IPMAT Indore 2024 — MCQ

The Setup: The terms of a geometric progression are real and positive, with the $p$-th term equal to $q$ and the $q$-th term equal to $p$. We need to find the logarithm of the first term.

Step 1: Define the sequence algebraically.
Let the first term be $a$ and the common ratio be $r$.
The general term formula gives:
$$T_p = a r^{p-1} = q$$
$$T_q = a r^{q-1} = p$$

Step 2: Apply logarithms to both equations.
Taking the log of both sides yields a system of linear equations:
$$\log(a) + (p-1)\log(r) = \log(q)$$
$$\log(a) + (q-1)\log(r) = \log(p)$$

Step 3: Eliminate $\log(r)$ to solve for $\log(a)$.
Subtract the second equation from the first:
$$(p-q)\log(r) = \log(q) - \log(p) \implies \log(r) = \frac{\log(q) - \log(p)}{p-q}$$
Substitute $\log(r)$ back into the first equation:
$$\log(a) = \log(q) - (p-1)\left(\frac{\log(q) - \log(p)}{p-q}\right)$$

Step 4: Simplify the expression.
Find a common denominator:
$$\log(a) = \frac{(p-q)\log(q) - (p-1)(\log(q) - \log(p))}{p-q}$$
$$\log(a) = \frac{p\log(q) - q\log(q) - p\log(q) + \log(q) + p\log(p) - \log(p)}{p-q}$$
$$\log(a) = \frac{(1-q)\log(q) - (1-p)\log(p)}{p-q}$$

Final Answer: $\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}$

The Setup: The shortest distance from a given point to a given circle is $4\text{ cm}$ and the longest distance is $9\text{ cm}$. We need to find the radius of the circle.

Step 1: Establish the distance relations based on the point's location.
Let the circle have center $O$ and radius $r$. Let the point be $P$.
The shortest and longest distances from a point to a circle lie along the straight line passing through the point and the center of the circle.

Step 2: Evaluate Case 1 (Point $P$ is outside the circle).
Shortest distance: $OP - r = 4$
Longest distance: $OP + r = 9$
Subtract the two equations to isolate $r$:
$$(OP + r) - (OP - r) = 9 - 4$$
$$2r = 5 \implies r = 2.5\text{ cm}$$

Step 3: Evaluate Case 2 (Point $P$ is inside the circle).
Shortest distance: $r - OP = 4$
Longest distance: $r + OP = 9$
Add the two equations to isolate $r$:
$$(r - OP) + (r + OP) = 4 + 9$$
$$2r = 13 \implies r = 6.5\text{ cm}$$

Final Answer: $2.5\text{ cm}$ or $6.5\text{ cm}$

The Setup: We are given the equations $|x+1|+(y+2)^{2}=0$ and $ax-3ay=1$. We must determine the value of $a$.

Step 1: Analyze the non-negative components of the first equation.
The absolute value function $|x+1|$ and the squared term $(y+2)^2$ must both be strictly greater than or equal to zero for all real numbers.
For their sum to exactly equal zero, each individual term must independently be zero:
$$|x+1| = 0 \implies x = -1$$
$$(y+2)^2 = 0 \implies y = -2$$

Step 2: Substitute the coordinates into the second linear equation.
Substitute $x = -1$ and $y = -2$ into $ax - 3ay = 1$:
$$a(-1) - 3a(-2) = 1$$
$$-a + 6a = 1$$
$$5a = 1 \implies a = \frac{1}{5}$$

Final Answer: $\frac{1}{5}$

The Setup: We need to find the number of real solutions for the absolute value quadratic equation $x^{2}-10|x|-56=0$.

Step 1: Use substitution to simplify the quadratic.
Recognize that for any real number, $x^2 = |x|^2$.
Rewrite the equation in terms of $|x|$:
$$|x|^2 - 10|x| - 56 = 0$$

Step 2: Factor the quadratic equation.
We look for two numbers that multiply to $-56$ and add to $-10$. These are $-14$ and $+4$:
$$(|x| - 14)(|x| + 4) = 0$$

Step 3: Solve for $|x|$ and determine real values for $x$.
This yields two potential cases for $|x|$:
Case 1: $|x| - 14 = 0 \implies |x| = 14 \implies x = 14 \text{ or } x = -14$
Case 2: $|x| + 4 = 0 \implies |x| = -4$
Because the absolute value of a real number cannot be negative, Case 2 produces no real solutions. This leaves exactly $2$ valid real solutions from Case 1.

Final Answer: 2

The Setup: We need to identify the greatest number among the expressions $2^{300}$, $3^{200}$, $4^{100}$, and $2^{100}+3^{100}$.

