Data Sufficiency — PPSC previous year questions

Each statement alone is one equation with two unknowns, so neither fixes the prices. Together, 2p + 3q = 10 and p + 2q = 5 give p = 5 and q = 0, so 3 pens and 4 pencils cost $15; both statements together are needed.

The angles of a triangle add to 180°. Statement 1 alone gives B = 60°, so A + C = 120° and A may or may not be 90°. Statement 2 alone gives C = 30°, so A + B = 150°, again undetermined. Using both, A = 180° − 60° − 30° = 90°, so the question is answered only when the statements are combined.

The stem already gives the daily output of 1000 chairs. Statement 1 adds the working hours, so the hourly rate is 1000 ÷ 8 = 125 chairs — sufficient on its own. Statement 2 states the hourly rate directly (125 chairs per hour) — also sufficient on its own.

Statement 1: x ÷ 2 being a whole number means x is even, so it is enough alone. Statement 2: x + 5 is odd only when x is even (even + odd = odd), so it is also enough alone. Each statement answers the question by itself.

Statement 1 gives only the ratio l = 2b. Statement 2 gives only 2(l + b) = 24. Neither alone fixes the two dimensions. Together they do: > 2(2b + b) = 24 → 6b = 24 → b = 4 cm, l = 8 cm > Area = 8 × 4 = 32 cm² So both statements together are sufficient.

Each statement alone has two unknowns, so neither gives x by itself. Adding the two equations gives 2y = 16, so y = 8 and x = 2, meaning both statements together are needed.

Statement I is a perfect square, so it fixes a on its own. > a² + 6a + 9 = (a + 3)² = 0 → a = −3 Statement II (a > 0) allows infinitely many values and settles nothing by itself, so Statement I alone is sufficient.

Statement 2 alone fixes the weekday: the third Saturday is the 18th, so the Saturdays fall on the 4th, 11th and 18th. > 14th = 18th − 4 days = Saturday − 4 = Tuesday Statement 1 gives only the weekday of the last day, and the month may have 28, 29, 30 or 31 days, so the 14th cannot be placed from it. Statement 2 alone is sufficient.

From statement I alone: Akash > Prabhakar > Manoj and Kamal, so Akash is the heaviest — sufficient. From statement II alone: Prabhakar and Akash are both heavier than Manoj, who is heavier than Kamal, but nothing compares Prabhakar with Akash, so the heaviest cannot be fixed — not sufficient.

Let the bigger number be b and the smaller be s. Statement I gives b = s + 6. Statement II gives 0.4s = 0.3b, so b = 4s/3. Statement III gives (b/2) : (s/3) = 2 : 1, so 3b = 4s, again b = 4s/3. So II and III carry the same information. Together they cannot fix the numbers. > I + II: s + 6 = 4s/3 → s = 18, b = 24, sum = 42 > I + III: same equation → sum = 42 I with either II or III is enough.