17 November 2025 · Shift 3

Average with correction Total marks first recorded $=20\times 55=1100$. One score was entered 30 short, so the corrected total =1100+(70-40)=1130. Corrected average $=\frac{1130}{20}=56.5$. The new student scores 20% more than this: $56.5\times \frac{120}{100}=67.8$. New total =1130+67.8=1197.8, now shared by 21 students. Final average $=\frac{1197.8}{21}=57.04$. Hence option (b) is the answer.

Percentage return The three shares are 32%, 28% and the rest, i.e. $100-32-28=40\%$, of ₹1,80,000. Mutual funds = ₹57,600; a 25% growth gives a gain of $0.25\times 57600=14400$, i.e. ₹14,400. Fixed deposits = ₹50,400; a 10% growth gives a gain of $0.10\times 50400=5040$, i.e. ₹5,040. Recurring deposits = ₹72,000; a 20% growth gives a gain of $0.20\times 72000=14400$, i.e. ₹14,400. Total gain = ₹14,400 + ₹5,040 + ₹14,400 = ₹33,840 on an investment of ₹1,80,000. Return $=\frac{33840}{180000}\times 100=18.8\%$. Hence option (a) is the answer.

Successive percentage change Let the original price be x. After the 25% rise the price becomes $x\times \frac{125}{100}=\frac{5x}{4}$. The 20% discount leaves 80% of that: $\frac{5x}{4}\times \frac{80}{100}=x$. So a 25% rise followed by a 20% discount cancel each other exactly, and the final price equals the original price. Final price is ₹84,000, so the original price was also ₹84,000. Hence option (d) is the answer.

Unit digit of squares If a number ends in 3, its square ends in 9, because $3\times 3=9$. All four options end in 9, so that test alone cannot decide — the number must also be a perfect square. The nearby squares are ${12}^2=144$, ${13}^2=169$ and ${14}^2=196$, so 129, 149 and 189 are not perfect squares at all. $169={13}^2$, and 13 itself ends in 3, so it satisfies both conditions. Hence option (b) is the answer.

Square root 3025 ends in 25, so its square root ends in 5. Drop the last two digits: 30 lies between $5^2=25$ and $6^2=36$, so the tens digit of the root is 5. Therefore $\sqrt{3025}=55$, and the check confirms it: $55\times 55=3025$. Hence option (c) is the answer.

Area of a circular ring External radius $R=\frac{54}{2}=27$ cm and internal radius $r=\frac{38}{2}=19$ cm. The cross-section is the ring between the two circles, so its area $=\pi R^2-\pi r^2=\pi \left(R^2-r^2\right)$. $R^2-r^2=\left(27+19\right)\left(27-19\right)=46\times 8=368$. Area $=\frac{22}{7}\times 368=\frac{8096}{7}=1156.57$ cm². Hence option (d) is the answer.

Difference of cubes Write both terms as cubes: $216x^3={\left(6x\right)}^3$ and $343y^3={\left(7y\right)}^3$. Use the identity $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$ with a=6x, b=7y: $216x^3-343y^3=\left(6x-7y\right)\left(36x^2+42xy+49y^2\right)$. Dividing by (6x-7y) leaves the quotient $36x^2+42xy+49y^2$. Compare it with $A{x}^2-B{y}^2+Cxy$: A=36, -B=49 so B=-49, and C=42. 2A+B-C=2(36)+(-49)-42=72-49-42=-19. Hence option (a) is the answer.

Completing the square Open the bracket and bring every term to one side: $x^2+y^2+z^2-8x-10y-6z+50=0$. Split 50 as 16+25+9 and group each variable with its own constant: $\left(x^2-8x+16\right)+\left(y^2-10y+25\right)+\left(z^2-6z+9\right)=0$. ${\left(x-4\right)}^2+{\left(y-5\right)}^2+{\left(z-3\right)}^2=0$. A sum of squares of real numbers is zero only when each square is zero, so x=4, y=5, z=3. $2\left(x+y\right)-3z=2\left(4+5\right)-3\times 3=18-9=9$. Hence option (d) is the answer.

