Cost per litre of X, Y, Z: $₹120, ₹125, ₹150$. Cost of Mixture P (3:2:5 ratio): Total parts = $3+2+5=10$. Cost for 10 parts = $3(120) + 2(125) + 5(150) = 360 + 250 + 750 = ₹1360$. Cost per litre of P = $1360/10 = ₹136$. Cost of Mixture Q (4:3:3 ratio): Total parts = $4+3+3=10$. Cost for 10 parts = $4(120) + 3(125) + 3(150) = 480 + 375 + 450 = ₹1305$. Cost per litre of Q = $1305/10 = ₹130.5$. Profit Calculation for 100 litres of P and 100 litres of Q sold at ₹136/litre: Profit from P = $100 \times (136 - 136) = ₹0$. Profit from Q = $100 \times (136 - 130.5) = 100 \times 5.5 = ₹550$. Total profit = $0 + 550 = ₹550$.

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<ul node="[object Object]"> <li node="[object Object]">Ratio of P's income to Q's income = $9:11$</li> <li node="[object Object]">Sum of both incomes = ₹$2,40,000$</li> <li node="[object Object]">Q's expenditure = ₹$84,000$</li> </ul> <hr class="my-[18px]" node="[object Object]"> Let P's income $=9x$, Let Q's income $=11x$ $9x + 11x = 2,40,000$ $20x = 2,40,000$ $x = 12,000$ Therefore: <ul node="[object Object]"> <li node="[object Object]">P's income = $9 \times 12,000$ = ₹$1,08,000$</li> <li node="[object Object]">Q's income = $11 \times 12,000$ = ₹$1,32,000$</li> </ul> <hr class="my-[18px]" node="[object Object]"> Using Income = Expenditure + Savings For Q: Q's savings = $1,32,000 - 84,000$ = ₹$48,000$ <hr class="my-[18px]" node="[object Object]"> P's savings are $\dfrac{100}{3}$% more than Q's savings P's savings = Q's savings + $\dfrac{100}{3}$% of Q's savings P's savings = $48,000 + \dfrac{100}{3} \times \dfrac{48,000}{100}$ P's savings = $48,000 + 16,000 = ₹64,000$ <hr class="my-[18px]" node="[object Object]"> For P: P's expenditure = P's income - P's savings P's expenditure = $1,08,000 - 64,000=₹44,000$

Let the original speed be $x$ km/h Reduced speed = $(x - 450)$ km/h Distance = $1400$ km Time increase = $60$ minutes = $1$ hour <hr class="my-[18px]" node="[object Object]"> Using Time = Distance ÷ Speed: Original time = $\dfrac{1400}{x}$ hours New time = $\dfrac{1400}{x-450}$ hours <hr class="my-[18px]" node="[object Object]"> Since new time is 1 hour more than original time: $\dfrac{1400}{x-450} - \dfrac{1400}{x} = 1$ <hr class="my-[18px]" node="[object Object]"> Multiplying by $x(x-450)$: $1400x - 1400(x-450) = x(x-450)$ $1400x - 1400x + 630000 = x^2 - 450x$ $630000 = x^2 - 450x$ $x^2 - 450x - 630000 = 0$ <hr class="my-[18px]" node="[object Object]"> Using quadratic formula: $x = \dfrac{450 \pm \sqrt{450^2 + 4 \times 630000}}{2}$ $x = \dfrac{450 \pm \sqrt{202500 + 2520000}}{2}$ $x = \dfrac{450 \pm \sqrt{2722500}}{2}$ $x = \dfrac{450 \pm 1650}{2}$ Taking positive value: $x = \dfrac{450 + 1650}{2} = 1050$ km/h <hr class="my-[18px]" node="[object Object]"> Reduced speed = $1050 - 450 = 600$ km/h <hr class="my-[18px]" node="[object Object]"> Percentage = $\dfrac{600}{1050} \times 100 = 57.14\%$

