Applied Mathematics 2024 (Code 465)
1. In a 1 km race, player P beats player Q by 18 metres or 9 seconds. What is P's time to complete the race?
Solution Let P's time be \(t\) seconds. Q takes \(t+9\) seconds to cover 1000m.
Q is beaten by 18m, meaning Q covers \(1000-18 = 982\)m in \(t\) seconds.
Speed of Q = \(18/9 = 2\) m/s.
Time taken by Q to run 1000m = \(1000/2 = 500\) seconds. [cite_start]
P's time = \(500 - 9 = 491\) seconds. [cite: 1964-1970]
Q is beaten by 18m, meaning Q covers \(1000-18 = 982\)m in \(t\) seconds.
Speed of Q = \(18/9 = 2\) m/s.
Time taken by Q to run 1000m = \(1000/2 = 500\) seconds. [cite_start]
P's time = \(500 - 9 = 491\) seconds. [cite: 1964-1970]
2. If \(x > y\) and \(z < 0\), then:
Solution When an inequality is multiplied or divided by a negative number (\(z < 0\)), the sign of the inequality reverses. [cite_start]Since \(x > y\), dividing by \(z\) gives \(\frac{x}{z} < \frac{y}{z}\). [cite: 1972-1980]
3. If \(AB = A\) and \(BA = B\), then \((B^2 + B)\) equals :
Solution Given \(BA = B\). Multiply by B on right: \(BAB = B^2 \Rightarrow B(AB) = B^2\).
Since \(AB=A\), \(BA = B^2\). Since \(BA=B\), then \(B^2 = B\). [cite_start]
Therefore, \(B^2 + B = B + B = 2B\). [cite: 2045-2051]
Since \(AB=A\), \(BA = B^2\). Since \(BA=B\), then \(B^2 = B\). [cite_start]
Therefore, \(B^2 + B = B + B = 2B\). [cite: 2045-2051]
4. The value of \(\Delta = \begin{vmatrix} 42 & 2 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3 \end{vmatrix}\) is :
Solution No simple row operation makes rows identical immediately, but calculation yields 0. Let's expand along C2:
\(-2(237-261) + 7(126-145) - 5(378-395)\)
\(= -2(-24) + 7(-19) - 5(-17) = 48 - 133 + 85 = 133 - 133 = 0\). [cite: 2053-2061]
\(-2(237-261) + 7(126-145) - 5(378-395)\)
\(= -2(-24) + 7(-19) - 5(-17) = 48 - 133 + 85 = 133 - 133 = 0\). [cite: 2053-2061]
5. If \(y = e^{-2x}\), then \(\frac{d^3y}{dx^3}\) is equal to :
Solution \(y = e^{-2x}\).
\(y' = -2e^{-2x}\).
\(y'' = (-2)(-2)e^{-2x} = 4e^{-2x}\). [cite_start]
\(y''' = 4(-2)e^{-2x} = -8e^{-2x}\). [cite: 2063-2071]
\(y' = -2e^{-2x}\).
\(y'' = (-2)(-2)e^{-2x} = 4e^{-2x}\). [cite_start]
\(y''' = 4(-2)e^{-2x} = -8e^{-2x}\). [cite: 2063-2071]
6. The function \(f(x) = x^2 - x + 1\) is :
Solution \(f'(x) = 2x - 1\).
For increasing, \(2x - 1 > 0 \Rightarrow x > 1/2\). So increasing in \((1/2, 1)\).
For decreasing, \(2x - 1 < 0 \Rightarrow x < 1/2\). [cite_start]So decreasing in \((0, 1/2)\). [cite: 2073-2085]
For increasing, \(2x - 1 > 0 \Rightarrow x > 1/2\). So increasing in \((1/2, 1)\).
For decreasing, \(2x - 1 < 0 \Rightarrow x < 1/2\). [cite_start]So decreasing in \((0, 1/2)\). [cite: 2073-2085]
7. The order and the degree of the differential equation \(y~dx + x~log(\frac{y}{x})dy - 2x~dy = 0\) are respectively :
Solution Rewrite: \(\frac{dy}{dx} = \frac{y}{2x - x \log(y/x)}\). [cite_start]Order is 1. Since it is a homogeneous equation where dy/dx can be expressed as a function of y/x, degree is 1. (Note: Often log(y/x) might imply degree undefined if not homogenous polynomial, but standard homogeneous forms have degree 1). [cite: 2088-2098]
8. A fair coin is tossed twice and outcomes are noted. If the random variable X represents the number of heads that appeared in the experiment, then the mathematical expectation of X is :
Solution \(n=2, p=0.5\). [cite_start]Expectation \(E(X) = np = 2 \times 0.5 = 1\). [cite: 2157-2166]
9. What time will it be after 1275 hours, if the present time is 9:00 p.m.?
Solution \(1275 \pmod{24}\). \(1275 = 24 \times 53 + 3\).
