25 November 2025 · Shift 2

Percentage of a number Let the number be x. Given $\frac{30}{100}\times x=120$. So $x=\frac{120\times 100}{30}=400$. Now $45\%$ of $400=\frac{45}{100}\times 400=180$. Hence option (b) is correct.

Successive percentage change Apply the two changes one after the other to the original price ₹300. After the $20\%$ rise the price is $300\times \frac{120}{100}=360$. The $5\%$ fall is taken on this new price of ₹360, not on the original ₹300. $360\times \frac{95}{100}=342$. So the final price is ₹342 and option (c) is correct.

Profit percentage Profit = Selling price − Cost price =195-150=45. Profit percentage is always taken on the cost price: Profit % $=\frac{45}{150}\times 100$. $\frac{45}{150}=\frac{3}{10}$, so the profit is $\frac{3}{10}\times 100=30\%$. Hence option (c) is correct.

Volume ratio of cylinders Volume of a cylinder $=\pi r^{2}h$. Let the first tank have radius r; then the second has radius 3r, and the height h is the same for both. $V_1=\pi r^{2}h$ and $V_2=\pi {\left(3r\right)}^{2}h=9\pi r^{2}h$. $V_1:V_2=\pi r^{2}h:9\pi r^{2}h=1:9$. Note that the radius is squared, so tripling it makes the volume 9 times, not 3 times; hence option (c) is correct.

Discount and marked price Let the cost price be x. A profit of $25\%$ means the profit is $\frac{25}{100}x$, and this profit is given as Rs. 120. $\frac{25}{100}x=120$, so $x=\frac{120\times 100}{25}=480$. Selling price =480+120=600. This selling price is what remains after a $20\%$ discount, so SP is $80\%$ of the marked price. $\frac{80}{100}\times MP=600$, hence $MP=\frac{600\times 100}{80}=750$. So the marked price is Rs. 750 and option (b) is correct.

Lateral surface of prism Only the lateral faces are painted, so the two hexagonal ends are left out. Lateral surface area of a right prism = (perimeter of base) × height. The base is a regular hexagon of side 8 cm, so its perimeter is $6\times 8=48$ cm. Lateral surface area $=48\times 14=672$ sq. cm. Cost $=672\times 12.50=8400$. The value of $\sqrt{3}$ would be needed only for the area of the hexagonal ends, which are not painted here. So the total painting cost is ₹8,400 and option (b) is correct.

Compound interest The money stays invested from age 21 to age 25, that is for 25-21=4 years. Formula: Amount $=P{\left(1+\frac{R}{100}\right)}^n$, with P=50000, R=10 and n=4. $A=50000\times {\left(\frac{11}{10}\right)}^4=50000\times \frac{14641}{10000}$. $A=5\times 14641=73205$. So she receives ₹73,205, and option (b) is correct.

Compound interest with repayment At $10\%$ compound interest the outstanding debt is multiplied by $\frac{11}{10}$ every year. After 1 year the debt is $10000\times \frac{11}{10}=11000$. He repays Rs. 5,000 there, so the balance carried forward is 11000-5000=6000. This Rs. 6,000 grows for the remaining 2 years: $6000\times {\left(\frac{11}{10}\right)}^2=6000\times \frac{121}{100}=7260$. Total money he actually pays =5000+7260=12260. Interest paid = total paid − loan taken =12260-10000=2260. Hence the total interest is Rs. 2,260 and option (a) is correct.

Combining ratios To link the three partners, Q's number must be the same in both ratios. P : Q = 2 : 3 and Q : R = 4 : 5; the LCM of 3 and 4 is 12. Multiply the first ratio by 4 and the second by 3: P : Q = 8 : 12 and Q : R = 12 : 15. Hence P : Q : R = 8 : 12 : 15, giving 8+12+15=35 equal parts. P's share $=\frac{8}{35}\times 70000=8\times 2000=16000$. So P gets Rs. 16,000 and option (a) is correct.

Ratio of numbers Let the three numbers be 3x, 5x and 7x. The largest is 7x and the smallest is 3x, so 7x-3x=24. 4x=24, hence x=6. The middle number is $5x=5\times 6=30$. Hence option (a) is correct.

Mixture and alligation The two liquids are mixed in equal proportions, so the milk percentage of the final mixture is simply the average of the two percentages. $\frac{40+m}{2}=55$ 40+m=110 m=110-40=70 So the second mixture contains 70% milk, and option (c) is correct.

Mixture and alligation The ratio 12:8 reduces to 3:2, so take 3 parts of the first mixture and 2 parts of the second. Acid from the first $=\frac{80}{100}\times 3=2.4$ parts. Acid from the second $=\frac{55}{100}\times 2=1.1$ parts. Total acid =2.4+1.1=3.5 parts in a total of 3+2=5 parts. Required percentage $=\frac{3.5}{5}\times 100=70$, so option (c) is correct.

Surds and indices Given $x=\sqrt{5}+\sqrt{3}$. Square both sides using ${\left(a+b\right)}^2=a^2+b^2+2ab$: $x^2=5+3+2\sqrt{5}\times \sqrt{3}=8+2\sqrt{15}$. Therefore $x^2-2\sqrt{15}=8+2\sqrt{15}-2\sqrt{15}=8$. Hence option (b) is correct.