Step 1: Normalize the exponents to a common power for direct comparison.
We can rewrite each single-term expression by factoring out a power of $100$ in the exponent:
$2^{300} = (2^3)^{100} = 8^{100}$
$3^{200} = (3^2)^{100} = 9^{100}$
$4^{100} = 4^{100}$

Step 2: Evaluate the additive term against the largest single base.
Compare the addition term $2^{100} + 3^{100}$ to the largest normalized term $9^{100}$:
Since $2^{100} < 3^{100}$, their sum satisfies $2^{100} + 3^{100} < 3^{100} + 3^{100} = 2 \times 3^{100}$.
Clearly, $2 \times 3^{100}$ is vastly smaller than $9^{100}$ (which is $3^{200}$).

Step 3: Conclude the greatest term.
Among the expressions $8^{100}$, $9^{100}$, and $4^{100}$, the term with the largest base is $9^{100}$, which corresponds back to $3^{200}$.

Final Answer: $3^{200}$

The Setup: An infinite geometric progression has a sum of $80$, and the sum of its first two terms is $35$. We need to find the number of terms $n$ for which the partial sum is closest to $100$.

Step 1: Establish equations for the GP.
Let the first term be $a$ and the common ratio be $r$.
$$S_\infty = \frac{a}{1-r} = 80 \implies a = 80(1-r)$$
$$S_2 = a + ar = a(1+r) = 35$$

Step 2: Solve for the common ratio $r$ and first term $a$.
Substitute the expression for $a$ into the second equation:
$$80(1-r)(1+r) = 35$$
$$80(1-r^2) = 35 \implies 1-r^2 = \frac{35}{80} = \frac{7}{16}$$
$$r^2 = \frac{9}{16} \implies r = \pm\frac{3}{4}$$
If $r = \frac{3}{4}$, then $a = 80(1 - 3/4) = 20$. The partial sum for positive terms converging to $80$ will never exceed $80$ or approach $100$.
If $r = -\frac{3}{4}$, then $a = 80(1 + 3/4) = 140$. This alternating series can exceed $80$ and approach $100$.

Step 3: Formulate the partial sum $S_n$ for $r = -3/4$.
$$S_n = \frac{140(1 - (-3/4)^n)}{1 - (-3/4)} = \frac{140}{7/4}(1 - (-3/4)^n) = 80(1 - (-3/4)^n)$$

Step 4: Test values of $n$ to find the sum closest to $100$.
For the sum to exceed $80$, $n$ must be odd (making $(-3/4)^n$ negative). Test odd integers:
If $n=3$: $S_3 = 80(1 - (-27/64)) \approx 113.75$ (Distance to 100 = $13.75$)
If $n=5$: $S_5 = 80(1 - (-243/1024)) \approx 98.98$ (Distance to 100 = $1.02$)
If $n=7$: $S_7 = 80(1 - (-2187/16384)) \approx 90.68$ (Distance to 100 = $9.32$)
The value $S_5$ yields the closest proximity to $100$.

Final Answer: 5

This question was cancelled by the exam board. No option was accepted as correct and all candidates were awarded full marks.

The Setup: There are $20$ identical balloons to be distributed among $5$ girls and $3$ boys such that everyone receives at least one balloon. A condition specifies that no girl receives fewer balloons than any boy does.

Step 1: Translate the logical constraints mathematically.
Let $G_i$ be the balloons received by girl $i$ and $B_j$ be the balloons received by boy $j$.
The condition implies $\min(G_i) \ge \max(B_j)$. Let $M = \max(B_j)$. Since $B_j \ge 1$, $M \ge 1$.
The boys must consume at least $M + 2$ balloons (one gets $M$, the others at least $1$).
The girls must consume at least $5M$ balloons.
Total bounded constraint: $5M + M + 2 \le 20 \implies 6M \le 18 \implies M \le 3$.

Step 2: Analyze valid configurations partitioned by $M$.
**Case 1: $M = 1$.**
All $3$ boys get $1$ balloon ($1$ way). The $5$ girls distribute the remaining $17$ balloons ($G_i \ge 1$).
Using stars and bars: $\binom{17-1}{5-1} = \binom{16}{4} = 1820$.
**Case 2: $M = 2$.**
Boy combinations with a max of $2$: $(2,2,2)$ [1 way], $(2,2,1)$ [3 ways], $(2,1,1)$ [3 ways].
Sum(B)=6, remaining 14 to girls ($G_i \ge 2$). Ways $= 1 \times 70 = 70$.
Sum(B)=5, remaining 15 to girls ($G_i \ge 2$). Ways $= 3 \times 126 = 378$.
Sum(B)=4, remaining 16 to girls ($G_i \ge 2$). Ways $= 3 \times 210 = 630$.
Total for Case 2 $= 70 + 378 + 630 = 1078$.
**Case 3: $M = 3$.**
Boys must sum to $\le 5$ since girls need at least $5 \times 3 = 15$. The only boy configuration is $(3,1,1)$ [3 ways].
Sum(B)=5, remaining 15 to girls ($G_i \ge 3$). Ways $= 3 \times 1 = 3$.

