XAT 2018 — QA & DI Question 16
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.
Answer & solution
- A
4
- B
5
- C
6
- D
8
10
Since 2 ≤ |x – 1| × |y + 3| ≤ 5, the value of |x – 1| × |y + 3| can be 2, 3, 4 or 5. Thus, we have the following cases.
Case 1: |x – 1| × |y + 3| = 2
⇒ |x – 1| = 2 and |y + 3| = 1
⇒ x = -1 and y = -4 or -2
Note: |x - 1| cannot be equal to 1 since x has to be a negative number.
∴ 2 possibilities.
Case 2: |x – 1| × |y + 3| = 3
⇒ |x – 1| = 3 and |y + 3| = 1
⇒ x = -2 and y = -4 or -2
∴ 2 possibilities.
Case 3: |x – 1| × |y + 3| = 4
⇒ |x – 1| = 4 and |y + 3| = 1
⇒ x = -3 and y = -4 or -2
⇒ |x – 1| = 2 and |y + 3| = 2
⇒ x = -1 and y = -1 or -5
∴ 4 possibilities.
Case 3: |x – 1| × |y + 3| = 5
⇒ |x – 1| = 5 and |y + 3| = 1
⇒ x = -4 and y = -4 or -2
∴ 2 possibilities.
∴ Total 2 + 2 + 4 + 2 = 10 possibilities.
Hence, option (e).