14 November 2025 · Shift 3

Profit and loss Selling price of 15 pens is ₹750, so the selling price of one pen $=\frac{750}{15}=50$, that is ₹50. Loss = selling price of 3 pens $=3\times 50=150$, that is ₹150. Cost price = selling price + loss, so the cost price of 15 pens =750+150=900, that is ₹900. Cost price of one pen $=\frac{900}{15}=60$, that is ₹60. Hence option (d) is correct.

Partnership Profit is divided in the ratio of (capital $\times$ time), and the whole period is 2 years, i.e. 24 months. A keeps ₹1,50,000 for 8 months and ₹1,80,000 for the remaining 16 months: $150000\times 8+180000\times 16=1200000+2880000=4080000$. B keeps ₹1,00,000 for the whole time: $100000\times 24=2400000$. C keeps ₹1,20,000 for 8 months and ₹1,00,000 for the remaining 16 months: $120000\times 8+100000\times 16=960000+1600000=2560000$. Ratio of shares =4080000:2400000:2560000=408:240:256=51:30:32, and 51+30+32=113. A's profit $=169500\times \frac{51}{113}=1500\times 51=76500$, that is ₹76,500. Hence option (a) is correct.

Trigonometric ratios $\cot \theta =\frac{12}{5}$ means that, in size, base : perpendicular =12:5. Hypotenuse $=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$, the familiar 5,12,13 triplet. So the magnitude of $\sin \theta =\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}=\frac{5}{13}$. In the third quadrant only $\tan \theta$ and $\cot \theta$ are positive, so $\sin \theta$ is negative there. Therefore $\sin \theta =-\frac{5}{13}$, and option (b) is correct.

Compound ratio The link between the two ratios is n, which stands as 8 in the first ratio and as 3 in the second. Make the two values of n equal by taking their LCM: $\mathrm{LCM}\left(8,3\right)=24$. Multiply the first ratio by 3: $m:n=7\times 3:8\times 3=21:24$. Multiply the second ratio by 8: $n:p=3\times 8:2\times 8=24:16$. Now n is 24 in both, so m:n:p=21:24:16, and option (a) is correct.

CI and SI difference Work with a sum of 100 and find both interests for 3 years at $10\%$ per annum. Simple interest $=\frac{100\times 10\times 3}{100}=30$, i.e. $30\%$ of the sum. Amount at compound interest $=100{\left(1+\frac{10}{100}\right)}^3=100\times \frac{1331}{1000}=133.1$, so the compound interest is 33.1, i.e. $33.1\%$ of the sum. Difference =33.1-30=3.1, so the difference is $3.1\%$ of the sum. Given that this difference is ₹31: $3.1\%$ of the sum =31. Sum $=\frac{31\times 100}{3.1}=1000$, that is ₹1,000. Hence option (b) is correct.

Compound interest The interest of the third year is earned on the amount standing at the end of the second year. Interest of the third year =5324-4840=484, that is ₹484. Rate $=\frac{484}{4840}\times 100=10\%$ per annum. The amount after 2 years is $P{\left(1+\frac{10}{100}\right)}^2=P\times \frac{121}{100}$. So $P\times \frac{121}{100}=4840$, giving $P=\frac{4840\times 100}{121}=4000$, that is ₹4,000. Hence option (b) is correct.

Average Sum of all 13 numbers $=13\times 51=663$. Sum of the first 7 numbers $=7\times 50=350$. Sum of the last 7 numbers $=7\times 51=357$. The two groups together contain 7+7=14 numbers, so every number is counted once except the 7th, which lies in both groups and is counted twice. Hence middle number =(350+357)-663=707-663=44. Hence option (a) is correct.

LCM and remainders A number that leaves the same remainder 6 on division by 8, 14 and 26 must be of the form (a multiple of their LCM) +6. $8=2^3$, $14=2\times 7$, $26=2\times 13$, so $\mathrm{LCM}=2^3\times 7\times 13=728$. Largest four-digit multiple of 728: $728\times 13=9464$, while $728\times 14=10192$ has five digits. So the required number =9464+6=9470. 9470 is still a four-digit number, so it is the largest one possible; note that 9464 itself leaves remainder 0, not 6. Hence option (d) is correct.

