Quantitative Aptitude

Fixing the wrong entry raises the class total by \(42 - 24 = 18\) marks, and this raises the average by \(31.4 - 30.8 = 0.6\). If the class has \(n\) students, the average rises by \(\frac{18}{n}\), so \(\frac{18}{n} = 0.6\) and \(n = 30\). Check: \(30 \times 30.8 = 924\) and \(924 + 18 = 942 = 30 \times 31.4\).

Answer: C

Let the distance from P to QR be \(h\). ST is \(\frac{h}{2}\) from QR, so it is also \(\frac{h}{2}\) from P. Because ST is parallel to QR, triangle PST is similar to triangle PQR with scale factor \(\frac{1}{2}\), so its area is \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\) of the area of PQR. The trapezium SQRT is the remaining \(\frac{3}{4}\), so the required ratio is \(\frac{1/4}{3/4} = \frac{1}{3}\).

Answer: A

In one roll, a 6 comes with probability \(\frac{1}{6}\) and any other face with probability \(\frac{5}{6}\), and the three rolls are independent. The single 6 can be on the first, second or third roll, so \(P = 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^2 = 3 \times \frac{25}{216} = \frac{75}{216}\).

Answer: A

Since the order does not matter, a purchase is fixed by how many scoops of each flavour are taken, that is, by whole numbers \(a, b, c, d \ge 0\) with \(a + b + c + d = 3\). By the stars and bars formula, the count is \(\binom{3 + 4 - 1}{3} = \binom{6}{3} = 20\). Check by cases: all three of one flavour gives 4 ways, two of one flavour and one of another gives \(4 \times 3 = 12\) ways, and three different flavours gives \(\binom{4}{3} = 4\) ways, a total of 20.

Answer: B

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

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

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

“Proportional” means \(x + y = k(x - y)\) for some fixed constant \(k\). Rearranging,

\[x + y = kx - ky \;\Rightarrow\; y(k + 1) = x(k - 1) \;\Rightarrow\; \frac{x}{y} = \frac{k + 1}{k - 1}.\]

The right-hand side involves only the constant \(k\), so \(\frac{x}{y}\) is the same for every pair that satisfies the condition: it is a constant. (Here \(k \neq \pm 1\), because \(k = 1\) would force \(y = 0\) and \(k = -1\) would force \(x = 0\).)

Answer: D

The median is the middle value once the data are sorted:

9, 10, 11, 11, 13, 14, 15, 17, 18, 69

There are ten values, an even count, so the median is the average of the 5th and 6th values: \(\frac{13 + 14}{2} = 13.5\).

The large value 69 drags the mean up to \(\frac{187}{10} = 18.7\) (option D), but the median depends only on the order of the values, so the outlier does not move it. Option C, 11, is the mode.

Answer: A

Take the coin counts as 5, 3 and 13; multiplying all three by the same factor would not change the percentage. The money held in each kind of coin is:

• ₹1 coins: 5 × ₹1 = ₹5
• ₹5 coins: 3 × ₹5 = ₹15
• ₹10 coins: 13 × ₹10 = ₹130

The total is ₹150, so the share in ₹5 coins is \(\frac{15}{150} \times 100 = 10\%\).

Option B, \(14\frac{2}{7}\%\), is the share of ₹5 coins by count (3 out of 21 coins), not by value.

Answer: C