14 November 2025 · Shift 2

Partnership Aman draws a salary of 3,000 every month, so his salary for the whole year $=3000\times 12=36000$. This salary is taken out of the total profit first; only the rest is shared in the ratio of the investments. Investment ratio =60000:80000:40000=3:4:2, i.e. 9 equal parts in all. Aman's total earning is 40,000, so his share of the remaining profit =40000-36000=4000. Aman gets 3 of the 9 parts, so 3 parts =4000 and the remaining profit $=\frac{4000}{3}\times 9=12000$. Total profit =36000+12000=48000. Hence, option (b) is the most appropriate answer.

Trigonometric ratios The expression $\sin A\cos B+\cos A\sin B$ is the expansion of $\sin \left(A+B\right)$, so all four ratios are needed. 8, 15, 17 is a Pythagorean triplet, since $8^2+{15}^2=64+225=289={17}^2$, so one right triangle gives every ratio. $\cos A=\frac{8}{17}\Rightarrow \sin A=\frac{15}{17}$, and $\sin B=\frac{15}{17}\Rightarrow \cos B=\frac{8}{17}$. $\sin A\cos B+\cos A\sin B=\frac{15}{17}\times \frac{8}{17}+\frac{8}{17}\times \frac{15}{17}$ $=\frac{120}{289}+\frac{120}{289}=\frac{240}{289}$ Hence, option (a) is the most appropriate answer.

Mixture, profit and loss Equal quantities are mixed, so take 1 kg of each type. First type: SP is ₹60/kg at 25% profit, so CP $=\frac{60\times 100}{125}=48$, i.e. ₹48/kg. Second type: SP is ₹36/kg at 25% loss, so CP $=\frac{36\times 100}{75}=48$, i.e. ₹48/kg. Cost of 2 kg of mixture =48+48=96, and it is sold for $48\times 2=96$. Selling price equals cost price, so there is neither profit nor loss. Hence, option (c) is the most appropriate answer.

Compound interest Under compound interest the amount is multiplied by the same factor every year. Interest earned during the 3rd year =5544-5040=504. This interest is earned on the amount standing at the end of the 2nd year, i.e. on ₹5040. Rate $=\frac{504}{5040}\times 100=10$. So the rate of interest is $10\%$ per annum. Hence, option (d) is the most appropriate answer.

Ratio and proportion To join the two ratios, the second number must carry the same value in both. First : Second =3:5 and Second : Third =4:7; the second number appears as 5 and as 4, whose LCM is 20. Multiply the first ratio by 4 and the second by 5: First : Second : Third =12:20:35. Total parts =12+20+35=67, and these 67 parts equal 190. Third number $=\frac{35}{67}\times 190=\frac{6650}{67}=99.25$ (approximately). Hence, option (b) is the most appropriate answer.

Averages Total spending of all 7 students $=7\times 150=1050$. Total spending of the 4 boys $=4\times 180=720$. Total spending of the 3 girls =1050-720=330. Average spending of a girl $=\frac{330}{3}=110$, i.e. ₹110. Hence, option (c) is the most appropriate answer.

HCF application A square tile can pave the courtyard exactly only if its side divides both sides of the courtyard, so the largest tile has side equal to the HCF of the two lengths. Convert to centimetres: 11.34 m =1134 cm and 26.25 m =2625 cm. $1134=2\times 3^4\times 7$ and $2625=3\times 5^4\times 7$. The common factors are 3 and 7, so HCF $=3\times 7=21$. Hence the largest square tile that can be used has side 21 cm. Hence, option (b) is the most appropriate answer.

Depreciation Depreciation behaves like compound interest with a negative rate, so the value is multiplied by the same factor every year. Formula: Value $=P{\left(1-\frac{R}{100}\right)}^n$. $=60000{\left(1-\frac{10}{100}\right)}^3=60000\times {\left(\frac{9}{10}\right)}^3$ $=60000\times \frac{729}{1000}=43740$ So the value of the motor after 3 years is ₹43,740. Hence, option (b) is the most appropriate answer.