Percentage and price Let the original price of one notebook be x rupees; after a fall of $20\%$ the new price is 0.8x rupees. Number bought for ₹300 at the old price $=\frac{300}{x}$. Number bought for ₹300 at the new price $=\frac{300}{0.8x}=\frac{375}{x}$. The new number exceeds the old by 4: $\frac{375}{x}-\frac{300}{x}=4$. $\frac{75}{x}=4$, so $x=\frac{75}{4}=18.75$. New price $=0.8\times 18.75=15$, that is ₹15. Hence option (c) is correct.

Speed, time and distance Let the distance between the home and the park be d km. Time taken going $=\frac{d}{4}$ hours and time taken returning $=\frac{d}{2}$ hours. Total time is 3 hours: $\frac{d}{4}+\frac{d}{2}=3$. $\frac{d+2d}{4}=3$, so $\frac{3d}{4}=3$. $d=\frac{3\times 4}{3}=4$, so the distance is 4 km. Hence option (a) is correct.

Time and work Take the total work as the LCM of 20 and 25, i.e. 100 units. One day's work of A $=\frac{100}{20}=5$ units and one day's work of B $=\frac{100}{25}=4$ units. Working together for 8 days they finish $8\times \left(5+4\right)=8\times 9=72$ units. Work still left =100-72=28 units. B alone needs $\frac{28}{4}=7$ more days, so option (b) is correct.

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$ Here P=10000, $R=6\%$ per annum and T=5 years. $SI=\frac{10000\times 6\times 5}{100}=\frac{300000}{100}$ SI=3000 So the total interest paid is ₹3,000, i.e. option (c).

Simple interest, equal time Second sum: SI=18000-15000=3000. $T=\frac{SI\times 100}{P\times R}=\frac{3000\times 100}{15000\times 10}=2$ years. The first sum also runs for 2 years, at $5\%$, so its interest $=\frac{x\times 5\times 2}{100}=\frac{x}{10}$. Amount = principal + interest, and the amount is ₹12,100: $x+\frac{x}{10}=12100\Rightarrow \frac{11x}{10}=12100$ $x=\frac{12100\times 10}{11}=11000$, so the sum is ₹11,000 — option (a).

Rationalising surds Rationalise the denominator by multiplying by the conjugate $\sqrt{169}+\sqrt{168}$: $x=\frac{1}{\sqrt{169}-\sqrt{168}}\times \frac{\sqrt{169}+\sqrt{168}}{\sqrt{169}+\sqrt{168}}=\frac{\sqrt{169}+\sqrt{168}}{169-168}$ $x=\sqrt{169}+\sqrt{168}$ $\sqrt{169}=13$, and since ${12.9}^2=166.41$ and ${13}^2=169$, $\sqrt{168}$ lies between 12.9 and 13 (it is about 12.96). So $x\approx 13+12.96=25.96$, which lies between 25 and 26. Hence the integer part of x is 25 — option (a).

Cone inside a sphere Cut the solid by the plane through the axis of the cone; the sphere gives a circle of diameter D and the cone gives a triangle inside it. In this section the base of the cone is the segment of length 2r through the centre O, and the height h is drawn perpendicular to it up to the vertex on the sphere. So the height h and the base diameter 2r are the two perpendicular sides, and the diameter D of the sphere closes the right triangle. By Pythagoras theorem: $h^2+{\left(2r\right)}^2=D^2$ $h^2=D^2-4r^2$ $h=\sqrt{D^2-4r^2}$, so option (a) is correct.

Mixture and alligation Water in the 50 litres $=10\%$ of $50=\frac{10}{100}\times 50=5$ litres, so milk =50-5=45 litres. Let x litres of pure water be added; the milk stays 45 litres. New water =(5+x) litres and new mixture =(50+x) litres. $\frac{5+x}{50+x}=\frac{25}{100}=\frac{1}{4}$ $4\left(5+x\right)=50+x\Rightarrow 20+4x=50+x$ $3x=30\Rightarrow x=10$ So 10 litres of pure water must be added — option (a).

