Basic Absolute Value Equations

[avinash.sunnasy]

Hello all,

According to MGMAT Eq, Ineq, and VICs guide, when attempting to solve an absolute value problem you should consider the positive and negative case.

E.g. 12+|w+4|=30
|w+4| = 18

Case 1: (+ve)
w-4 = 16
w = 22

Case 2:(-ve)
-w+4=18
w=-14

Please can you help me to solve the following problem using the above techniques. If they are not applicable please explain why and what to look for to differentiate the following absolute value problem from those which use the above technique.

If y is an integer, then the least possible value of |23-5y| is
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Thanks,

A.

[gokul_nair1984]

You are missing out on the basics:

|a|=3 means ,

a=+3(when a is positive)
a=-3(when a is negative)

Therefore for your explained case,

12+|w+4|=30
|w+4| = 18

Case 1: (+ve)
w+4=18 ..=>w=14

Case 2:(-ve)
w+4=-18..=>w=-22

I am quite sure that MGMAT books might not have explained as you have perceived..

Now coming to your Question,

If y is an integer, then the least possible value of |23-5y| is
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

The least possible value can be cannot be less than 0 as the term is within a modulus sign.

Therefore we can equate it to 0..
=> 23-5y=0
or y=4.6...Rounding it off to 5,

we get |23-5*5|=|-2|=2 Thus B is the answer.

I hope you are clear with modulus basics.

[tim]

Thanks, Gokul. You can also try plugging in the answer choices on this one..