ISI MS(QMBA) Chapter-wise PYQs

### Method Explanation To find the last digit of a number raised to a large power, we observe the pattern of the last digits of the successive powers of the base number. These patterns are cyclic. Once the cycle length is identified, we divide the given exponent by the cycle length. The remainder of this division will correspond to the position in the cycle, which then gives us the last digit of the number. If the remainder is $0$, it means the last digit is the same as the last digit of the term corresponding to the cycle length. **Given:** The expression $7^{300}$. **To Find:** The last digit of $7^{300}$. **Solution:** Step 1: Determine the pattern of the last digits of powers of $7$. We calculate the first few powers of $7$ and note their last digits: * $7^1 = 7$ (Last digit is $7$) * $7^2 = 49$ (Last digit is $9$) * $7^3 = 7^2 \times 7 = 49 \times 7 = 343$ (Last digit is $3$) * $7^4 = 7^3 \times 7 = 343 \times 7 = 2401$ (Last digit is $1$) * $7^5 = 7^4 \times 7 = 2401 \times 7 = 16807$ (Last digit is $7$) Step 2: Identify the cycle length of the last digits. From Step 1, the pattern of the last digits is $7, 9, 3, 1, 7, \dots$. This pattern repeats every $4$ powers. Therefore, the cycle length is $4$. Step 3: Divide the exponent by the cycle length and find the remainder. The exponent is $300$. We divide $300$ by the cycle length $4$: The remainder of this division is $0$. Step 4: Determine the last digit based on the remainder. When the remainder is $0$, it means the last digit is the same as the last digit of the term corresponding to the cycle length. In this case, it's the $4^{th}$ digit in the cycle. The cycle of last digits is $(7, 9, 3, 1)$. The $4^{th}$ digit in this cycle is $1$. Therefore, the last digit of $7^{300}$ is $1$. **Conclusion:** The last digit in $7^{300}$ is $1$. The final answer is $\boxed{1}$

### Method Explanation The problem asks us to find the sum of a series of function values. The function is given as $f(x) = \frac{9^x}{9^x + 3}$. This type of function often has a special property related to symmetry, specifically $f(x) + f(1-x) = C$ for some constant $C$. We will first verify this property for the given function. Once the property is established, we will group the terms in the series into pairs that sum to this constant $C$. We will then count the number of such pairs and determine if there is a middle term that does not form a pair. Finally, we will sum the contributions from the pairs and the middle term (if any) to find the total sum. **Given:** The function $f(x) = \frac{9^x}{9^x + 3}$. The series to be summed: $S = f\left(\frac{1}{1996}\right) + f\left(\frac{2}{1996}\right) + \dots + f\left(\frac{1995}{1996}\right)$. **To Find:** The value of the sum $S$. **Solution:** **Step 1: Verify the property $f(x) + f(1-x)$.** Let's evaluate $f(x) + f(1-x)$: We can rewrite $9^{1-x}$ as $\frac{9}{9^x}$. Substitute this into the second term: To simplify the second fraction, multiply its numerator and denominator by $9^x$: Now, we can factor out 3 from the denominator of the second term: Since the denominators are the same, we can add the numerators: This property holds for the given function. **Step 2: Identify the terms that form pairs in the sum.** The sum is $S = f\left(\frac{1}{1996}\right) + f\left(\frac{2}{1996}\right) + \dots + f\left(\frac{1995}{1996}\right)$. We can group terms such that their arguments sum to 1. For a term $f\left(\frac{k}{1996}\right)$, its pair will be $f\left(1 - \frac{k}{1996}\right) = f\left(\frac{1996-k}{1996}\right)$. Each such pair sums to 1: For example: $f\left(\frac{1}{1996}\right) + f\left(\frac{1995}{1996}\right) = 1$ $f\left(\frac{2}{1996}\right) + f\left(\frac{1994}{1996}\right) = 1$ and so on. **Step 3: Determine the number of terms and pairs.** The sum contains terms from $k=1$ to $k=1995$. The total number of terms is $N = 1995$. Since $N=1995$ is an odd number, there will be a middle term that does not form a pair. The position of the middle term is $\frac{N+1}{2} = \frac{1995+1}{2} = \frac{1996}{2} = 998$. So, the middle term is $f\left(\frac{998}{1996}\right)$. The number of pairs is $\frac{N-1}{2} = \frac{1995-1}{2} = \frac{1994}{2} = 997$. **Step 4: Calculate the value of the middle term.** The middle term is $f\left(\frac{998}{1996}\right)$. This simplifies to $f\left(\frac{1}{2}\right)$. Substitute $x = \frac{1}{2}$ into the function $f(x)$: **Step 5: Calculate the total sum.