24 November 2025 · Shift 2

Mean proportional The mean proportional between two numbers a and b is $\sqrt{ab}$, because it is the number x for which a:x=x:b. Here a=9 and b=25, so $x=\sqrt{9\times 25}=\sqrt{225}$. $\sqrt{225}=15$. Check: 9:15 and 15:25 both reduce to 3:5. Hence option (a) is correct.

Average of even numbers The ten consecutive even numbers starting from 14 are 14,16,18,20,22,24,26,28,30,32. They form an arithmetic progression, and for an AP the average equals the mean of the first and the last term. Average $=\frac{14+32}{2}=\frac{46}{2}$. =23. Hence option (b) is correct.

Successive percentage change For two successive changes of $x\%$ and $y\%$, the net change is $x+y+\frac{xy}{100}$ per cent. Here x=+15 and y=-15, so the net change $=15-15+\frac{15\times \left(-15\right)}{100}$. $=-\frac{225}{100}=-2.25$, that is a decrease of $2.25\%$. Verify with 100: $100\to 115\to 115\times 0.85=97.75$, a fall of 2.25 on 100. An equal rise and fall never cancel out, so option (a) is wrong; hence option (b) is correct.

Successive percentage change Use the net change formula $x+y+\frac{xy}{100}$ with x=+25 and y=-25. Net change $=25-25+\frac{25\times \left(-25\right)}{100}=-\frac{625}{100}$. =-6.25, that is a decrease of $6.25\%$. Check with 100: $100\to 125\to 125\times 0.75=93.75$, which is 6.25 below 100. In general a rise and fall of the same $x\%$ always leaves a loss of $\frac{x^2}{100}\%$, so option (a) is correct.

Discount percentage Discount = marked price - selling price =1800-1440=360. Discount per cent is always taken on the marked price, so it is $\frac{360}{1800}\times 100$. $\frac{360}{1800}=\frac{1}{5}$, so the discount is $\frac{1}{5}\times 100=20$. Hence option (c) is correct.

Speed, time and distance Distance = speed $\times$ time. $=45\times 6$. =270, so the car covers 270 km. Hence option (c) is correct.

Volume of pyramid and cuboid The monument is a cuboid of height 0.8 m carrying a square-based pyramid of height 1.8 m, both on the same 1.4 m by 1.4 m base. Total height =0.8+1.8=2.6 m. Volume of the cuboid $=1.4\times 1.4\times 0.8=1.568$ cubic metres. Base area of the pyramid $=1.4\times 1.4=1.96$ square metres. Volume of the pyramid $=\frac{1}{3}\times 1.96\times 1.8=\frac{3.528}{3}=1.176$ cubic metres. Total volume $=1.568+1.176=2.744\approx 2.74$ cubic metres. So the monument is 2.6 m tall with a volume of about 2.74 cubic metres, and option (c) is correct.

Number systems The number sets sit one inside the other: natural numbers are inside integers, integers inside rational numbers, and rational numbers inside real numbers. That makes options (a), (b) and (d) true statements. But real numbers also contain the irrational numbers, such as $\sqrt{2}$ and $\pi$, which cannot be written as $\frac{p}{q}$ with integers p and q, $q\ne 0$. So 'every real number is rational' is the false statement, and option (c) is correct.

Volume ratio of solids Let the side of the cube be a, so the cube's height is also a. The cylinder has the same height a and radius $r=\frac{a}{3}$. Volume of the cube $=a^3$. Volume of the cylinder $=\pi r^{2}h=\pi {\left(\frac{a}{3}\right)}^2\times a=\frac{\pi a^3}{9}$. Ratio cylinder : cube $=\frac{\pi a^3}{9}:a^3$, and cancelling $a^3$ gives $\frac{\pi }{9}:1=\pi :9$. Hence option (d) is correct.

