21 November 2025 · Shift 2

Average, excluded number Formula: Sum = Average $\times$ Number of terms. Sum of the 12 numbers $=35\times 12=420$ After one number is removed, 11 numbers are left, so their sum $=33\times 11=363$ Excluded number =420-363=57 Hence, option (b) is the most appropriate answer.

Surds, square root value The value of $\sqrt{35}$ is given, so change the required surd into a form that contains $\sqrt{35}$. Multiply the numerator and the denominator inside the root by 5: $\sqrt{\frac{7}{5}}=\sqrt{\frac{7\times 5}{5\times 5}}=\frac{\sqrt{35}}{5}$ Substituting $\sqrt{35}=5.916$: $=\frac{5.916}{5}=1.1832$ Hence, option (a) is the most appropriate answer.

Rationalisation, algebraic identity Rationalise x by multiplying the numerator and the denominator by $\sqrt{3}+\sqrt{2}$: $x=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\frac{3+2+2\sqrt{6}}{3-2}=5+2\sqrt{6}$ In the same way $y=\frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2}{3-2}=5-2\sqrt{6}$. So x+y=10 and $xy={\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2=25-24=1$. Identity: $x^2+y^2={\left(x+y\right)}^2-2xy$ $={\left(10\right)}^2-2\times 1=100-2=98$ Hence, option (b) is the most appropriate answer.

Partnership, profit sharing When the money stays invested for the same time, the profit is divided in the ratio of the investments. Ratio =3:5:7, so the total number of equal parts =3+5+7=15. Value of one part $=\frac{45000}{15}=3000$ Q's share $=5\times 3000=15000$ So Q gets ₹15,000. Hence, option (b) is the most appropriate answer.

Partnership, profit sharing The profit is shared in the ratio of the investments 4:5:6. Total number of equal parts =4+5+6=15 Value of one part $=\frac{60000}{15}=4000$ X's share $=4\times 4000=16000$ So X gets ₹16,000. Hence, option (a) is the most appropriate answer.

Combining two ratios Q is common to both ratios, so make the number standing for Q the same in each of them. In P:Q=3:4 the value of Q is 4, and in Q:R=5:6 it is 5; the LCM of 4 and 5 is 20. Multiply each term of the first ratio by 5: P:Q=15:20 Multiply each term of the second ratio by 4: Q:R=20:24 Now Q is 20 in both, so they can be joined: P:Q:R=15:20:24. Hence, option (a) is the most appropriate answer.

Average, missing term Formula: Average $=\frac{\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{terms}}{\mathrm{number}\mathrm{of}\mathrm{terms}}$ $\frac{11+14+x+23}{4}=18$ $48+x=18\times 4=72$ x=72-48=24 Hence, option (c) is the most appropriate answer.

Average of grouped numbers Sum of all four numbers $=30\times 4=120$ Sum of the first and the second $=25\times 2=50$ Sum of the third and the fourth $=35\times 2=70$, and 50+70=120, so the data are consistent. The first number is 20, so the second number =50-20=30. Hence, option (c) is the most appropriate answer.

Percentage, finding the whole 84 marks stand for 70% of the maximum marks. $\frac{70}{100}\times \mathrm{Maximum}=84$ Maximum $=\frac{84\times 100}{70}=120$ So the paper carried 120 maximum marks. Hence, option (b) is the most appropriate answer.

Profit percentage Cost price =400 and selling price =500, so Profit =500-400=100. Profit percentage is always taken on the cost price, not on the selling price. Formula: Profit $\%=\frac{\mathrm{Profit}}{\mathrm{Cost}\mathrm{price}}\times 100$ $=\frac{100}{400}\times 100=25$ So the profit is 25%. Hence, option (c) is the most appropriate answer.

Profit percentage Profit is the excess of the selling price over the cost price, and the profit percentage is always taken on the cost price. Formula: profit % $=\frac{SP-CP}{CP}\times 100$. Here CP = ₹800 and SP = ₹960, so the profit =960-800=160, i.e. ₹160. Profit % $=\frac{160}{800}\times 100=\frac{1}{5}\times 100=20$. Hence the profit percentage is $20\%$, so option (c) is correct.

Successive discounts Two successive discounts act one after the other, so take a convenient marked price of ₹100 and apply them in turn. After the first discount of $30\%$: $100\times \frac{70}{100}=70$. After the second discount of $20\%$: $70\times \frac{80}{100}=56$. The price falls from ₹100 to ₹56, so the total fall is ₹44 on ₹100, that is $44\%$. The formula gives the same value: $a+b-\frac{ab}{100}=30+20-\frac{30\times 20}{100}=44$. Hence the single equivalent discount is $44\%$, so option (d) is correct.

Mixture and alligation Two solutions of different strengths are mixed about the mean strength, and alligation gives the mixing ratio at once. Rule: (quantity of the weaker solution) : (quantity of the stronger solution) = (stronger − mean) : (mean − weaker). Here the two strengths are $20\%$ and $60\%$, and the mean strength wanted is $50\%$. Ratio =(60-50):(50-20)=10:30=1:3. Hence the $20\%$ solution and the $60\%$ solution must be mixed in the ratio 1:3, so option (d) is correct.