Volume of hemispherical shell Inner radius r=6 cm and thickness =1 cm, so outer radius R=6+1=7 cm. Volume of iron = outer hemisphere - inner hemisphere $=\frac{2}{3}\pi R^3-\frac{2}{3}\pi r^3$ $=\frac{2}{3}\pi \left(R^3-r^3\right)=\frac{2}{3}\pi \left(7^3-6^3\right)$ $=\frac{2}{3}\pi \left(343-216\right)=\frac{2}{3}\pi \times 127$ $=\frac{254\pi }{3}$, so the iron used is $\frac{254\pi }{3}$ cm${}^3$ — option (d).

Recurring decimal to fraction Let $x=0.545454\dots$; the repeating block is 54, which has 2 digits. Multiply by 100: $100x=54.545454\dots$ Subtract: $100x-x=54.545454\dots -0.545454\dots =54$ $99x=54\Rightarrow x=\frac{54}{99}$ Divide numerator and denominator by 9: $x=\frac{6}{11}$, so option (b) is correct.

Laws of exponents Law of exponents: ${\left(a^m\right)}^n=a^{m\times n}$. ${\left(7^3\right)}^2=7^{3\times 2}=7^6$ So $7^x=7^6$; when the bases are equal the powers must be equal. x=6, so option (c) is correct.

Orthocentre of a triangle A(0,0) and B(9,0) both lie on the x-axis, so AB is along the x-axis. A(0,0) and C(0,12) both lie on the y-axis, so AC is along the y-axis. The axes are perpendicular, so $AB\perp AC$ and the triangle is right angled at A. In a right-angled triangle the two legs are themselves altitudes, and they meet at the vertex holding the right angle, so that vertex is the orthocentre. Hence the orthocentre is A(0,0) — option (a) is correct.

Similar triangles, area ratio In two similar triangles the ratio of the areas equals the square of the ratio of any pair of corresponding lengths — sides, altitudes, medians or perimeters. $\frac{A_1}{A_2}={\left(\frac{h_1}{h_2}\right)}^2=\frac{144}{169}$ $\frac{h_1}{h_2}=\sqrt{\frac{144}{169}}=\frac{12}{13}$ So the corresponding altitudes are in the ratio 12:13 — option (b) is correct.

Equation of a parallel line Parallel lines have equal slopes. Comparing y=5x-7 with y=mx+c gives slope m=5. Point-slope form: $y-y_1=m\left(x-x_1\right)$ with $\left(x_1,y_1\right)=\left(3,5\right)$. y-5=5(x-3) y-5=5x-15 y=5x-10, so option (c) is correct.

Similar figures, scale Park : model =100:1, so every length in the model is $\frac{1}{100}$ of the corresponding length in the park. Length of the path in the model $=\frac{25}{100}=0.25$ m 0.25 m $=0.25\times 100=25$ cm, so option (b) is correct.

Identity a³+b³+c³ Identity: if a+b+c=0, then $a^3+b^3+c^3=3abc$. Numerator: a=144-121=23, b=121-81=40, c=81-144=-63 and 23+40-63=0. So the numerator $=3\times 23\times 40\times \left(-63\right)$. Denominator: a=12-11=1, b=11-9=2, c=9-12=-3 and 1+2-3=0. So the denominator $=3\times 1\times 2\times \left(-3\right)$. Required value $=\frac{3\times 23\times 40\times \left(-63\right)}{3\times 1\times 2\times \left(-3\right)}=\frac{23\times 40\times 63}{6}$ $=23\times 420=9660$, so option (a) is correct.

Trigonometric identity Write both terms in terms of sine and cosine: $\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$ $=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }$ (since ${\sin }^2\theta +{\cos }^2\theta =1$) Given $\tan \theta +\cot \theta =3$, so $\frac{1}{\sin \theta \cos \theta }=3$ $\sin \theta \cos \theta =\frac{1}{3}$, so option (c) is correct.