Quantitative Aptitude

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

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

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

A number is divisible by 3 exactly when its digit sum is. The five digits add up to \(1 + 3 + 4 + 6 + 7 = 21\), and a 4-digit number with distinct digits from this set leaves out exactly one digit, so its digit sum is 21 minus the digit left out.

That is a multiple of 3 only when the digit left out is a multiple of 3, i.e. 3 or 6. Each choice leaves four distinct non-zero digits, which can be arranged in \(4! = 24\) ways.

Total \(= 2 \times 24 = 48\).

Answer: B

After the leading 2, the terms are two geometric series interleaved: the powers of \(\tfrac{1}{2}\) (\(\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \tfrac{1}{16}, \ldots\)) and the powers of \(\tfrac{1}{3}\) (\(\tfrac{1}{3}, \tfrac{1}{9}, \tfrac{1}{27}, \ldots\)). All terms are positive, so they may be regrouped. With \(\frac{a}{1 - r}\) for each series:

\[\frac{1/2}{1 - 1/2} = 1, \qquad \frac{1/3}{1 - 1/3} = \frac{1}{2}.\]

Sum \(= 2 + 1 + \tfrac{1}{2} = \tfrac{7}{2}\).

Answer: B

The chart shows shares of the valid votes, so remove the invalid ones first: 1,15,000 − 5,000 = 1,10,000 valid votes.

B and C together hold 25% + 20% = 45% of these: 0.45 × 1,10,000 = 49,500.

Option (C), 51,750, is the trap of taking 45% of all 1,15,000 votes cast, including the invalid ones.

Answer: B

Each of the three children is independently a boy or a girl with probability \(\tfrac{1}{2}\), so all \(2^{3} = 8\) ordered outcomes are equally likely.

Two girls and one boy happens as GGB, GBG or BGG: \(\binom{3}{1} = 3\) outcomes, one for each choice of which child is the boy. The probability is \(\tfrac{3}{8}\).

The tempting \(\tfrac{1}{8}\) counts only one particular birth order.

Answer: A

Let the annual return of fund A be \(r\). Each person invests ₹50,000 in total, so it is enough to compare the returns.

• Person 1: \(10{,}000r + 0.15 \times 40{,}000 = 10{,}000r + 6{,}000\)
• Person 2: \(20{,}000r + 0.15 \times 30{,}000 = 20{,}000r + 4{,}500\)

Person 1 gets ₹500 more, so \((10{,}000r + 6{,}000) - (20{,}000r + 4{,}500) = 500\), which gives \(1{,}500 - 10{,}000r = 500\) and \(r = 0.10\), i.e. 10%.

Answer: B