If -12

Question

Competition Mode Question

If  -12 \leq x \leq 1  and  -4 \leq y \leq 10 , sum of minimum and maximum possible value of  x - |y|  is

A. -22
B. -21
C. -17
D. 21
E. 22

shameekv1989 · Accepted Answer

If −12≤x≤1 and −4≤y≤10, sum of minimum and maximum possible value of x−|y|x−|y| is

A. -22
B. -21
C. -17
D. 21
E. 22

Answer - B ->

x min value = -12
|y| min value = 0 (|y| will get rid of negative values of y)

x max value = 1
y max value = 10

min value of (x - |y|) = -12-10 = -22
max value of (x-|y|) = 1 - 0 = 1

Therefore sum of min and max values of (x-|y|) = -22+1 = -21

rajatchopra1994 · Answer

Explanation:

for minimum value, x=-12, y=10
x−|y| = -12-10 = -22
for maximum value, x=1 , y =0
x−|y| = 1-0 = 1

Sum of maximum & minimum = -22 + 1 = -21

IMO-B

CareerGeek · Answer

Maximum value of (x, |y|) = (1, 10) & minimum value of (x, |y|) = (-12, 0)

Maximum value of x - |y| = 1 - |0| = 1
Minimum value of x - |y| = -12 - |10| = -12 - 10 = -22

Sum of maximum and minimum value = 1 - 22 = -21

Option B

hardikajmani · Answer

Values of (x, |y|) 
Max ->  (1, 10) 
Min ->  (-12, 0)

Max value of  x - |y| = 1 - |0| = 1 
Min value of  x - |y| = -12 - |10| = -22

Sum = 1 - 22 = -21

Option B

madgmat2019 · Answer

minimum value of x−|y| is -12-10 = -22
maximum value of x−|y| is 1-0 = 1
sum = -22+1 = -21

OA:B

joohwangie · Answer

If −12≤x≤1 and −4≤y≤10, sum of minimum and maximum possible value of x−|y| is

minimum:
x=-12
y=10

-12-10=-22

maximum:
x=1
y=0
1-0=1

sum:
-22+1
=-21

B

Archit3110 · Answer

max value will be at x=1&y=0 ; i.e 1 and min value x=-12 and y = 10 ; -22
so sum ; -22+1 ;-21
IMO B

If −12≤x≤1 and −4≤y≤10, sum of minimum and maximum possible value of x−|y| is

A. -22
B. -21
C. -17
D. 21
E. 22

exc4libur · Answer

min: -12-|10|=-22
max: 1-|0|=1
sum: -21

Ans (B)

GMATSheep · Answer

For the smallest possible value let x = -12 and y = 10 ; -12 - |10| = -22
For the largest possible value let x = 1 and y = 0 ; 1 - |0| = 1
Sum of numbers: -22 + 1 = -21
ANS) B

ostrick5465 · Answer

x - |y|  get the minimum value when x minimum and  |y|  maximum
=> x = -12, y = 10=> Max  x - |y| = -12 - |10| = -22

x - |y|  get the maximum value when x maximum and  |y|  minimum
=> x = 1, y = 0 => Min  x - |y| = 1 - |0| = 1

=> sum of minimum and maximum value of  x - |y|  is  -21

=> Choice B

lacktutor · Answer

If −12≤x≤1 and −4≤y≤10, sum of minimum and maximum possible value of x−|y| is ?

The max =x−|y|= 1-|0|= 1
The min= x−|y|= -12 -|-10|= -12-10= -22

--> The sum = 1-22= -21

The answer is B.

eakabuah · Answer

We are given the ranges -12<=x<=1 and -4<=y<=10 and we are to find the sum of the minimum and maximum possible values of x-|y|
We know the |y| will always be +ve, hence to get the maximum value of x-|y|, |y| must be minimum, and from the given range, y must equal zero while x must be the maximum value of x and this is 1.
so Maximum value of x-|y| = 1-|0| = 1.

For the minimum, |y| must be maximum possible value in the range while x must be the minimum value possible.
hence Minimum value of x-|y| = -12-|10| = -22
Maximum+minimum = 1-22=-21

B is the answer.

unraveled · Answer

If −12≤x≤1 and −4≤y≤10, sum of minimum and maximum possible value of x−|y| is

A. -22
B. -21
C. -17
D. 21
E. 22

x−|y|(max) = x(max) − |y|(min) = 1 − |-4| = - 3
x−|y|(min) = x(min) − |y|(max) = -12 − |10| = - 22

Also, 
−12≤x≤1 and 
−4≤y≤10 = 4≤|y|≤10 = -10 ≤ -|y| ≤ -4

Adding 2 equations gives 
-12-10 ≤ x -|y| ≤ 1-4
-22 ≤ x -|y| ≤ -3

Sum of maximum and minimum = - 22 - 3 = -25

unraveled · Answer

What a pity mistake i have made. |y| can take '0' as minimum value and x -12

And maximum value of |y| is 10 and x is -12. Hence -12 - 10 + 1 - 0 = -21

EtaCarnia · Answer

Hey  Bunuel

I am totally off the grid in this one. I am getting -25. My solution is like this.

Max value of X and Y : 1 and 10
Min Value of X and Y : -12 and -4 .

while look at the solutions some have take minimum value of Y as 0. I am not getting it how .
Can you help me with is please.?

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If -12 <= x <= 1 and -4 <= y <= 10, sum of minimum and maximum possibl

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