General Aptitude

The sentence contrasts Rehman’s early hesitation with something that “never wavered”. The blank therefore needs a noun meaning firm determination, one that can be followed by “to contribute”.

• resolve (as a noun) means exactly that: his resolve to contribute never wavered.
• ambivalence means mixed feelings, the opposite of something unwavering.
• satisfaction is contentment, not a commitment, and does not take “to contribute”.
• revolve is a verb meaning to turn around, so it cannot follow “Rehman’s”.

Answer: C

A bird builds a nest and lives in it. The structure that bees build and live in is a hive, so Bee : Hive keeps the same relation.

The other words do not fit: a kennel is a shelter for a dog, a lair is the den of a wild animal such as a fox or a lion, and a hammock is a hanging bed for people.

Answer: C

Multiply both sides by \(e^{x}\) to get \(Pe^{2x} = Q\), which must hold for every real \(x\).

At \(x = 0\) this gives \(P = Q\), and at \(x = \frac{1}{2}\ln 2\) it gives \(2P = Q\). Together these force \(P = 0\), and then \(Q = 0\). In short, \(e^{2x}\) keeps changing, so \(Pe^{2x}\) can stay equal to the constant \(Q\) only when \(P = 0\).

• B holds only at \(x = 0\).
• C would make the positive number \(e^{x}\) equal the negative number \(-e^{-x}\).
• D needs \(P = 0\) with \(Q \neq 0\), but \(P = 0\) forces \(Q = 0\) as well.

Answer: A

Fold the net with the open triangle at the front. The filled circle above it becomes the top, the open circle below it the bottom, and the blank squares on either side the left and right faces. The filled triangle, two squares to the right, lands at the back.

So the opposite pairs are: open triangle and filled triangle, filled circle and open circle, and the two blank faces.

• B and D show the filled triangle next to the open triangle, but those faces are opposite.
• C shows two blank faces sharing an edge, but the blank faces are opposite.
• A shows the open triangle with the filled circle above it and a blank face beside it, exactly as folded.

Answer: A

The product \(p_{1}p_{2}\) is divisible by \(p_{1}\), which is neither 1 nor the product itself (because \(p_{2} \ge 2\)). So \(p_{1}p_{2}\) always has a third divisor and is never prime, even when \(p_{1} = p_{2}\). Statement B holds for all primes.

One counterexample rules out each other option:

• A: \(2 + 3 = 5\) is prime.
• C: \(3 + 5 + 1 = 9\) is not prime.
• D: \(3 \times 5 + 1 = 16\) is not prime.

Answer: B

Ramya’s sentence is a counterfactual: “if I had known …, I would not have …”. This form describes a situation that did not happen, so it tells us that Ramya did not, in fact, know that Josephine was in the hospital. Her point is that knowing would not have changed anything.

• A contradicts that meaning.
• C and D may or may not be true, but nothing in the sentence mentions a past friendship or the cause of the hospital stay.

Answer: B

Compare IMAGE with FHBNJ. Moving each letter of IMAGE one step forward in the alphabet gives J N B H F, and writing that in reverse order gives FHBNJ. FIELD follows the same rule: one step forward gives G J F M E, which reversed is EMFJG.

For BEACH, one step forward gives C F B D I, and reversing it gives IDBFC. None of the other options fits this rule.

Answer: B

Test the first step, \(i = 1\), where \(I(1) = -4\):

• \(X\): \(20 + (-4) = 16\) matches, while \(20 \times (-4) = -80\) does not, so \(X\) is a running sum. This rules out B and C.
• \(Y\): \(20 \times (-4) = -80\) matches, so \(Y\) is multiplied by the current input. D would give \(20 \times 20 = 400\) instead.

The rule in A also fits the later columns: \(16 + 10 = 26\), \(26 + 15 = 41\), \(-80 \times 10 = -800\) and \(-800 \times 15 = -12000\).

Answer: A

Find the area of triangle PLS in two ways.

• With base PS: PS = 2 cm, and L lies on QR, which is 2 cm from PS. So the area is \(\frac{1}{2} \times 2 \times 2 = 2\) cm².
• With base PL: S lies on MN, which is parallel to PL at a distance LM. So the area is \(\frac{1}{2} \times PL \times LM\), half the rectangle.

Hence the rectangle’s area is \(2 \times 2 = 4\) cm², wherever L sits on QR.

Answer: D

Treat each zone as a point and each river segment as a possible bridge between the two zones on its banks: P (Z1–Z2), R (Z2–Z3), Q (Z1–Z3), S (Z1–Z5), T (Z3–Z5), V (Z3–Z4) and U (Z4–Z5).

Joining 5 zones needs at least 4 bridges, and 4 are enough only if they reach every zone without forming a loop.

• A has only 3 bridges and leaves Z4 cut off.
• B also leaves Z4 cut off, and its bridges on Q, S and T join Z1, Z3 and Z5 in a loop.
• C joins Z1, Z2, Z4 and Z5 each to Z3, so all zones are connected with 4 bridges.
• D connects every zone but uses 5 bridges.

Answer: C