Test Level 2: Inequalities - 2 CAT MCQs & solutions - Free

x = 0, y = ±7
Or x = ±1, y = ±6
Similarly, (±2, ±5)
(±3, ±4), (±4, ±3), (±5, ±2), (±6, ±1), (±7, 0)
(0, ±7) means (0, 7) or (0, -7) → 2 pairs
(±1, ±6) means (1, 6), (1, -6), (-1, 6), (-1, - 6) → 4 pairs
So, total number of possible pairs = 2 + 6 × 4 + 2 = 28

(m - n)² + (m - 1)² + (n - 1)² ≥ 0 [Any square is always non-negative]
⇒ m² - 2mn + n² + m² - 2m + 1 + n² - 2n + 1 ≥ 0
⇒ 2m² + 2n² + 2 ≥ 2mn + 2m + 2n
⇒ m² + n² + 1 ≥ mn + m + n {Dividing both sides by 2}

m cannot be < -3 because that will result in (-ve)(-ve)/(-ve), which will be negative and thus < 0.
m cannot be -2 or -3 because that will result in 0, which is not > 0.
m cannot be -1, 0 or 1 because that will result in (+ve)(+ve)/(-ve), which is < 0.
m cannot be 2 because that will result in dividing by 0.
Thus, m can only be 3, 4, 5, 6, ......., 99.
So, required sum is 

2x + y = 40
Possible pairs (x, y) are = (1, 38) (2, 36) ……….. (13, 14)
And (14, 12) is not possible because x ≤ y.
So, there are a total of 13 pairs.

2y - x > 2x - y ⇒ y > x
i.e. y = 0, x < 0 ⇒ No pair exists.
y = 1, x = 0, ⇒ Number of pairs = 1
y = 2, x = 0, 1 ⇒ Number of pairs = 2
y = 3, x = 0, 1, 2 ⇒ Number of pairs = 3
…
…
…
If y = 20, x = 0 to 19 ⇒ Number of pairs
= 20
Total number of pairs = 1 + 2 + 3 + …… + 20 = 210

Since 5 and 2 are both greater than zero, we can multiply the first inequality by 10 and retain the sign >.
2(2x – 3) > 5(3x – 2)
4x – 6 > 15x – 10

2nd inequality:

Both inequalities  will be satisfied if 

a is an even integer and b is an odd integer.
So, b - a will be odd.
Since b - a > 5 and c > b.
So, c - a will be minimum if c and b both are consecutive odd integers.
So, c - a > 7
Hence, minimum value of c - a will be 9 as c is odd and a is an even integer.

Let x be the smaller of the two consecutive odd positive integers. Then, the other odd integer is x + 2.
It is given that both the integers are smaller than 18 and their sum is more than 20.
Therefore,
x + 2 < 18 and x + (x + 2) > 20
⇒ x < 16 and 2x + 2 > 20
⇒ x < 16 and 2x > 18
⇒ x < 16 and x > 9
⇒ 9 < x < 16
⇒ x = 11, 13, 15 [∵ x is an odd integer]
Hence, the required pairs of odd integers are (11, 13), (13, 15) and (15, 17).

If these are the three sides of a triangle, then
(1 + 3c) + 1 > 1 - c ... (i) (∵ Sum of two sides must be greater than the third side)
Also, 1 + 3c + 1 - c > 1 ... (ii)
And 1 + 1 - c > 1 + 3c ... (iii)
From (i), we get

From (ii), we get

From (iii), we get

Combining the three inequalities,

If we put x = 3,
Then |3 − 1| + |3 − 2|+|3 − 3| = 3 ≤ 6
Therefore option (a), (c), (d) are not correct.
Hence only option (b) is correct.