19 November 2025 · Shift 2

Algebraic identity Identity: ${\left(a+b\right)}^2=a^2+b^2+2ab$. So $a^2+b^2={\left(a+b\right)}^2-2ab$. $a^2+b^2=131-2\times 8$ $a^2+b^2=131-16=115$ Hence, option (c) is correct.

Partnership In a partnership the profit is divided in the ratio of (capital × time). A invests for the whole year: $36000\times 12=432000$. B joins after x months, so B's money works for (12-x) months: $24000\times \left(12-x\right)$. $\frac{432000}{24000\left(12-x\right)}=\frac{2}{1}$ $432000=48000\left(12-x\right)$ $12-x=9\Rightarrow x=3$ So B joined after 3 months, i.e. option (b).

Trigonometric ratios ${\cos }^2\theta =\frac{1}{2}$ with $0^{\circ }<\theta <{90}^{\circ }$ gives $\cos \theta =\frac{1}{\sqrt{2}}$, i.e. $\theta ={45}^{\circ }$. Then ${\sin }^2\theta =1-{\cos }^2\theta =1-\frac{1}{2}=\frac{1}{2}$. Complementary angles: $\sec \left({90}^{\circ }-\theta \right)=\csc \theta$, so ${\sec }^2\left({90}^{\circ }-\theta \right)={\csc }^2\theta =\frac{1}{{\sin }^2\theta }=2$. ${\sin }^2\theta +{\sec }^2\left({90}^{\circ }-\theta \right)=\frac{1}{2}+2=\frac{5}{2}$ Hence, option (b) is correct.

Compound interest Formula: $A=P{\left(1+\frac{r}{100}\right)}^n$, where A is the amount and P the principal. $26450=P{\left(1+\frac{15}{100}\right)}^2=P\times {\left(\frac{23}{20}\right)}^2$ $26450=P\times \frac{529}{400}$ $P=\frac{26450\times 400}{529}=20000$ So the principal is ₹20,000, i.e. option (a).

Angles of a triangle BC is produced to D, so $\angle ACD$ is the exterior angle of $△ABC$ at C. Exterior angle theorem: $\angle ACD=\angle BAC+\angle ABC$. Let $\angle ABC=5k$; then $\angle BAC=\frac{4}{5}\times 5k=4k$. $4k+5k={135}^{\circ }$ $9k={135}^{\circ }\Rightarrow k={15}^{\circ }$ $\angle BAC=4k=4\times {15}^{\circ }={60}^{\circ }$ Hence, option (b) is correct.

Replacement of mixture Formula: if 5 litres are drawn out and replaced by water each time, then after n operations the wine left $=x{\left(1-\frac{5}{x}\right)}^n$, where x litres is the capacity of the cask. The operation is carried out three times in all, and the wine finally left is $\frac{27}{64}$ of the original quantity. ${\left(1-\frac{5}{x}\right)}^3=\frac{27}{64}={\left(\frac{3}{4}\right)}^3$ $1-\frac{5}{x}=\frac{3}{4}$ $\frac{5}{x}=\frac{1}{4}\Rightarrow x=20$ So the cask originally held 20 litres of wine, i.e. option (d).

Trigonometric identities $\tan \theta +\cot \theta =2$ means $\tan \theta +\frac{1}{\tan \theta }=2$, i.e. ${\tan }^2\theta -2\tan \theta +1=0$. ${\left(\tan \theta -1\right)}^2=0\Rightarrow \tan \theta =1$, and as $\theta$ is acute, $\theta ={45}^{\circ }$. At ${45}^{\circ }$, $\sin \theta =\cos \theta =\frac{1}{\sqrt{2}}$. ${\sin }^3\theta +{\cos }^3\theta ={\left(\frac{1}{\sqrt{2}}\right)}^3+{\left(\frac{1}{\sqrt{2}}\right)}^3=\frac{2}{2\sqrt{2}}$ $=\frac{1}{\sqrt{2}}$ Hence, option (b) is correct.

Races on a circular track On a circular track two runners moving in the same direction meet for the first time when the faster one gains exactly one full lap on the slower one. Relative speed =45-30=15 m/s. Time of first meeting $=\frac{4500}{15}=300$ seconds. Hence, option (c) is correct.

