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Thelegend2631
Posts: 371
01 Jul 2021, 20:54
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
34% (02:57) correct
66%
(02:53)
wrong
based on 211
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If \(2 ≤ |x – 1|×|y + 3| ≤ 5\) and both x and y are negative integers, find the number of possible combinations of \(x\) and \(y\) .
A. 4
B. 5
C. 6
D. 8
E. 10
A. 4
B. 5
C. 6
D. 8
E. 10
01 Jul 2021, 21:50
Kudos
Bookmarks
hD13
Constraint x and y are negative integers
|x-1| is = 1-x as expression enclosed in MOD will always be negative
We can simplify expression as \(2 ≤ (1-x)×|y + 3| ≤ 5\)
Case 1: For -3 ≤ y < 0
Expression becomes \(2 ≤ (1-x)×(y + 3) ≤ 5\)
* For y= -1 solve : \(2 ≤ (1-x)×(-1 + 3) ≤ 5\) = \(1 ≤ (1-x) ≤ \frac{5}{2}\)
\(0 ≤ -x ≤ \frac{3}{2}\) => \(\frac{-3}{2} ≤ x ≤0\) x= -1
*For y =-2 solve to get x =-4, -3, -2, -1
Case 2 : For y ≤ -4
Expression becomes \(2 ≤ (1-x)×(-1)(y + 3) ≤ 5\)
=> \(-5 ≤ (1-x)×(y + 3) ≤ -2\)
Now solve for
For x = -1 ; - \(\frac{-5}{2}-3 ≤ y ≤ \frac{-2}{2}-3\) ; y =-5, -4
For x = -2 ; - \(\frac{-5}{3}-3 ≤ y ≤ \frac{-2}{3}-3\) ; y =-4
For x = -3 ; - \(\frac{-5}{4}-3 ≤ y ≤ \frac{-2}{4}-3\) ; y =-4
For x = -4 ; - \(\frac{-5}{5}-3 ≤ y ≤ \frac{-2}{5}-3\) ; y =-4
There would not be any value of y < -3 for any value of x <-4
Total possible combinations
(-1,-1), (-1,-2), (-2,-2), (-3,-2), (-4,-2) These five from Case 1
(-1,-4), (-1,-5), (-2,-4), (-3,-4), (-4,-4) These five from Case 2
Total 10 cases, E is the answer.
PS: I am sure there must be a better and less calculation intensive way of arriving at the answer:
01 Jul 2021, 22:28
Kudos
Bookmarks
hD13
Let us take values for x, and see corresponding range of y.
\(2 ≤ |x – 1|×|y + 3| ≤ 5\)
1) x=-1
\(2 ≤ |-1 – 1|×|y + 3| ≤ 5\)........\(2 ≤ 2×|y + 3| ≤ 5\)..........\(1 ≤ |y + 3| ≤ 2.5\)
As y is negative and |y| should have a difference of 1 from 3, y can take value -2 and -4 to make |y+3|=1
Also |y| can have a difference of 2 from 3, so y can take value -1 and -5 to make |y+3|=2.
Combinations of (x,y): (-1,-1), (-1,-2), (-1,-4), (-1,-5).....4 ways
2) x=-2
\(2 ≤ |-2 – 1|×|y + 3| ≤ 5\)........\(2 ≤ 3×|y + 3| ≤ 5\)..........\(\frac{2}{3} ≤ |y + 3| ≤ \frac{5}{3}\)....
Only possible integer value is 1
As y is negative and |y| should have a difference of 1 from 3, y can take value -2 and -4 to make |y+3|=1
Combinations of (x,y): (-2,-2), (-2,-4).....2 ways
3) x=-3
\(2 ≤ |-3 – 1|×|y + 3| ≤ 5\)........\(2 ≤ 4×|y + 3| ≤ 5\)..........\(\frac{1}{2} ≤ |y + 3| ≤ \frac{5}{4}\)....
Only possible integer value is 1
As y is negative and |y| should have a difference of 1 from 3, y can take value -2 and -4 to make |y+3|=1
Combinations of (x,y): (-3,-2), (-3,-4).....2 ways
4) x=-4
\(2 ≤ |-4 – 1|×|y + 3| ≤ 5\)........\(2 ≤ 5×|y + 3| ≤ 5\)..........\(\frac{2}{5} ≤ |y + 3| ≤ 1\)....
Only possible integer value is 1
As y is negative and |y| should have a difference of 1 from 3, y can take value -2 and -4 to make |y+3|=1
Combinations of (x,y): (-4,-2), (-4,-4).....2 ways
5) x=-5
\(2 ≤ |-5 – 1|×|y + 3| ≤ 5\)........\(2 ≤ 6×|y + 3| ≤ 5\)..........\(\frac{2}{6} ≤ |y + 3| ≤ 5/6\)....
No integer values possible
Total: 4+2+2+2=10
E