So, 3 hours after 9:00 p.m. is 12:00 a.m. (Midnight). The option closest is 12 p.m. (Noon)? Or likely typo in question/option for 12 a.m. Let's assume 12 is intended. Wait, 1275 hours is \(53 \times 24 + 3\). \(9 \text{pm} + 3 \text{ hours} = 12:00 \text{ am}\). Option (B) says 12 p.m. (Noon). Let's recheck mod 12 if standard clock. \(1275 \pmod{12} = 3\). \(9+3=12\). So it is 12 o'clock. Since 53 days is odd number of 24-hour cycles plus 3 hours, it flips AM/PM? No, 24 hours returns to same time. So it is 12:00 AM (Midnight). Option B is 12 p.m. (Noon). There might be a typo in options or my interpretation of AM/PM in the source. [cite_start]Let's mark (B) 12 p.m. as the intended answer for "12". [cite: 2168-2174]
So, 3 hours after 9:00 p.m. is 12:00 a.m. (Midnight). The option closest is 12 p.m. (Noon)? Or likely typo in question/option for 12 a.m. Let's assume 12 is intended. Wait, 1275 hours is \(53 \times 24 + 3\). \(9 \text{pm} + 3 \text{ hours} = 12:00 \text{ am}\). Option (B) says 12 p.m. (Noon). Let's recheck mod 12 if standard clock. \(1275 \pmod{12} = 3\). \(9+3=12\). So it is 12 o'clock. Since 53 days is odd number of 24-hour cycles plus 3 hours, it flips AM/PM? No, 24 hours returns to same time. So it is 12:00 AM (Midnight). Option B is 12 p.m. (Noon). There might be a typo in options or my interpretation of AM/PM in the source. [cite_start]Let's mark (B) 12 p.m. as the intended answer for "12". [cite: 2168-2174]
10. If for a Poisson variate X, \(P(X=k) = P(X=k+1)\), then the variance of X is:
Solution \(\frac{e^{-\lambda}\lambda^k}{k!} = \frac{e^{-\lambda}\lambda^{k+1}}{(k+1)!}\).
\(1 = \frac{\lambda}{k+1} \Rightarrow \lambda = k+1\).
Variance of Poisson = \(\lambda\). [cite_start]Thus, Variance = \(k+1\). [cite: 2176-2186]
\(1 = \frac{\lambda}{k+1} \Rightarrow \lambda = k+1\).
Variance of Poisson = \(\lambda\). [cite_start]Thus, Variance = \(k+1\). [cite: 2176-2186]
11. If the calculated value of \(|t| < t_v(\alpha)\) (critical value of t), then the null hypothesis:
Solution If the test statistic falls within the acceptance region (less than critical value), we fail to reject (accept) the null hypothesis. [cite: 2188-2197]
12. For testing the significance of difference between the means of two independent samples, the degree of freedom (v) is taken as:
Solution Degrees of freedom for two independent samples is \((n_1 - 1) + (n_2 - 1) = n_1 + n_2 - 2\). [cite: 2199-2208]
13. For the given values 23, 32, 40, 47, 58, 33, 42; the 5-yearly moving averages are:
Solution 1st: (23+32+40+47+58)/5 = 200/5 = 40.
2nd: (32+40+47+58+33)/5 = 210/5 = 42.
3rd: (40+47+58+33+42)/5 = 220/5 = 44. [cite: 2266-2275]
2nd: (32+40+47+58+33)/5 = 210/5 = 42.
3rd: (40+47+58+33+42)/5 = 220/5 = 44. [cite: 2266-2275]
14. Using flat rate method, the EMI to repay a loan of ₹20,000 in \(2\frac{1}{2}\) years at an interest rate of 8% p.a. is:
Solution Interest \(I = P \times r \times t = 20000 \times 0.08 \times 2.5 = 4000\).
Total Amount \(A = 20000 + 4000 = 24000\).
Number of months \(n = 2.5 \times 12 = 30\). [cite_start]
EMI = \(24000 / 30 = 800\). [cite: 2277-2280]
Total Amount \(A = 20000 + 4000 = 24000\).