** The total sum $S$ is the sum of all the pairs plus the middle term: **Conclusion:** The sum of the terms $f\left(\frac{1}{1996}\right) + f\left(\frac{2}{1996}\right) + \dots + f\left(\frac{1995}{1996}\right)$ is $997.5$. The final answer is $\boxed{\text{997.5}}$

### Method Explanation The problem states that three terms are in an Arithmetic Progression (AP). The fundamental property of an AP is that if $A, B, C$ are three consecutive terms in an AP, then the middle term $B$ is the arithmetic mean of the other two, i.e., $2B = A+C$. We will apply this property to the given logarithmic terms. First, we will set up the equation using the AP property. Since the logarithms are in different bases (base 2 and base 4), we will convert all logarithmic terms to a common base, preferably base 2, using the change of base formula $\log_b a = \frac{\log_c a}{\log_c b}$. After converting to a common base, we will use logarithm properties such as $\log_b M + \log_b N = \log_b (MN)$ and $k \log_b M = \log_b (M^k)$ to simplify the equation. Once the equation is in the form $\log_2(\text{expression 1}) = \log_2(\text{expression 2})$, we can equate the arguments of the logarithms. This will lead to an algebraic equation, which we will solve for $x$. Finally, it's crucial to check if the obtained value(s) of $x$ satisfy the domain requirements for the original logarithmic expressions (i.e., the arguments of the logarithms must be strictly positive). **Given:** The three terms $\log_2(5 \cdot 2^x + 1)$, $\log_4(2^{1-x} + 1)$ and $1$ are in AP. **To Find:** The value of $x$. **Solution:** Step 1: Apply the condition for terms in AP. If $A, B, C$ are in AP, then $2B = A+C$. Here, $A = \log_2(5 \cdot 2^x + 1)$, $B = \log_4(2^{1-x} + 1)$, and $C = 1$. So, we have: Step 2: Convert all logarithms to a common base (base 2). We use the change of base formula: $\log_b a = \frac{\log_c a}{\log_c b}$. For the term $\log_4(2^{1-x} + 1)$: Since $\log_2 4 = 2$, we get: Substitute this back into the AP equation: Step 3: Simplify the equation using logarithm properties. We know that $1$ can be written as $\log_2 2$. So, the equation becomes: Using the logarithm property $\log_b M + \log_b N = \log_b (MN)$: Step 4: Equate the arguments of the logarithms. Since the bases are the same, we can equate the arguments: Step 5: Solve the algebraic equation for $x$. Let $y = 2^x$. Then $2^{1-x} = 2^1 \cdot 2^{-x} = \frac{2}{2^x} = \frac{2}{y}$. Substitute $y$ into the equation: Multiply both sides by $y$ to clear the denominator: Rearrange into a standard quadratic form $ay^2+by+c=0$: Use the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: This gives two possible values for $y$: Step 6: Substitute back $y = 2^x$ and check for valid solutions. Since $y = 2^x$, $y$ must be strictly positive ($2^x > 0$ for all real $x$). The value $y_2 = -\frac{1}{2}$ is not possible, so we discard it. For $y_1 = \frac{2}{5}$: To solve for $x$, take $\log_2$ on both sides: Using the logarithm property $\log_b (M/N) = \log_b M - \log_b N$: Step 7: Verify the domain of the original logarithmic terms. The arguments of the logarithms must be positive: 1. $5 \cdot 2^x + 1$: Since $2^x > 0$ for any real $x$, $5 \cdot 2^x + 1$ is always positive. 2. $2^{1-x} + 1$: Since $2^{1-x} > 0$ for any real $x$, $2^{1-x} + 1$ is always positive. Both arguments are always positive, so the solution $x = 1 - \log_2 5$ is valid. **Conclusion:** The value of $x$ is $1 - \log_2 5$. The final answer is $\boxed{1 - \log_2 5}$

### Method Explanation The problem asks us to determine the relationship between $a, b, c$ given a specific algebraic equation. The core strategy is to manipulate the given equation algebraically to see if it simplifies into one of the standard conditions for Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP). Since the equation involves reciprocals and differences, it hints towards a connection with Harmonic Progression, which is defined by the reciprocals of its terms forming an AP. We will aim to transform the given equation into the characteristic condition for three numbers to be in HP, which is $b = \frac{2ac}{a+c}$ or its equivalent $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$. The condition $b \neq a+c$ will also be considered to ensure validity of steps. **Given:** The equation $\frac{1}{a} + \frac{1}{c} + \frac{1}{a-b} + \frac{1}{c-b} = 0$. Also, $b \neq a+c$. The numbers $a, b, c$ are distinct non-zero real numbers (implied by the presence of reciprocals and terms like $a-b, c-b$). **To Find:** Which of the following statements is true: 1. $a, b, c$ are in GP 2. $a, b, c$ are in AP 3. $a, b, c$ are in HP 4. None of these **Solution:** **Step 1: Rearrange the given equation.