27 November 2025 · Shift 3

Percentage of a number Let the number be x. $\frac{25}{100}\times x=35$ $x=35\times 4=140$ Now $80\%$ of the number $=\frac{80}{100}\times 140$. =112, so option (b) is correct.

Profit percentage Let the cost price of one article be ₹1, so the cost price of 15 articles is ₹15. The gain on selling those 15 articles equals the cost price of 5 articles, i.e. ₹5. Profit percentage $=\frac{\mathrm{profit}}{\mathrm{cost}\mathrm{price}}\times 100=\frac{5}{15}\times 100$. $=\frac{100}{3}=33.33\%$ Hence option (b) is correct.

Alternate-day work Take the total work as the LCM of 12 and 18, i.e. 36 units. A does $\frac{36}{12}=3$ units a day and B does $\frac{36}{18}=2$ units a day. Working alternately, one full cycle of two days (A then B) finishes 3+2=5 units. Seven such cycles, that is 14 days, finish $7\times 5=35$ units. The 15th day is A's turn and only 36-35=1 unit is left, which A does in $\frac{1}{3}$ day. Total time $=14+\frac{1}{3}=14\frac{1}{3}$ days, so option (c) is correct.

Speed, time and distance Let the original speed be x km/h, so the reduced speed is (x-8) km/h. The time at the reduced speed exceeds the time at the original speed by 1 hour. $\frac{160}{x-8}-\frac{160}{x}=1$ 160x-160(x-8)=x(x-8) $1280=x^2-8x$ $x^2-8x-1280=0$, that is (x-40)(x+32)=0. A speed cannot be negative, so x=40 km/h. Hence option (b) is correct.

Simple interest Let the sum (principal) be P; the rate is $10\%$ per annum and the time is 5 years. Simple interest $=\frac{P\times 10\times 5}{100}=\frac{P}{2}$. Amount $=P+\frac{P}{2}=\frac{3P}{2}$. $\frac{3P}{2}=6000$ $P=\frac{6000\times 2}{3}=4000$ The original sum is ₹4,000, so option (c) is correct.

Volume of a cuboid A rectangular parallelepiped is a cuboid, whose volume is length × breadth × height. Volume $=4\times 1.8\times 1.5=10.8$ cubic metres. One cubic metre holds 1000 litres. Capacity $=10.8\times 1000=10800$ litres, so option (c) is correct.

Volume of a pyramid The base is a right triangle whose two legs are 5 cm and 12 cm. Area of the base $=\frac{1}{2}\times 5\times 12=30$ sq cm. Volume of a pyramid $=\frac{1}{3}\times \mathrm{base}\mathrm{area}\times \mathrm{height}$. $=\frac{1}{3}\times 30\times 15=150$ cubic cm. Hence option (b) is correct.

Angle bisector theorem PM makes $\angle RPM=\angle QPM$, so PM is the bisector of $\angle P$ and meets QR at M. By the angle bisector theorem, $\frac{QM}{MR}=\frac{PQ}{PR}$. $\frac{QM}{MR}=\frac{21}{14}=\frac{3}{2}$ So QR is cut in the ratio 3:2 and $MR=\frac{2}{3+2}\times QR$. $MR=\frac{2}{5}\times 15.5=6.2$ cm, so option (b) is correct.

Percentage and ratio Students who failed $=40\%$ of 400=160. Students who passed =400-160=240. Passed with distinction $=40\%$ of 240=96. Passed without distinction =240-96=144. Required ratio =96:144=2:3, so option (a) is correct.

RHS congruence rule The RHS criterion states that if the hypotenuse and one side of a right triangle equal the hypotenuse and the corresponding side of another right triangle, the two triangles are congruent. Assertion (A) says exactly this, so A is true. Reason (R) says that the RHS rule applies to right-angled triangles, which is also true. R names the very rule that makes A true, so R is the correct explanation of A. Hence option (a) is correct.

Perimeter of rectangle The same wire is bent into the rectangle, so the perimeter of the rectangle equals the circumference of the circular loop. Circumference $=2\pi r=2\times \frac{22}{7}\times 21=132$ cm. Let the length and breadth be 7x and 4x, so 2(7x+4x)=132. $22x=132\Rightarrow x=6$, hence the length $=7\times 6=42$ cm. Diameter of the loop $=2\times 21=42$ cm. Required sum =42+42=84 cm, so option (a) is correct.

Trigonometric ratios $\csc \theta =\frac{2}{\sqrt{3}}$ gives $\sin \theta =\frac{\sqrt{3}}{2}$, so $\theta ={60}^{\circ }$. For $\theta ={60}^{\circ }$, $\cos \theta =\frac{1}{2}$. $\sec \theta =\frac{1}{\cos \theta }=\frac{1}{\frac{1}{2}}=2$. So option (b) is correct.

Partnership profit share All three invest for the same period, so the profit is divided in the ratio of the investments. 40000:50000:60000=4:5:6, and the ratio total is 4+5+6=15. Let the total profit be P; then R's share $=\frac{6}{15}P$. $\frac{6}{15}P=6000\Rightarrow P=6000\times \frac{15}{6}=15000$. The total profit is ₹15,000, so option (b) is correct.

