 or <0 to understand the sign of multiplier. Here, however all numbers are +ve and no error was captured.

assume a=2 and b=3 is the lower ceiling, then ab>6 and ab/3>2
assume a=3 and b=4 is the upper ceiling, then ab<12 and ab/3<4

2<ab/3<4
LalaB wrote:
pemdas wrote:2<a<3<b<4

this is similar question from Kaplan which asks to solve for (1/3)ab

care to solve this question by assessing min/max values for a and b?
I don't mean inequalities are must here, but they help clear the way
2<a<3
3<b<4
6<ab<12

2<1/3abd<4

using a calculator and my previous way of problem solving,we can find out whether the solution above is right.


@pemdas, is my answ ok?
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by LalaB » Mon Dec 26, 2011 8:46 am
pemdas wrote:in general an answer found is correct, but the way two inequalities were multiplied was aberrant. Instead, i would separate the signs and put >0 or <0 to understand the sign of multiplier. Here, however all numbers are +ve and no error was captured.

assume a=2 and b=3 is the lower ceiling, then ab>6 and ab/3>2
assume a=3 and b=4 is the upper ceiling, then ab<12 and ab/3<4

2<ab/3<4
hm, pemdas show me the difference with my approach )) i have done the same. multiplied one side of the inequity with the side of another inequity.
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by LalaB » Mon Dec 26, 2011 8:49 am
btw, if we are asked (b-a), then the answer will be -

2<a<3
3<b<4

3-3<b-a<4-2
0<b-a<2

i mean now we subtract opposite sides
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by pemdas » Mon Dec 26, 2011 9:44 am
multiplication operation with inequalities in this form isn't right
LalaB wrote:2<a<3
3<b<4
here all numbers are +ve, BUT in case the sign were different
-2<a<3
3<b<4
according to your method we would get -6<ab<12?

Nevertheless, 'a' could be -7/4 and 'b' could be 15/4, then ab=-105/16 which is less (and not greater) than -6 (namely, it's -6.56)
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by LalaB » Mon Dec 26, 2011 9:52 am
pemdas wrote:multiplication operation with inequalities in this form isn't right
LalaB wrote:2<a<3
3<b<4
here all numbers are +ve, BUT in case the sign were different
-2<a<3
3<b<4
according to your method we would get -6<ab<12?

Nevertheless, 'a' could be -7/4 and 'b' could be 15/4, then ab=-105/16 which is less (and not greater) than -6 (namely, it's -6.56)
good point :)
pemdas, let me exlain what I am trying to do.
to find the limits of two equaties I try to find the max and the min values.

-2<a<3
3<b<4

in this case the minimum value will be -2*4(I multiply min of one equity with max of another one) , the max - 3*3 (I multiply min of one equity with max of another one)
I mean i always try to minimize one side, and maximize another side

so i got

-8<ab<9
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by rijul007 » Mon Dec 26, 2011 10:08 am
karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?

3
4
5
6
7
3<x<6<y<10

3 < x < 6
6 < y < 10

For smallest value of y-x, subtract largest value of x from smallest value of y
y-x = 6.000000.....1 - 5.99999.....9 = 0.0000000......2

For largestvalue of y-x, subtract smallest possible value of x from largest possible value of y
y-x = 9.99999.....9 - 3.000000........1 = 6.9999999999.......8

Integer value of y-x
0 < y-x < 7

Greatest possible positive integer = 6
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by pemdas » Mon Dec 26, 2011 10:14 am
why you restrict the RHS to 9? The point is a multiplication of inequalities containing more than one inequality sign may not be admissible. We have to test each inequality.
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