The event-wise number of participants are as follows: Track: 2124 Field: 2064 Swimming: 2364 Racket Sports: 960 Boxing and Wrestling: 2004

Total number of participants from the five states = $1800 + 1752 + 1992 + 2160 + 1812 = 9516$ Now, $1800 \div 9516 \approx 0.1892 \approx 18.9\%$

The event-wise number of participants from the five states are as follows: Track: 2124 Field: 2064 Swimming: 2364 Hence, the required combined number of participants will be $5.75 \times 2124 + 5.25 \times 2064 + 4 \times 2364 = 12213 + 10836 + 9456 = 32505$

The event-wise number of participants from the five states are as follows: Track: 2124 Field: 2064 Swimming: 2364 Hence, the required combined number of participants will be 5.75 × 2124 + 5.25 × 2064 + 4 × 2364 = 12213 + 10836 + 9456 = 32505 Total female participants from Kerala = 2160 × 0.675 = 1458 528 × 0.5625 + 576 × 7/9 + 504 × 9/(9+5) = 297 + 448+ 324 = 1069 If all 144 participants from Kerala in racket sports were female, then the value of y will be the lowest. 1069+144 = 1213 1458 – 1213 = 245 = minimum possible number of female participants from Kerala in boxing and wrestling events. Hence, maximum possible number of male participants from Kerala in boxing and wrestling events = 408 – 245 =163 Now, 245 ÷ 163 ≈ 1.5031 ⇒ 1.5 is the correct answer.

Let the general term be $T(x) = \frac{x^2+5x+4}{x^2+5x+6}$. Using factorization, $x^2+5x+4 = (x+1)(x+4)$ and $x^2+5x+6 = (x+2)(x+3)$. So, $T(x) = \frac{(x+1)(x+4)}{(x+2)(x+3)}$. The given expression is the product of $T(x)$ for $x$ from 90 to 98. $\frac{m}{n} = \prod_{x=90}^{98} \frac{(x+1)(x+4)}{(x+2)(x+3)}$ $\frac{m}{n} = \frac{(90+1)(90+4)}{(90+2)(90+3)} \times \frac{(91+1)(91+4)}{(91+2)(91+3)} \times ... \times \frac{(98+1)(98+4)}{(98+2)(98+3)}$ $\frac{m}{n} = \frac{91 \times 94}{92 \times 93} \times \frac{92 \times 95}{93 \times 94} \times \frac{93 \times 96}{94 \times 95} \times ... \times \frac{99 \times 102}{100 \times 101}$ Rearranging the terms to see the telescoping product: $\frac{m}{n} = \left(\frac{91}{92} \times \frac{92}{93} \times \frac{93}{94} \times ... \times \frac{99}{100}\right) \times \left(\frac{94}{93} \times \frac{95}{94} \times \frac{96}{95} \times ... \times \frac{102}{101}\right)$ The first part simplifies to $\frac{91}{100}$. The second part simplifies to $\frac{102}{93}$. So, $\frac{m}{n} = \frac{91}{100} \times \frac{102}{93}$. Simplify the fraction: $\frac{102}{93} = \frac{3 \times 34}{3 \times 31} = \frac{34}{31}$. $\frac{m}{n} = \frac{91}{100} \times \frac{34}{31} = \frac{91 \times 34}{100 \times 31} = \frac{3106}{3100}$. To find $m$ and $n$ as co-primes, simplify the fraction $\frac{3106}{3100}$. Divide numerator and denominator by their greatest common divisor, which is 2. $\frac{3106 \div 2}{3100 \div 2} = \frac{1553}{1550}$. 1553 and 1550 are co-primes. Thus, $m = 1553$ and $n = 1550$. The value of $|m - n| = |1553 - 1550| = 3$.