Simplification of surds Use the identity ${\left(x+y\right)}^2-{\left(x-y\right)}^2=4xy$ with $x=\sqrt{7}$ and $y=\sqrt{3}$. Numerator $=4\times \sqrt{7}\times \sqrt{3}=4\sqrt{21}$. Denominator $=\sqrt{7}\times \sqrt{3}=\sqrt{21}$. $\frac{4\sqrt{21}}{\sqrt{21}}=4$ Hence, option (d) is correct.

Area of a trapezium Formula: area of a trapezium $=\frac{1}{2}\times$ (sum of the parallel sides) $\times$ distance between them. Area $=\frac{1}{2}\times \left(9.4+11.8\right)\times 8.5$ $=\frac{1}{2}\times 21.2\times 8.5$ $=10.6\times 8.5=90.1$ So the required area is $90.1{cm}^2$, i.e. option (b).

Square: diagonal and area In a square of side a, the diagonal is $d=a\sqrt{2}$. So $a=\frac{d}{\sqrt{2}}=\frac{2\sqrt{44}}{\sqrt{2}}=2\sqrt{22}$ cm. Area of the square $=a^2={\left(2\sqrt{22}\right)}^2=4\times 22=88$ sq cm. A quicker route is to use the diagonal directly: area $=\frac{d^2}{2}=\frac{4\times 44}{2}=88$ sq cm. Hence, option (b) is the correct answer.

Ratio and average of amounts Let the amount with P be x. Q has 3.5 times the amount of P, so Q=3.5x. S has 50% less than Q, i.e. half of Q, so $S=\frac{3.5x}{2}=1.75x$. R equals the average of those three persons P, Q and S: $R=\frac{x+3.5x+1.75x}{3}=\frac{6.25x}{3}$. Given P+R=6000, so $x+\frac{6.25x}{3}=6000$, i.e. $\frac{9.25x}{3}=6000$. $x=\frac{18000}{9.25}=\frac{72000}{37}$ $R=6000-x=\frac{222000-72000}{37}=\frac{150000}{37}=4054.05\dots$ Hence R has about Rs. 4054, option (c).

Profit percentage Selling price of 12 kg is ₹576, so the SP of 1 kg $=\frac{576}{12}=48$, i.e. ₹48. Profit = SP of 4 kg $=4\times 48=192$, i.e. ₹192. Cost price of 12 kg = SP − Profit = ₹576 − ₹192 = ₹384. Profit % $=\frac{192}{384}\times 100=50$, i.e. 50%. Shortcut: the profit on 12 kg equals the SP of 4 kg, so the CP of 12 kg equals the SP of 12-4=8 kg, giving profit % $=\frac{4}{8}\times 100=50$. Hence, option (b) is the correct answer.

Equal gain and loss percent When two articles are sold at the same price, one at a gain of x% and the other at a loss of the same x%, the deal always ends in a loss. Loss % $=\frac{x^2}{100}=\frac{{35}^2}{100}=\frac{1225}{100}=12.25$, i.e. 12.25%. Check: $C{P}_1=\frac{800\times 100}{135}=592.59$ and $C{P}_2=\frac{800\times 100}{65}=1230.77$, so the total CP =1823.36 against a total SP of 1600. Loss =1823.36-1600=223.36 and $\frac{223.36}{1823.36}\times 100=12.25$. Hence, option (b) is the correct answer.

Ratio after percentage change New profit = old profit × (1 + rate of increase), applied term by term. New ratio $=3\times \frac{140}{100}:7\times \frac{150}{100}:6\times \frac{160}{100}$ =4.2:10.5:9.6 Multiplying every term by 10: 42:105:96. Dividing every term by the common factor 3: 14:35:32. Hence, option (b) is the correct answer.

Average age Total age of the 22 students $=22\times 16=352$ years. With the first teacher there are 23 people and the average becomes 16+2=18 years, so the total $=23\times 18=414$ years. Age of the first teacher =414-352=62 years. With the second teacher there are 24 people and the average rises by 3 more, to 18+3=21 years, so the total $=24\times 21=504$ years. Age of the second teacher =504-414=90 years. Hence, option (c) is the correct answer.

