27 November 2025 · Shift 1

Algebraic identity The expression has the form $x^2-y^2$ with $x=a^2+b^2$ and y=2ab, so use $x^2-y^2=\left(x-y\right)\left(x+y\right)$. ${\left(a^2+b^2\right)}^2-{\left(2ab\right)}^2=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)$ $={\left(a-b\right)}^2{\left(a+b\right)}^2={\left(\left(a-b\right)\left(a+b\right)\right)}^2={\left(a^2-b^2\right)}^2$ $=a^4-2a^2{b}^2+b^4$ Hence option (a) is the correct answer.

Comparing surds Write every value as a single square root; then the one with the largest number inside is the greatest. $\sqrt{15}$ stays as it is, $3\sqrt{5}=\sqrt{3^2\times 5}=\sqrt{45}$, and $\sqrt{30}$ stays as it is. Since 45>30>15, we get $3\sqrt{5}>\sqrt{30}>\sqrt{15}$. So the greatest value is $3\sqrt{5}$, and option (b) is the correct answer.

Income and expenditure ratio Let the incomes be 5x and 4x, and the expenditures be 3y and 2y. Saving = Income − Expenditure, and each of them saves ₹3000: 5x-3y=3000 … (i) $4x-2y=3000\Rightarrow 2x-y=1500\Rightarrow y=2x-1500$ … (ii) Putting (ii) into (i): 5x-3(2x-1500)=3000 $5x-6x+4500=3000\Rightarrow x=1500$ A's income $=5x=5\times 1500=7500$, i.e. ₹7500, so option (a) is the correct answer.

Percentage Required percentage $=\frac{\mathrm{Part}}{\mathrm{Whole}}\times 100=\frac{75}{500}\times 100$ $=\frac{75}{5}=15$ So Rs. 75 is $15\%$ of Rs. 500, and option (b) is the correct answer.

False weight profit He charges the price of 1000 gm but actually hands over only 920 gm, so his cost corresponds to 920 gm while his receipt corresponds to 1000 gm. Gain % $=\frac{\mathrm{True}\mathrm{weight}-\mathrm{False}\mathrm{weight}}{\mathrm{False}\mathrm{weight}}\times 100$ $=\frac{1000-920}{920}\times 100=\frac{80}{920}\times 100$ $=\frac{8000}{920}=8.6956\dots$ Rounded to two decimal places the gain is $8.70\%$, so option (b) is the correct answer.

Speed, time and distance Let the original speed be s km/h. The distance is the same in both cases. Distance = Speed $\times$ Time, so $4s=\left(s+5\right)\times 3$ 4s=3s+15 s=15 His original speed is 15 km/h, so option (c) is the correct answer.

Trains moving towards each other The train from B is the slower one; let its speed be 5x. The train from A is 20% faster, so its speed is $\frac{120}{100}\times 5x=6x$. Moving towards each other, their relative speed is 5x+6x=11x. Together they cover the whole 660 km in 5 hours: $11x\times 5=660$ $11x=132\Rightarrow x=12$ Speed of the slower train $=5x=5\times 12=60$ km/h, so option (b) is the correct answer.

Compound interest rate Formula: Amount $=P{\left(1+\frac{R}{100}\right)}^n$, with P=7290, n=2 and Amount =8410. ${\left(1+\frac{R}{100}\right)}^2=\frac{8410}{7290}=\frac{841}{729}$ Both are perfect squares: $\frac{841}{729}=\frac{{29}^2}{{27}^2}$, so $1+\frac{R}{100}=\frac{29}{27}$ $\frac{R}{100}=\frac{29}{27}-1=\frac{2}{27}$ $R=\frac{200}{27}=7.407\dots \approx 7.40$ So the rate is $7.40\%$ per annum, and option (b) is the correct answer.

Time and work Take the total work as LCM(20,25)=100 units. A's one day's work $=\frac{100}{20}=5$ units and B's one day's work $=\frac{100}{25}=4$ units. In 10 days A and B together do $\left(5+4\right)\times 10=90$ units. Remaining work =100-90=10 units, which C does in 10 days, so C's one day's work $=\frac{10}{10}=1$ unit. A, B and C together do 5+4+1=10 units in one day. Time taken $=\frac{100}{10}=10$ days, so option (a) is the correct answer.