Sum of odd place digits = $8 + 5 + 3 + 1 = 17$ Sum of even place digits = $7 + k + 2 = 9 + k$ Hence Sum of odd place digits - Sum of even place digits = $0, 11, 22,....$ Hence $17 - (9 + k) = 0, 11, 22,....$ $8 - k = 0, 11, 22,....$ Since k is a digit (0-9), the possible values for $8 - k$ are between $8-9=-1$ and $8-0=8$. The only multiple of 11 in this range is 0. Hence $8 - k = 0$ Possible value of $k = 8$ And this is the greatest as well as least possible value of k Hence the difference between the greatest value and smallest possible values = $8 - 8 = 0$

Let cost price of article A =$100x$ Cost price of Article B =$100y$ From the given statements, $18x-10y=85$ $8y-11x = -34$ Solving this equations, we get $x=10$, $y=9.5$ Cost price of article A = $100x$ = Rs.$1000$ Cost price of article B = $100y$ = Rs.$950$

Time for 1 idol to craft by A, B and c are 15,12, and 20 days Taking the LCM of (15,12,20) = 60 units Efficiency of A = 60/15 = 4 units/day Efficiency of B = 60/12 = 5 units/day Efficiency of C = 60/20 = 3 units/day Work done in every 2 days = 9 + 8 =17 units. So, the number of cycles will be 3 and the work done during 3 cycles = 3 x 17 = 51 units. Remaining work on 7th day = 60-51 = 9 units. Work is completed on 7th day by A and B. Therefore, 7 days are required to craft the 1 idol. Hence, First Idol constructed in 7 days. On the 8th day B and C will work. On the 14th day remaining work to be done will be 9/60 to complete the 2nd idol, but on the 14th day work will be done by B and C and their capacity is only of 8/60. Hence 2nd idol will be completed on 15th day.

Hints : nth term=$ar^{n-1}$ n=24 Sum=$a(r^n-1)/r-1$ =$1(2^{24}-1)/(2-1)$ =$2^{24}-1$

Number of ways of selecting (4 consonants out of 7) and (2 vowels out of 4) = $^7C_4 \times ^4C_2 = 35 \times 6 = 210$ Number of groups, each having 4 consonants and 2 vowels = 210. Each group contains 6 letters. Number of ways 6 letters can arrange themselves = $6! = 720$ $\therefore$ Required number of ways = $(210 \times 720) = 151200$.

Hints : Breadth = 396/33=12 Hence maximum 3 rooms (each of width 4meter) can be constructed. Thus, number of sheets required=2 [Hence, Option 3 should be the correct answer.]

We need to find the Principal (original amount invested). Given: Simple Interest = ₹243, Rate = 7.5%, Time = 4 years <hr class="my-[18px]" node="[object Object]"> Using the Simple Interest formula: $SI = \dfrac{P \times R \times T}{100}$ Substituting the known values: $243 = \dfrac{P \times 7.5 \times 4}{100}$ <hr class="my-[18px]" node="[object Object]"> Simplifying: $243 = \dfrac{P \times 30}{100}$ $243 = \dfrac{30P}{100}$ <hr class="my-[18px]" node="[object Object]"> Solving for P: $243 \times 100 = 30P$ $24,300 = 30P$ $P = \dfrac{24,300}{30}$ $P = 810$ <hr class="my-[18px]" node="[object Object]"> Verification: $SI = \dfrac{810 \times 7.5 \times 4}{100} = \dfrac{24,300}{100} = 243$ ✓ The sum invested was ₹810.

Circumference of the wheel = $2\pi r = 2 \times \frac{22}{7} \times 70 = 440$ cm. Number of revolutions per minute = 250. Distance covered in one minute = Circumference $\times$ Revolutions = $440 \times 250 = 110000$ cm = 1100 m = 1.1 km. Speed of the bus = Distance per minute $\times$ 60 minutes/hour = $1.1 \times 60 = 66$ km/hr. Usual time taken = 3 hours. Total distance to the institute = Speed $\times$ Time = $66 \times 3 = 198$ km. One day, she left 1 hour late, so the available time to reach on time = $3 - 1 = 2$ hours. Required speed to cover 198 km in 2 hours = $\frac{198}{2} = 99$ km/hr. Increase in speed required = Required speed - Usual speed = $99 - 66 = 33$ km/hr.

