Applied Mathematics (Suppl. 2024)
1. In what ratio must water be mixed with milk to gain \(16\frac{2}{3}\%\) on selling the mixture at cost price?
Solution Let CP of milk be 1 per litre. To gain \(16\frac{2}{3}\%\) (which is \(50/3 \%\)), the profit comes entirely from water.
Profit % = \(\frac{\text{Water}}{\text{Milk}} \times 100\).
\(\frac{50}{3} = \frac{W}{M} \times 100 \Rightarrow \frac{W}{M} = \frac{50}{300} = \frac{1}{6}\).
Profit % = \(\frac{\text{Water}}{\text{Milk}} \times 100\).
\(\frac{50}{3} = \frac{W}{M} \times 100 \Rightarrow \frac{W}{M} = \frac{50}{300} = \frac{1}{6}\).
2. In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. By how much can A beat C in the same race?
Solution When A runs 100m, B runs 75m. (Ratio A:B = 100:75 = 4:3).
When B runs 100m, C runs 96m. (Ratio B:C = 100:96 = 25:24).
Combine ratios: A:C = \(\frac{A}{B} \times \frac{B}{C} = \frac{4}{3} \times \frac{25}{24} = \frac{100}{72}\).
When A runs 100m, C runs 72m. Margin = \(100 - 72 = 28\) m.
When B runs 100m, C runs 96m. (Ratio B:C = 100:96 = 25:24).
Combine ratios: A:C = \(\frac{A}{B} \times \frac{B}{C} = \frac{4}{3} \times \frac{25}{24} = \frac{100}{72}\).
When A runs 100m, C runs 72m. Margin = \(100 - 72 = 28\) m.
3. If \(A = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}\) and \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), then \((A^2 - 6A)\) is equal to:
Solution \(A^2 = \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}\).
\(6A = \begin{bmatrix} 24 & 6 \\ 18 & 12 \end{bmatrix}\).
\(A^2 - 6A = \begin{bmatrix} 19-24 & 6-6 \\ 18-18 & 7-12 \end{bmatrix} = \begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix} = -5I\).
\(6A = \begin{bmatrix} 24 & 6 \\ 18 & 12 \end{bmatrix}\).
\(A^2 - 6A = \begin{bmatrix} 19-24 & 6-6 \\ 18-18 & 7-12 \end{bmatrix} = \begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix} = -5I\).
4. If \(A = \begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix}\) and \(A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}\), then value of x is:
Solution We know \(A A^{-1} = I\).
\(\begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2x & 0 \\ 0 & 2x \end{bmatrix}\).
For this to be identity \(I\), \(2x = 1 \Rightarrow x = 1/2\).
\(\begin{bmatrix} 2x & 0 \\ x & x \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2x & 0 \\ 0 & 2x \end{bmatrix}\).
For this to be identity \(I\), \(2x = 1 \Rightarrow x = 1/2\).
5. \(\int 2^{2x} \cdot 3^x dx\) is equal to:
Solution \(2^{2x} = (2^2)^x = 4^x\).
Integrand becomes \(4^x \cdot 3^x = (4 \cdot 3)^x = 12^x\).
\(\int 12^x dx = \frac{12^x}{\ln 12} + C\).
Integrand becomes \(4^x \cdot 3^x = (4 \cdot 3)^x = 12^x\).
\(\int 12^x dx = \frac{12^x}{\ln 12} + C\).
6. If \(x + y = 8\), then the maximum value of \((xy)\) is:
Solution For a given sum, the product is maximum when numbers are equal. \(x = y = 8/2 = 4\).
Max Product = \(4 \times 4 = 16\).
Max Product = \(4 \times 4 = 16\).
7. The demand curve for a monopolist is given by \(x = 100 - 4p\). The value of x for which \(MR = 0\), is:
Solution \(x = 100 - 4p \Rightarrow 4p = 100 - x \Rightarrow p = 25 - 0.25x\).
Revenue \(R = px = 25x - 0.25x^2\).
\(MR = \frac{dR}{dx} = 25 - 0.5x\). Set \(MR = 0 \Rightarrow 0.5x = 25 \Rightarrow x = 50\).
Revenue \(R = px = 25x - 0.25x^2\).
\(MR = \frac{dR}{dx} = 25 - 0.5x\). Set \(MR = 0 \Rightarrow 0.5x = 25 \Rightarrow x = 50\).
8. A random variable X takes values -1, 0, 1. If its mean is 0.6 and \(P(X=0)=0.2\), then \(P(X=1)\) is:
Solution Let \(P(-1)=a, P(1)=b\).
Sum of probs: \(a + 0.2 + b = 1 \Rightarrow a+b=0.8\).
Mean: \((-1)a + (0)(0.2) + (1)b = 0.6 \Rightarrow b-a=0.6\).
Adding equations: \(2b=1.4 \Rightarrow b=0.7\). So \(P(X=1)=0.7\).
Sum of probs: \(a + 0.2 + b = 1 \Rightarrow a+b=0.8\).
Mean: \((-1)a + (0)(0.2) + (1)b = 0.6 \Rightarrow b-a=0.6\).