** The given equation is: Group the terms: **Step 2: Combine the fractions within each group.** For the first group: For the second group: Substitute these back into the rearranged equation: **Step 3: Manipulate the equation to isolate terms.** Move the second term to the right side: **Step 4: Analyze the condition $b \neq a+c$.** If $a+c=0$, then $c=-a$. The equation from Step 3 becomes: This implies $2b=0$, so $b=0$. If $a+c=0$ and $b=0$, then the given condition $b \neq a+c$ becomes $0 \neq 0$, which is a contradiction. Therefore, $a+c \neq 0$. This allows us to safely divide by $a+c$ if needed. **Step 5: Continue simplifying the equation.** Since $a+c \neq 0$, we can rewrite the equation from Step 3 as: Now, cross-multiply: Expand the left side: Rearrange terms: Combine like terms: This expansion is getting complicated. Let's try to achieve the HP condition $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ directly. From Step 3: We know that for $a,b,c$ to be in HP, $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must be in AP. This means $\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$, or equivalently $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$. So, if $a,b,c$ are in HP, then $\frac{a+c}{ac} = \frac{2}{b}$. Let's substitute $\frac{a+c}{ac}$ with $\frac{2}{b}$ into the equation from Step 3: Cross-multiply: Subtract $2b^2$ from both sides: Move terms to one side to simplify: Factor out $b$ from the right side: Divide by $(a+c)$ (which we established is not zero): **Step 6: Conclude the relationship between $a, b, c$.** The derived condition $b = \frac{2ac}{a+c}$ is the definition of the Harmonic Mean of $a$ and $c$. For three numbers $a, b, c$, if the middle term $b$ is the Harmonic Mean of $a$ and $c$, then $a, b, c$ are in Harmonic Progression. Therefore, $a, b, c$ are in HP. **Conclusion:** The given equation, along with the condition $b \neq a+c$, implies that $a, b, c$ are in Harmonic Progression. The final answer is $\boxed{\text{3. } a, b, c \text{ are in HP}}$

### Method Explanation The problem requires proving a compound inequality involving three positive numbers $a, b, c$ such that $a < b < c$. We will tackle this by proving each part of the inequality separately. The general strategy will be to manipulate each inequality algebraically, moving all terms to one side, and then using the given conditions ($a < b < c$ and $a, b, c > 0$) to show that the resulting expression is positive. Since $a, b, c$ are positive, the denominators $c(a+b+c)$ and $a(a+b+c)$ will always be positive, allowing us to cross-multiply without changing the direction of the inequality. **Given:** $a, b, c$ are positive numbers such that $a < b < c$. **To Find:** Show that $\frac{a^2}{c} < \frac{a^2+b^2+c^2}{a+b+c} < \frac{c^2}{a}$. **Solution:** We need to prove two separate inequalities: 1. $\frac{a^2}{c} < \frac{a^2+b^2+c^2}{a+b+c}$ 2. $\frac{a^2+b^2+c^2}{a+b+c} < \frac{c^2}{a}$ Let's prove the first inequality: **Part 1: Prove $\frac{a^2}{c} < \frac{a^2+b^2+c^2}{a+b+c}$** Step 1: Rewrite the inequality by cross-multiplication. Since $a, b, c > 0$, both $c$ and $a+b+c$ are positive. We can cross-multiply without changing the direction of the inequality: Step 2: Expand both sides of the inequality. Step 3: Rearrange the terms to show the difference is positive. Subtract $a^2c$ from both sides: Now, move all terms to one side to show the expression is positive: Step 4: Group terms and use the given conditions ($a < b < c$ and $a, b, c > 0$). We can group the terms as follows: Let's analyze each group: - For the term $(cb^2 - a^2b)$: Factor out $b$: $b(cb - a^2)$. Since $a < b < c$ and all are positive: $c > b \implies cb > b^2$. $b > a \implies b^2 > a^2$. Therefore, $cb > b^2 > a^2$, which implies $cb - a^2 > 0$. Since $b > 0$, it follows that $b(cb - a^2) > 0$. - For the term $(c^3 - a^3)$: Since $c > a$ and both are positive, $c^3 > a^3$. Therefore, $c^3 - a^3 > 0$. Since both $(cb^2 - a^2b)$ and $(c^3 - a^3)$ are positive, their sum is also positive. Thus, $cb^2 + c^3 - a^3 - a^2b > 0$, which proves the first part of the inequality. Now, let's prove the second inequality: **Part 2: Prove $\frac{a^2+b^2+c^2}{a+b+c} < \frac{c^2}{a}$** Step 1: Rewrite the inequality by cross-multiplication. Since $a, b, c > 0$, both $a$ and $a+b+c$ are positive. We can cross-multiply without changing the direction of the inequality: Step 2: Expand both sides of the inequality. Step 3: Rearrange the terms to show the difference is positive. Subtract $ac^2$ from both sides: Now, move all terms to one side to show the expression is positive: Step 4: Group terms and use the given conditions ($a < b < c$ and $a, b, c > 0$). We can group the terms as follows: Let's analyze each group: - For the term $(bc^2 - ab^2)$: Factor out $b^2$: $b^2(c - a)$. Since $a < b < c$ and all are positive: $c > a \implies c - a > 0$. Since $b > 0$, $b^2 > 0$. Therefore, $b^2(c - a) > 0$. - For the term $(c^3 - a^3)$: Since $c > a$ and both are positive, $c^3 > a^3$. Therefore, $c^3 - a^3 > 0$. Since both $(bc^2 - ab^2)$ and $(c^3 - a^3)$ are positive, their sum is also positive. Thus, $bc^2 + c^3 - a^3 - ab^2 > 0$, which proves the second part of the inequality. Since both inequalities have been proven, the compound inequality holds true. **Conclusion:** Given $a, b, c$ are positive numbers such that $a < b < c$, we have shown that $\frac{a^2}{c} < \frac{a^2+b^2+c^2}{a+b+c}$ and $\frac{a^2+b^2+c^2}{a+b+c} < \frac{c^2}{a}$. Therefore, $\frac{a^2}{c} < \frac{a^2+b^2+c^2}{a+b+c} < \frac{c^2}{a}$ is proven. The final answer is $\boxed{\text{The inequality is proven.}}$

### Method Explanation This problem describes a scenario where an initial value (starting elevation) is given, and a quantity (elevation) changes at a constant rate over time. This is a classic application of an Arithmetic Progression (AP) or a linear function. We will model the balloon's elevation as a function of time. The initial elevation will serve as the starting point (analogous to the first term or the constant in a linear equation), and the constant rate of rise will be the common difference (analogous to the slope in a linear equation). **Given:** * Initial elevation of the balloon (at time $t=0$) = $24$ feet above sea level. * Rate of rise = $45$ feet/minute. **To Find:** * The equation that models the balloon's elevation $h$ as a function of time $t$. **Solution:** **Step 1: Identify the initial value and the rate of change.** The balloon starts at an elevation of $24$ feet. This is the elevation at time $t=0$. The balloon rises at a constant rate of $45$ feet per minute. This means for every minute that passes, the elevation increases by $45$ feet. **Step 2: Relate to the concept of Arithmetic Progression or a linear function.** An Arithmetic Progression is a sequence where each term after the first is obtained by adding a constant difference to the preceding term. In this context, the elevation at each minute mark forms an AP. Let $h(t)$ be the elevation of the balloon at time $t$ minutes. * At $t=0$ minutes (initial time), the elevation is $h(0) = 24$ feet. * At $t=1$ minute, the elevation will be $h(1) = h(0) + 45 = 24 + 45$ feet. * At $t=2$ minutes, the elevation will be $h(2) = h(1) + 45 = (24 + 45) + 45 = 24 + 2 \times 45$ feet. * At $t=3$ minutes, the elevation will be $h(3) = h(2) + 45 = (24 + 2 \times 45) + 45 = 24 + 3 \times 45$ feet. This pattern shows that the elevation $h(t)$ increases by $45$ feet for each minute $t$. **Step 3: Formulate the equation.** Following the pattern from Step 2, for any time $t$ (in minutes), the elevation $h$ can be expressed as: Substituting the given values: It is conventional to write the term with the variable first: **Step 4: Compare with the general $n^{th}$ term formula of an AP (optional but good for conceptual alignment).** The $n^{th}$ term of an AP is given by $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. If we consider the elevation at $t=0$ as the "zeroth" term, or if we adjust the index: Let $a$ be the elevation at $t=0$, so $a=24$. Let $d$ be the common difference (rate of rise), so $d=45$. The elevation at time $t$ can be thought of as the $(t+1)^{th}$ term if we start counting from $t=0$ as the first term. So, $T_{t+1} = a + ((t+1)-1)d = a + td$. Substituting $a=24$ and $d=45$: $h(t) = 24 + 45t$. This confirms the derived equation. **Step 5: Check the given options.