22 November 2025 · Shift 1

Average of numbers Average $=\frac{\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{observations}}{\mathrm{number}\mathrm{of}\mathrm{observations}}$. Sum = 17+27+37+47+57 = 185. Average $=\frac{185}{5}=37$. The five numbers form an AP with common difference 10, and for an odd number of terms in AP the average is simply the middle term, which is 37 — a useful shortcut. Hence, option (d) is the most appropriate answer.

Time and work Take the whole job as 1 unit. One day's work of A $=\frac{1}{30}$ and one day's work of B $=\frac{1}{40}$. Working together, one day's work $=\frac{1}{30}+\frac{1}{40}=\frac{4+3}{120}=\frac{7}{120}$. In 10 days they finish $10\times \frac{7}{120}=\frac{70}{120}=\frac{7}{12}$ of the job. Work left $=1-\frac{7}{12}=\frac{5}{12}$. Hence, option (b) is the most appropriate answer.

Decimal expansion of rationals A rational number is any number that can be written as $\frac{p}{q}$ with p and q integers and $q\ne 0$. If the denominator in lowest terms has only 2 and 5 as prime factors, the division comes to an end, e.g. $\frac{3}{8}=0.375$ — a terminating decimal. For any other denominator the remainders must start repeating (only finitely many remainders are possible), e.g. $\frac{1}{3}=0.333\dots$ — a recurring decimal. So the decimal expansion of a rational number is either terminating or recurring; a non-terminating, non-recurring decimal such as $\sqrt{2}=1.41421356\dots$ belongs to an irrational number. Options (c) and (d) name only one of the two possibilities each, so they are incomplete. Hence, option (a) is the most appropriate answer.

Profit, loss and marked price Let the marked price be M and the cost price be C. Selling at 50% of the marked price gives a loss of 20%, so $\frac{50}{100}M=\frac{80}{100}C$. Therefore $M=\frac{80}{50}C=1.6C$. If it is sold at 90% of the marked price, $SP=\frac{90}{100}\times 1.6C=1.44C$. Profit $\%=\frac{1.44C-C}{C}\times 100=44\%$. Hence, option (a) is the most appropriate answer.

Discount and profit Cost price is ₹11,000 and marked price is ₹14,000. A gain of 20% fixes the selling price: $SP=11000\times \frac{120}{100}=13200$, that is ₹13,200. Discount = MP - SP = 14000 - 13200 = 800, that is ₹800. Discount $\%=\frac{800}{14000}\times 100=\frac{40}{7}\%$. $\frac{40}{7}=5\frac{5}{7}$, so the rate of discount is $5\frac{5}{7}\%$. Hence, option (a) is the most appropriate answer.

Discount on marked price Sale price = marked price $\times \left(1-\frac{\mathrm{discount}\%}{100}\right)$. Here the discount is $62\frac{1}{2}\%=\frac{125}{2}\%$, so the customer pays $100-62\frac{1}{2}=37\frac{1}{2}\%=\frac{75}{2}\%$ of the marked price. Sale price $=8000\times \frac{75}{200}=8000\times \frac{3}{8}=3000$. So the shoe is sold for ₹3,000. Hence, option (b) is the most appropriate answer.

Average and correction of data Total marks of the 10 students $=10\times 73.5=735$. One score was read as 45 instead of 54, so the total is short by 54-45 = 9. Correct total of the 10 students = 735 + 9 = 744. After removing one student, the 9 students left average 76.2, so their total $=9\times 76.2=685.8$. Marks of the removed student = 744 - 685.8 = 58.2. Hence, option (c) is the most appropriate answer.

Volume of a triangular prism First check the base triangle: ${60}^2+{91}^2=3600+8281=11881={109}^2$. So 60, 91, 109 is a Pythagorean triplet and the base is right-angled, with the two legs 60 cm and 91 cm. Area of the base $=\frac{1}{2}\times 60\times 91=2730{\mathrm{cm}}^2$. Volume of the prism $=\mathrm{area}\mathrm{of}\mathrm{base}\times \mathrm{height}=2730\times 10=27300{\mathrm{cm}}^3$. The rate is 5 paise per ${\mathrm{cm}}^3$, that is ₹0.05 per ${\mathrm{cm}}^3$. Total cost $=27300\times 0.05=1365$, that is ₹1365. Hence, the total cost of filling the box is ₹1365.