Curved surface of a cone Base circumference $=2\pi r=22$, so $r=\frac{22\times 7}{2\times 22}=3.5$ cm. Curved surface area of a cone $=\pi rl$, where l is the slant height. $=\frac{22}{7}\times 3.5\times 7$ =77 cm². Hence option (b) is the answer.

Trigonometric identity The Pythagorean identity for cosecant is $1+{\cot }^{2}A={\mathrm{cosec}}^{2}A$. Rearranging it gives ${\mathrm{cosec}}^{2}A-{\cot }^{2}A=1$ for every angle A, so the value of $\sin A$ is not even needed. Check it with the given data: $\sin A=\frac{5}{13}$ means opposite =5, hypotenuse =13, so base $=\sqrt{{13}^2-5^2}=\sqrt{144}=12$. Then $\mathrm{cosec}A=\frac{13}{5}$ and $\cot A=\frac{12}{5}$, so $\frac{169}{25}-\frac{144}{25}=\frac{25}{25}=1$. Hence option (d) is the answer.

Trigonometric identity Open the numerator with the identity ${\left(1+a\right)}^2+{\left(1-a\right)}^2=2\left(1+a^2\right)$. So the numerator $=2\left(1+{\sin }^{2}x\right)$. Here $\sin x=\frac{5}{13}$, so ${\sin }^{2}x=\frac{25}{169}$ and ${\cos }^{2}x=1-\frac{25}{169}=\frac{144}{169}$. Numerator $=2\left(1+\frac{25}{169}\right)=2\times \frac{194}{169}=\frac{388}{169}$. Required value $=\frac{388}{169}÷\frac{144}{169}=\frac{388}{144}=\frac{97}{36}$. Hence option (c) is the answer.

Simple interest rate Take the amount at the end of 2 years as 117 units; then the amount at the end of 5 years is 147 units. The extra 3 years bring simple interest =147-117=30 units, so interest of 1 year $=\frac{30}{3}=10$ units. Interest of 2 years $=2\times 10=20$ units, so the principal =117-20=97 units. Rate = (interest of one year ÷ principal) × 100 $=\frac{10}{97}\times 100$ $=\frac{1000}{97}=10.30\%$ (approximately). Hence option (c) is the answer.

Tangential quadrilateral When a circle is inscribed in a quadrilateral, the two tangents drawn from each vertex are equal in length. Adding those equal tangent lengths gives the standard result: AB+CD=BC+AD. 10+12=13+AD AD=22-13=9 cm. Hence option (c) is the answer.

Ratio and profit share A:B=4:5 and B:C=6:7; make B's number the same in both ratios. Multiply the first ratio by 6 and the second by 5: A:B=24:30 and B:C=30:35. So A:B:C=24:30:35 and the total is 24+30+35=89 parts. A's share $=\frac{24}{89}\times 89000=24\times 1000=$ Rs 24,000. Hence option (b) is the answer.

Time and work efficiency The same work is done both times, so 16(8B+12G)=20(6B+14G), where B and G are one day's work of a boy and a girl. 128B+192G=120B+280G $8B=88G\Rightarrow 1B=11G$, i.e. one boy works as fast as 11 girls. Total work $=16\left(8\times 11G+12G\right)=16\times 100G=1600$ girl-days. Time taken by 20 girls $=\frac{1600}{20}=80$ days. Hence option (a) is the answer.

Right triangle ratios $\angle Q={90}^{\circ }$, so PR is the hypotenuse and $QR=\sqrt{{25}^2-{24}^2}=\sqrt{625-576}=\sqrt{49}=7$ cm. For angle P the opposite side is QR=7, the adjacent side is PQ=24 and the hypotenuse is PR=25. $\sin P=\frac{7}{25}$ and $\cos P=\frac{24}{25}$. $\sin P+\cos P=\frac{7}{25}+\frac{24}{25}=\frac{31}{25}$ Hence option (b) is the answer.