Looking at the signs: $+ - - + - - + - - + \ldots$ The pattern repeats every 3 terms, so we group accordingly. <hr class="my-[18px]" node="[object Object]"> Group the terms: $S = \left(1 - \dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{8} - \dfrac{1}{16} - \dfrac{1}{32}\right) + \left(\dfrac{1}{64} - \dfrac{1}{128} - \dfrac{1}{256}\right) + \ldots$ <hr class="my-[18px]" node="[object Object]"> Calculate each group: Group 1: $1 - \dfrac{1}{2} - \dfrac{1}{4}$ Convert to same denominator: $\dfrac{4}{4} - \dfrac{2}{4} - \dfrac{1}{4} = \dfrac{4-2-1}{4} = \dfrac{1}{4}$ Group 2: $\dfrac{1}{8} - \dfrac{1}{16} - \dfrac{1}{32}$ $\dfrac{4}{32} - \dfrac{2}{32} - \dfrac{1}{32} = \dfrac{4-2-1}{32} = \dfrac{1}{32}$ Group 3: $\dfrac{1}{64} - \dfrac{1}{128} - \dfrac{1}{256}$ $\dfrac{4}{256} - \dfrac{2}{256} - \dfrac{1}{256} = \dfrac{4-2-1}{256} = \dfrac{1}{256}$ <hr class="my-[18px]" node="[object Object]"> Rewrite as a new series: $S = \dfrac{1}{4} + \dfrac{1}{32} + \dfrac{1}{256} + \ldots$ Notice that: $\dfrac{1}{32} = \dfrac{1}{4} \times \dfrac{1}{8}$ $\dfrac{1}{256} = \dfrac{1}{4} \times \dfrac{1}{64} = \dfrac{1}{4} \times \left(\dfrac{1}{8}\right)^2$ So: $S = \dfrac{1}{4}\left(1 + \dfrac{1}{8} + \dfrac{1}{64} + \ldots\right)$ <hr class="my-[18px]" node="[object Object]"> The series inside parentheses is geometric with: <ul node="[object Object]"> <li node="[object Object]">First term: $a = 1$</li> <li node="[object Object]">Common ratio: $r = \dfrac{1}{8}$</li> </ul> Using the infinite geometric series formula $\dfrac{a}{1-r}$: $\text{Sum inside parentheses} = \dfrac{1}{1-\dfrac{1}{8}} = \dfrac{1}{\dfrac{7}{8}} = \dfrac{8}{7}$ <hr class="my-[18px]" node="[object Object]"> Final calculation: $S = \dfrac{1}{4} \times \dfrac{8}{7} = \dfrac{8}{28} = \dfrac{2}{7}$

Let the total work be $60$ units (LCM of $30$ and $20$ for clean calculations). <hr class="my-[18px]" node="[object Object]"> Find the work rates: $(A + C)$ complete $60$ units in $30$ days, so $(A + C) = \dfrac{60}{30} = 2$ units/day $(A + B + C)$ complete $60$ units in $20$ days, so $(A + B + C) = \dfrac{60}{20} = 3$ units/day Therefore, $B = (A + B + C) - (A + C) = 3 - 2 = 1$ unit/day <hr class="my-[18px]" node="[object Object]"> Since $B$ is $20\%$ more efficient than $C$: $B = 1.2C$ $1 = 1.2C$ $C = \dfrac{1}{1.2} = \dfrac{5}{6}$ units/day $A = (A + C) - C = 2 - \dfrac{5}{6} = \dfrac{12 - 5}{6} = \dfrac{7}{6}$ units/day <hr class="my-[18px]" node="[object Object]"> Phase 1: $C$ completes $\dfrac{1}{4}$ of the work Work done by $C = \dfrac{1}{4} \times 60 = 15$ units Time taken by $C = \dfrac{15}{\dfrac{5}{6}} = 15 \times \dfrac{6}{5} = 18$ days <hr class="my-[18px]" node="[object Object]"> Phase 2: $A$ and $B$ complete remaining work Remaining work $= 60 - 15 = 45$ units Combined rate of $A$ and $B = \dfrac{7}{6} + 1 = \dfrac{7}{6} + \dfrac{6}{6} = \dfrac{13}{6}$ units/day Time taken by $A$ and $B = \dfrac{45}{\dfrac{13}{6}} = 45 \times \dfrac{6}{13} = \dfrac{270}{13}$ days <hr class="my-[18px]" node="[object Object]"> Total time $= 18 + \dfrac{270}{13} = \dfrac{234}{13} + \dfrac{270}{13} = \dfrac{504}{13} = 38\dfrac{10}{13}$ days