Corrected average Total attendance of the first two events $=2\times 168.5=337$. Total attendance of the next two events $=2\times 182.5=365$. Recorded total for all four events =337+365=702. One of the first two attendances was actually 15 more than recorded, and one of the last two was 7 less, so the total changes by +15-7=+8. Corrected total =702+8=710. Corrected average $=\frac{710}{4}=177.5$. Hence option (b) is correct.

Cube and trailing zeros Separate the zeros before cubing: $700=7\times {10}^2$. Cube both factors: ${700}^3=7^3\times {\left({10}^2\right)}^3=343\times {10}^6$. So ${700}^3=343000000$, which ends in 6 zeros. The rule is that cubing multiplies the count of trailing zeros by 3, i.e. $2\times 3=6$, so option (c) is correct.

Complementary angle identity Use the complementary-angle relation $\sin \left({90}^{\circ }-\theta \right)=\cos \theta$. Squaring it gives ${\sin }^2\left({90}^{\circ }-\theta \right)={\cos }^2\theta$. Hence ${\sin }^2\theta +{\sin }^2\left({90}^{\circ }-\theta \right)={\sin }^2\theta +{\cos }^2\theta$. By the identity the radar system uses, ${\sin }^2\theta +{\cos }^2\theta =1$. So the required value is 1 for every $\theta$, and option (a) is correct.

Simple interest and amount From 15 February 2025 to 15 June 2025 is exactly 4 months, so $T=\frac{4}{12}=\frac{1}{3}$ year. Formula: $SI=\frac{P\times R\times T}{100}$. $SI=\frac{30000\times 18\times \frac{1}{3}}{100}=\frac{30000\times 18}{100}\times \frac{1}{3}$. $SI=5400\times \frac{1}{3}=1800$. Amount repaid =P+SI=30000+1800=31800. So the repayment is ₹31,800 and option (b) is correct.

SI and CI ratio Simple interest on ₹(x+400) at $32\%$ for 4 years $=\frac{\left(x+400\right)\times 32\times 4}{100}=\frac{128\left(x+400\right)}{100}$. Compound interest on ₹(x-400) at $20\%$ for 2 years $=\left(x-400\right)\left({\left(\frac{6}{5}\right)}^2-1\right)$. ${\left(\frac{6}{5}\right)}^2-1=\frac{36}{25}-1=\frac{11}{25}$, so $CI=\frac{44\left(x-400\right)}{100}$. Given ratio: $\frac{128\left(x+400\right)}{44\left(x-400\right)}=\frac{49}{11}$. Since $\frac{128}{44}=\frac{32}{11}$, this reduces to $\frac{32\left(x+400\right)}{11\left(x-400\right)}=\frac{49}{11}$, i.e. 32(x+400)=49(x-400). $32x+12800=49x-19600\Rightarrow 17x=32400$. $x=\frac{32400}{17}=1905.88\dots$, which is ₹1,906 to the nearest rupee, so option (a) is correct.

Difference of SI and CI Simple interest: $SI=\frac{20000\times 10\times 3}{100}=6000$. Compound interest: $CI=P\left({\left(1+\frac{R}{100}\right)}^3-1\right)=20000\left({\left(\frac{11}{10}\right)}^3-1\right)$. ${\left(\frac{11}{10}\right)}^3=\frac{1331}{1000}$, so $CI=20000\times \frac{331}{1000}=6620$. Difference =6620-6000=620. So the two schemes differ by ₹620 and option (c) is correct.

Mixture and replacement Let the initial mixture be 5k litres, so juice =4k litres and water =k litres. Taking out 50 litres of the mixture does not change the ratio, so what remains is (5k-50) litres with water $=\frac{5k-50}{5}=k-10$ litres. After pouring in 20 litres of water: total =5k-50+20=5k-30 and water =k-10+20=k+10. Water is $40\%$ of the new mixture: $k+10=\frac{40}{100}\left(5k-30\right)=2k-12$. $\Rightarrow k=22$. Initial quantity $=5k=5\times 22=110$ litres, so option (c) is correct.

