Step 1: Understand the trapezium structure.
We are given $PQRS$ with $PS \parallel QR$. Lines $PQ$ and $SR$ are extended to meet at point $A$ above the trapezium. We want to calculate $\angle PAS$. Step 2: Using Statement I.
- From the figure, $\triangle AQR \sim \triangle APS$ (because $PS \parallel QR$).
- This similarity gives ratio:
\[ \frac{AQ}{AP} = \frac{QR}{PS} = \frac{AR}{AS} = k \tag{1} \] Now, we are given $PQ=3$, $RS=4$, and $\angle QPS=60^\circ$. Let us drop a perpendicular from $Q$ onto $PS$, meeting at $M$. In right $\triangle PQM$: \[ \sin 60^\circ = \frac{QM}{PQ} = \frac{QM}{3} \] \[ QM = 3\cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \] Similarly, drop a perpendicular from $R$ onto $PS$, meeting at $N$. Then $RN = QM = \tfrac{3\sqrt{3}}{2}$. Now, in right $\triangle RSN$: \[ \sin \angle RSN = \frac{RN}{RS} = \frac{\tfrac{3\sqrt{3}}{2}}{4} = \frac{3\sqrt{3}}{8} \] Hence, \[ \angle RSN = \sin^{-1}\!\left(\frac{3\sqrt{3}}{8}\right) \] Finally, in $\triangle APS$, \[ \angle PAS = 180^\circ - (\angle APS + \angle PSA) \] But $\angle APS = 120^\circ$ (straight line property at $P$ since $\angle QPS=60^\circ$) and $\angle PSA = \angle RSN$. Thus, \[ \angle PAS = 120^\circ - \sin^{-1}\!\left(\frac{3\sqrt{3}}{8}\right) \] Therefore, Statement I alone gives us a clear value of $\angle PAS$.
Step 3: Using Statement II.
If $PS=10$ and $QR=5$, then the ratio of parallel sides is $2:1$. But without knowing any angle (like $\angle QPS$), the trapezium can scale differently, so $\angle PAS$ cannot be uniquely determined.
Final Decision:
- Statement I alone $\Rightarrow$ sufficient.
- Statement II alone $\Rightarrow$ insufficient.
- Combined use is not required since I alone works.
\[ \boxed{\text{Answer: (A)}} \]
Step 1: Observe the recurrence.
The rule is: \[ A_{n+1} = A_n^2+1 \] This means once the initial term is fixed, the entire sequence is uniquely determined.
Step 2: Using Statement I ($A_0=1$).
Start with $A_0=1$: \[ A_1 = 1^2+1=2, \quad A_2=2^2+1=5, \quad A_3=5^2+1=26, \ldots \] Thus, the sequence is completely determined. With a computer or theoretical methods, $\gcd(A_{900}, A_{1000})$ can be found.
Hence, Statement I alone is sufficient.
Step 3: Using Statement II ($A_1=2$).
From the recurrence: \[ A_1 = A_0^2+1 \] So $2=A_0^2+1 \Rightarrow A_0=1$. Thus Statement II actually implies Statement I, and again the sequence is uniquely fixed. Therefore, Statement II alone is also sufficient.
Step 4: Conclude sufficiency.
Since either statement alone fixes the entire sequence and hence allows computation of $\gcd(A_{900}, A_{1000})$, each statement alone is sufficient. \[ \boxed{\text{Answer: (D)}} \]





Light Chemicals is an industrial paint supplier with presence in three locations: Mumbai, Hyderabad and Bengaluru. The sunburst chart below shows the distribution of the number of employees of different departments of Light Chemicals. There are four departments: Finance, IT, HR and Sales. The employees are deployed in four ranks: junior, mid, senior and executive. The chart shows four levels: location, department, rank and gender (M: male, F: female). At every level, the number of employees at a location/department/rank/gender are proportional to the corresponding area of the region represented in the chart.
Due to some issues with the software, the data on junior female employees have gone missing. Notice that there are junior female employees in Mumbai HR, Sales and IT departments, Hyderabad HR department, and Bengaluru IT and Finance departments. The corresponding missing numbers are marked u, v, w, x, y and z in the diagram, respectively.
It is also known that:
a) Light Chemicals has a total of 210 junior employees.
b) Light Chemicals has a total of 146 employees in the IT department.
c) Light Chemicals has a total of 777 employees in the Hyderabad office.
d) In the Mumbai office, the number of female employees is 55.

An investment company, Win Lose, recruit's employees to trade in the share market. For newcomers, they have a one-year probation period. During this period, the employees are given Rs. 1 lakh per month to invest the way they see fit. They are evaluated at the end of every month, using the following criteria:
1. If the total loss in any span of three consecutive months exceeds Rs. 20,000, their services are terminated at the end of that 3-month period,
2. If the total loss in any span of six consecutive months exceeds Rs. 10,000, their services are terminated at the end of that 6-month period.
Further, at the end of the 12-month probation period, if there are losses on their overall investment, their services are terminated.
Ratan, Shri, Tamal and Upanshu started working for Win Lose in January. Ratan was terminated after 4 months, Shri was terminated after 7 months, Tamal was terminated after 10 months, while Upanshu was not terminated even after 12 months. The table below, partially, lists their monthly profits (in Rs. ‘000’) over the 12-month period, where x, y and z are masked information.
Note:
• A negative profit value indicates a loss.
• The value in any cell is an integer.
Illustration: As Upanshu is continuing after March, that means his total profit during January-March (2z +2z +0) ≥
Rs.20,000. Similarly, as he is continuing after June, his total profit during January − June ≥
Rs.10,000, as well as his total profit during April-June ≥ Rs.10,000.