Let me break this down into simple steps that anyone can follow! We have $13$ guests total. We're putting $5$ guests at one round table and $8$ guests at another round table. We need to find ALL possible ways to do this. <hr class="my-[18px]" node="[object Object]"> First, we need to decide which $5$ guests (out of $13$) sit at the first table. Number of ways to choose $5$ guests from $13$ = $\binom{13}{5}$ Why do we use combinations here? Because at this step, we only care about WHO sits at which table, not WHERE they sit at the table. $\binom{13}{5} = \dfrac{13!}{5! \times 8!} = \dfrac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287$ The remaining $8$ guests automatically go to the second table. <hr class="my-[18px]" node="[object Object]"> Now we arrange the people around each circular table. Key concept: For a round table with $n$ people, we fix one person's position (to avoid counting rotations as different arrangements) and arrange the remaining $(n-1)$ people. This gives us $(n-1)!$ arrangements. First table ($5$ people): $(5-1)! = 4! = 24$ ways Second table ($8$ people): $(8-1)! = 7! = 5040$ ways <hr class="my-[18px]" node="[object Object]"> Since each choice is independent, we multiply: Total arrangements = (Ways to choose groups) $\times$ (Ways to arrange table $1$) $\times$ (Ways to arrange table $2$) Total = $\binom{13}{5} \times 4! \times 7!$ $Total = \dfrac{13!}{5! \times 8!} \times 4! \times 7!$ <hr class="my-[18px]" node="[object Object]"> Let me show you a shortcut: $\dfrac{13!}{5! \times 8!} \times 4! \times 7! = \dfrac{13! \times 4! \times 7!}{5! \times 8!}$ Since $5! = 5 \times 4!$ and $8! = 8 \times 7!$: $= \dfrac{13! \times 4! \times 7!}{(5 \times 4!) \times (8 \times 7!)} = \dfrac{13!}{5 \times 8} = \dfrac{13!}{40}$ Final Answer: $\dfrac{13!}{40}$

We need to find the area enclosed by four boundary lines. Let's rewrite each line in slope-intercept form: $x = 6$ is a vertical line $y = 0$ is the x-axis $3x - 2y = 0$ gives us $y = \dfrac{3x}{2}$ $x - 4y + 10 = 0$ gives us $y = \dfrac{x + 10}{4}$ <hr class="my-[18px]" node="[object Object]"> Find where each pair of lines intersects: Intersection of $y = 0$ and $y = \dfrac{3x}{2}$: Setting equal: $0 = \dfrac{3x}{2}$ gives us $x = 0$ Point: $(0, 0)$ Intersection of $y = 0$ and $x = 6$: Point: $(6, 0)$ Intersection of $x = 6$ and $y = \dfrac{x + 10}{4}$: Substituting: $y = \dfrac{6 + 10}{4} = \dfrac{16}{4} = 4$ Point: $(6, 4)$ Intersection of $y = \dfrac{3x}{2}$ and $y = \dfrac{x + 10}{4}$: Setting equal: $\dfrac{3x}{2} = \dfrac{x + 10}{4}$ Cross multiply: $12x = 2x + 20$ Solving: $10x = 20$, so $x = 2$ Then $y = \dfrac{3(2)}{2} = 3$ Point: $(2, 3)$ <hr class="my-[18px]" node="[object Object]"> Split the quadrilateral into two triangles: Triangle 1: $(0,0)$, $(2,3)$, $(6,0)$ Triangle 2: $(2,3)$, $(6,4)$, $(6,0)$ Using the triangle area formula: Area = $\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ Triangle 1 area: $\dfrac{1}{2}|0(3-0) + 2(0-0) + 6(0-3)| = \dfrac{1}{2}|0 + 0 - 18| = 9$ Triangle 2 area: $\dfrac{1}{2}|2(4-0) + 6(0-3) + 6(3-4)| = \dfrac{1}{2}|8 - 18 - 6| = \dfrac{1}{2} \times 16 = 8$ Total Area = $9 + 8 = 17$ square units

