General Aptitude

Courage and bravery are synonyms: both name the strength to face danger or fear without giving in. The second pair must therefore match yearning with a word of the same meaning. Yearning is a deep, often wistful desire for something, which is exactly what longing means.

Yelling (shouting), yawning (opening the mouth wide from tiredness) and glaring (staring angrily) only resemble yearning in sound or spelling, not in meaning.

Answer: A

The playing was already in progress when another past event, the rain, interrupted it. An action that continued up to a moment in the past takes the past perfect continuous, had been + verb-ing: “We had been playing tennis in the lawn when it suddenly started to rain.”

• Have been playing is present perfect continuous; it ties the action to the present and clashes with the past tense “started”.
• Would have been playing and could be playing express a guess or a possibility, not an action that actually took place.

Answer: B

The condition \(U \le 4\) includes 4, so count every pixel with value 0, 1, 2, 3 or 4, row by row:

• Row 1 (0, 1, 0, 2): all 4 pixels
• Row 2 (4, 7, 3, 3): 3 pixels
• Row 3 (5, 5, 4, 4): 2 pixels
• Row 4 (6, 7, 3, 2): 2 pixels

The total is \(4 + 3 + 2 + 2 = 11\). As a check, only five pixels are above 4 (5, 5, 6, 7, 7), and \(16 - 5 = 11\). Leaving out the three pixels equal to 4 would give 8, option B.

Answer: C

Each region carries the sum of the numbers of the shapes it lies in. The given label 4 confirms the rule: that region is inside the rectangle and the ellipse but outside the triangle, and \(1 + 3 = 4\).

• P is inside all three shapes: \(P = 1 + 2 + 3 = 6\).
• Q is inside the triangle and the ellipse only: \(Q = 2 + 3 = 5\).
• R is inside the rectangle and the triangle only: \(R = 1 + 2 = 3\).

Answer: A

The new width is \(1.1W\), while the length stays \(L\).

• Area: \(L \times 1.1W = 1.1LW\), exactly 10% more for every \(L\) and \(W\).
• Perimeter: it grows by \(0.2W\), which is less than 10% of \(2(L + W)\) because the length does not grow.
• Diagonal: \(\sqrt{L^{2} + 1.21W^{2}} \lt 1.1\sqrt{L^{2} + W^{2}}\), so it also grows by less than 10%.
• Square: only if \(1.1W = L\), which holds for one particular ratio, not for all values.

Answer: C

Each response uses a phrasal verb with “up”, and each verb suits exactly one statement:

• P – 3: a house in a mess needs to be cleared up, that is, tidied.
• Q – 4: a complaint about marks is taken up (raised) with the teacher.
• R – 1: to someone who avoids politics, Kanishk promises not to bring it up (mention it).
• S – 2: an unknown word can be looked up in a dictionary.

This matching is P – 3; Q – 4; R – 1; S – 2.

Answer: B

Let \(w\) be the common weight function, taken as strictly increasing up to age 50 and strictly decreasing after it, so its single peak is at 50.

• If both ages are at most 50, then \(a_{1} \lt a_{2}\) gives \(w(a_{1}) \lt w(a_{2})\).
• If both ages are at least 50, then \(w(a_{1}) \gt w(a_{2})\).

Either way the weights differ. These two cases cover options A and C, and also D, because an age of exactly 50 falls into one of them. Equal weights are therefore possible only with the ages on opposite sides of the peak: \(a_{1} \lt 50 \lt a_{2}\).

Answer: B

Joining the centre to the vertices gives three kinds of triangles:

• 1, 4, 7: a 60° angle at the centre, so equilateral with side \(r\); area \(\frac{\sqrt{3}}{4}r^{2}\).
• 2, 3, 5, 6, 8, 9: sides \(r\) and \(r\) with 30° between them; area \(\frac{1}{2}r^{2}\sin 30^\circ = \frac{r^{2}}{4}\).
• 10, 11, 12: sides \(d\) and \(d\) with the dodecagon’s 150° angle; area \(\frac{d^{2}}{4} = \frac{(2 - \sqrt{3})r^{2}}{4}\), since \(d^{2} = 2r^{2}(1 - \cos 30^\circ)\).

The total area is \(3r^{2}\), so 3 squares of side \(r\). One equilateral, two 30° and one small triangle add up to exactly \(r^{2}\), and they fit: the equilateral triangle stands on one side of the square, the 30° triangles fill the spaces beside it, and the small triangle closes the gap at the opposite side. So each square uses 4 triangles.

Answer: A

Write the right side as a power of 3: \(27 \times 9^{x} = 3^{3} \times 3^{2x} = 3^{2x + 3}\). Equating exponents gives \(x^{2} = 2x + 3\), that is, \(x^{2} - 2x = 3\).

The required value is \(\dfrac{2^{x^{2}}}{(2^{x})^{2}} = \dfrac{2^{x^{2}}}{2^{2x}} = 2^{x^{2} - 2x} = 2^{3}\).

There is no need to find \(x\): both roots, \(x = 3\) and \(x = -1\), satisfy \(x^{2} - 2x = 3\). Adding the exponents instead of subtracting them (with \(x = 3\)) would give the wrong option \(2^{15}\).

Answer: C

From the chart, 20 shifts had 5 patients, 40 had 6, 30 had 7 and 10 had 8, a total of 100 shifts. They saw \(5(20) + 6(40) + 7(30) + 8(10) = 630\) patients, so the mean per shift is \(\bar{X} = \frac{630}{100} = 6.3\).

The earning \(1000(X - 0.2)\) is a linear function of \(X\), so its average is the same function of the average: \(1000(6.3 - 0.2) = 6100\), that is, ₹ 6,100. Forgetting the \(-0.2\) gives 6,300, option B.

Answer: A