Adding equations: \(2b=1.4 \Rightarrow b=0.7\). So \(P(X=1)=0.7\).
9. 100 identical coins (prob p of heads) are tossed. If \(P(50 \text{ heads}) = P(51 \text{ heads})\), then p is:
Solution \(^{100}C_{50} p^{50} q^{50} = ^{100}C_{51} p^{51} q^{49}\).
\(\frac{100!}{50!50!} q = \frac{100!}{51!49!} p \Rightarrow \frac{q}{50} = \frac{p}{51} \Rightarrow 51(1-p) = 50p \Rightarrow p = \frac{51}{101}\).
\(\frac{100!}{50!50!} q = \frac{100!}{51!49!} p \Rightarrow \frac{q}{50} = \frac{p}{51} \Rightarrow 51(1-p) = 50p \Rightarrow p = \frac{51}{101}\).
10. Probability a bomb hits target is \(4/5\). Probability that exactly 2 out of 6 bombs hit is:
Solution \(P(X=2) = ^6C_2 (4/5)^2 (1/5)^4 = 15 \cdot \frac{16}{25} \cdot \frac{1}{625} = \frac{240}{15625} = \frac{48}{3125} = \frac{48}{5^5}\).
11. A specific characteristic of a sample is known as a:
Solution A characteristic of a population is a parameter; a characteristic of a sample is a statistic.
12. The test statistic for a one sample t-test, denoted by t, is defined as:
Solution The standard formula for t-statistic uses the standard error of the mean, which is \(S/\sqrt{n}\).
13. If \(n=6, \Sigma y=84, \Sigma xy=108, \Sigma x^2=70, \Sigma x=0\), the straight line trend is:
Solution \(a = \Sigma y / n = 84/6 = 14\).
\(b = \Sigma xy / \Sigma x^2 = 108/70 \approx 1.54\).
Equation: \(y = a + bx = 14 + 1.54x\).
\(b = \Sigma xy / \Sigma x^2 = 108/70 \approx 1.54\).
Equation: \(y = a + bx = 14 + 1.54x\).
14. At what rate will present value of a perpetuity of ₹500 payable quarterly be ₹40,000?
Solution \(PV = R/i \Rightarrow 40000 = 500/i \Rightarrow i = 500/40000 = 0.0125\).
This 'i' is per quarter. Annual rate = \(0.0125 \times 4 = 0.05 = 5\%\).
This 'i' is per quarter. Annual rate = \(0.0125 \times 4 = 0.05 = 5\%\).
15. If nominal rate r% is compounded k times a year, the effective rate \(r_e\) is:
Solution Effective rate formula is \((1 + \frac{r}{100k})^k - 1\).
16. Annual depreciation is ₹40,000, scrap value ₹50,000, life 15 years. Original cost is:
Solution Depreciation = \(\frac{C-S}{n} \Rightarrow 40000 = \frac{C-50000}{15}\).
\(600000 = C - 50000 \Rightarrow C = 650000\).
\(600000 = C - 50000 \Rightarrow C = 650000\).
17. Solutions to LPP minimizing \(z=3x+2y\) subject to \(x+y \ge 8, 3x+5y \le 15, x, y \ge 0\):
Solution Constraint \(x+y \ge 8\) defines area far from origin. Constraint \(3x+5y \le 15\) defines area near origin (intercepts 5 and 3).
There is no common region (infeasible). Thus, zero solutions.
There is no common region (infeasible). Thus, zero solutions.
18. Investment of ₹10,000 grows to ₹60,000 in 4 years. CAGR is: (Given \(6^{1/4} = 1.565\))
Solution \(CAGR = (\frac{EV}{SV})^{1/n} - 1 = (\frac{60000}{10000})^{1/4} - 1 = 6^{0.25} - 1\).
\(1.565 - 1 = 0.565 = 56.5\%\).
\(1.565 - 1 = 0.565 = 56.5\%\).
19. Assertion (A): The degree of \((\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^2 + \sin(\frac{dy}{dx}) + 1 = 0\) is 3.
Reason (R): Highest power of highest order derivative in a polynomial diff eq is its degree.
Reason (R): Highest power of highest order derivative in a polynomial diff eq is its degree.
Solution Due to the term \(\sin(dy/dx)\), the equation is not a polynomial in derivatives, so degree is not defined. Assertion is False. Reason is the correct definition of degree.
20. Assertion (A): Minor of element \(a_{13}\) in matrix \(\begin{bmatrix} 0 & 2 & 6 \\ 1 & 2 & -1 \\ 2 & 1 & 3 \end{bmatrix}\) is \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\).
Reason (R): Minor is obtained by deleting \(j^{th}\) row and \(i^{th}\) column.
Reason (R): Minor is obtained by deleting \(j^{th}\) row and \(i^{th}\) column.
Solution Assertion: Removing row 1 and col 3 leaves \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\). This is True.
Reason: Minor is obtained by deleting \(i^{th}\) row and \(j^{th}\) column (not j-th row and i-th column). Reason is False.
Reason: Minor is obtained by deleting \(i^{th}\) row and \(j^{th}\) column (not j-th row and i-th column). Reason is False.