Volume of a trapezoidal prism The base of each tank is a trapezium with parallel sides 20 cm and 16 cm and the distance between them 12 cm. Area of the trapezium $=\frac{1}{2}\times \left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{distance}\mathrm{between}\mathrm{them}$. $=\frac{1}{2}\times \left(20+16\right)\times 12=\frac{1}{2}\times 36\times 12=216{\mathrm{cm}}^2$. Volume of one tank $=216\times 60=12960{\mathrm{cm}}^3$. Volume of 20 tanks $=20\times 12960=259200{\mathrm{cm}}^3=2,59,200{\mathrm{cm}}^3$. Hence, option (d) is the most appropriate answer.

Area of an annulus The jogging path is a ring: the track of radius 18.5 m is the inner circle and the 4.5 m band lies outside it. Inner radius $r=18.5m$ and outer radius $R=18.5+4.5=23m$. Area of the ring $=\pi R^2-\pi r^2=\pi \left(R^2-r^2\right)$. $R^2-r^2={23}^2-{18.5}^2=529-342.25=186.75$. Area $=3.1416\times 186.75=586.6938m^2\approx 586.69m^2$. Hence, option (a) is the most appropriate answer.

Remainder theorem Remainder theorem: when a polynomial f(x) is divided by a linear divisor, the remainder is the value of f(x) at the zero of that divisor. Here $3x-6=0\Rightarrow x=2$, so the required remainder is f(2). $f\left(2\right)=4{\left(2\right)}^3+3{\left(2\right)}^2-12\left(2\right)+1$ =32+12-24+1=21 Hence, option (a) is the correct answer.

Centres of a triangle An altitude of a triangle is the perpendicular drawn from a vertex to the opposite side. The three altitudes of a triangle are concurrent, and their common point is called the orthocentre. The other centres come from other cevians: medians meet at the centroid, perpendicular bisectors of the sides at the circumcentre, and angle bisectors at the incentre. Hence, option (c) is the correct answer.

Trigonometric identity Add the two fractions by taking the LCM of the denominators. $\frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}=\frac{\sin x\left(\left(1+\cos x\right)+\left(1-\cos x\right)\right)}{\left(1-\cos x\right)\left(1+\cos x\right)}$ $=\frac{2\sin x}{1-{\cos }^{2}x}=\frac{2\sin x}{{\sin }^{2}x}=\frac{2}{\sin x}$ So $\frac{2}{\sin x}=4\Rightarrow \sin x=\frac{1}{2}\Rightarrow x={30}^{\circ }$, which lies in $0<x<{90}^{\circ }$. $\sin x+\cos x=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2}$ Hence, option (a) is the correct answer.

Circle and square mensuration Area of the circle $=\pi r^2=9856$ $r^2=9856\times \frac{7}{22}=3136\Rightarrow r=56$ cm The circle is inscribed in the square, so the side of the square equals the diameter: a=2r=112 cm. Side of the equilateral triangle $=\frac{112}{2}=56$ cm Sum of the perimeters $=4\times 112+3\times 56=448+168=616$ cm Hence, option (a) is the correct answer.

Pythagoras theorem In $△ABC$ the right angle is at B, so AC is the hypotenuse. $BC=\sqrt{A{C}^2-A{B}^2}=\sqrt{8^2-{\left(4\sqrt{3}\right)}^2}=\sqrt{64-48}=\sqrt{16}=4$ cm R is the mid-point of BC, so $BR=\frac{4}{2}=2$ cm. $△ABR$ is also right-angled at B, so $AR=\sqrt{A{B}^2+B{R}^2}$. $AR=\sqrt{{\left(4\sqrt{3}\right)}^2+2^2}=\sqrt{48+4}=\sqrt{52}=2\sqrt{13}$ cm Hence, option (c) is the correct answer.

