karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?
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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.
dkarthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?
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not sure, that it is a good idea, but still if I have no time, I will think this way-karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?
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LalaB wrote:not sure, that it is a good idea, but still if I have no time, I will think this way-karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?
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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
karthikpandian19 wrote:@pemdas
After seeing the explanations, this problem cannot be solved with Inequalities.
OA is 6LalaB wrote:not sure, that it is a good idea, but still if I have no time, I will think this way-karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?
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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
2<a<3pemdas 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
LalaB wrote:2<a<3pemdas 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
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?
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.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
here all numbers are +ve, BUT in case the sign were differentLalaB wrote:2<a<3
3<b<4
according to your method we would get -6<ab<12?-2<a<3
3<b<4
good pointpemdas wrote:multiplication operation with inequalities in this form isn't righthere all numbers are +ve, BUT in case the sign were differentLalaB wrote:2<a<3
3<b<4according to your method we would get -6<ab<12?-2<a<3
3<b<4
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)
3<x<6<y<10karthikpandian19 wrote:If 3<x<6<y<10, then what is the greatest possible positive integer difference of x and y?
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