Speed, time and distance Let the distance between the two stations be d km. At 120 km/h the train is 2 minutes early and at 90 km/h it is 3 minutes late, so the two journey times differ by 2+3=5 minutes $=\frac{5}{60}=\frac{1}{12}$ hour. $\frac{d}{90}-\frac{d}{120}=\frac{1}{12}$ $d\left(\frac{4-3}{360}\right)=\frac{1}{12}$ $\frac{d}{360}=\frac{1}{12}\Rightarrow d=30$ Hence the distance between the two stations is 30 km, option (c).

Successive percentage increase Let the original salary be 100 and let each increment be r%. $100{\left(1+\frac{r}{100}\right)}^2=148.84$ ${\left(1+\frac{r}{100}\right)}^2=1.4884$ $1+\frac{r}{100}=\sqrt{1.4884}=1.22$ $\frac{r}{100}=0.22\Rightarrow r=22$ Hence each increment was 22%, option (c).

Population growth The birth rate adds 20% and the death rate removes 16%, so the net growth is 20-16=4% per year. Population after 3 years $=250000{\left(1+\frac{4}{100}\right)}^3=250000{\left(\frac{26}{25}\right)}^3$ $=250000\times \frac{17576}{15625}=16\times 17576$ =281216 Hence the population at the end of 3 years is 2,81,216, option (b).

Time and work together A's one day's work $=\frac{1}{12}$ and B's one day's work $=\frac{1}{15}$. Working together, their one day's work $=\frac{1}{12}+\frac{1}{15}=\frac{5+4}{60}=\frac{9}{60}=\frac{3}{20}$. Time taken together $=\frac{20}{3}=6\frac{2}{3}$ days. Hence, option (c) is the correct answer.

Successive discounts After a discount of 30% the price becomes $\frac{70}{100}$ of the marked price, and a further discount of 40% makes it $\frac{60}{100}$ of that. SP $=3200\times \frac{70}{100}\times \frac{60}{100}$ $=3200\times \frac{42}{100}=1344$ Hence the final selling price is ₹1,344, option (b).

Effective discount percent Take the marked price as 100. After 35% off it becomes 65, and after a further 20% off it becomes $65\times \frac{80}{100}=52$. So 100 finally costs 52, and the effective discount =100-52=48, i.e. 48%. By formula, effective discount $=a+b-\frac{ab}{100}=35+20-\frac{35\times 20}{100}=55-7=48$. The marked price ₹1,200 is not needed for the percentage; it only fixes the price paid, $1200\times 0.52=624$, i.e. ₹624. Hence, option (b) is the correct answer.

Perimeter of equilateral triangle The rope goes right round the plot, so the length of rope needed is the perimeter of the equilateral triangle. Perimeter $=3\times \mathrm{side}=3\times 85=255$ m. Total expense $=255\times 20=5100$, i.e. ₹5,100. Hence, option (d) is the correct answer.

Mixture and profit Let the milk cost ₹c per litre, and suppose m litres of milk are mixed with w litres of water. The seller pays only for the milk, so CP =mc; he sells the whole (m+w) litres at the same ₹c per litre, so SP =(m+w)c. Profit =(m+w)c-mc=wc, which is exactly the value of the free water. Profit % $=\frac{wc}{mc}\times 100=\frac{w}{m}\times 100=20$ $\frac{w}{m}=\frac{20}{100}=\frac{1}{5}$ Hence water : milk =1:5, option (b).

Comparing surds To compare surds of different orders, bring them to a common index. The LCM of 3, 2 and 4 is 12. $\sqrt[3]{11}={11}^{\frac{4}{12}}=\sqrt[12]{{11}^4}=\sqrt[12]{14641}$ $\sqrt{7}=7^{\frac{6}{12}}=\sqrt[12]{7^6}=\sqrt[12]{117649}$ $\sqrt[4]{12}={12}^{\frac{3}{12}}=\sqrt[12]{{12}^3}=\sqrt[12]{1728}$ Since 1728<14641<117649, we get $\sqrt[4]{12}<\sqrt[3]{11}<\sqrt{7}$, which the decimal values confirm: 1.86<2.22<2.65. So the three surds line up as $\sqrt[4]{12},\sqrt[3]{11},\sqrt{7}$. Hence, option (b) is the correct answer.