CAT Previous Year Questions - Inequalities & Logarithms (July 15) - CAT MCQ

Case I)
If both x and y are both positive.
Then,  is always true, since We are increasing the numerator and decreasing the denominator.

Case II)
If both x and y are both negative.

Case III)
x is positive and y is negative.

This is always true since we are increasing the numerator and decreasing the denominator.
Case IV)
x is negative and y is positive.

Observe that k is always positive.


So, the given condition holds good when both x & y are positive or x is positive but y is negative or x is negative, y is positive and y < −x Since y is negative in the third option, −x < y , implies that x > |y|, that is x is positive.
We know that when y is negative and x is positive the condition always holds good.

Let the total amount be equal to T.
T = n × 352
“However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330”

So, the maximum value that n can take is 16.

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

We have ∣n−60∣ < ∣n−100∣ <∣n−20∣
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.
Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be ≤ 40, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference. Thus we have the condition that n > 40.
The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be ≤ 60, as then it would be closer to 20 than 100. Thus we have the condition that n > 60.
The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be \ge80≥80, as then it would be closer to 100 than 60. Thus we have the condition that n<80.
The number which satisfies the conditions are 61, 62, 63, 64......79.
Thus, a total of 19 numbers.

log_x (x² + 12) = 4
This means, x² + 12 = x⁴
Let k = x²
k + 12 = k²
k² – k – 12 = 0
k² – 4k + 3k – 12 = 0
k(k – 4) + 3(k – 4) = 0
k = 4 or k = -3
But since k = x², k is always non-negative.
∴ k = 4
∴ x² = 4
Since x is the base of the log function, it can should always be positive.
∴ x = 2
3 log_yx = 1
log_yx³ = 1
x³ = y¹
y = x³ = 2³ = 8
∴ x + y = 2 + 8 = 10

But, since the argument of the log can't be negative, x = 4
∴ The common ratio of the G.P is 

∴ The common difference of the AP = (7/2)

For any value of n < 16, the numerator is positive. For any value of n > 16, it is negative.
For any value of n < 64, the denominator is positive. For any value of n > 64, it is negative.
For a fraction to be negative, the numerator and the denominator must be of opposite signs.
In this case, n should be between 16 and 64.
The number of values that n can take = 63 - 16 = 47.