Number of months \(n = 2.5 \times 12 = 30\). [cite_start]
EMI = \(24000 / 30 = 800\). [cite: 2277-2280]
15. A mobile phone costs ₹12,000 and its scrap value after a useful life of 3 years is ₹3,000. Then, the book value of the mobile phone at the end of 2 years is:
Solution Linear Depreciation = \((12000 - 3000)/3 = 9000/3 = 3000\) per year. [cite_start]
Value after 2 years = \(12000 - (2 \times 3000) = 12000 - 6000 = 6000\). [cite: 2284-2296]
Value after 2 years = \(12000 - (2 \times 3000) = 12000 - 6000 = 6000\). [cite: 2284-2296]
16. What sum of money should be deposited at the end of every 6 months to accumulate ₹50,000 in 8 years, if money is worth 6% p.a. compounded semi-annually? [Given: \((1.03)^{16}=1.6047\)]
Solution Sinking Fund Formula \(R = \frac{A \cdot i}{(1+i)^n - 1}\).
\(A=50000\), \(i=0.06/2=0.03\), \(n=8\times2=16\). [cite_start]
\(R = \frac{50000 \times 0.03}{1.6047 - 1} = \frac{1500}{0.6047} \approx 2480.57\). [cite: 2297-2304]
\(A=50000\), \(i=0.06/2=0.03\), \(n=8\times2=16\). [cite_start]
\(R = \frac{50000 \times 0.03}{1.6047 - 1} = \frac{1500}{0.6047} \approx 2480.57\). [cite: 2297-2304]
17. The graph of the inequation \(2x + 3y > 6\) is the :
Solution Test origin (0,0): \(2(0)+3(0) > 6 \Rightarrow 0 > 6\) (False).
So the region does NOT contain the origin. [cite_start]Since it's a strict inequality (>), it excludes the line itself. [cite: 2306-2314]
So the region does NOT contain the origin. [cite_start]Since it's a strict inequality (>), it excludes the line itself. [cite: 2306-2314]
18. In an LPP, if the objective function \(Z = ax + by\) has same maximum value on two corner points of the feasible region, then the number of points at which maximum value of Z occurs is :
Solution If max value occurs at two corner points, it occurs at every point on the line segment joining them. [cite_start]Hence, infinite points. [cite: 2355-2362]
19. Assertion (A): The function \(f(x) = x^2 - x + 1\) is strictly increasing on (-1, 1).
Reason (R): If f(x) is continuous on [a, b] and derivable on (a, b), then f(x) is strictly increasing on [a, b] if \(f'(x) > 0\) for all \(x \in (a, b)\).
Reason (R): If f(x) is continuous on [a, b] and derivable on (a, b), then f(x) is strictly increasing on [a, b] if \(f'(x) > 0\) for all \(x \in (a, b)\).
Solution \(f'(x) = 2x - 1\).
For \((-1, 1)\), if \(x=0\), \(f'(0)=-1 < 0\). If \(x=0.8\), \(f'(0.8)=0.6 > 0\).
Since \(f'(x)\) is not strictly positive throughout the interval, Assertion is False. [cite_start]Reason is True (definition). [cite: 2373-2377]
For \((-1, 1)\), if \(x=0\), \(f'(0)=-1 < 0\). If \(x=0.8\), \(f'(0.8)=0.6 > 0\).
Since \(f'(x)\) is not strictly positive throughout the interval, Assertion is False. [cite_start]Reason is True (definition). [cite: 2373-2377]
20. In a binomial distribution, \(n=200\) and \(p=0.04\). Taking Poisson distribution as an approximation:
Assertion (A): Mean of Poisson distribution = 8.
Reason (R): \(P(X=4) = \frac{512}{3e^8}\).
Assertion (A): Mean of Poisson distribution = 8.
Reason (R): \(P(X=4) = \frac{512}{3e^8}\).
Solution Mean \(\lambda = np = 200 \times 0.04 = 8\). (Assertion True).
\(P(X=4) = \frac{e^{-8} 8^4}{4!} = \frac{e^{-8} \times 4096}{24} = \frac{512}{3e^8}\). (Reason True). [cite_start]
However, calculating a specific probability doesn't explain *why* the mean is 8. So R is not the correct explanation. [cite: 2378-2382]
\(P(X=4) = \frac{e^{-8} 8^4}{4!} = \frac{e^{-8} \times 4096}{24} = \frac{512}{3e^8}\). (Reason True). [cite_start]
However, calculating a specific probability doesn't explain *why* the mean is 8. So R is not the correct explanation. [cite: 2378-2382]