18 November 2025 · Shift 2

Discount and marked price Let the marked price be M. Discount is always reckoned on the marked price, and the cost price does not change, so whatever the selling price gains the profit gains too. Extra profit $=15\%\mathrm{of}M-7\%\mathrm{of}M=8\%\mathrm{of}M$ $\frac{8M}{100}=768$ $M=\frac{768\times 100}{8}=9600$ So the marked price is ₹9,600 and option (a) is the answer; the cost price ₹3,000 is extra data not needed here.

Discount on a ratio Let the marked prices be 3x (pen) and 5x (book), so the two together are marked at 8x. Total discount $=12\%\mathrm{of}8x=0.96x$ Discount on the pen $=17\%\mathrm{of}3x=0.51x$ Discount on the book =0.96x-0.51x=0.45x Rate of discount on the book $=\frac{0.45x}{5x}\times 100=9\%$ Hence option (c) is the answer.

Weighted average The sections have different strengths, so a weighted average is needed: total marks divided by total students. Total marks $=72\times 20+68\times 15+75\times 25$ =1440+1020+1875=4335 Total students =20+15+25=60 Overall average $=\frac{4335}{60}=72.25$, so option (a) is the answer.

Perfect cubes A four-digit number lies between 1000 and 9999, so find the biggest whole number whose cube stays inside that range. ${21}^3=9261$, which is a four-digit number, while ${22}^3=10648$ already has five digits. So the largest four-digit perfect cube is 9261. The remaining options 9271, 9281 and 9361 lie between ${21}^3$ and ${22}^3$, so none of them is the cube of a whole number; option (a) is the answer.

Partnership ratio In a partnership the profit is divided in the ratio of (capital $\times$ time), because money that stays longer earns more. A's capital worked for the whole year: $50000\times 12=600000$ B joined after 6 months, so B's capital worked for only 6 months: $30000\times 6=180000$ A:B=600000:180000=10:3 Hence option (a) is the answer.

Compound angle formula Split ${75}^{\circ }$ into two standard angles: ${75}^{\circ }={45}^{\circ }+{30}^{\circ }$. Using the given expansion with $A={45}^{\circ }$ and $B={30}^{\circ }$: $\cos {75}^{\circ }=\cos {45}^{\circ }\cos {30}^{\circ }-\sin {45}^{\circ }\sin {30}^{\circ }$ $=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$ $=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ Hence option (c) is the answer.

Ratio after percentage rise Take the numbers as $2x,7x,5x,3x$. Increasing a quantity by $r\%$ multiplies it by $\left(1+\frac{r}{100}\right)$. New values: $2x\times 1.2=2.4x$, $7x\times 1.3=9.1x$, $5x\times 1.4=7x$, $3x\times 1.1=3.3x$ New ratio =2.4:9.1:7:3.3, and multiplying every term by 10 clears the decimals. =24:91:70:33, so option (d) is the answer.

Mixture removed The ratio 5:4 has 5+4=9 parts, so one part $=\frac{720}{9}=80$ litres. Milk $=5\times 80=400$ litres and water $=4\times 80=320$ litres. What is drawn off is mixture, not pure milk, so the 180 litres removed also carry milk and water in the ratio 5:4. Milk removed $=\frac{5}{9}\times 180=100$ litres Milk left =400-100=300 litres, so option (a) is the answer.

Mixture - adding water The ratio 1:3 has 1+3=4 parts, so one part $=\frac{80}{4}=20$ litres. Wine =20 litres and water $=3\times 20=60$ litres. Only water is poured in, so the wine stays fixed at 20 litres - use it as the anchor. For the new ratio 1:5 the water must be $5\times 20=100$ litres. Water to be added =100-60=40 litres, so option (d) is the answer.

Trigonometric identity Given: $\sec \theta +\tan \theta =6$ ... (i) Identity: ${\sec }^2\theta -{\tan }^2\theta =1$, which factorises as $\left(\sec \theta +\tan \theta \right)\left(\sec \theta -\tan \theta \right)=1$. Putting (i) into it: $6\times \left(\sec \theta -\tan \theta \right)=1$, so $\sec \theta -\tan \theta =\frac{1}{6}$ ... (ii) Adding (i) and (ii): $2\sec \theta =6+\frac{1}{6}=\frac{37}{6}$ $\sec \theta =\frac{37}{12}$ $\cos \theta =\frac{1}{\sec \theta }=\frac{12}{37}$, so option (b) is the answer.

Relative speed - chase In a chase both vehicles move the same way, so the bike gains only at the relative speed 95-50=45 km/hr. Time to close the 24 km gap $=\frac{24}{45}$ hr $=\frac{24}{45}\times 60=32$ minutes, so t=32. Therefore t-15=32-15=17 minutes. Distance covered by the bike in 17 minutes $=95\times \frac{17}{60}=\frac{1615}{60}=26.9$ km (nearly). This is nearest to 27 km, so option (c) is correct.

Surds - square identity Here x and y are of the form a+b and a-b with $a=\sqrt{13}$ and $b=\sqrt{5}$. Use the identity ${\left(a+b\right)}^2+{\left(a-b\right)}^2=2\left(a^2+b^2\right)$; the cross terms cancel, so no surd survives. Hence $x^2+y^2=2\left({\left(\sqrt{13}\right)}^2+{\left(\sqrt{5}\right)}^2\right)=2\left(13+5\right)$. $=2\times 18=36$. So option (a) is correct.