Simple interest rate Simple interest is $SI=\frac{P\times R\times T}{100}$, so the rate is $R=\frac{SI\times 100}{P\times T}$. Here P=8500, SI=2805 and T=4 years. $R=\frac{2805\times 100}{8500\times 4}=\frac{280500}{34000}$. $\frac{280500}{34000}=8.25$, so the rate comes to about $8.25\%$ per annum. Among the four choices only $8.38\%$ lies this close: at $7.25\%$, $8.75\%$ or $9.25\%$ the four-year interest would be ₹2,465, ₹2,975 or ₹3,145, all far from ₹2,805. Hence option (b) is the most appropriate answer.

Surface area of hemisphere Only the outer curved surface of a hemispherical bowl is polished, and its area is $2\pi r^2$. With r=7 cm: area $=2\times \frac{22}{7}\times 7\times 7$. $=2\times 22\times 7=308$ square cm. Cost of polishing $=308\times 0.50=154$. Hence the polishing costs ₹154, so option (a) is correct.

Interior angle sum The interior angles of any polygon of n sides add up to $\left(n-2\right)\times {180}^{\circ }$. Here the panel is a decagon, so n=10 and the sum $=\left(10-2\right)\times {180}^{\circ }$. $=8\times {180}^{\circ }={1440}^{\circ }$. Hence the total of all interior angles is ${1440}^{\circ }$, so option (c) is correct.

Sides of regular polygon At every vertex of a polygon the interior angle and the exterior angle together make a straight angle. Exterior angle $={180}^{\circ }-{160}^{\circ }={20}^{\circ }$. In a regular polygon all exterior angles are equal and add up to ${360}^{\circ }$, so $n=\frac{360}{20}$. n=18. Hence the polygon has 18 sides, so option (c) is correct.

Volume of cone and sphere Equal volumes of material mean that the volume of the cone equals the volume of the sphere. Volume of the cone $=\frac{1}{3}\pi r^{2}h$ and volume of the sphere $=\frac{4}{3}\pi r^3$. So $\frac{1}{3}\pi r^{2}h=\frac{4}{3}\pi r^3$. Cancelling $\frac{1}{3}\pi r^2$ from both sides gives h=4r. Hence the height of the cone is 4r, so option (c) is correct.

Melting and recasting Melting changes the shape but not the amount of metal, so the volume of the big ball equals the total volume of the 125 small balls. $\frac{4}{3}\pi R^3=125\times \frac{4}{3}\pi r^3$, which gives $R^3=125r^3$. With R=10: $1000=125r^3$, so $r^3=8$. $r=\sqrt[3]{8}=2$. Hence the radius of each smaller ball is 2 m, so option (b) is correct.

Recurring decimal to fraction In $0.272727\dots$ the block '27' repeats from the very first decimal place, so it is a pure recurring decimal with a two-digit period. Rule: a pure recurring decimal equals the repeating block divided by as many 9s as there are digits in that block. So $0.27=\frac{27}{99}$. $\frac{27}{99}=\frac{27÷9}{99÷9}=\frac{3}{11}$. A check confirms it: $3÷11=0.272727\dots$ Hence the required fraction is $\frac{3}{11}$, so option (c) is correct.

Slab-wise fare and ratio The fare is charged in two slabs, so work out each part separately and then add. First 50 km at ₹18 per km: $50\times 18=900$. The remaining 125-50=75 km at ₹14 per km: $75\times 14=1050$. Total fare =900+1050=1950, that is ₹1,950. The ratio 2:3 has 2+3=5 parts, so B's share is $\frac{3}{5}$ of the total fare. B pays $\frac{3}{5}\times 1950=1170$. Hence passenger B paid ₹1,170, so option (c) is correct.

Double angle formula $\tan \theta =\frac{5}{12}$ means the side opposite $\theta$ is 5 units and the side adjacent to it is 12 units. Hypotenuse $=\sqrt{5^2+{12}^2}=\sqrt{169}=13$, the familiar triplet (5,12,13). So $\sin \theta =\frac{5}{13}$ and $\cos \theta =\frac{12}{13}$. Identity: $\sin 2\theta =2\sin \theta \cos \theta$. $\sin 2\theta =2\times \frac{5}{13}\times \frac{12}{13}=\frac{120}{169}$. Hence $\sin 2\theta =\frac{120}{169}$, so option (d) is correct.

Combining ratios y occurs in both ratios, so make its value the same in each before joining them. Given x:y=5:6 and y:z=3:7. Multiply the second ratio throughout by 2 so that y becomes 6: y:z=6:14. Now the chain gives x:y:z=5:6:14. Hence x:z=5:14, so option (b) is correct.

Sum doubling at simple interest If a sum doubles, the interest earned is equal to the sum itself, that is SI=P. Formula: $SI=\frac{P\times R\times T}{100}$. Substituting SI=P and T=12: $P=\frac{P\times R\times 12}{100}$. Cancelling P from both sides: 100=12R, so $R=\frac{100}{12}$. R=8.33 approximately. Hence the required rate is about $8.33\%$ per annum, so option (b) is correct.

Trigonometric identity Treat 2A as one single angle and use the identity ${\sin }^{2}2A+{\cos }^{2}2A=1$. ${\sin }^{2}2A=1-{\left(\frac{12}{13}\right)}^2=1-\frac{144}{169}=\frac{25}{169}$. $\sin 2A=\sqrt{\frac{25}{169}}=\frac{5}{13}$, the positive value being taken as 2A is acute here. The triplet (5,12,13) shows the same thing: with adjacent side 12 and hypotenuse 13, the opposite side is 5. Hence $\sin 2A=\frac{5}{13}$, so option (d) is correct.