28 November 2025 · Shift 1

Average of numbers Sum = average $\times$ count, so the sum of all 9 numbers $=9\times 83=747$. Sum of the first 4 numbers $=4\times 89=356$. Sum of the last 4 numbers $=4\times 80=320$. The first four and the last four together cover every number except the 5th, and they do not overlap. 5th number =747-356-320=747-676=71. So option (b) is the correct answer.

Percentage change Formula: if the price rises by $R\%$, the consumption must fall by $\frac{R}{100+R}\times 100\%$ to keep the expenditure unchanged. Here R=50, so the reduction $=\frac{50}{100+50}\times 100\%=\frac{50}{150}\times 100\%$. $=\frac{100}{3}\%=33.33\%$ Check: let the price be ₹100 and the consumption 3 units, so the expenditure is ₹300; after the rise the price is ₹150 and ₹300 buys only 2 units, a drop of 1 unit in 3, i.e. $33.33\%$. So option (d) is the correct answer.

Discount and price A discount of 30% is taken off the original price, so what is paid is $100\%-30\%=70\%$ of it. Therefore $70\%$ of the original price =700. Original price $=\frac{700\times 100}{70}=1000$, i.e. ₹1,000. Check: $30\%$ of 1000=300, and 1000-300=700, as given. So option (d) is the correct answer.

Mixture and dilution Acid in the 40 litres $=60\%$ of $40=\frac{60}{100}\times 40=24$ litres. So the water already present =40-24=16 litres. Only water is poured in, so the 24 litres of acid never changes; the whole change is in the water. The acid has to end up as $40\%$ measured against the water, i.e. acid : water =40:100=2:5. So $24=\frac{40}{100}\times \left(\mathrm{final}\mathrm{water}\right)$, giving final water $=\frac{24\times 100}{40}=60$ litres. Water to be added =60-16=44 litres, so option (a) is the correct answer.

Time and work Let the whole work be $\mathrm{LCM}\left(40,60\right)=120$ units. A's one day's work $=\frac{120}{40}=3$ units and B's one day's work $=\frac{120}{60}=2$ units. Working together they finish 3+2=5 units a day, so in 10 days they finish $5\times 10=50$ units. Work left =120-50=70 units. As a fraction of the whole work this is $\frac{70}{120}=\frac{7}{12}$, so option (c) is the correct answer.

Packing cuboids Match each edge of the carton with an edge of the box so that no space is wasted. $60÷15=4$, $35÷7=5$ and $30÷10=3$ — each division is exact, so the boxes fill the carton completely. Number of boxes $=4\times 5\times 3=60$. The volumes confirm it: $\frac{60\times 35\times 30}{15\times 10\times 7}=\frac{63000}{1050}=60$. So option (c) is the correct answer.

Similar triangles In two similar triangles the corresponding angles are equal, so the second sign must carry exactly the same three angles as the first. The three angles of the first sign are ${40}^{\circ }$, ${80}^{\circ }$ and ${60}^{\circ }$. One angle of the similar sign is given as ${60}^{\circ }$, so the other two can only be ${80}^{\circ }$ and ${40}^{\circ }$. The check on the angle sum holds: ${60}^{\circ }+{80}^{\circ }+{40}^{\circ }={180}^{\circ }$. So option (a) is the correct answer.

Interior angle of polygon Formula: each interior angle of a regular polygon of n sides $=\frac{\left(n-2\right)\times {180}^{\circ }}{n}$. A regular hexagon has n=6, so each interior angle $=\frac{\left(6-2\right)\times {180}^{\circ }}{6}$. $=\frac{4\times {180}^{\circ }}{6}=\frac{{720}^{\circ }}{6}={120}^{\circ }$ The exterior angle checks it: $\frac{{360}^{\circ }}{6}={60}^{\circ }$ and ${180}^{\circ }-{60}^{\circ }={120}^{\circ }$. So option (b) is the correct answer.

Mixture and profit Let the quantity bought at ₹28 per kg be x kg; then the quantity bought at ₹20 per kg is 2x kg. The two parts make up the whole purchase, so 2x+x=600, i.e. 3x=600 and x=200. So 400 kg came at ₹20 and 200 kg at ₹28, and the cost price $=400\times 20+200\times 28=8000+5600=13600$. Selling price $=600\times 25=15000$, so the profit =15000-13600=1400, exactly the ₹1,400 stated. Hence the quantity bought at ₹28 per kg is 200 kg, and option (d) is the correct answer.