Step 3: Sum the cases.
Total exact valid distributions $= 1820 + 1078 + 3 = 2901$.

Note: All four options give ranges from $6000$ upward, so none of them contains $2901$. This question was cancelled by the exam board and all candidates were awarded full marks. The value $2901$ is what the stated constraints actually yield.

Final Answer: 2901

The Setup: We are provided a complex logarithmic expression for a constant $a$, and we need to evaluate $5^a$.

Step 1: Simplify the numerator and denominator using logarithmic properties.
Expression: $a = \frac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_7 25(\log_7 8 - \log_7 4)}$
Using the quotient rule $\log_x(y) - \log_x(z) = \log_x(y/z)$:
Numerator: $(\log_7 4)(\log_7 5 - \log_7 2) = (2\log_7 2)(\log_7 5 - \log_7 2)$
Denominator: $(\log_7 25)(\log_7 8 - \log_7 4) = (2\log_7 5)(\log_7 (8/4)) = (2\log_7 5)(\log_7 2)$

Step 2: Expand terms to locate cancellations.
$$a = \frac{2\log_7 2 (\log_7 5 - \log_7 2)}{2\log_7 5 \log_7 2}$$
Cancel the common factor $2\log_7 2$:
$$a = \frac{\log_7 5 - \log_7 2}{\log_7 5}$$

Step 3: Separate the fraction and change bases.
$$a = 1 - \frac{\log_7 2}{\log_7 5}$$
Using the change of base formula:
$$a = 1 - \log_5 2$$

Step 4: Evaluate $5^a$.
$$5^a = 5^{1 - \log_5 2} = \frac{5^1}{5^{\log_5 2}} = \frac{5}{2}$$

Final Answer: $\frac{5}{2}$

The Setup: We need to find the smallest possible total number of students in a class, given the percentage of girls is strictly between $48\%$ and $50\%$.

Step 1: Formulate the inequality constraint.
Let $G$ be the integer number of girls and $N$ be the total integer number of students.
$$48\% < \frac{G}{N} < 50\%$$
$$\frac{12}{25} < \frac{G}{N} < \frac{1}{2}$$

Step 2: Test values for $N$ starting from the lower bound constraints.
For the fraction to sit strictly between $0.48$ and $0.50$, $N$ must be minimally larger than $25$.
If $N = 25$: $\frac{12}{25} = 0.48$ (Fails the strict greater-than inequality).
If $N = 26$: The maximum integer $G$ strictly less than half of $26$ is $12$. $\frac{12}{26} \approx 0.461$ (Fails, as it is less than $0.48$).
If $N = 27$: The maximum integer $G$ strictly less than half of $27$ is $13$. $\frac{13}{27} \approx 0.48148$ (Valid).

Step 3: Conclude the minimum total.
Since $0.48 < 0.48148 < 0.50$, the minimum total class size $N$ that supports a valid integer amount of girls is $27$.

Final Answer: 27

The Setup: In triangle $ABC$, the side $AB$ is $c$, the median $BD$ has length $k$, and the angle $\angle BDA = \theta$ where $\theta < 90^\circ$. We need to find the area of the entire triangle $ABC$.

Step 1: Establish the area formula using the median.
A median divides a triangle into two smaller triangles of identical area.
$$\text{Area}(ABC) = 2 \times \text{Area}(ABD)$$
Using the base $AD$ and the altitude from $B$ to $AD$ (which is $k\sin\theta$):
$$\text{Area}(ABD) = \frac{1}{2} \times AD \times (k\sin\theta)$$
$$\text{Area}(ABC) = 2 \left( \frac{1}{2} \times AD \times k\sin\theta \right) = AD \cdot k\sin\theta$$

Step 2: Use the Cosine Rule to express $AD$.
In $\triangle ABD$, apply the Law of Cosines:
$$c^2 = AD^2 + k^2 - 2(AD)(k)\cos\theta$$
Rearrange into a quadratic equation in terms of $AD$:
$$AD^2 - (2k\cos\theta)AD + (k^2 - c^2) = 0$$
Solve for $AD$ via the quadratic formula (taking the positive valid geometric root):
$$AD = \frac{2k\cos\theta + \sqrt{4k^2\cos^2\theta - 4(k^2 - c^2)}}{2}$$
$$AD = k\cos\theta + \sqrt{c^2 - k^2(1 - \cos^2\theta)} = k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta}$$

Step 3: Substitute $AD$ back into the Area equation.
$$\text{Area}(ABC) = \left( k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta} \right) k\sin\theta$$
$$\text{Area}(ABC) = k^2\sin\theta\cos\theta + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta}$$
Using the double-angle identity $2\sin\theta\cos\theta = \sin(2\theta)$:
$$\text{Area}(ABC) = \frac{k^2\sin 2\theta}{2} + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta}$$

Final Answer: $\dfrac{k^{2}\sin 2\theta}{2}+k\sin\theta\sqrt{c^{2}-k^{2}\sin^{2}\theta}$

The Setup: In a triangle $ABC$ with $AB=AC$, $D$ is a point on $BC$ creating $\angle BAD=30^\circ$, and $E$ is a point on $AC$ creating $AD=AE$. We need to determine the angle $\angle CDE$.

