Combining Inequalities

[adhirajkohli]

Hi,

I had this doubt pertaining to combining inequalities.
If we have 2 inequalities with the same 'sign' , can we combine them by addition or subtraction ?

Consider a "remixed" OG problem ( to avoid copyright issues )

If 2p < 2a < 2q and 2r < 2b < 2t is a < b ?

(I) p < r
(II) q < r

if we look at (I) logically we know that even though p < r we cannot assume q < r hence this statement is INSUFFICIENT
looking at (II) we know that if q < r and r < b then it follows that a < b therefore (II) alone is SUFFICIENT.

But if i were to try solving this by combining inequalities :-
(I) p < r
p < a
subtract
0 < r - a
=> a < r
also r < b
adding the two
a + r < b + r
=> a < b hence sufficient

(II) q < r
r < b
add
q + r < r + b
=> q < b
but a < q
adding the two
a + q < b + q
hence a < b therefore sufficient

where am i going wrong with this approach ?

Thanks

[debmalya.dutta]

addition of inequalities with the same sign is ok .
Subtract is tricky

for example
3 > 0 - inequality 1
5 > 0 - inequality 2
This does not mean that inequality 1- inequality 2 =>3-5 is greater than zero.. Infact it is less than zero

So when you are subtracting 2 inequalities ,you should know which one is greater and then apply the sign changes wherever applicable


2p < 2a < 2q and 2r < 2b < 2t

Statement 1
since p < r does not mean anything for the relationship between a and b
for example
2 < 2a (say a=6) < 10
4 < 2b (say b=6) < 10
hence insufficient

Statement 2
q < r
2p < 2a < 2q and 2r < 2b < 2t can be condensed to
2p < 2a < 2q< 2r < 2b < 2t
we can see that a < b
hence sufficient

Hence option is B

[jnelson0612]

Thank you debmalya. I agree with your explanation. It is much safer on such problems to pick numbers. You can quickly see that statement 1 is insufficient but that statement 2 is sufficient.