Given the system of equations: (1) $x - 2y + 4z = 6$ (2) $3x + y + 5z = 4$ (3) $2x - y + 3z = 4$ Add (2) and (3): $(3x+y+5z) + (2x-y+3z) = 4+4$ $5x + 8z = 8$ (Equation A) Multiply (3) by 2: $4x - 2y + 6z = 8$ (Equation 3') Subtract (1) from (3'): $(4x - 2y + 6z) - (x - 2y + 4z) = 8 - 6$ $3x + 2z = 2$ (Equation B) From (B), $2z = 2 - 3x$. Substitute this into (A): $5x + 4(2z) = 8$ $5x + 4(2 - 3x) = 8$ $5x + 8 - 12x = 8$ $-7x = 0 \Rightarrow x = 0$ Substitute $x=0$ into (B): $3(0) + 2z = 2 \Rightarrow 2z = 2 \Rightarrow z = 1$. Substitute $x=0$ and $z=1$ into (2): $3(0) + y + 5(1) = 4$ $y + 5 = 4 \Rightarrow y = -1$. Now, calculate $2x - 3y + 5z$: $2(0) - 3(-1) + 5(1) = 0 + 3 + 5 = 8$.

Given functions $f(x) = 2x - 3$ and $g(x) = x^3 + 5$. First, find the composite function $(fog)(x)$: $(fog)(x) = f(g(x)) = f(x^3 + 5)$ Substitute $x^3+5$ into $f(x)$: $2(x^3 + 5) - 3$ $(fog)(x) = 2x^3 + 10 - 3 = 2x^3 + 7$ Now, we need to find $(fog)^{-1}(-9)$. Let $(fog)^{-1}(-9) = k$. By definition of inverse functions, this means $(fog)(k) = -9$. Using the expression for $(fog)(x)$ we found: $2k^3 + 7 = -9$ Subtract 7 from both sides: $2k^3 = -9 - 7$ $2k^3 = -16$ Divide by 2: $k^3 = -8$ Take the cube root of both sides: $k = \sqrt[3]{-8}$ $k = -2$ Therefore, $(fog)^{-1}(-9) = -2$.

Let the total number of registered voters = $V$ <hr class="my-[18px]" node="[object Object]"> People who actually voted: $100\% - 18\% = 82\%$ of total voters Votes polled = $0.82V$ <hr class="my-[18px]" node="[object Object]"> Valid votes from the polled votes: $100\% - 15\% = 85\%$ of polled votes Valid votes = $0.85 \times 0.82V = 0.697V$ <hr class="my-[18px]" node="[object Object]"> Votes received by each candidate: B received $56\%$ of valid votes = $0.56 \times 0.697V = 0.39032V$ A received remaining valid votes = $44\%$ of valid votes = $0.44 \times 0.697V = 0.30668V$ <hr class="my-[18px]" node="[object Object]"> Using the majority information: B won by $6273$ votes, so: $0.39032V - 0.30668V = 6273$ $0.08364V = 6273$ $V = \dfrac{6273}{0.08364} = 75000$ <hr class="my-[18px]" node="[object Object]"> Valid votes received by A: A's votes $= 0.30668 \times 75000 = 23001$ Therefore, A received $23001$ valid votes.

When we insert $6$ arithmetic means between $3$ and $\dfrac{13}{2}$, we create an arithmetic progression with $8$ terms total. <ul node="[object Object]"> <li node="[object Object]">First term: $a_1 = 3$</li> <li node="[object Object]">Last term: $a_8 = \dfrac{13}{2}$</li> <li node="[object Object]">The $6$ arithmetic means are: $a_2, a_3, a_4, a_5, a_6, a_7$</li> </ul> <hr class="my-[18px]" node="[object Object]"> In an AP: $a_n = a_1 + (n-1)d$ So: $a_8 = a_1 + 7d$ $\dfrac{13}{2} = 3 + 7d$ $\dfrac{13}{2} - 3 = 7d$ $\dfrac{13}{2} - \dfrac{6}{2} = 7d$ $\dfrac{7}{2} = 7d$ $d = \dfrac{1}{2}$ <hr class="my-[18px]" node="[object Object]"> First arithmetic mean: $a_2 = 3 + (1)\left(\dfrac{1}{2}\right) = \dfrac{7}{2}$ Fifth arithmetic mean: $a_6 = 3 + (5)\left(\dfrac{1}{2}\right) = 3 + \dfrac{5}{2} = \dfrac{11}{2}$ <hr class="my-[18px]" node="[object Object]"> Ratio = Fifth mean : First mean = $\dfrac{11}{2} : \dfrac{7}{2} = 11 : 7$ Answer: $11:7$