Circular track meetings When two runners move in the same direction, the relative speed is the difference of their speeds. Relative speed =20-15=5 m/s. They cross each other every time the faster runner gains one full lap of 500 m on the slower one. Time for one crossing $=\frac{500}{5}=100$ s. 10 minutes $=10\times 60=600$ s. Number of crossings $=\frac{600}{100}=6$, so option (c) is correct.

Corrected average Total marks first taken $=50\times 80=4000$. Only one entry is wrong, and the two marks differ by 92-39=53. Replacing that entry by the correct value lowers the total by 53: 4000-53=3947. Correct average $=\frac{3947}{50}=78.94$. So option (b) is correct.

Ratio of salaries Let the three salaries be 9x, 7x and 5x. The supervisor's salary is 7x=35000, so $x=\frac{35000}{7}=5000$. Manager's salary $=9x=9\times 5000=45000$. So the manager earns 45,000 and option (c) is correct.

Tangent and radius The radius drawn to the point of contact is perpendicular to the tangent, so $\angle OTP={90}^{\circ }$. In right triangle OTP, $O{P}^2=O{T}^2+P{T}^2=7^2+{24}^2=49+576=625$. $OP=\sqrt{625}=25$ units. The point of the circle nearest to P lies on the line PO, at a distance of one radius from O. $PN=OP-\mathrm{radius}=25-7=18$ units, so option (a) is correct.

Polygon angles At each vertex the interior and exterior angles form a linear pair, so their sum is ${180}^{\circ }$. Let the exterior angle be x; then the interior angle is 8x. $8x+x={180}^{\circ }\Rightarrow 9x={180}^{\circ }\Rightarrow x={20}^{\circ }$. For a regular polygon the exterior angles add up to ${360}^{\circ }$, so $n=\frac{{360}^{\circ }}{{20}^{\circ }}=18$. The polygon has 18 sides, so option (d) is correct.

Algebraic identities From xy=15 we get $y=\frac{15}{x}$. Putting this in $2x^2+y^2=59$ gives $2x^2+\frac{225}{x^2}=59$. Multiplying by $x^2$: $2x^4-59x^2+225=0$. $x^2=\frac{59\pm \sqrt{{59}^2-4\times 2\times 225}}{4}=\frac{59\pm \sqrt{1681}}{4}=\frac{59\pm 41}{4}$. This gives $x^2=25$ or $x^2=\frac{9}{2}$; the whole-number solution is x=5, and then $y=\frac{15}{5}=3$. 2y+x=6+5=11 and 2y-x=6-5=1. So option (b) is correct.

Laws of indices Write each base as a power of a prime: $32=2^5$ and $243=3^5$. ${\left(\frac{32}{243}\right)}^{\frac{2}{5}}={\left(\frac{2^5}{3^5}\right)}^{\frac{2}{5}}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$. $\sqrt[5]{32}={\left(2^5\right)}^{\frac{1}{5}}=2$. ${\left(\frac{243}{32}\right)}^{\frac{3}{5}}={\left(\frac{3}{2}\right)}^3=\frac{27}{8}$. Product $=\frac{4}{9}\times 2\times \frac{27}{8}=\frac{216}{72}=3$. So option (a) is correct.

Trigonometric identity Change every ratio into sine and cosine: $\cot \theta =\frac{\cos \theta }{\sin \theta }$. $\cot \theta \times \cos \theta +\sin \theta =\frac{{\cos }^2\theta }{\sin \theta }+\sin \theta$. Taking $\sin \theta$ as the common denominator: $=\frac{{\cos }^2\theta +{\sin }^2\theta }{\sin \theta }$. Since ${\sin }^2\theta +{\cos }^2\theta =1$, this equals $\frac{1}{\sin \theta }=\csc \theta$. So option (b) is correct.

Common tangents The number of common tangents depends only on how the two circles are placed, not on their radii. Two circles that cut each other at two points overlap, so no tangent can pass between them; only the two direct (external) common tangents exist. Equal radii merely make these two tangents parallel to the line joining the centres; the count stays the same. Hence 2 common tangents are possible, so option (c) is correct.

Trigonometric ratios $\cos 3A=\frac{7}{25}$ means, for the angle 3A, base =7 and hypotenuse =25. Perpendicular $=\sqrt{{25}^2-7^2}=\sqrt{625-49}=\sqrt{576}=24$, the familiar 7, 24, 25 triplet. $\sin 3A=\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}=\frac{24}{25}$. $\cot 3A=\frac{\mathrm{base}}{\mathrm{perpendicular}}=\frac{7}{24}$. So option (a) is correct.

Compound interest For annual compounding, Amount $=P{\left(1+\frac{r}{100}\right)}^n$. $6050=P{\left(1+\frac{10}{100}\right)}^2=P{\left(\frac{11}{10}\right)}^2=P\times \frac{121}{100}$. $P=6050\times \frac{100}{121}=5000$. The required sum is ₹5,000, so option (a) is correct.

Discount and marked price '150 percent more than the discount' means the marked price is the discount plus $150\%$ of it, i.e. 2.5 times the discount. Take the discount as 100 units; then the marked price is $2.5\times 100=250$ units. Selling price = marked price - discount =250-100=150 units. Required percentage $=\frac{150}{250}\times 100=60$. The selling price is 60 percent of the marked price, so option (b) is correct.