13 November 2025 · Shift 3

Profit and loss A loss of $20\%$ means the selling price is $80\%$ of the cost price. Cost price $=\frac{2000\times 100}{80}=2500$, i.e. ₹2,500. For a gain of $10\%$ the new selling price $=2500\times \frac{110}{100}$. =2750, so the bicycle must be sold for ₹2,750 and option (d) is correct.

Partnership In a partnership the capital is counted month by month, so multiply every amount by the months it stayed in the business. P $=10000\times 12=120000$. Q $=12000\times 6+8000\times 6=72000+48000=120000$. R $=16000\times 6+24000\times 6=96000+144000=240000$. Ratio of the shares =120000:120000:240000=1:1:2. R's share $=84000\times \frac{2}{4}=42000$, so R received ₹42,000 and option (c) is correct.

Trigonometric identity Write each part with sine and cosine only: $cosecx=\frac{1}{\sin x}$ and $\sec x=\frac{1}{\cos x}$. So the expression $=\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}$. Now $\frac{\cos x}{\sin x}=\cot x$ and $\frac{\sin x}{\cos x}=\tan x$. Hence the expression $=\cot x+\tan x$, so option (d) is correct.

Ratio and proportion Let the number of shoes be 5x and the number of socks be 3x. 5x=50 gives x=10. Socks $=3x=3\times 10=30$, so option (a) is correct.

Compound interest Interest is compounded yearly, so the full first year takes the rate $20\%$. The extra 6 months take simple interest at $20\%$ per annum on the new amount, that is $10\%$. Amount $=6000\times \frac{120}{100}\times \frac{110}{100}=6000\times 1.2\times 1.1$. =7920. Compound interest =7920-6000=1920, so the interest is ₹1,920 and option (a) is correct.

Compound interest Interest of the first year $=250000\times \frac{20}{100}=50000$. He withdraws $40\%$ of it $=50000\times \frac{40}{100}=20000$. Balance left to work in the second year =250000+50000-20000=280000. Interest of the second year $=280000\times \frac{20}{100}=56000$. Balance at the end =280000+56000=336000, i.e. ₹3,36,000, so option (c) is correct.

Ratio and LCM Let the numbers be 5x and 3x; since 5 and 3 have no common factor, x is their HCF. LCM $=5\times 3\times x=15x$. 15x=180 gives x=12. So the numbers are $5\times 12=60$ and $3\times 12=36$. Sum =60+36=96, so option (a) is correct.

Average Total age of the 25 students $=25\times 16=400$ years. After the 5 join there are 30 students of average 18, so their total age $=30\times 18=540$ years. Total age of the 5 new students =540-400=140 years. Their average $=\frac{140}{5}=28$ years, so option (c) is correct.

Percentage Work each option out as a number. $16\%$ of $350=\frac{16}{100}\times 350=56$. $25\%$ of 400=100, $28\%$ of 300=84 and $30\%$ of 340=102. The smallest of 56, 100, 84 and 102 is 56, so option (a) is correct.

Time, speed and distance Let the distance between the home and the shop be d km. Time going $=\frac{d}{7}$ hours and time returning $=\frac{d}{5}$ hours. $\frac{d}{7}+\frac{d}{5}=6$. $\frac{5d+7d}{35}=6\Rightarrow \frac{12d}{35}=6$. $d=\frac{6\times 35}{12}=17.5$, so the distance is 17.5 km and option (c) is correct.

Time and work Take the total work as the LCM of 16, 24 and 32, that is 96 units. Then P+Q+R do $\frac{96}{16}=6$ units a day and P+Q do $\frac{96}{24}=4$ units a day. So one day's work of R alone =6-4=2 units. This fits the third fact too: Q+R do $\frac{96}{32}=3$ units a day, so Q alone does 3-2=1 unit and P alone 4-1=3 units. Time taken by R alone $=\frac{96}{2}=48$ days. Hence option (c) is correct.

Profit and discount Priyanka buys three pants, i.e. more than two items, so the 10% discount applies and ₹2,700 is the discounted selling price. Marked price $=2700\times \frac{100}{90}=3000$, i.e. ₹3,000. The 25% profit is earned on the ₹2,700 actually received, so cost price $=2700\times \frac{100}{125}=2160$, i.e. ₹2,160. At full price the three pants would fetch the marked price, ₹3,000. Profit =3000-2160=840, i.e. ₹840. Hence option (b) is correct.

Simple interest — rate Let the principal be P. Tripling means the amount becomes 3P, so the interest earned is 3P-P=2P. Formula: $SI=\frac{P\times R\times T}{100}$. Substituting $SI=2P$ and T=4: $2P=\frac{P\times R\times 4}{100}$. $R=\frac{2\times 100}{4}=50$ So the rate is 50% per annum; hence option (d) is correct.

Cuboid — total surface area Let the breadth be b=3 cm; then length l=6b=18 cm and height h=4b=12 cm. Formula: total surface area =2(lb+bh+hl). $=2\left(18\times 3+3\times 12+12\times 18\right)$ $=2\left(54+36+216\right)=2\times 306=612$ So the total surface area is $612{cm}^2$; hence option (b) is correct.