Surds - simplification $a+b=\left(\sqrt{7}+\sqrt{5}\right)+\left(\sqrt{7}-\sqrt{5}\right)=2\sqrt{7}$. $a-b=\left(\sqrt{7}+\sqrt{5}\right)-\left(\sqrt{7}-\sqrt{5}\right)=2\sqrt{5}$. So $\frac{a+b}{a-b}=\frac{2\sqrt{7}}{2\sqrt{5}}=\sqrt{\frac{7}{5}}=\frac{\sqrt{35}}{5}$. Numerically $\frac{\sqrt{35}}{5}=\frac{5.916}{5}=1.183$ (nearly). The given options are worth $\frac{3\sqrt{6}}{5}=1.47$, $\frac{2\sqrt{6}}{3}=1.63$ and $\frac{2\sqrt{3}}{7}=0.49$, none of which equals 1.183. Hence option (d) is correct.

Area - tiling a floor Let the tile be 5x cm long and 4x cm broad. Perimeter of the tile: $2\left(5x+4x\right)=288\Rightarrow 18x=288\Rightarrow x=16$. So the tile measures $5\times 16=80$ cm by $4\times 16=64$ cm, and its area $=80\times 64=5120$ cm². Floor: 25.6 m =2560 cm, 14.4 m =1440 cm, so its area $=2560\times 1440=3686400$ cm². Number of tiles $=\frac{3686400}{5120}=720$. Hence option (a) is correct.

Rectangle - diagonal and area The diagonal of a rectangle divides it into two right triangles whose legs are the two sides. By Pythagoras, other side $=\sqrt{{15}^2-9^2}=\sqrt{225-81}=\sqrt{144}=12$ cm. Area = length $\times$ breadth $=12\times 9=108$ cm². So the other side is 12 cm and the area 108 cm², i.e. option (b) is correct.

Average of consecutive numbers For an odd count of equally spaced numbers the average is the middle term, so the 4th of the seven even numbers is 50. The seven numbers are therefore 44, 46, 48, 50, 52, 54, 56 and their sum $=7\times 50=350$. The next 5 consecutive even numbers are 58, 60, 62, 64, 66, with sum $=\frac{5}{2}\left(58+66\right)=\frac{5}{2}\times 124=310$. New average $=\frac{350+310}{7+5}=\frac{660}{12}=55$. Hence option (d) is correct.

Successive percentage change Let the original population be 100. After a fall of 15% it becomes 100-15=85. The rise now has to be measured on the reduced base 85, not on 100 - that is why the answer is more than 15%. Increase needed =100-85=15, so required percentage $=\frac{15}{85}\times 100$. $=\frac{300}{17}=17.647\dots \approx 17.6\%$. Hence option (b) is correct.

Markup and discount The markup is calculated on the cost price, the discount on the marked price. Marked price $=600\times \frac{120}{100}=720$, i.e. ₹720. Selling price $=720\times \frac{90}{100}=648$, i.e. ₹648. In one step: $600\times \frac{120}{100}\times \frac{90}{100}=600\times 1.08=648$, so the net gain is only 8%. Hence option (b) is correct.

Compound interest rate For compound interest, amount $=P{\left(1+\frac{r}{100}\right)}^n$. Here the sum becomes 25 times in 2 years, so $P{\left(1+\frac{r}{100}\right)}^2=25P$. ${\left(1+\frac{r}{100}\right)}^2=25\Rightarrow 1+\frac{r}{100}=5$ (the positive root). $\frac{r}{100}=4\Rightarrow r=400\%$ per annum. Check: $P\to 5P\to 25P$ in two years, which fits. Hence option (d) is correct.

Time, speed and distance Let the normal speed be v km/h, so the increased speed is (v+20) km/h. Starting 40 minutes late, the faster journey must save $\frac{40}{60}=\frac{2}{3}$ hour over the same 160 km. So $\frac{160}{v}-\frac{160}{v+20}=\frac{2}{3}$. $160\times \frac{\left(v+20\right)-v}{v\left(v+20\right)}=\frac{2}{3}\Rightarrow \frac{3200}{v^2+20v}=\frac{2}{3}$. $9600=2v^2+40v\Rightarrow v^2+20v-4800=0$. (v+80)(v-60)=0, and speed cannot be negative, so v=60 km/h. Hence option (a) is correct.

Percentage - win ratio Let x matches be played before the losing streak. Matches won $=60\%$ of $x=\frac{3x}{5}$; since the next 10 matches are all losses, the number of wins stays $\frac{3x}{5}$. After the streak the matches played become x+10 and the winning percentage is $50\%$. So $\frac{3x/5}{x+10}=\frac{1}{2}\Rightarrow \frac{6x}{5}=x+10$. $6x=5x+50\Rightarrow x=50$. Check: 30 wins in 50 matches is 60%, and 30 wins in 60 matches is 50%. Hence option (b) is correct.

Time and work One day's work of A and B together $=\frac{1}{9}$. One day's work of A alone $=\frac{1}{18}$. So one day's work of B $=\frac{1}{9}-\frac{1}{18}=\frac{2-1}{18}=\frac{1}{18}$. B therefore needs 18 days alone - A and B are equally efficient here. Hence option (a) is correct.

Discount on marked price A discount of 36% leaves $100-36=64\%$ of the original price. Final price $=3060\times \frac{64}{100}$. $=30.6\times 64=1958.4$, i.e. ₹1,958.4. Hence option (c) is correct.

Perimeter and cost of fencing Fencing is laid along the boundary, so the length of fence needed is the perimeter of the triangle. Perimeter =70+105+132=307 m. Total expense $=307\times 19$. $=307\times 20-307=6140-307=5833$, i.e. ₹5,833. Hence option (a) is correct.

Ratio and perimeter Let the sides be 2x, 5x and 9x metres. Perimeter is the sum of the sides: 2x+5x+9x=16x=480. So $x=\frac{480}{16}=30$. The sides are $2\times 30=60$ m, $5\times 30=150$ m and $9\times 30=270$ m. Hence option (a) is correct.