Cosec-cot identity Use ${\csc }^2\theta -{\cot }^2\theta =1$, which factorises as $\left(\csc \theta +\cot \theta \right)\left(\csc \theta -\cot \theta \right)=1$. Here $\csc \theta +\cot \theta =5$, so $\csc \theta -\cot \theta =\frac{1}{5}$. Adding the two equations: $2\csc \theta =5+\frac{1}{5}=\frac{26}{5}$. $\csc \theta =\frac{13}{5}$, and $\sin \theta$ is its reciprocal, so $\sin \theta =\frac{5}{13}$. $\theta$ lies in the first quadrant, so this positive value is the admissible one. Hence option (a) is the answer.

Tangent to a circle A line can meet a circle at two points, at exactly one point, or at no point at all. The line that meets it at exactly one point is the tangent, and the radius drawn to that point of contact is perpendicular to it. A secant cuts the circle at two points, a chord is the segment joining two points of the circle, and a diameter is the chord through the centre — each of these meets the circle twice. Hence option (c) is the answer.

Parallel chords of a circle The perpendicular from the centre to a chord bisects that chord, so each chord gives a right triangle with the radius. For the 16 cm chord: half-chord =8, so its distance $=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ cm. For the 30 cm chord: half-chord =15, so its distance $=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8$ cm. If both chords lie on the same side of the centre, the distance between them =15-8=7 cm. If they lie on opposite sides of the centre, the distance =15+8=23 cm. Hence option (a) is the answer.

Speed, time and distance Speed of the train $=\frac{180}{6}=30$ km/h. Speed of the car $=30\times \frac{120}{100}=36$ km/h. Speed of the bike $=36\times \frac{125}{100}=45$ km/h. Distance to be covered $=180\times \frac{130}{100}=234$ km. Time by car $=\frac{234}{36}=6.5$ hours and time by bike $=\frac{234}{45}=5.2$ hours. Difference =6.5-5.2=1.3 hours $=1.3\times 60=78$ minutes. Hence option (a) is the answer.

Angle between two chords When two chords cut each other inside a circle, the angle between them equals half the sum of the two arcs they intercept. Arc AC subtends $\angle AOC={60}^{\circ }$ at the centre and arc BD subtends $\angle BOD={70}^{\circ }$, so those arcs measure ${60}^{\circ }$ and ${70}^{\circ }$. $\angle APC=\frac{1}{2}\left({60}^{\circ }+{70}^{\circ }\right)=\frac{{130}^{\circ }}{2}={65}^{\circ }$ Hence option (b) is the answer.

Chord and central angle Join O to P and to Q; then OP=OQ=r and $\angle POQ={120}^{\circ }$. The perpendicular from O to PQ bisects both the chord and the angle, so it makes ${60}^{\circ }$ with each radius and the half-chord $=r\sin {60}^{\circ }$. So $PQ=2r\sin {60}^{\circ }=2r\times \frac{\sqrt{3}}{2}=r\sqrt{3}$. $r\sqrt{3}=8\sqrt{3}\Rightarrow r=8$ cm. Hence option (c) is the answer.

Surface area of a solid Volume of a hemisphere $=\frac{2}{3}\pi r^3$, so use the given volume to find the common radius. $\frac{2}{3}\pi r^3=\frac{250\pi }{3}\Rightarrow r^3=125\Rightarrow r=5$ cm. The cone has h=12 cm, so its slant height $l=\sqrt{r^2+h^2}=\sqrt{25+144}=\sqrt{169}=13$ cm. The circular face where the two solids join is hidden, so the outer surface is only the curved surface of the hemisphere plus the curved surface of the cone. Total surface area $=2\pi r^2+\pi rl=2\pi {\left(5\right)}^2+\pi \left(5\right)\left(13\right)$ $=50\pi +65\pi =115\pi$ cm². Hence option (b) is the answer.

Squaring surds Square each surd separately and then add. $a^2={\left(\sqrt{7}+\sqrt{3}\right)}^2=7+3+2\sqrt{21}=10+2\sqrt{21}$ $b^2={\left(\sqrt{7}-\sqrt{3}\right)}^2=7+3-2\sqrt{21}=10-2\sqrt{21}$ On adding, the surd terms cancel: $a^2+b^2=10+10=20$. Hence option (d) is the answer.

Slope of a line The slope of the line joining $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $m=\frac{y_2-y_1}{x_2-x_1}$. $m=\frac{6-3}{4-2}=\frac{3}{2}$ Hence option (d) is the answer.