Given Information: <ul node="[object Object]"> <li node="[object Object]">Triangle $ABC$ with angle bisector $AD$ meeting $BC$ at $D$</li> <li node="[object Object]">$AB = 30$ cm</li> <li node="[object Object]">$BC = 20$ cm</li> <li node="[object Object]">$BD$ is $4$ cm less than $DC$</li> </ul> <hr class="my-[18px]" node="[object Object]"> Since $BD$ is $4$ cm less than $DC$: <ul node="[object Object]"> <li node="[object Object]">Let $DC = x$ cm</li> <li node="[object Object]">Then $BD = (x - 4)$ cm</li> </ul> Since $D$ lies on $BC$: $BD + DC = BC$ $(x - 4) + x = 20$ $2x - 4 = 20$ $2x = 24$ $x = 12$ Therefore: $DC = 12$ cm and $BD = 8$ cm <hr class="my-[18px]" node="[object Object]"> When a line bisects an angle of a triangle, it divides the opposite side in the ratio of the adjacent sides. The Angle Bisector Theorem states: $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ Substituting our values: $\dfrac{8}{12} = \dfrac{30}{AC}$ <hr class="my-[18px]" node="[object Object]"> Cross multiply: $8 \times AC = 12 \times 30$ $8 \times AC = 360$ $AC = \dfrac{360}{8}$ $AC = 45$ cm Answer: $AC = 45$ cm

Understanding what we need to find: We have 4 partners (P, Q, R, S) sharing profit in certain ratios. We need to find the difference between (P's share + S's share) vs (Q's share + R's share). <hr class="my-[18px]" node="[object Object]"> Convert all ratios to a common form. The ratios are given separately, but we need one combined ratio to compare all four partners. Given ratios: $\dfrac{P}{Q} = \dfrac{2}{3}$; $\dfrac{Q}{R} = \dfrac{4}{5}$; $\dfrac{R}{S} = \dfrac{5}{12}$ By making the numbers the same, we get: $P : Q : R : S = 8 : 12 : 15 : 36$ <hr class="my-[18px]" node="[object Object]"> Calculate individual shares: Total ratio parts = $8 + 12 + 15 + 36 = 71$ parts Since total profit = ₹$24,850$ Each part = $\dfrac{24850}{71} = ₹350$ Individual shares: <ul node="[object Object]"> <li node="[object Object]">P's share = $8 \times 350 = ₹2,800$</li> <li node="[object Object]">Q's share = $12 \times 350 = ₹4,200$</li> <li node="[object Object]">R's share = $15 \times 350 = ₹5,250$</li> <li node="[object Object]">S's share = $36 \times 350 = ₹12,600$</li> </ul> <hr class="my-[18px]" node="[object Object]"> Find the required difference: Combined share of P and S = $2,800 + 12,600 = ₹15,400$ Combined share of Q and R = $4,200 + 5,250 = ₹9,450$ Difference = $15,400 - 9,450 = ₹5,950$

What we know: Perimeter = $148$ cm A rhombus has 4 equal sides, so: Side length = $\dfrac{148}{4} = 37$ cm <hr class="my-[18px]" node="[object Object]"> What we know: <ul node="[object Object]"> <li node="[object Object]">One diagonal = $70$ cm</li> <li node="[object Object]">Side length = $37$ cm</li> </ul> The diagonals of a rhombus cut each other at right angles and split each other in half. This creates right-angled triangles with: <ul node="[object Object]"> <li node="[object Object]">One leg = $\dfrac{70}{2} = 35$ cm</li> <li node="[object Object]">Other leg = $\dfrac{d_2}{2}$ (unknown)</li> <li node="[object Object]">Hypotenuse = $37$ cm</li> </ul> Using Pythagoras theorem: $35^2 + \left(\dfrac{d_2}{2}\right)^2 = 37^2$ $1225 + \left(\dfrac{d_2}{2}\right)^2 = 1369$ $\left(\dfrac{d_2}{2}\right)^2 = 144$ $\dfrac{d_2}{2} = 12$ So $d_2 = 24$ cm <hr class="my-[18px]" node="[object Object]"> Area of rhombus = $\dfrac{1}{2} \times d_1 \times d_2$ Area = $\dfrac{1}{2} \times 70 \times 24 = 840$ cm² <hr class="my-[18px]" node="[object Object]"> For the trapezium: <ul node="[object Object]"> <li node="[object Object]">Parallel sides are $x$ cm and $50$ cm</li> <li node="[object Object]">Height = $14$ cm</li> </ul> Area of trapezium = $\dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$ Area = $\dfrac{1}{2} \times (x + 50) \times 14 = 7(x + 50)$ <hr class="my-[18px]" node="[object Object]"> Since both areas are equal: $840 = 7(x + 50)$ $\dfrac{840}{7} = x + 50$ $120 = x + 50$ $x = 70$ cm

