Latest Elitmus Aptitude Question SOLUTION: minimum value of |x|+|x-1|+|x-2| if |x|=x ,x>=0 |x|=-x,x

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(#M40030576) ELITMUS Numerical Ability question quant Keep an EYE

minimum value of |x|+|x-1|+|x-2|
if |x|=x ,x>=0
|x|=-x,x

Asked In Elitmus keshav (10 years ago)

Unsolved Read Solution (10)

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RAKESH  (10 years ago)

draw the graph,it can be seen that
minimum value of |x|+|x-1|+|x-2| will be at x=1
and value is =1+0+1=2
ans- 2

Is this solution Helpfull? Yes (51) | No (3)  

ankit chauhan  (10 years ago)

if we take x=0 then we get |x|=x so |x|+|x-1|+|x-2|=x+x-1+x-2=3x-3 X=0,,so minmum value is -3.

Is this solution Helpfull? Yes (3) | No (9)  

animesh jian  (10 years ago)

best way to solve is eliminate yhe option.

Is this solution Helpfull? Yes (1) | No (1)  

Shaibal Dutta  (10 years ago)

min value is 2. for x=1 , bcz |-2|=2 nd |-3|=3 for x=0 ;

Is this solution Helpfull? Yes (1) | No   

kunan  (10 years ago)

6 use differnciation for both case

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keshav  (10 years ago)

|X|= -X, IF X

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senaritra Das  (10 years ago)

can u plz send me the graph of dis prob @rakesh email-id= senaelitmus@gmail.com

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vishal rathore  (10 years ago)

= -3x + 3 if x

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vishal rathore  (10 years ago)

-3x+3 if x

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vivek  (8 years ago)

Here the each term of series are in ap..such that..
1) 6,11,16...,996 2) 7,12,17....,997. 3) 8,13,18,....,998 4) 9,14,19,.....,999.
Here if we add each term of each series in position wise then we get a number which is divisible by 10.
So take any series we get similar ans.
a/c to a.p,
996=6+(n-1)5
198=(n-1)
So, n =199. We can't take 1,2,3,4 as it contain 1.
so required ans is 199.

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