Volume of a prism The water a pipe holds is its volume, and for a right prism volume = base area $\times$ length. The base is an equilateral triangle of side 8 cm, whose area $=\frac{\sqrt{3}}{4}\times 8^2=\frac{\sqrt{3}}{4}\times 64=16\sqrt{3}$ sq cm. Volume $=16\sqrt{3}\times 30=480\sqrt{3}$ cubic cm. Taking $\sqrt{3}=1.732$, volume $=480\times 1.732=831.36$ cubic cm. So option (c) is the correct answer.

Inradius of a triangle The sprinkler touches all three sides, so its circle is the incircle of the triangle and the radius asked for is the inradius r. Semi-perimeter $s=\frac{26+28+30}{2}=\frac{84}{2}=42$ m. By Heron's formula the area $=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\sqrt{42\times 16\times 14\times 12}$. $=\sqrt{112896}=336$ square metres. For any triangle the inradius is area divided by the semi-perimeter, so $r=\frac{336}{42}=8$ m. So the radius of the sprinkler is 8 m, and option (a) is the correct answer.

Combining ratios b occurs in both ratios, so make the value of b the same in each of them. In a:b=8:3 multiply both parts by 5, giving a:b=40:15. In b:c=5:13 multiply both parts by 3, giving b:c=15:39. Now the two chains join: a:b:c=40:15:39. Dropping the middle term, a:c=40:39, so option (b) is the correct answer.

Angle between altitude and bisector Let the base angles be $\angle B$ and $\angle C$; then $\angle BAC=180°-\angle B-\angle C$. AS bisects $\angle BAC$, so $\angle BAS=\frac{180°-\angle B-\angle C}{2}=90°-\frac{\angle B+\angle C}{2}$. In the right triangle ABT, $\angle BAT=180°-90°-\angle B=90°-\angle B$. Since $\angle B>\angle C$, the foot T lies between B and S, so $\angle TAS=\angle BAS-\angle BAT$. $\angle TAS=\left(90°-\frac{\angle B+\angle C}{2}\right)-\left(90°-\angle B\right)=\angle B-\frac{\angle B+\angle C}{2}$. $\angle TAS=\frac{2\angle B-\angle B-\angle C}{2}=\frac{\angle B-\angle C}{2}$. So option (b) is the correct answer.

Mixture percentage The mixture is syrup plus water, so its volume is 4+n litres while the syrup in it stays 4 litres. The syrup is $10\%$ of the mixture, so $\frac{4}{4+n}=\frac{10}{100}=\frac{1}{10}$. Cross-multiplying, $4+n=4\times 10=40$. n=40-4=36. So 36 litres of water are added, and option (c) is the correct answer.

Angle at centre and circumference The angle a chord subtends at the centre is twice the angle it subtends at any point on the remaining arc of the circle. Here the chord AB subtends $\angle APB=55°$ at the circumference. So $\angle AOB=2\times \angle APB=2\times 55°=110°$. The angle at the centre is 110°, so option (c) is the correct answer.

Data sufficiency: trains To cross a platform a train must cover its own length plus the platform, so the distance needed here is 600+1800=2400 m; the only unknown is the speed. Statement I: crossing a 1,400 m platform means covering 600+1400=2000 m in 100 s, so the speed is $\frac{2000}{100}=20$ m/s and the required time is $\frac{2400}{20}=120$ s. Hence statement I alone is sufficient. Statement II: crossing a pole means covering just the length of the train, so the speed is $\frac{600}{30}=20$ m/s and again the time is $\frac{2400}{20}=120$ s. Hence statement II alone is also sufficient. Either statement alone answers the question, so option (c) is the correct answer.

Rationalising a surd To clear the surd from the denominator, multiply the top and the bottom by the conjugate $5-\sqrt{3}$. $\frac{7}{5+\sqrt{3}}=\frac{7\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}$. The denominator is a difference of squares: $5^2-{\left(\sqrt{3}\right)}^2=25-3=22$. The numerator is $7\times 5-7\times \sqrt{3}=35-7\sqrt{3}$. So the expression equals $\frac{35-7\sqrt{3}}{22}$, and option (a) is the correct answer.