Let the speed of the boat in still water be $V$ km/h and the speed of the stream be $4.5$ km/h. For the same distance $D$: <ul node="[object Object]"> <li node="[object Object]">Time in still water = $\dfrac{D}{V}$</li> <li node="[object Object]">Time upstream = $\dfrac{D}{V - 4.5}$</li> </ul> Given that time in still water is $20\%$ less than time upstream: $\dfrac{D}{V} = 0.8 \times \dfrac{D}{V - 4.5}$ <hr class="my-[18px]" node="[object Object]"> Cancel $D$ from both sides: $\dfrac{1}{V} = \dfrac{0.8}{V - 4.5}$ $V - 4.5 = 0.8V$ $0.2V = 4.5$ $V = 22.5$ km/h <hr class="my-[18px]" node="[object Object]"> Calculate actual speeds: <ul node="[object Object]"> <li node="[object Object]">Upstream speed = $V - 4.5 = 22.5 - 4.5 = 18$ km/h</li> <li node="[object Object]">Downstream speed = $V + 4.5 = 22.5 + 4.5 = 27$ km/h</li> </ul> <hr class="my-[18px]" node="[object Object]"> Time for $60$ km downstream: $\dfrac{60}{27} = \dfrac{20}{9}$ hours Time for $30$ km upstream: $\dfrac{30}{18} = \dfrac{5}{3}$ hours <hr class="my-[18px]" node="[object Object]"> Total time = $\dfrac{20}{9} + \dfrac{5}{3} = \dfrac{20}{9} + \dfrac{15}{9} = \dfrac{35}{9}$ hours

The solution will be added shortly to this PYQ!

The solution will be added shortly to this PYQ!

iimb-bbadbe 2025 Complete Paper Solution | QA Slot 1

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iimb-bbadbe 2025 Complete Paper Solution | QA Slot 1

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<div><div class="latex-wrapper">If $1, \omega, \omega^2$ are the cube roots of unity, then the value of $\dfrac{1+9+11\omega^2}{\omega^2+9+11\omega} + \dfrac{3+5\omega+7\omega^2}{3\omega+5\omega^2+7}$ is:<br /></div></div>

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<div><div class="latex-wrapper">What is the sum of the infinite series: $1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \dots?$<br /></div></div>

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<div><div class="latex-wrapper">A person hosts a party of 13 guests and places 5 of them on one table and the remaining on the other table, the tables being round. In how many ways can he arrange the guests?<br /></div></div>

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<div><div class="latex-wrapper">Find the area (in square units) bounded by $x=6, y=0, 3x-2y=0$, and $x-4y+10=0$.<br /></div></div>

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<div><div class="latex-wrapper">If $x - 2y + 4z = 6$, $3x + y + 5z = 4$ and $2x - y + 3z = 4$, then what is the value of $(2x – 3y + 5z)$?<br /></div></div>

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<div><div class="latex-wrapper">If f: R -&gt; R is defined by f(x) = 2x – 3 and g: R -&gt; R is defined by g(x) = x³ + 5, then (fog)⁻¹(-9) is equal to:<br /></div></div>

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<div><div class="latex-wrapper">If six arithmetic means are inserted between 3 and $\frac{13}{2}$, the ratio of the fifth and the first mean is:<br /></div></div>

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<div><div class="latex-wrapper">In $\triangle ABC$, $AD$ is the bisector of $\angle A$ meeting $BC$ at $D$. If $AB = 30$ cm, $BC = 20$ cm and the length of $BD$ is 4 cm less than $DC$, then the length (in cm) of $AC$ is<br /></div></div>

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<div><div class="latex-wrapper">What was the number of literate population in the year 2019 in town 1?</div></div>

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<div><div class="latex-wrapper">What is the ratio between total population and literate population in 2019 in town 3?</div></div>

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<div><div class="latex-wrapper">Number of literate population of town 2 in year 2019 is approximately what percent of literate population of town 1 in the year 2021?</div></div>

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<div><div class="latex-wrapper">What was the approximate literate population in town 2 in 2020?</div></div>