Test Level 3: Inequalities CAT MCQs & solutions - Free

Total money with Vijay and Manju is less than Rs. 5.
If Vijay is near to 0, then Manju is near to 5.
If Manju is near to 0, then Vijay is near to 5.
Option (1):
3V + 4M < 24
Let V = 1, then M < 5
3 × 1 + 4 × 4 < 24 [True]
Option (2):
3V + 4M < 18
3 + 16 < 18 [False]

x = 1, 2y ≤ 16 ⇒ y ∈ [1, 2, 3, 4, 5, 6, 7, 8] ⇒ 8 pairs 
x = 2, 2y ≤ 8 ⇒ y ∈ [1, 2, 3, 4] ⇒ 4 pairs
x = 3, y = 0 ⇒ 0 pair
So, a total of 12 pairs are there.

If a < b and c < 0, let us give some values to these variables such as a = 2, b = 3 and c = -1 and put in the options.
We will have only option 3 true, i.e.


 which is true.

u is always negative; hence, for us to have a minimum value of vz/u , vz should be positive.
Also, for the least value, the numerator has to be the maximum positive value and the denominator has to be the smallest negative value. In other words, vz has to be 2 and u has to be -0.5.
Hence, minimum value of vz/u = 2/-0.5 = -4
For us to get the maximum value, vz has to be the smallest negative value and u has to be the highest negative value.
Thus, vz has to be -2 and u has to be -0.5.
Hence, maximum value of vz/u = -2/-0.5 = 4

Since a is always negative and b², c and d² are always positive. So ab²/cd² is always negative.
So, the maximum value is negative of minimum magnitude.
So, a and b² should be minimum in magnitude and c and d² should be maximum.
So, take a = -1, b = 2 and c = 4 and d = 3.

So, maximum value = 

Consider p = 1, q = 2, r = 3, s = 4
Case 1: p + q < r + s
which implies, 3 < 7, which holds the inequality.
Case 2: (p + q) (rs) = 36
pq (r + s) = 14
36 is not less than 14
Hence, the second option is incorrect
Case 3: (p + q) (r + s) = 21
pq + rs = 14
Hence, option 3 is incorrect.

If 7a - 2 ≥ 0, then, 
If 2a + 3 ≥ 0, then 
So, three different intervals of a are possible.
Case 1:
 then inequality becomes,
2a + 3 > 7a - 2
5 > 5a
1 > a
So, for case 1,

Case 2:

2a + 3 > 2 - 7a
9a > - 1

So, for case 2,

Case 3:
  then inequality becomes,
-2a - 3 > 2 - 7a
5a > 5
a > 1, which is not possible as the given inequality is formed only if

So, combining case 1 and case 2,
 is the correct interval of a.

Total number of kids surveyed = 100 + 200 + 100 = 400
25% of 400 = 100
Number of kids who chose brand C = 20 + 40 + k = 60 + k
Not more than 25% of the kids chose brand C.
This implies that at the most, 25% of the kids chose brand C.
Thus, at the most, 100 kids chose brand C.
Thus, 60 + k ≤ 100
Or, k ≤ 40
Hence, answer option (3) is correct.

For the first pair, use the equation, remainder of (7/3)x = remainder of (123/3).
Thus, the first pair is (0, 41). Since only positive integers are permitted, we reject this pair.
But the other pairs can be derived from the above pair.
Now, every third value of x after 0 will give an integral value of y i.e. for x = 3, 6, 9, 12, 15, 18.. and so on, we get integral values of y as 34, 27, 20, 13, 6, - 1,……. and so on.
Thus, only five pairs satisfy our condition: (3, 34), (6, 27), (9, 20), (12, 13) and (15, 6).

2 < x < 3 and -8 < y < -7
Since, the value of xy² is also positive and the least value of x + y is -8 + 2 = -6      ....(i)
And since 2 < x < 3,
⇒ 4 < x² < 9
And - 8 < y < -7
⇒ 4y < x²y < 9y
⇒ -63 < x²y < -32      ....(ii)
While -105 < 5xy < -80    ....(iii)
Hence, from (i), (ii) and (iii), 5xy is the least because xy² is positive.