26 November 2025 · Shift 2

Simplification of surds Use the identity ${\left(a+b\right)}^2=a^2+2ab+b^2$ with $a=\sqrt{5}$ and $b=\sqrt{3}$. ${\left(\sqrt{5}+\sqrt{3}\right)}^2={\left(\sqrt{5}\right)}^2+2\times \sqrt{5}\times \sqrt{3}+{\left(\sqrt{3}\right)}^2$ $=5+2\sqrt{5\times 3}+3$ $=8+2\sqrt{15}$ The surd $\sqrt{15}$ cannot be simplified further, so the exact value is $8+2\sqrt{15}$ and option (b) is correct.

Percentage increase A rise of 20% multiplies the old price by $1+\frac{20}{100}=\frac{6}{5}$. New price $=800\times \frac{6}{5}=960$ Checking the other way, the increase itself is $\frac{20}{100}\times 800=160$, and 800+160=960. So the new price is Rs. 960 and option (b) is correct.

Profit and loss Let the cost price be x. At a profit of 20% the selling price is $\frac{120}{100}x=1.2x$. At a profit of 50% it would be $\frac{150}{100}x=1.5x$, and this is Rs. 600 more than the first selling price. 1.5x-1.2x=600 0.3x=600, so $x=\frac{600}{0.3}=2000$ The cost price is Rs. 2,000, so option (d) is correct.

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$, where P=1500, R=12 and T=4. $SI=\frac{1500\times 12\times 4}{100}$ $=\frac{72000}{100}=720$ The simple interest is Rs. 720, so option (b) is correct.

Efficiency in work P finishes $\frac{1}{4}$ of the work in 10 days, so P alone needs $10\times 4=40$ days for the whole work. Q finishes $\frac{1}{5}$ of the work in 12 days, so Q alone needs $12\times 5=60$ days. Efficiency is inversely proportional to the time taken, so P : Q $=\frac{1}{40}:\frac{1}{60}=3:2$. P has the higher efficiency, and it exceeds Q's by $\frac{3-2}{2}\times 100=50\%$. So P is 50% more efficient than Q and option (a) is correct.

Central angle of a polygon Join the centre of the regular polygon to each of its vertices; this cuts the complete angle of 360° at the centre into equal parts, one for every side. A regular pentagon has 5 sides, so there are 5 equal central angles. Central angle $=\frac{360°}{5}=72°$ So the angle subtended at the centre by two adjacent vertices is 72° and option (a) is correct.

Right pyramid height In a right square pyramid the vertical height, half of one base side and the slant height form a right-angled triangle, the slant height being the hypotenuse. Half the base side $=\frac{16}{2}=8$ m. $l^2=h^2+{\left(\frac{a}{2}\right)}^2$ ${17}^2=h^2+8^2$ $h^2=289-64=225$ $h=\sqrt{225}=15$ The vertical height is 15 m, so option (c) is correct.

Volume of a cone A cut parallel to the base leaves a small cone at the tip that is similar to the whole cone, so its height and radius are reduced in the same ratio. The cut is at half the height, so the small cone has height $\frac{18}{2}=9$ cm and radius $\frac{14}{2}=7$ cm. Volume $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 9$ $=\frac{1}{3}\times \frac{22}{7}\times 49\times 9=22\times 7\times 3$ =462 The volume of the smaller cone is $462{\mathrm{cm}}^3$, so option (b) is correct.

Surface area of a sphere The surface area of a sphere is $4\pi r^2$, so it varies as the square of the radius. New area : old area $=9^2:8^2=81:64$ Increase in area =81-64=17 parts out of the original 64 parts. Percentage increase $=\frac{17}{64}\times 100$ $=\frac{1700}{64}=26.5625\%\approx 26.56\%$ So the surface area rises by about 26.56% and option (b) is correct.

Area of a circular path Let the radius of the lawn be r m; the path is 2 m wide all round, so the outer radius is (r+2) m. Given: $\pi {\left(r+2\right)}^2=2\times \pi r^2$ Cancelling $\pi$, ${\left(r+2\right)}^2=2r^2$ Taking the positive square root, $r+2=\sqrt{2}r$ $r\left(\sqrt{2}-1\right)=2$, so $r=\frac{2}{\sqrt{2}-1}=\frac{2\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\left(\sqrt{2}+1\right)$ Putting $\sqrt{2}=1.41$, $r=2\times 2.41=4.82$ The radius of the lawn is about 4.82 m, so option (c) is correct.

Algebraic identities Let the three numbers be a, b and c, so a+b+c=30, $a^2+b^2+c^2=350$ and abc=1000. Formula: ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)$. 900=350+2(ab+bc+ca), so $ab+bc+ca=\frac{550}{2}=275$. Identity: $a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$. $a^3+b^3+c^3=30\left(350-275\right)+3\times 1000=30\times 75+3000$. $a^3+b^3+c^3=2250+3000=5250$, so option (b) is correct.

Trigonometric ratios The given relation $7\sin \theta -9\cos \theta =0$ means $7\sin \theta =9\cos \theta$. For an acute angle $\cos \theta \ne 0$, so both sides may be divided by $\cos \theta$. $7\times \frac{\sin \theta }{\cos \theta }=9$, that is $7\tan \theta =9$. $\tan \theta =\frac{9}{7}$, so option (b) is correct.

Surface area of a pyramid The tent is a square pyramid, so the canvas covers the square base together with the four triangular slant faces. Area of the square base $=12\times 12=144$ sq m. Each slant face is a triangle whose base is the side 12 m and whose height is the slant height 8 m. Area of one face $=\frac{1}{2}\times 12\times 8=48$ sq m, so the four faces give $4\times 48=192$ sq m. Total canvas =144+192=336 sq m, so option (b) is correct.

