30 November 2025 · Shift 2

Square root of a decimal Write the decimal as a fraction: $0.25=\frac{25}{100}$. $\sqrt{0.25}=\sqrt{\frac{25}{100}}=\frac{\sqrt{25}}{\sqrt{100}}=\frac{5}{10}=0.5$. A quick check confirms it: $0.5\times 0.5=0.25$. Note that a decimal with 2 places has a square root with 1 place, which rules out 0.05 and 0.005. Hence option (c) is correct.

Powers and multiplication Evaluate the power first: $3^4=3\times 3\times 3\times 3=81$. Now $111\times 81$. $111\times 81=111\times 80+111\times 1=8880+111=8991$. Hence option (b) is correct.

Loss percentage A loss of 20% means the article is sold for $100-20=80\%$ of its cost price. Formula: S.P. = C.P. $\times \frac{100-loss\%}{100}$. S.P. $=500\times \frac{80}{100}=400$. Check: loss =500-400=100, and $\frac{100}{500}\times 100=20\%$. Hence option (a) is correct, ₹400.

Profit percentage Let the cost price of one item be x and its selling price be y. Given: 20y=25x, so $\frac{y}{x}=\frac{25}{20}=\frac{5}{4}$. Take C.P. =4 and S.P. =5 per item; since S.P. exceeds C.P., it is a profit. Profit % $=\frac{S.P.-C.P.}{C.P.}\times 100=\frac{5-4}{4}\times 100=25\%$. Hence option (b) is correct.

Profit percentage A profit of 20% means the selling price is $100+20=120\%$ of the cost price. Formula: S.P. = C.P. $\times \frac{100+profit\%}{100}$. S.P. $=800\times \frac{120}{100}=960$. Check: profit =960-800=160, and $\frac{160}{800}\times 100=20\%$. Hence option (a) is correct, ₹960.

Discount and marked price Discount is always reckoned on the marked price, so after a 12% discount the price paid is $100-12=88\%$ of it. So M.P. $\times \frac{88}{100}=880$. M.P. $=880\times \frac{100}{88}=1000$. Check: 12% of 1000 is 120, and 1000-120=880. Hence option (a) is correct, ₹1,000.

Discount A discount of 25% leaves $100-25=75\%$ of the marked price to be paid. S.P. $=3200\times \frac{75}{100}=2400$. Check: discount $=\frac{25}{100}\times 3200=800$, and 3200-800=2400. Hence option (d) is correct, ₹2,400.

Speed, distance and time Formula: speed $=\frac{distance}{time}$. Speed $=\frac{90}{1.8}$ km/h. Clear the decimal by multiplying numerator and denominator by 10: $\frac{900}{18}=50$. Check: $50\times 1.8=90$ km, exactly the distance given. Hence option (c) is correct, 50 km/h.

Simple interest rate Formula: $S.I.=\frac{P\times R\times T}{100}$, so $R=\frac{S.I.\times 100}{P\times T}$. Here P=5960, $S.I.=876.60$ and T=3 years. $R=\frac{876.60\times 100}{5960\times 3}=\frac{87660}{17880}$. $\frac{87660}{17880}=4.902\dots$, i.e. about $4.90\%$. Hence option (a) is correct.

Simple interest rate Formula: $R=\frac{S.I.\times 100}{P\times T}$. Here P=7500, $S.I.=2604$ and T=4 years. $R=\frac{2604\times 100}{7500\times 4}=\frac{260400}{30000}$. $\frac{260400}{30000}=8.68$, so the rate is $8.68\%$. Check: $\frac{7500\times 8.68\times 4}{100}=2604$, the interest given. Hence option (d) is correct.

Train crossing a platform To cross a platform completely, the train must cover its own length plus the length of the platform. Distance =180+220=400 m Speed $=\frac{400}{80}=5$ m/s Converting to km/h: $5\times \frac{18}{5}=18$ km/h Hence option (d) is correct.

Time and work The wall is the same, so the total work (workers $\times$ days) stays constant. Work $=16\times 20=320$ worker-days With 8 workers: days $=\frac{320}{8}=40$ Shortcut: the number of workers is halved, so the time doubles — $20\times 2=40$ days. Hence option (a) is correct.

Percentage change in volume For a cone, $V=\frac{1}{3}\pi r^{2}h$, so the volume depends on $r^{2}h$ — three linear dimensions multiplied together. Heating raises every dimension by 10%, so both r and h are multiplied by 1.1. New volume $=\frac{1}{3}\pi {\left(1.1r\right)}^2\left(1.1h\right)={\left(1.1\right)}^3\times \frac{1}{3}\pi r^{2}h$ ${\left(1.1\right)}^3=1.331$, so the new volume is 1.331 times the old one. Increase $=\left(1.331-1\right)\times 100=33.1\%$ The figures 25 cm and 34 cm are never needed — a uniform scaling changes the volume by the same percentage whatever the size. Hence option (c) is correct.

Curved surface area of cone The lateral surface of a cone is its curved surface, whose area is $\pi rl$, where l is the slant height. Slant height: $l=\sqrt{r^2+h^2}=\sqrt{5^2+{12}^2}=\sqrt{25+144}=\sqrt{169}=13$ m Area affected $=\pi rl=\pi \times 5\times 13=65\pi$ square metres The base is on the ground and is not exposed, so $\pi r^2$ is not added. Hence option (b) is correct.