Algebraic identity Use the identity ${\left(a+b\right)}^2-{\left(a-b\right)}^2=4ab$; it saves squaring the surds separately. Here $a=\sqrt{10}$ and $b=\sqrt{2}$. $4ab=4\times \sqrt{10}\times \sqrt{2}=4\sqrt{20}$. Since $\sqrt{20}=2\sqrt{5}$, the value is $4\times 2\sqrt{5}=8\sqrt{5}$. Hence option (d) is correct.

Volume of sphere and hemisphere Volume of the solid sphere $=\frac{4}{3}\pi R^3=\frac{4}{3}\pi {\left(12\right)}^3=\frac{4}{3}\pi \times 1728=2304\pi$ cm³. The bowl is hollow, so its internal radius =4-1=3 cm. Metal in one bowl $=\frac{2}{3}\pi \left(R^3-r^3\right)=\frac{2}{3}\pi \left(4^3-3^3\right)$. $=\frac{2}{3}\pi \left(64-27\right)=\frac{2}{3}\pi \times 37=\frac{74\pi }{3}$ cm³. Number of bowls $=\frac{2304\pi }{\frac{74\pi }{3}}=\frac{2304\times 3}{74}=\frac{6912}{74}\approx 93.4$. Only complete bowls can be cast, so about 93 bowls are made and option (c) is correct.

Rational and irrational numbers Check each option against all three conditions. 3.5 is a terminating decimal, so it is rational and condition (i) already fails. $\sqrt{11}\approx 3.32$ and $\sqrt{15}\approx 3.87$ are irrational and lie between 3 and 4, but their squares are 11 and 15, which are rational, so condition (iii) fails. $\sqrt{3}+2\approx 1.732+2=3.732$ is irrational and lies between 3 and 4. Its square is ${\left(\sqrt{3}+2\right)}^2=3+4\sqrt{3}+4=7+4\sqrt{3}$, which is again irrational. All three conditions hold only for $\sqrt{3}+2$, so option (c) is correct.

Fractions First change the mixed numbers into improper fractions: $7\frac{1}{2}=\frac{15}{2}$ kg and $3\frac{3}{5}=\frac{18}{5}$ kg. Rice left $=\frac{15}{2}-\frac{18}{5}=\frac{75-36}{10}=\frac{39}{10}$ kg. This is shared equally among 3 bags: $\frac{39}{10}÷3=\frac{39}{10}\times \frac{1}{3}=\frac{13}{10}$ kg. So each bag holds $\frac{13}{10}$ kg, and option (b) is correct.

Partnership In a partnership the profit is shared in the ratio of capital $\times$ time, so first find each partner's capital-months. P: $40,000\times 12=4,80,000$; Q: $60,000\times 12=7,20,000$; R joined 6 months late, so R: $1,20,000\times 6=7,20,000$. Ratio $=4,80,000:7,20,000:7,20,000=2:3:3$, which makes 2+3+3=8 equal parts. R's share $=\frac{3}{8}\times 4,00,000=1,50,000$. So R receives Rs. 1,50,000 and option (b) is correct.

Circular track meetings They run in opposite directions, so the speeds add up: relative speed =10+8=18 km/h. Starting together, they meet once for every 3 km of relative distance covered, that is once every $\frac{3}{18}=\frac{1}{6}$ hour =10 minutes. Time up to the 9th meeting $=9\times 10=90$ minutes =1 hour 30 minutes. Adding this to 6:00 a.m. gives 7:30 a.m., so option (b) is correct.

Average of real numbers Average =0 means the total of the 150 numbers is $0\times 150=0$. If every one of them were negative the total would be negative, so at least one number has to be positive. A single positive number is enough, provided it is large enough to cancel the whole negative total. For example, take 149 numbers each equal to -1 and one number equal to +149; the sum is -149+149=0. So the largest possible count of negative numbers is 150-1=149, and option (d) is correct.

Successive discount Successive discounts act one after the other, each on the price left after the previous one. After the first discount: $4000\times \frac{90}{100}=3600$. After the second: $3600\times \frac{80}{100}=2880$. After the third: $2880\times \frac{75}{100}=2160$. So the selling price is ₹2,160 and option (c) is correct.

Average correction Recorded total =45+50+65+55+60=275. The wrong entry 50 has to be replaced by 80, so the total rises by 80-50=30. Correct total =275+30=305. Correct average $=\frac{305}{5}=61$. Hence option (c) is correct.

Algebraic simplification Let x=20. Then $800=2x^2$ and 21=x+1, so the expression becomes $\frac{x^4+2x^2+1-x^2}{x^2+x+1}$. The numerator simplifies to $x^4+2x^2+1-x^2=x^4+x^2+1$. Use the standard factorisation $x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)$. So the fraction $=\frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2+x+1}=x^2-x+1$. Putting x=20: 400-20+1=381. Hence option (a) is correct.

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$. Here P=1800, R=3 and T=6. $SI=\frac{1800\times 3\times 6}{100}=\frac{32400}{100}$. SI=324, so the simple interest is Rs. 324 and option (b) is correct.

Pipes and cisterns Work with one-minute rates: $P+Q+R=\frac{1}{10}$, $Q+R=\frac{1}{15}$ and $R+P=\frac{1}{20}$. Q's own rate is found by removing R+P from the rate of all three together. $Q=\frac{1}{10}-\frac{1}{20}=\frac{2-1}{20}=\frac{1}{20}$. Q fills $\frac{1}{20}$ of the tank in one minute, so alone it needs 20 minutes. Hence option (b) is correct.