Recurring decimal to fraction For a purely recurring decimal, write the repeating block over as many nines as it has digits. The block here is 54, which has two digits, so $0.545454\dots =\frac{54}{99}$. Divide numerator and denominator by 9: $\frac{54}{99}=\frac{6}{11}$, which is the simplest form. Difference $=\frac{6}{11}-\frac{1}{4}=\frac{24-11}{44}=\frac{13}{44}$. So the fraction is $\frac{6}{11}$ and the difference is $\frac{13}{44}$, and option (c) is correct.

Trigonometric ratios $\sec \theta =\frac{\mathrm{hypotenuse}}{\mathrm{base}}=\frac{5}{4}$, so take hypotenuse =5 and base =4. By Pythagoras, perpendicular $=\sqrt{5^2-4^2}=\sqrt{25-16}=3$. $\tan \theta =\frac{\mathrm{perpendicular}}{\mathrm{base}}=\frac{3}{4}$. In the first quadrant every trigonometric ratio is positive, so $\tan \theta =\frac{3}{4}$ and option (c) is correct.

Partnership profit sharing Both partners invest for the same period, so the profit is divided in the ratio of their capitals. A:B=15000:25000=3:5. Total parts =3+5=8, so one part $=\frac{80000}{8}=10000$. A's share $=3\times 10000=30000$, that is Rs. 30,000, so option (a) is correct.

Combining two ratios B carries the same number 4 in both ratios, so the two ratios can be joined without changing anything. A:B=3:4 and B:C=4:11 give A:B:C=3:4:11. Total parts =3+4+11=18. B's share $=\frac{4}{18}\times 2200=\frac{8800}{18}=488.88$. So B gets Rs. 488.88, and option (a) is correct.

Degree to radian conversion Formula: radian measure = degree measure $\times \frac{\pi }{180}$, because ${180}^{\circ }=\pi$ radians. Radian measure $=140\times \frac{\pi }{180}=\frac{140\pi }{180}$. Divide numerator and denominator by 20: $\frac{140\pi }{180}=\frac{7\pi }{9}$. Hence option (b) is correct.

Compound interest, changing rates When the rate changes from year to year, the two years together act like a single rate $r_1+r_2+\frac{r_1{r}_2}{100}$. Effective rate $=6+12+\frac{6\times 12}{100}=18+0.72=18.72\%$. So $18.72\%$ of the principal is the total interest: $\frac{18.72}{100}\times P=3744$. $P=\frac{3744\times 100}{18.72}=20000$. The original principal was ₹20,000, so option (b) is correct.

Average and total marks Total = average $\times$ number of students, so work with totals rather than averages. Total marks of the 49 students $=49\times 75=3675$. Total marks of all 50 students at the new average $=50\times 65=3250$. The change caused by the 50th student =3250-3675=-425. Its numerical value is 425, which is the figure asked for, so option (d) is correct.

Rise in water level The water rises by exactly the volume of the submerged sphere, so the starting height of 12 cm is extra information. Volume of the sphere $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(4\right)}^3=\frac{256\pi }{3}$ cubic cm. Base area of the cylinder $=\pi R^2=\pi {\left(8\right)}^2=64\pi$ sq cm. Rise in level $=\frac{\mathrm{volume}\mathrm{added}}{\mathrm{base}\mathrm{area}}=\frac{256\pi }{3}÷64\pi =\frac{4}{3}$. $\frac{4}{3}=1.33$ cm, so the water level rises by 1.33 cm and option (b) is correct.

Mixture and alligation Both solutions have parts totalling 2+5=7, so take 7 units of A and 7 units of B, which is the ratio 1:1. Alcohol =2+5=7 units and water =5+2=7 units, so the new mixture has alcohol : water =1:1. In 600 ml of this mixture, alcohol =300 ml and water =300 ml. Let x ml of water be added; alcohol stays 300 ml, so $\frac{300}{300+x}=\frac{3}{4}$. 1200=900+3x, giving 3x=300 and x=100. So 100 ml of water must be added, and option (a) is correct.

Algebraic identities Square the given relation: ${\left(a-\frac{1}{a}\right)}^2=a^2-2+\frac{1}{a^2}$. So $a^2+\frac{1}{a^2}=8^2+2=64+2=66$. Square once more: ${\left(a^2+\frac{1}{a^2}\right)}^2=a^4+2+\frac{1}{a^4}={66}^2=4356$. $a^4+\frac{1}{a^4}=4356-2=4354$, so option (b) is correct.

Trains moving towards each other From 8 AM to 10 AM only the first train is moving, that is for 2 hours. Distance it covers alone $=50\times 2=100$ km. Distance still between the two trains at 10 AM =490-100=390 km. After 10 AM they approach each other, so the relative speed =50+80=130 km/h. Time taken to meet $=\frac{390}{130}=3$ hours. 3 hours after 10 AM is 1:00 PM, so option (a) is correct.

Marked price and discount Let the discount be 2 units, a convenient choice because 150 percent of it is a whole number. Marked price $=150\%$ of $2=\frac{150}{100}\times 2=3$ units. Selling price = marked price - discount =3-2=1 unit. Required percentage $=\frac{\mathrm{MP}}{\mathrm{SP}}\times 100=\frac{3}{1}\times 100=300\%$. So the marked price is 300 percent of the selling price, and option (b) is correct.