Time and work Total work for one job $=19\times 10=190$ man-days. Four times the work $=4\times 190=760$ man-days. With 10 men, time $=\frac{760}{10}=76$ days. So 10 men need 76 days and option (b) is correct.

tan-cot identity Let $\tan \theta =t$, so $\cot \theta =\frac{1}{t}$ and the condition is $t+\frac{1}{t}=2$. Squaring: $t^2+\frac{1}{t^2}={\left(t+\frac{1}{t}\right)}^2-2=2^2-2=2$. Squaring once more: $t^4+\frac{1}{t^4}={\left(t^2+\frac{1}{t^2}\right)}^2-2=2^2-2=2$. The same result follows directly, since $t+\frac{1}{t}=2$ forces t=1, i.e. $\theta ={45}^{\circ }$, and then ${\tan }^4\theta +{\cot }^4\theta =1+1=2$. So the value is 2 and option (c) is correct.

Surface area of a combined solid The cone and the hemisphere are stuck together along their flat circular faces, so both of those flat faces lie inside the solid and are not part of its surface. Only the two curved surfaces are exposed. Curved surface area of the hemisphere $=2\pi r^2$. Curved surface area of the cone $=\pi rl$, where l is the slant height. Total surface area $=2\pi r^2+\pi rl$, so option (c) is correct.

Total cost and profit Pages printed $=25\times 300=7500$, so printing cost $=7500\times 20=$ ₹1,50,000. Binding $=25\times 50=$ ₹1,250 and supply $=25\times 40=$ ₹1,000. Advertising is a one-time expense of ₹3,000. Total cost price =150000+1250+1000+3000= ₹1,55,250. Total selling price $=25\times 6500=$ ₹1,62,500. Profit =162500-155250= ₹7,250, so option (d) is correct.

Combining two ratios To join A:B and B:C the value of B must be the same in both. Here B=5 in A:B=7:5 and B=5 in B:C=5:6, so no adjustment is needed. Writing them together, A:B:C=7:5:6. Hence option (a) is correct.

Partnership profit share All three invested for the same period, so the profit is shared in the ratio of the investments, 8:6:6. Total number of parts =8+6+6=20. Q's share $=\frac{6}{20}\times 160000=48000$. So Q receives ₹48,000 and option (a) is correct.

Principal from compound interest For 2 years, $CI=P\left({\left(1+\frac{R}{100}\right)}^2-1\right)$. With $R=15\%$, ${\left(1+\frac{15}{100}\right)}^2={\left(\frac{23}{20}\right)}^2=\frac{529}{400}$. So $CI=P\left(\frac{529}{400}-1\right)=\frac{129P}{400}$. $\frac{129P}{400}=3150\Rightarrow P=\frac{3150\times 400}{129}=\frac{1260000}{129}$. $P=9767.44\dots \approx 9767.4$. So the principal is ₹9,767.4 and option (a) is correct.

Square root of a surd Try to write $21+8\sqrt{5}$ as a perfect square ${\left(a+b\right)}^2=a^2+b^2+2ab$. Match the surd part: $2ab=8\sqrt{5}$, so $ab=4\sqrt{5}$; take a=4 and $b=\sqrt{5}$. Check the rational part: $a^2+b^2=16+5=21$, which is exactly the constant term. Hence $21+8\sqrt{5}={\left(4+\sqrt{5}\right)}^2$. Therefore $\sqrt{21+8\sqrt{5}}=4+\sqrt{5}$, so option (b) is correct.

Rise in water level The ball is completely submerged, so the water level rises by exactly the volume of the ball. Volume of the ball $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(14\right)}^3$. If the level rises by h, then $\pi {\left(14\right)}^{2}h=\frac{4}{3}\pi {\left(14\right)}^3$. $h=\frac{4}{3}\times 14=\frac{56}{3}=18.66$ cm. New height =20+18.66=38.66 cm. Since 38.66>28, the ball really is under water, so option (d) is correct.