Let's solve this step by step: Area of floor = $346.5$ $m^2$ Floor is circular, so using area formula: $346.5 = \pi r^2$ Radius of floor = $r = \sqrt{\frac{346.5}{\pi}} = 10.5$ m Height of tent = $h = 14$ m Slant height using Pythagoras theorem: $l = \sqrt{h^2 + r^2} = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5$ m Area of curved surface = $\pi r l$ $= \pi × 10.5 × 17.5 = 577.5$ m$^2$ Width of canvas = $1.1$ m Length of canvas required = $\frac{577.5}{1.1} = 525$ m The answer is $525$ m.

Given: <ul node="[object Object]"> <li node="[object Object]">Players: 11 <ul node="[object Object]"> <li node="[object Object]">Team average: 28 years</li> <li node="[object Object]">Three groups of 3 players: $\frac{25+28+30}{3}$ years</li> <li node="[object Object]">Captain's age = Youngest player's age + 11</li> </ul> </li> </ul> Let $x =$ youngest player's age Average of 9 players $= \frac{3×25 + 3×28 + 3×30}{9} = \frac{249}{9}$ Using average formula: $\frac{9 × \frac{249}{9} + x + (x+11)}{11} = 28$ $\frac{249 + 2x + 11}{11} = 28$ $249 + 2x + 11 = 308$ $2x = 48$ $x = 24$ Captain's age $= 24 + 11 = 35$ years

Angle between the hour and minute hand $= |30H - \frac{11}{2}M|$ where $H$ is the hour hand, and $M$ is the minute hand So, $H=5$ $M = 15$ Substituting: $= |30(5) - \frac{11}{2}(15)|$ $= |150 - \frac{165}{2}|$ $= |150 - 82.5|$ $= |67.5|$ Therefore, the angle between the hands is $67.5°$

Let the annual rent be $R$ Purchase price = $40R$ Rate of return = $\frac{\text{Annual Return}}{\text{Investment}} × 100\%$ $= \frac{R}{40R} × 100\%$ $= \frac{1}{40} × 100\%$ $= 2.50\%$ Therefore, the rate of return is 2.50% per annum.

Given: Diagonal 1 $(d_1)$ = 40m Diagonal 2 $(d_2)$ = 30m In a rhombus, using diagonals to find side $(s)$: $s^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2$ $s^2 = (\frac{40}{2})^2 + (\frac{30}{2})^2$ $s^2 = 400 + 225$ $s^2 = 625$ $s = 25$ meters Perimeter of rhombus = $4s$ $= 4 × 25$ $= 100$ meters Therefore, the perimeter is $100$ meters.

For fractions $\dfrac{a_1}{b_1}, \dfrac{a_2}{b_2}, ..., \dfrac{a_n}{b_n}$, the HCF formula is $\dfrac{HCF(a_1, a_2, ..., a_n)}{LCM(b_1, b_2, ..., b_n)}$ Given fractions are $\frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \frac{21}{40}$. HCF of numerators (9, 12, 18, 21) = 3, and LCM of denominators (10, 25, 35, 40) = 1400. Therefore, the HCF: $\frac{3}{1400}$

The series goes like this $=(\text{the number}\times n)+n$ to give the next number, where $n$ is consecutive first natural numbers. $(15\times1)+1=16$ $(16\times2)+2=34$ $(34\times3)+3=105$ $(105\times4)+4=424$ $(424\times5)+5=2125$ Hence, $2124$ is the odd term out in the above series.

Let total length of pole be $x$ cm Initially, $\frac{4}{7}x$ is in mud. After pulling out $\frac{1}{3}$ of the pole, $250$ cm remains in mud. Therefore: $\frac{4}{7}x - \frac{1}{3}x = 250$ $\frac{12}{21}x - \frac{7}{21}x = 250$ $\frac{5}{21}x = 250$ $x = 250 × \frac{21}{5}$ $x = 1050$ cm Therefore, the total length of the pole is 1050 cm.

Given: <ul node="[object Object]"> <li node="[object Object]">Total profit = ₹$9600$</li> <li node="[object Object]">A's capital = ₹$12000$</li> <li node="[object Object]">B's capital = ₹$20000$</li> <li node="[object Object]">A gets $10\%$ of profit for managing</li> <li node="[object Object]">Rest divided in proportion to capital</li> </ul> <ol node="[object Object]"> <li node="[object Object]">A's managing share = $10\%$ of $9600$ = ₹$960$</li> </ol> <ol start="2" node="[object Object]"> <li node="[object Object]">Remaining profit = $9600 - 960$ = ₹$8640$</li> </ol> <ol start="3" node="[object Object]"> <li node="[object Object]">Proportion for remaining profit:</li> </ol> A : B = $12000 : 20000 = 3 : 5$ <ol start="4" node="[object Object]"> <li node="[object Object]">A's share from remaining = $\frac{3}{8} × 8640$ = ₹$3240$</li> </ol> <ol start="5" node="[object Object]"> <li node="[object Object]">Total money A receives = Managing share + Proportional share</li> </ol> = $960 + 3240$ $=$ ₹$4200$ Therefore, A will receive ₹$4200$.

Let filling capacity = $x$ $m^3$/minute Then emptying capacity = $(x + 10)$ $m^3$/minute Given tank volume = $2400$ $m^3$ For filling: $\frac{2400}{x} = t$ minutes For emptying: $\frac{2400}{x+10} = (t-8)$ minutes Therefore: $\frac{2400}{x} - \frac{2400}{x+10} = 8$ $\frac{2400(x+10) - 2400x}{x(x+10)} = 8$ $\frac{24000}{x(x+10)} = 8$ $24000 = 8x(x+10)$ $24000 = 8x^2 + 80x$ $8x^2 + 80x - 24000 = 0$ $x^2 + 10x - 3000 = 0$ Solving quadratic: $x = 50$ or $x = -60$ (reject negative) Therefore, filling capacity is 50 $m^3$/minute.

jipmat 2021 Complete Paper Solution | QA

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jipmat 2021 Complete Paper Solution | QA

Question 1.

<div><div class="latex-wrapper">The length of canvas 1.1 m wide required to build a conical tent of height 14m and the floor area 346.5 $m^{2}$ is<br /></div></div>

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<div><div class="latex-wrapper">At what angle the hands of a clock are inclined at 15 minutes past 5?<br /></div></div>

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<div><div class="latex-wrapper">A man pays 40 times the annual rent to purchase a building. The rate % per annum he derives from his investment is<br /></div></div>

Question 5.

<div><div class="latex-wrapper">In a rhombus, the length of the two diagonals are 40m and 30m respectively. Its perimeter will be<br /></div></div>

Question 6.

<div><div class="latex-wrapper">The HCF of $(\frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \frac{21}{40})$ is<br /></div></div>

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<div><div class="latex-wrapper">The odd term out in the series 15, 16, 34, 105, 424, 2124, 12756 is<br /></div></div>

Question 8.

<div><div class="latex-wrapper">$(\frac{4}{7})$ of a pole is in the mud. When $(\frac{1}{3})$ of it is pulled out, 250 cm of the pole is still in the mud. The length of the pole is<br /></div></div>

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