Simple interest At 10% per annum simple interest for 2 years, the interest is $2\times 10\%=20\%$ of the sum. So Amount $=100\%+20\%=120\%$ of the sum, and this equals ₹9000. Sum $=\frac{9000}{120}\times 100=75\times 100=7500$ The required sum is ₹7500, so option (c) is the correct answer.

Area of a trapezium Formula: area of a trapezium $=\frac{1}{2}\times$ (sum of the parallel sides) $\times$ height. Here the parallel sides (bases) are 12 cm and 16 cm and the area is $98{\mathrm{cm}}^2$. Substituting: $98=\frac{1}{2}\times \left(12+16\right)\times h$. $98=\frac{1}{2}\times 28\times h=14h$. $h=\frac{98}{14}=7$ cm, so option (b) is the correct answer.

Surface area of a cuboid The outer surface of a closed cuboid is its total surface area =2(lb+bh+hl). Here l=4 m, b=2.5 m and h=1.5 m. $lb=4\times 2.5=10$, $bh=2.5\times 1.5=3.75$, $hl=1.5\times 4=6$. Total surface area $=2\left(10+3.75+6\right)=2\times 19.75=39.5m^2$. Cost $=39.5\times 12=474$. So the painting costs ₹474 and option (c) is the correct answer.

Volume of a pyramid The base is an equilateral triangle of side 10 cm, so base area $=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times {10}^2=25\sqrt{3}{\mathrm{cm}}^2$. Formula: volume of a pyramid $=\frac{1}{3}\times$ base area $\times$ height. $=\frac{1}{3}\times 25\sqrt{3}\times 9=75\sqrt{3}{\mathrm{cm}}^3$. Taking $\sqrt{3}=1.732$, volume $=75\times 1.732=129.9{\mathrm{cm}}^3$. So option (a) is the correct answer.

Right triangle, altitude $\angle QPR=\angle QRP$, so $△PQR$ is isosceles and the sides opposite these equal angles are equal: QR=PQ=10 cm. By Pythagoras, $PR=\sqrt{P{Q}^2+Q{R}^2}=\sqrt{{10}^2+{10}^2}=10\sqrt{2}$ cm. QM is the altitude to the hypotenuse, and area of $△PQR$ counted two ways gives $\frac{1}{2}\times PQ\times QR=\frac{1}{2}\times PR\times QM$. $QM=\frac{PQ\times QR}{PR}=\frac{10\times 10}{10\sqrt{2}}=5\sqrt{2}$ cm, which indeed satisfies PR=2QM. $P{R}^2-Q{M}^2={\left(10\sqrt{2}\right)}^2-{\left(5\sqrt{2}\right)}^2=200-50=150$. So option (a) is the correct answer.

Centroid of a triangle A median joins a vertex to the mid-point of the opposite side, so every triangle has three medians. The three medians are always concurrent, and their common point is called the centroid (usually written G). The centroid divides each median in the ratio 2:1 measured from the vertex, and it is the centre of gravity of a triangular lamina. The other options come from different sets of lines: perpendicular bisectors meet at the circumcentre, angle bisectors at the incentre and altitudes at the orthocentre. So option (b) is the correct answer.

Volume of cone and cylinder Let the cone have radius r=12 cm and height h; then the cylinder has radius R and height 2h. Given: volume of cone $=\frac{1}{2}\times$ volume of cylinder. $\frac{1}{3}\pi r^{2}h=\frac{1}{2}\times \pi R^2\left(2h\right)=\pi R^{2}h$. Cancelling $\pi h$ from both sides: $R^2=\frac{r^2}{3}=\frac{{12}^2}{3}=\frac{144}{3}=48$. $R=\sqrt{48}=\sqrt{16\times 3}=4\sqrt{3}$ cm. So the radius of the cylinder is $4\sqrt{3}$ cm and option (b) is the correct answer.

Common tangents to circles The number of common tangents depends only on how d compares with $r_1+r_2$ and $\left(r_1-r_2\right)$. The given condition $\left(r_1-r_2\right)<d<r_1+r_2$ is exactly the condition for the two circles to cut each other at two distinct points. Two intersecting circles overlap, so no line can pass between them; the internal (transverse) tangents therefore do not exist. Only the two direct (external) common tangents can be drawn, one on each side. For reference: $d>r_1+r_2$ gives 4 tangents, $d=r_1+r_2$ gives 3, and $d=\left(r_1-r_2\right)$ gives 1. So the required number is 2 and option (b) is the correct answer.