Partnership with withdrawal Profit in a partnership is divided in the ratio of capital multiplied by the time for which it stayed in the business. A keeps ₹20,000 for the first 6 months and only ₹10,000 for the next 6 months, while B keeps ₹30,000 for the whole 12 months. A's capital-months $=20000\times 6+10000\times 6=120000+60000=180000$. B's capital-months $=30000\times 12=360000$. So A:B=180000:360000=1:2, and the profit splits into 1+2=3 equal parts. A's share $=\frac{1}{3}\times 90000=30000$, that is ₹30,000. So option (b) is the correct answer.

Simple interest principal Under simple interest the same interest is added every year, so the amount rises by a fixed step each year. Interest of 9-6=3 years =19000-16000=3000. Interest of 1 year $=\frac{3000}{3}=1000$. Interest of 6 years $=6\times 1000=6000$. Principal = amount after 6 years - interest of 6 years =16000-6000=10000. So the original principal is ₹10,000, and option (b) is the correct answer.

Algebraic identity Take 2 common from the expression: $2a^2+2b^2-4ab=2\left(a^2-2ab+b^2\right)$. The bracket is the identity ${\left(a-b\right)}^2$, so the expression is $2{\left(a-b\right)}^2$. Only a-b is needed, and it is given as a-b=3; the value of a+b is not required. $2{\left(a-b\right)}^2=2\times 3^2=2\times 9=18$. So option (a) is the correct answer.

Sum of fractions Evaluate each option and keep the one whose total is exactly 100. (a) $\frac{200}{4}+\frac{3}{4}=50+\frac{3}{4}=50.75$, which is far below 100. (b) $\frac{250}{4}+\frac{1}{3}=62.5+0.33\dots =62.83\dots$, again short of 100. (c) $\frac{75}{4}+\frac{350}{3}=18.75+116.66\dots =135.41\dots$, past 100. (d) $\frac{150}{3}+\frac{200}{4}=50+50=100$, exactly the required sum. So option (d) is the correct answer.

Simplifying surds First bring both surds to the same form: $\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}$. So $x=5\sqrt{3}+\sqrt{3}=6\sqrt{3}$. Squaring, $x^2={\left(6\sqrt{3}\right)}^2=6^2\times {\left(\sqrt{3}\right)}^2=36\times 3$. $x^2=108$. So option (b) is the correct answer.

Compound interest ratio Write the rate as a fraction: $14\frac{2}{7}\%=\frac{100}{7}\%=\frac{100}{7\times 100}=\frac{1}{7}$ of the sum each year. So in one year the amount becomes $1+\frac{1}{7}=\frac{8}{7}$ times what it was at the start of that year. Amount at the end of the first year $=P\times \frac{8}{7}$. Amount at the end of the second year $=P\times {\left(\frac{8}{7}\right)}^2=P\times \frac{64}{49}$. Required ratio $=\frac{8P}{7}:\frac{64P}{49}=\frac{8}{7}:\frac{64}{49}$. Multiplying both parts by $\frac{49}{8P}$ gives 7:8. So option (b) is the correct answer.

Average of consecutive odds An odd number of consecutive terms is symmetric about its middle term, so for 7 such numbers the middle (4th) term is the average. Hence the 4th number is 39, and the seven numbers are 33, 35, 37, 39, 41, 43, 45. The seven consecutive odd numbers just before these start two below 33 and run backwards: 19, 21, 23, 25, 27, 29, 31. The newly added numbers are these seven, and their middle term is 25. So their average is 25, and option (c) is the correct answer.

Time and work, leaving Take the total work as the LCM of 16, 36 and 48, which is 144 units. Then P does $\frac{144}{16}=9$ units, Q does $\frac{144}{36}=4$ units and R does $\frac{144}{48}=3$ units in a day. For the first 6 days all three work: $6\times \left(9+4+3\right)=6\times 16=96$ units. For the last 4 days R works alone: $4\times 3=12$ units. So the middle stretch, worked by Q and R together, covers 144-96-12=36 units. Q and R together do 4+3=7 units a day, so that stretch lasts $\frac{36}{7}$ days. Total time $=6+\frac{36}{7}+4=10+5\frac{1}{7}=15\frac{1}{7}$ days. So option (b) is the correct answer.