Time and work Take the total work as the LCM of 18 and 36, i.e. 36 units. A's one day work $=\frac{36}{18}=2$ units and B's one day work $=\frac{36}{36}=1$ unit. Working together for 6 days they complete $\left(2+1\right)\times 6=18$ units. Work still left =36-18=18 units. B alone does 1 unit a day, so he needs $\frac{18}{1}=18$ more days. Hence, option (a) is the most appropriate answer.

Marked price and discount Let the cost price be ₹100. Marked price $=100+20\%$ of 100=120. A discount of 15% is given on the marked price, so SP $=120\times \frac{85}{100}=102$. Profit =102-100=2, so profit percentage $=\frac{2}{100}\times 100=2\%$. Hence, option (a) is the most appropriate answer.

Simple interest, two rates Let the money put in Account A be x; then Account B gets (20000-x). Formula: $SI=\frac{P\times R\times T}{100}$ $\frac{x\times 5\times 4}{100}+\frac{\left(20000-x\right)\times 7\times 4}{100}=5200$ 0.2x+0.28(20000-x)=5200 5600-0.08x=5200 $0.08x=400\Rightarrow x=5000$ So ₹5,000 was invested in Account A, and option (c) is correct.

Algebraic identity Identity: ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)$ Substituting the given values: ${\left(18\right)}^2=a^2+b^2+c^2+2\times 80$ $324=a^2+b^2+c^2+160$ $a^2+b^2+c^2=324-160=164$ Hence option (c) is correct.

LCM and perfect square A number divisible by both 18 and 54 must be a multiple of their LCM. $18=2\times 3^2$ and $54=2\times 3^3$, so $LCM=2\times 3^3=54$ In a perfect square every prime must carry an even power, so $2^1$ has to be raised to $2^2$ and $3^3$ to $3^4$. Required number $=2^2\times 3^4=4\times 81=324$ Check: $324={18}^2$, $324÷18=18$ and $324÷54=6$, both exact. Hence option (a) is correct.

Ratio and mixture The ratio 5 : 6 has 5+6=11 parts, and 44 litres is shared in these parts. Milk $=\frac{5}{11}\times 44=20$ litres Water $=\frac{6}{11}\times 44=24$ litres Only milk is added, so water stays at 24 litres while milk becomes 20+8=28 litres. New ratio $=28:24=\frac{28}{4}:\frac{24}{4}=7:6$ Hence option (c) is correct.

Circle inscribed in square When a circle is inscribed in a square it touches all four sides, so its diameter equals the side of the square. Diameter =18 cm, therefore radius $r=\frac{18}{2}=9$ cm Area of circle $=\pi r^2=3.14\times {\left(9\right)}^2$ $=3.14\times 81=254.34{\mathrm{cm}}^2$ Hence option (b) is correct.

Sphere recast into cylinder Melting and recasting does not change the quantity of metal, so volume of sphere = volume of cylinder. Radius of cylinder =3R, height =H. $\frac{4}{3}\pi R^3=\pi {\left(3R\right)}^{2}H$ $\frac{4}{3}\pi R^3=9\pi R^{2}H$ $H=\frac{4R^3}{3\times 9R^2}=\frac{4R}{27}$ $\frac{H}{R}=\frac{4}{27}$, that is H:R=4:27 Hence option (c) is correct.

Hemisphere TSA and cylinder volume Total surface area of a solid hemisphere = curved surface + flat circular face $=2\pi r^2+\pi r^2=3\pi r^2$ Volume of a cylinder $=\pi r^{2}h$ Both solids have the same radius r, and the two quantities are numerically equal. $3\pi r^2=\pi r^{2}h$ Cancelling $\pi r^2$ from both sides: h=3 Hence the height is 3 units — option (d) is correct.

