**Q1. ‘X’ is a perfect square less than 40, and ‘Y’ is a perfect cube less than 100.**

**Quantity I** – What will be probability of X > Y.

**Quantity II** – What will be probability that the sum of X & Y is an odd number.

(a) Quantity I > Quantity II

(b) Quantity I ≥ Quantity II

(c) Quantity II > Quantity I

(d) Quantity II ≥ Quantity I

(e) Quantity I = Quantity II or Relation cannot be established

**Q2. If 2x ^{2} + 3x – 6 < 0 and 3y^{2} + 16y + 18 < 0, and ‘x’ and ‘y’ are integers, then**

**Quantity I**– Value of ‘x’

**Quantity II**– Value of ‘y’

(a) Quantity I > Quantity II

(b) Quantity I ≥ Quantity II

(c) Quantity II > Quantity I

(d) Quantity II ≥ Quantity I

(e) Quantity I = Quantity II or Relation cannot be established

**Q3. ‘Q’ is two-digit number divisible by 16, and when 4 added into the number then it is divisible by 4 & 6.**

**Quantity I** – Maximum possible value of (Q – 6).

**Quantity II** – Minimum possible value of 250% of (Q + 4)

(a) Quantity I > Quantity II

(b) Quantity I < Quantity II

(c) Quantity I ≥ Quantity II

(d) Quantity I ≤ Quantity II

(e) Quantity I = Quantity II or No relation

**Q4. ‘A’ type of 5P taps can fill a tank in P/2 hours, while ‘A’ type of 2Q taps can fill the same tank in 4Q/5 hours. **

**Quantity I:** Value of ‘Q+20’.

**Quantity II:** Value of ‘1.25P’

(a) Quantity I ≥ Quantity II

(b) Quantity I = Quantity II or No relation

(c) Quantity I > Quantity II

(d) Quantity I < Quantity II

(e) Quantity I ≤ Quantity II

**Q5. A container contains 160 lit of pure milk and P liters of water. Ratio of 25% of mixture to the initial quantity of water is 3 : 4 and when Q liters of water is added to it ratio of total mixture to water becomes 9 : 4.**

**Quantity I** – Find the P + Q.

**Quantity II** – Find the 4(P – Q).

(a) Quantity I > Quantity II

(b) Quantity I ≥ Quantity II

(c) Quantity II > Quantity I

(d) Quantity II ≥ Quantity I

(e) Quantity I = Quantity II or Relation cannot be established

**Solutions**

. . .