Averages of groups Turn each average into a total by multiplying by 2. Science + Mathematics $=1320\times 2=2640$ Mathematics + Literature $=1480\times 2=2960$ Adding the two: Science $+2\times$ Mathematics + Literature =2640+2960=5600 But Science + Mathematics + Literature =4050, so subtracting leaves one extra Mathematics. Mathematics =5600-4050=1550 Hence option (b) is correct.

Surface area of prism A right prism on a rectangular base is simply a cuboid of dimensions $5.5\times 3.5\times 12$ cm. Formula: TSA $=2\times \left(\mathrm{base}\mathrm{area}\right)+\left(\mathrm{base}\mathrm{perimeter}\right)\times \left(\mathrm{height}\right)$ Base area $=5.5\times 3.5=19.25$ sq cm Base perimeter $=2\left(5.5+3.5\right)=2\times 9=18$ cm TSA $=2\times 19.25+18\times 12=38.5+216=254.5$ Hence option (c), 254.5 sq cm, is correct.

Trigonometric ratios of 45° In a ${45}^{\circ }$–${45}^{\circ }$–${90}^{\circ }$ triangle the two legs are equal, so the ratio of perpendicular to base is 1. Therefore $\tan {45}^{\circ }=1$, and $\cot {45}^{\circ }=\frac{1}{\tan {45}^{\circ }}=1$. $\tan A+\cot B=\tan {45}^{\circ }+\cot {45}^{\circ }=1+1=2$ Hence option (d) is correct.

Trigonometric ratios $\sin x=\frac{5}{13}$ means perpendicular =5 and hypotenuse =13. Base $=\sqrt{{13}^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ (the 5–12–13 triplet). So $\tan x=\frac{5}{12}$ and $cosecx=\frac{13}{5}$. $2\tan x+cosecx=2\times \frac{5}{12}+\frac{13}{5}=\frac{5}{6}+\frac{13}{5}$ $=\frac{25+78}{30}=\frac{103}{30}$ Hence option (a) is correct.

Simple interest Let the principal be P. In both cases the time is the same, 6 years. At $r\%$: $P+\frac{P\times r\times 6}{100}=12860$ At $\left(r+5\right)\%$: $P+\frac{P\times \left(r+5\right)\times 6}{100}=14420$ Subtracting, only the extra 5% for 6 years survives: $\frac{P\times 5\times 6}{100}=14420-12860=1560$ $\frac{30P}{100}=1560$ $P=\frac{1560\times 100}{30}=5200$ Hence option (c), Rs. 5,200, is correct.

Mixture and ratio Milk : Water =2:3, so the 45 litres are made of 2+3=5 parts. 1 part $=\frac{45}{5}=9$ litres, so milk $=2\times 9=18$ L and water $=3\times 9=27$ L. Only water is poured in, so the milk stays fixed at 18 litres. For the new ratio 1:4, water must be $4\times 18=72$ litres. Water to be added =72-27=45 litres. Hence option (c) is correct.

Time and work efficiency Equal times mean equal efficiencies, so Amita's one-day work = Bina's + Sita's. Take the total work as 175 units (the LCM of 25 and 35). Sita alone in 35 days: Sita $=\frac{175}{35}=5$ units/day. Amita and Bina in 25 days: Amita + Bina $=\frac{175}{25}=7$ units/day. From the first line, Amita - Bina = Sita =5 units/day. Adding the last two: $2\times$ Amita =7+5=12, so Amita =6 and Bina =7-6=1 unit/day. Time for Bina alone $=\frac{175}{1}=175$ days. Hence option (b) is correct.

Average after removal Sum of the 8 numbers $=8\times 25=200$. Statement I: if the remaining 7 average 24, their sum $=7\times 24=168$. Removed number =200-168=32, exactly as claimed, so I is correct. Statement II: if 18 is removed, the remaining sum =200-18=182. Average $=\frac{182}{7}=26$, exactly as claimed, so II is correct too. Since both statements hold, neither of them is incorrect. Hence option (d) is correct.

Fundamental trigonometric identity Identity: ${\sin }^2\theta +{\cos }^2\theta =1$ ${\sin }^2\theta =1-{\left(0.6\right)}^2=1-0.36=0.64$ $\sin \theta =\sqrt{0.64}=0.8$ (positive, since $\theta$ is acute) The sides are in the ratio 0.6:0.8:1, i.e. the familiar 3:4:5 triangle. Hence option (a) is correct.

Trigonometric simplification The expression has the form $\left(a-b\right)\left(a+b\right)=a^2-b^2$ with a=1 and $b=\cos x$. $\left(1-\cos x\right)\left(1+\cos x\right)=1^2-{\cos }^{2}x=1-{\cos }^{2}x$ From ${\sin }^{2}x+{\cos }^{2}x=1$ we get $1-{\cos }^{2}x={\sin }^{2}x$. So the expression simplifies to ${\sin }^{2}x$; note that $2\sin x$ and $1-\sin x$ come from wrongly expanding as a sum. Hence option (a) is correct.

Equivalent discount Under the scheme the customer takes home 5 articles but pays for only 4. Let the marked price of one article be ₹100, so 5 articles are marked at ₹500. Amount actually paid $=4\times 100=$ ₹400 Discount =500-400= ₹100 on a marked price of ₹500. Discount % $=\frac{100}{500}\times 100=20\%$ The tempting 25% comes from taking the free article as a fraction of the 4 paid for, but discount is always measured on the marked price. Hence option (b) is correct.