Linear equation on a fraction Let the numerator be x; the denominator is 4 more, so the fraction is $\frac{x}{x+4}$. Adding 5 to both parts: $\frac{x+5}{x+4+5}=\frac{x+5}{x+9}$ $\frac{x+5}{x+9}=\frac{6}{7}$ Cross-multiplying: 7(x+5)=6(x+9) 7x+35=6x+54 x=19 Original fraction $=\frac{19}{19+4}=\frac{19}{23}$ Hence option (b) is correct.

Powers of x minus 1/x Given $x-\frac{1}{x}=4$. Climb to the sixth power in two jumps: first cube, then square. Identity: $x^3-\frac{1}{x^3}={\left(x-\frac{1}{x}\right)}^3+3\left(x-\frac{1}{x}\right)$ $x^3-\frac{1}{x^3}={\left(4\right)}^3+3\times 4=64+12=76$ Identity: $x^6+\frac{1}{x^6}={\left(x^3-\frac{1}{x^3}\right)}^2+2$ $x^6+\frac{1}{x^6}={\left(76\right)}^2+2$ =5776+2=5778 Hence option (a) is correct.

Simplifying surds Take the perfect squares out of each surd first. $\sqrt{180}=\sqrt{36\times 5}=6\sqrt{5}$ $\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$ $\frac{6\sqrt{5}+2\sqrt{5}}{\sqrt{5}}=\frac{8\sqrt{5}}{\sqrt{5}}=8$ Hence option (a) is correct.

Trigonometric identity Use the identity $1+{\tan }^{2}x={\sec }^{2}x$ to clear the denominator. $\frac{{\sec }^{2}x+{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{{\sec }^{2}x+{\tan }^{2}x}{{\sec }^{2}x}$ $=1+\frac{{\tan }^{2}x}{{\sec }^{2}x}$ Now $\frac{{\tan }^{2}x}{{\sec }^{2}x}=\frac{{\sin }^{2}x}{{\cos }^{2}x}\times {\cos }^{2}x={\sin }^{2}x$ So the expression equals $1+{\sin }^{2}x$. Option (b) would need $\sin x=0$ for every x, which is not true, so option (c) is correct.

Pythagorean triplet in trigonometry Treat 2A as a single angle $\theta$, so $\cos \theta =\frac{5}{13}=\frac{\mathrm{base}}{\mathrm{hypotenuse}}$. Perpendicular $=\sqrt{{13}^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ $\sin \theta =\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}=\frac{12}{13}$ So $\sin 2A=\frac{12}{13}$ and option (a) is correct.

Ratios from tan A $\tan A=4=\frac{4}{1}$, so in a right triangle the perpendicular is 4 and the base is 1. Hypotenuse $=\sqrt{4^2+1^2}=\sqrt{16+1}=\sqrt{17}$ $\sin A=\frac{4}{\sqrt{17}}$ and $\cos A=\frac{1}{\sqrt{17}}$ Check with the given identity: $\frac{16}{17}+\frac{1}{17}=\frac{17}{17}=1$ Options (b), (c) and (d) fail this check, so option (a) is correct.

Slope of a line Formula: slope $m=\frac{y_2-y_1}{x_2-x_1}$ Taking $\left(x_1,y_1\right)=\left(x,3\right)$ and $\left(x_2,y_2\right)=\left(4,-5\right)$: $4=\frac{-5-3}{4-x}=\frac{-8}{4-x}$ 4(4-x)=-8 16-4x=-8 $4x=24\Rightarrow x=6$ Hence option (c) is correct.

Escalator, relative speed Let the length of the escalator be L, the man's own speed m and the escalator's speed e. Walking up while the escalator moves up, the two speeds add: $m+e=\frac{L}{20}$ Walking down while the escalator still moves up, the escalator opposes him: $m-e=\frac{L}{80}$ Adding the two equations removes e: $2m=\frac{L}{20}+\frac{L}{80}=\frac{4L+L}{80}=\frac{5L}{80}=\frac{L}{16}$ $m=\frac{L}{32}$ On a stationary escalator only his own speed acts, so time $=\frac{L}{m}=\frac{L}{L/32}=32$ seconds Hence option (a) is correct.