Step 1: Assign variables to the base angles of the primary triangle.
Let $\angle B = \angle C = \alpha$ (since $AB = AC$).
The vertex angle of the large triangle is $\angle BAC = 180^\circ - 2\alpha$.
The remaining upper angle inside is $\angle DAC = \angle BAC - \angle BAD = (180^\circ - 2\alpha) - 30^\circ = 150^\circ - 2\alpha$.

Step 2: Determine the base angles of the internal isosceles triangle.
In $\triangle ADE$, since $AD = AE$, the base angles are equal:
$$\angle ADE = \angle AED = \frac{180^\circ - \angle DAC}{2}$$
$$\angle ADE = \frac{180^\circ - (150^\circ - 2\alpha)}{2} = \frac{30^\circ + 2\alpha}{2} = 15^\circ + \alpha$$

Step 3: Use the exterior angle theorem to map $\angle CDE$.
Look at $\triangle ABD$. The exterior angle at $D$ is $\angle ADC$, equaling the sum of the remote interior angles:
$$\angle ADC = \angle B + \angle BAD = \alpha + 30^\circ$$
Geometrically, angle $\angle ADC$ is comprised of two adjacent components:
$$\angle ADC = \angle ADE + \angle CDE$$
Substitute the known expressions:
$$\alpha + 30^\circ = (15^\circ + \alpha) + \angle CDE$$
$$\angle CDE = 30^\circ - 15^\circ = 15^\circ$$

Final Answer: $15^{\circ}$

The Setup: We are provided the logarithmic equations $\log_{4}x=a$ and $\log_{25}x=b$. We need to determine the value of $\log_{x} 10$ expressed in terms of variables $a$ and $b$.

Step 1: Change the base of the given logarithms to $x$.
Using the base-change inversion rule $\log_y x = \frac{1}{\log_x y}$:
$$\log_4 x = a \implies \log_x 4 = \frac{1}{a}$$
$$\log_{25} x = b \implies \log_x 25 = \frac{1}{b}$$

Step 2: Simplify the bases to prime numbers.
Using the logarithm power rule $\log_x(y^c) = c \log_x(y)$:
$$\log_x(2^2) = \frac{1}{a} \implies 2 \log_x 2 = \frac{1}{a} \implies \log_x 2 = \frac{1}{2a}$$
$$\log_x(5^2) = \frac{1}{b} \implies 2 \log_x 5 = \frac{1}{b} \implies \log_x 5 = \frac{1}{2b}$$

Step 3: Calculate $\log_x 10$.
Since $10 = 2 \times 5$, we can separate the target logarithm using the product rule:
$$\log_x 10 = \log_x(2 \times 5) = \log_x 2 + \log_x 5$$

Step 4: Substitute the derived fractions and combine.
$$\log_x 10 = \frac{1}{2a} + \frac{1}{2b}$$
Find a common denominator to combine the rational expressions:
$$\log_x 10 = \frac{b + a}{2ab} = \frac{a+b}{2ab}$$

Final Answer: $\dfrac{a+b}{2ab}$

The Setup: There are $5$ boys and $3$ girls sitting randomly around a circular table. We must find the probability that there is at least one boy sitting between any two girls.

Step 1: Calculate the total number of unrestricted circular arrangements.
For $n$ distinct individuals sitting in a circle, the number of unique arrangements is $(n - 1)!$.
Total people $= 5 + 3 = 8$.
$$\text{Total Arrangements} = (8 - 1)! = 7! = 5040$$

Step 2: Calculate the number of restricted arrangements (girls separated).
First, arrange the $5$ boys in a circle.
$$\text{Ways to seat boys} = (5 - 1)! = 4! = 24$$
Seating the $5$ boys creates exactly $5$ gaps between them. To ensure no two girls sit adjacent to each other, we place the $3$ girls into these distinct gaps.
Choose $3$ gaps out of $5$, and arrange the girls:
$$\text{Ways to seat girls} = \binom{5}{3} \times 3! = 10 \times 6 = 60$$
Total valid arrangements $= 24 \times 60 = 1440$.

Step 3: Calculate the probability.
$$\text{Probability} = \frac{1440}{5040} = \frac{144}{504} = \frac{12}{42} = \frac{2}{7}$$