** 1. $h = 24t + 45$: This would imply an initial elevation of $45$ and a rise of $24$ feet/minute, which is incorrect. 2. $t = 24h + 45$: This incorrectly expresses time as a function of height and swaps the roles of initial value and rate. 3. $h = 45t + 24$: This matches our derived equation, with $24$ as the initial elevation and $45$ as the rate of rise per minute. 4. $t = 45h + 24$: This incorrectly expresses time as a function of height. **Conclusion:** The equation that correctly models the balloon's elevation $h$ as a function of time $t$ is $h = 45t + 24$. The final answer is $\boxed{\text{h = 45t + 24}}$

### Method Explanation This problem involves calculating probabilities in a game where two players, A and B, take turns. The game ends when one player throws a '6'. Since the game can theoretically go on indefinitely, we will use the concept of an infinite Geometric Progression (GP) to sum the probabilities of each player winning on their respective turns. We will first define the probabilities of success (throwing a '6') and failure (not throwing a '6') in a single throw. Then, we will identify the sequence of events that lead to A winning and B winning. Each sequence of events will have a specific probability, and these probabilities will form a Geometric Progression. Finally, we will use the formula for the sum of an infinite GP, $S_\infty = \frac{a}{1-r}$ (where $a$ is the first term and $r$ is the common ratio), to find the total probability for each player. **Given:** * A standard six-sided die is used. * Players A and B throw the die alternatively. * The first player to throw a '6' wins. * Player A starts the game. **To Find:** * The probability that A wins the game. * The probability that B wins the game. **Solution:** Step 1: Define the probabilities of success and failure in a single throw. Let $p$ be the probability of throwing a '6' in a single throw. For a standard die, there is one '6' out of six faces. Let $q$ be the probability of *not* throwing a '6' in a single throw. Step 2: Calculate the probability of A winning. A can win on their 1st turn, 2nd turn, 3rd turn, and so on. * **A wins on 1st turn:** A throws a '6'. Probability = $p = \frac{1}{6}$. * **A wins on 2nd turn:** A fails, B fails, then A throws a '6'. Probability = $q \cdot q \cdot p = q^2 p = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$. * **A wins on 3rd turn:** A fails, B fails, A fails, B fails, then A throws a '6'. Probability = $q \cdot q \cdot q \cdot q \cdot p = q^4 p = \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}$. This forms an infinite Geometric Progression for A's winning probabilities: The first term of this GP is $a_A = p$. The common ratio is $r_A = q^2$. Since $q = 5/6$, $q^2 = (5/6)^2 = 25/36$. As $|r_A| = 25/36 < 1$, the sum converges. Using the formula for the sum of an infinite GP, $S_\infty = \frac{a}{1-r}$: Substitute the values of $p$ and $q$: Step 3: Calculate the probability of B winning. B can win on their 1st turn, 2nd turn, 3rd turn, and so on. * **B wins on 1st turn:** A fails, then B throws a '6'. Probability = $q \cdot p = \frac{5}{6} \cdot \frac{1}{6}$. * **B wins on 2nd turn:** A fails, B fails, A fails, then B throws a '6'. Probability = $q \cdot q \cdot q \cdot p = q^3 p = \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}$. * **B wins on 3rd turn:** A fails, B fails, A fails, B fails, A fails, then B throws a '6'. Probability = $q \cdot q \cdot q \cdot q \cdot q \cdot p = q^5 p = \left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}$. This forms an infinite Geometric Progression for B's winning probabilities: The first term of this GP is $a_B = qp$. The common ratio is $r_B = q^2$. As calculated before, $q^2 = 25/36 < 1$, so the sum converges. Using the formula for the sum of an infinite GP, $S_\infty = \frac{a}{1-r}$: Substitute the values of $p$ and $q$: Step 4: Verify the sum of probabilities. The sum of the probabilities of A winning and B winning should be $1$, as one of them must win. This confirms our calculations are consistent. **Conclusion:** The probability that A wins the game is $\frac{6}{11}$. The probability that B wins the game is $\frac{5}{11}$. The final answer is $\boxed{\text{A wins with probability } \frac{6}{11}, \text{ B wins with probability } \frac{5}{11}}$.