Sphere: area and volume For a sphere, surface area $=4\pi r^2\propto r^2$ and volume $=\frac{4}{3}\pi r^3\propto r^3$. Let the surface area go from 100 to 100+69=169 units. $\frac{r22}{r12}=\frac{169}{100}\Rightarrow \frac{r_2}{r_1}=\frac{13}{10}$ $\frac{V_2}{V_1}={\left(\frac{13}{10}\right)}^3=\frac{2197}{1000}=2.197$ Increase in volume $=\left(2.197-1\right)\times 100\%=119.7\%$ Hence, option (a) is the correct answer.

Partnership Both partners invested for the same period (one year), so the profit is divided in the ratio of the capitals. A:B=70000:80000=7:8 Total of the ratio parts =7+8=15 B's share $=\frac{8}{15}\times 300000=160000$ Hence, option (a) is the correct answer. So B receives ₹1,60,000 out of the total profit of ₹3,00,000.

Trigonometric simplification Write every ratio in terms of $\sin x$ and $\cos x$: $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$. $\frac{\sin x\cdot \sec x}{\tan x}=\frac{\sin x\times \frac{1}{\cos x}}{\frac{\sin x}{\cos x}}$ $=\frac{\sin x}{\cos x}\times \frac{\cos x}{\sin x}=1$ Hence, option (b) is the correct answer.

Standard trigonometric values Recall the standard values: $\sin {30}^{\circ }=\frac{1}{2}$ and $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$. $\sin {30}^{\circ }\times \cos {30}^{\circ }=\frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$ Hence, option (b) is the correct answer.

HCF application Every row holds trees of one variety only and all rows hold the same number, so that number must divide 26, 39 and 52 exactly. The largest such number is the HCF of the three counts. $26=2\times 13$, $39=3\times 13$, $52=2^2\times 13$, so the only common factor is 13. $\mathrm{HCF}\left(26,39,52\right)=13$ Hence, option (b) is the correct answer.

Average with new members Let the class have n students whose average weight is A kg, so their total weight is nA kg. First condition: $\frac{nA+60}{n+1}=A+2$ $nA+60=nA+A+2n+2\Rightarrow A+2n=58$ So every option fixes a class size: $n=\frac{58-A}{2}$. Only A=42 kg, which gives n=8, raises the average by about 3.5 kg once both 60 kg students have joined: $\frac{8\times 42+120}{10}=45.6$ kg. Hence, option (a) is the correct answer.

Time and work Let the total work be the LCM of the three times: $\mathrm{LCM}\left(12,16,24\right)=48$ units. One day's work of A, B and C together =4+3+2=9 units. Work finished in 2 days $=9\times 2=18$ units, so the work left =48-18=30 units. After A leaves, B and C together do 3+2=5 units per day. Time needed $=\frac{30}{5}=6$ days Hence, option (a) is the correct answer.

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$, where P=50000, T=5 and SI=25000. $25000=\frac{50000\times R\times 5}{100}$ 25000=2500R $R=\frac{25000}{2500}=10$ So the rate of interest is 10% per annum. Hence, option (d) is the correct answer.

Trigonometric ratios $\cot \theta =1\Rightarrow \tan \theta =1\Rightarrow \theta ={45}^{\circ }$ In such a triangle the base and the perpendicular are equal, so the hypotenuse $=\sqrt{1^2+1^2}=\sqrt{2}$ and $\sin \theta =\frac{1}{\sqrt{2}}$. ${\sin }^2\theta ={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}$ Hence, option (c) is the correct answer.

Trigonometric ratios $\cot A=\frac{\mathrm{base}}{\mathrm{perpendicular}}=\sqrt{3}$, so take base $=\sqrt{3}$ and perpendicular =1. Hypotenuse $=\sqrt{{\left(\sqrt{3}\right)}^2+1^2}=\sqrt{4}=2$ $cosecA=\frac{\mathrm{hypotenuse}}{\mathrm{perpendicular}}=\frac{2}{1}=2$ $\sec A=\frac{\mathrm{hypotenuse}}{\mathrm{base}}=\frac{2}{\sqrt{3}}$ These are exactly the values of $A={30}^{\circ }$, since $\cot {30}^{\circ }=\sqrt{3}$. Hence, option (d) is the correct answer.