Trigonometric identity Write both functions in terms of sine and cosine: $\cot x=\frac{\cos x}{\sin x}$ and $\mathrm{cosec}x=\frac{1}{\sin x}$. $\frac{\cot x}{\mathrm{cosec}x}=\frac{\cos x/\sin x}{1/\sin x}$. Dividing by $\frac{1}{\sin x}$ is the same as multiplying by $\sin x$: $=\frac{\cos x}{\sin x}\times \sin x=\cos x$. A quick check with $x={60}^{\circ }$: $\frac{\cot {60}^{\circ }}{\mathrm{cosec}{60}^{\circ }}=\frac{1/\sqrt{3}}{2/\sqrt{3}}=\frac{1}{2}=\cos {60}^{\circ }$. So option (a) is the correct answer.

Partnership and profit Let the total profit be x. 10% of it goes to charity, so the amount actually divided between P and Q is $90\%$ of $x=\frac{9x}{10}$. The investments are in the ratio 5:3, so P's share is $\frac{5}{5+3}=\frac{5}{8}$ of the divided amount. $\frac{5}{8}\times \frac{9x}{10}=900$ $\frac{9x}{16}=900\Rightarrow x=900\times \frac{16}{9}=1600$. So the total profit is ₹1,600 and option (d) is the correct answer.

Successive discounts Successive discounts act one after the other, the second one on the already reduced price. After 10% off, the price is 90% of ₹1,500: $1500\times \frac{90}{100}=1350$. After 20% off that, the price is 80% of ₹1,350: $1350\times \frac{80}{100}=1080$. Shortcut: the single equivalent discount is $10+20-\frac{10\times 20}{100}=28\%$, and $1500\times \frac{72}{100}=1080$. So the final price is ₹1,080 and option (b) is the correct answer.

Ratio and proportion Let the two prices be 7x and 4x, so that their ratio stays 7:4. The adult ticket is the 7-part: $7x=280\Rightarrow x=\frac{280}{7}=40$. Child ticket $=4x=4\times 40=160$. So option (c) is the correct answer.

Average of marks Formula: total = average $\times$ number of tests. For an average of 52 over 5 tests, the required total is $5\times 52=260$. Total of the first four tests =40+55+60+45=200. Marks needed in the 5th test =260-200=60. So option (c) is the correct answer.

Internal common tangent Formula: length of the internal (transverse) common tangent $=\sqrt{d^2-{\left(r_1+r_2\right)}^2}$, where d is the distance between the centres. Here $r_1+r_2=3+7=10$ and the tangent length is 24. $24=\sqrt{d^2-{10}^2}$ Squaring: $576=d^2-100\Rightarrow d^2=676$. $d=\sqrt{676}=26$ unit. So option (d) is the correct answer.

Trigonometric ratios $\mathrm{cosec}x=\frac{\mathrm{hypotenuse}}{\mathrm{perpendicular}}=\frac{5}{3}$, so take perpendicular =3 and hypotenuse =5. Base $=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$ (the familiar 3-4-5 triangle). $\sec x=\frac{\mathrm{hypotenuse}}{\mathrm{base}}=\frac{5}{4}$. So option (b) is the correct answer.

Trigonometric identities $\cot \theta =\frac{3}{4}$ means base =3 and perpendicular =4, so $\tan \theta =\frac{4}{3}$. Using ${\sec }^2\theta =1+{\tan }^2\theta$: ${\sec }^2\theta =1+\frac{16}{9}=\frac{25}{9}$. Using ${\mathrm{cosec}}^2\theta =1+{\cot }^2\theta$: ${\mathrm{cosec}}^2\theta =1+\frac{9}{16}=\frac{25}{16}$. ${\sec }^2\theta -{\mathrm{cosec}}^2\theta =\frac{25}{9}-\frac{25}{16}=25\times \frac{16-9}{144}=\frac{25\times 7}{144}$. $=\frac{175}{144}$, so option (a) is the correct answer.