Surd — perfect square The last number added is one more than the middle number, so use the identity $n^2+n+\left(n+1\right)=n^2+2n+1={\left(n+1\right)}^2$. Here n=1236, so the expression under the root becomes ${\left(1236+1\right)}^2={\left(1237\right)}^2$. Therefore its value $=\sqrt{{\left(1237\right)}^2}=1237$. Hence option (c) is correct.

Mixture — repeated replacement Formula for repeated replacement: juice left $=A{\left(1-\frac{x}{A}\right)}^n$, where A=60, x=6 and n=2. $=60{\left(1-\frac{6}{60}\right)}^2=60{\left(\frac{9}{10}\right)}^2$ $=60\times \frac{81}{100}=48.6$ Check: after the first operation 60-6=54 litres of juice are left; the second draw of 6 litres of mixture removes $\frac{54}{60}\times 6=5.4$ litres of juice, leaving 54-5.4=48.6 litres. So 48.6 litres of pure juice remain; hence option (c) is correct.

Perimeter and area Perimeter of the rectangle =2(l+b)=2(10+8)=36 cm. The square has the same perimeter, so 4a=36, giving side a=9 cm. Area of the square $=a^2=9^2=81$ So the area is $81{cm}^2$; hence option (c) is correct.

Time and distance The speed is constant, so the time taken is directly proportional to the distance covered. $\frac{3}{7}$ of the journey takes 6 hours, so $\frac{1}{7}$ of it takes $\frac{6}{3}=2$ hours. The remaining journey is $1-\frac{3}{7}=\frac{4}{7}$ of the whole. Time needed for it $=4\times 2=8$ hours. Hence option (b) is correct.

Centroid of a triangle Formula: for vertices $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)$ the centroid is $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$. x-coordinate $=\frac{2+3+4}{3}=\frac{9}{3}=3$ y-coordinate $=\frac{3+6+9}{3}=\frac{18}{3}=6$ So the centroid is (3,6); hence option (d) is correct.

Similar triangles — ratios In similar triangles the ratio of the areas is the square of the ratio of the corresponding sides. So $\frac{s_1}{s_2}=\sqrt{\frac{36}{64}}=\frac{6}{8}$. Every side of the larger triangle is that same multiple of the matching side, so the perimeters are in the ratio of the sides. Hence the perimeters are in the ratio 6:8, and option (a) is correct.

Slope of a line Formula: slope $m=\frac{y_2-y_1}{x_2-x_1}$. Here $\left(x_1,y_1\right)=\left(3,2\right)$ and $\left(x_2,y_2\right)=\left(5,12\right)$. $m=\frac{12-2}{5-3}=\frac{10}{2}=5$ So the slope is 5; hence option (d) is correct.

Quadratic equation Multiply throughout by y(y+1) to clear the denominators: $y\cdot y+\left(y+1\right)=4y\left(y+1\right)$. $y^2+y+1=4y^2+4y$ $3y^2+3y-1=0$ Formula: $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ with a=3, b=3 and c=-1. $y=\frac{-3\pm \sqrt{9+12}}{6}=\frac{-3\pm \sqrt{21}}{6}$ The root offered in the options is $y=\frac{-3+\sqrt{21}}{6}$. Hence option (a) is correct.

Similar triangles — altitude In similar triangles every pair of corresponding lengths — sides, altitudes, medians, angle bisectors — is in the same ratio. So $\frac{15}{h}=\frac{5}{8}$, where h is the required altitude. $h=\frac{15\times 8}{5}=24$ So the corresponding altitude of the larger triangle is 24 cm; hence option (c) is correct.

Laws of indices Write the base as a power of 5: $25=5^2$. ${25}^{\frac{3}{2}}={\left(5^2\right)}^{\frac{3}{2}}=5^3=125$ The second term carries the negative index, so ${25}^{-\frac{3}{2}}=\frac{1}{{25}^{\frac{3}{2}}}=\frac{1}{125}$. Adding: $125+\frac{1}{125}=\frac{125\times 125+1}{125}=\frac{15625+1}{125}=\frac{15626}{125}$ Hence option (b) is correct.

Trigonometric identity Add the two fractions: $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\left(1+\cos \theta \right)}^2}{\sin \theta \left(1+\cos \theta \right)}$. Numerator $={\sin }^2\theta +{\cos }^2\theta +2\cos \theta +1=1+2\cos \theta +1=2\left(1+\cos \theta \right)$. So the left side becomes $\frac{2\left(1+\cos \theta \right)}{\sin \theta \left(1+\cos \theta \right)}=\frac{2}{\sin \theta }$. Hence $\frac{2}{\sin \theta }=\frac{4}{\sqrt{3}}$, so $\sin \theta =\frac{\sqrt{3}}{2}$ and $\theta ={60}^{\circ }$. $\tan {60}^{\circ }+\sec {60}^{\circ }=\sqrt{3}+2$ ${\left(\tan \theta +\sec \theta \right)}^{-1}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=2-\sqrt{3}$ Hence option (a) is correct.