Hints :The roots are real and positive if $D\ge0$, sum of roots $\ge 0$and product of roots $&gt; 0$.(I) $D \ge 0 \implies 4m^2-20m (m-3) \ge 0 \implies -4m^2 + 15m \ge 0$$\implies 4m^2-15m \le 0 \implies m \in [0,3.75]$(II) $\rightarrow$ Sum of roots $\gt 0$:$\frac{2m}{m-3} \gt 0 \implies m \gt 0 \quad m \in (-\infty,0) \cup (3,\infty)$(III) $\rightarrow$ product of roots $\gt 0$:$\frac{5m}{m-3} \gt 0 \implies \frac{m}{m-3} \gt 0 \implies m \in (-\infty,0) \cup (3,\infty)$Combining (I), (II) and (III)Hence $m \in (3,3.75]$

Let the numbers be $x$ and $x+2$. $x+x+2 &gt; 22$ $2x+2 &gt; 22$ $2x &gt; 20$ $x &gt; 10$ Both numbers are smaller than 19, so $x+2 &lt; 19$, which means $x &lt; 17$. Since $x$ is an odd positive integer, possible values for $x$ are 11, 13, 15. The pairs are (11, 13), (13, 15), (15, 17). Total pairs = 3.

Intena Angle = 2 x exte. ayle ant. + ext. = 180 2eat. + eat. = 180 =&gt; ext. ayle=60° No. of beles = $\frac{360}{60} = 6$

$19 \times 3 - 21.25 \times 2 = 57 - 42.5 = 14.5$

Let the cost per jar be Rs y and the cost of each chocolate be Rs z. Hence, we have y + 40z = 640 and 2y + 100z = 1540 ⇒ y + 50z = 770 Hence, we have 10z = 130 ⇒ 5z = 65 ⇒ 35z = 455 and 40z = 520 Hence, y = 640 – 520 = 120 ⇒ y + 35z = 120 + 455 = 575

Since PQRS is a cyclic quadrilateral, $\angle SQR = \angle SPR = 72^\circ \Rightarrow \angle SPQ = 72^\circ + 28^\circ = 100^\circ$. Hence, $\angle QRS = 180^\circ -100^\circ = 80^\circ$.

iimb-bbadbe 2024 Complete Paper Solution | QA Slot 2

Instructions

Question 1.

Question 2.

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Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

iimb-bbadbe 2024 Complete Paper Solution | QA Slot 2

Instructions

Question 1.

<div><div class="latex-wrapper">Among the five types of events, which one saw maximum participation from athletes of the five given states, taken together?<br /></div></div>

Question 2.

<div><div class="latex-wrapper">Approximately what percentage of the total number of participants from the five states taken together was equal to the total number of<br />participants from Punjab?<br /></div></div>

Question 3.

Question 4.

Question 5.

Question 6.

<div><div class="latex-wrapper">If the number 123k578 is divisible by 11, then what is the value of k?<br /></div></div>

Question 7.

Question 8.

Question 9.

Question 10.

<div><div class="latex-wrapper">Out of 7 different consonants and 4 different vowels, how many words (not necessarily meaningful) of 4 consonants and 2 vowels can be formed?<br /></div></div>

Question 11.

Question 12.

<div><div class="latex-wrapper">The simple interest on a certain sum of money invested at 7.5% per annum for 4 years is ₹243. What was the sum (in ₹) invested?<br /></div></div>

Question 13.

Question 14.

<div><div class="latex-wrapper">If the equation $(m - 3)x^2 - 2mx + 5m = 0$ has real and positive roots, then the possible value(s) of m lies in:</div></div>

Question 15.

<div><div class="latex-wrapper">A pair of consecutive odd positive integers, both of which are smaller than 19, have a sum that is more than 22. How many such pairs will be there, if we allow a number to be in more than one pair?<br /></div></div>

Question 16.

<div><div class="latex-wrapper">The measure of each exterior angle of a regular polygon having ‘p' sides is half of the measure of each of its interior angles. The value of 'p' is:<br /></div></div>

Question 17.

<div><div class="latex-wrapper">The average of three numbers is 19 while the average of two of these numbers is 21.25. What is the value of the third number?<br /></div></div>

Question 18.

<div><div class="latex-wrapper">Regarding the function $f(x)=x^{2}-5 x+1$ and $g(x)=1-x-x^{3}$ on the set of real numbers, which of the following statements is not true?<br /></div></div>

Question 19.

Question 20.

<div><div class="latex-wrapper">PQRS is a cyclic quadrilateral whose diagonals PR and QS intersect at the point T. $\angle SQR$ measures $72^\circ$, and $\angle QPR$ measures $28^\circ$. What is the measure of $\angle QRS$?<br /></div></div>