17 November 2025 · Shift 2

Overall loss percentage The two selling prices are equal, but the two cost prices are not, so find each cost price first. Gain of $25\%$: CP $=\frac{840\times 100}{125}=672$. Loss of $25\%$: CP $=\frac{840\times 100}{75}=1120$. Total CP =672+1120=1792, total SP $=2\times 840=1680$, so there is a loss of 1792-1680=112. Loss $\%=\frac{112}{1792}\times 100=6.25\%$. Shortcut worth remembering: equal selling prices with the same gain and loss of $x\%$ always end in a loss of $\frac{x^2}{100}\%=\frac{625}{100}=6.25\%$. Hence, option (b) is the most appropriate answer.

Successive discounts In successive discounts the second discount is worked out on the price left after the first one, never on the marked price. Price after the first discount $=750\times \frac{75}{100}=562.5$. This ₹562.50 falls to ₹450, so the second reduction is 562.5-450=112.5. Second discount $\%=\frac{112.5}{562.5}\times 100=20\%$. Hence, option (d) is the most appropriate answer.

Discount and profit No amount is given, so take the cost price as ₹100 — percentages will not change. Marked price $=100+30\%$ of 100=130. Discount of $12\%$ on the marked price leaves SP $=130\times \frac{88}{100}=114.4$. Profit =114.4-100=14.4 on a cost of 100, so profit $\%=14.4\%$. The same result comes from the successive-change formula: $30-12-\frac{30\times 12}{100}=18-3.6=14.4$. Hence, option (c) is the most appropriate answer.

Average of groups Total of the five old departments $=5\times 50=250$ employees. The sixth department adds 80 more, so the company now has 250+80=330 employees. The transfer of $5\times 20=100$ employees only moves people from one department to another — the company total stays 330. New average $=\frac{330}{6}=55$ employees per department. Hence, option (c) is the most appropriate answer.

Compound interest, varying rate When the rate changes every year, multiply the principal by one growth factor for each year. Formula: Amount $=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)$. After year 1: $20000\times \frac{120}{100}=24000$. After year 2: $24000\times \frac{125}{100}=30000$. After year 3: $30000\times \frac{110}{100}=33000$. Total interest =33000-20000=13000, i.e. ₹13,000. Hence, option (b) is the most appropriate answer.

Square of a number Bring the number close to a round figure and use ${\left(a-b\right)}^2=a^2-2ab+b^2$ with a=120, b=1. ${119}^2={\left(120-1\right)}^2={120}^2-2\times 120\times 1+1^2$. =14400-240+1=14161. Check: ${119}^2$ must be a little less than ${120}^2=14400$, and 14161 is the only option that fits. Hence, option (b) is the most appropriate answer.

Perfect squares A perfect square is the square of a whole number, so list the squares lying near the given range. $6^2=36$ (below 40), $7^2=49$ (between 40 and 60), $8^2=64$ (above 60). So 49 is the only perfect square between 40 and 60; 51 and 59 are not squares at all and 64 lies outside the range. Hence, option (c) is the most appropriate answer.

Partnership with added capital Profits are shared in the ratio of capital $\times$ time, so add up each partner's month-wise capital over the 12 months. P: $2000\times 6+4000\times 6=12000+24000=36000$. Q: $3000\times 6+\left(3000+y\right)\times 6=18000+18000+6y=36000+6y$. Therefore $\frac{36000}{36000+6y}=\frac{12}{19}$. Cross-multiplying: $36000\times 19=12\left(36000+6y\right)$, i.e. 684000=432000+72y. 72y=252000, so y=3500. Hence, option (c) is the most appropriate answer.

Partnership profit share Profit is divided in the ratio of capital $\times$ number of months for which it worked. P kept ₹20,000 for the whole year: $20000\times 12=240000$. Q joined 4 months late, so his money worked for only 8 months: $30000\times 8=240000$. Ratio of shares =240000:240000=1:1. P's share $=70000\times \frac{1}{2}=35000$, i.e. ₹35,000. Hence, option (c) is the most appropriate answer.

Partnership profit share Each partner's share is proportional to capital $\times$ time. P: $3000\times 6=18000$; Q: $2000\times 12=24000$; R: $4000\times 9=36000$. Ratio =18000:24000:36000=3:4:6, which makes 3+4+6=13 equal parts. P's share $=390000\times \frac{3}{13}=30000\times 3=90000$, i.e. ₹90,000. Hence, option (b) is the most appropriate answer.

Percentage change in area The area of a circle is $A=\pi r^2$, so the area depends on the square of the radius; the actual value 20 cm is not needed. New radius $=r\left(1+\frac{1.6}{100}\right)=1.016r$. New area $=\pi {\left(1.016r\right)}^2=1.032256\pi r^2$. Increase $=\left(1.032256-1\right)\times 100=3.2256\%\approx 3.22\%$. Shortcut: if the radius rises by $x\%$, the area rises by $\left(2x+\frac{x^2}{100}\right)\%=\left(3.2+0.0256\right)\%=3.2256\%$. Hence, option (c) is the most appropriate answer.

Algebraic identities Square the first equation: ${\left(x+y\right)}^2=x^2+2xy+y^2=7^2=49$. Put $x^2+y^2=25$ in it: 25+2xy=49, so 2xy=24 and xy=12. Use the identity $x^4+x^2{y}^2+y^4={\left(x^2+y^2\right)}^2-x^2{y}^2$, since ${\left(x^2+y^2\right)}^2=x^4+2x^2{y}^2+y^4$. Substituting: $={\left(25\right)}^2-{\left(12\right)}^2=625-144=481$. Hence, option (c) is the most appropriate answer.

Square identities The two squares differ only in the middle term: ${\left(4a+3b\right)}^2={\left(4a-3b\right)}^2+4\times \left(4a\right)\left(3b\right)={\left(4a-3b\right)}^2+48ab$. From 24ab=48 we get ab=2, so $48ab=48\times 2=96$. Therefore ${\left(4a+3b\right)}^2=25+96=121$. Required value $={\left(4a+3b\right)}^2+12ab=121+12\times 2=121+24=145$. Hence, option (a) is the most appropriate answer.

Trigonometric identities Identity: ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =1$, i.e. $\left(\mathrm{cosec}\theta -\cot \theta \right)\left(\mathrm{cosec}\theta +\cot \theta \right)=1$. Given $\mathrm{cosec}\theta -\cot \theta =\frac{1}{\sqrt{3}}$, so $\mathrm{cosec}\theta +\cot \theta =\sqrt{3}$. Adding the two: $2\mathrm{cosec}\theta =\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}$, so $\mathrm{cosec}\theta =\frac{2}{\sqrt{3}}$. Subtracting them: $2\cot \theta =\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$, so $\cot \theta =\frac{1}{\sqrt{3}}$. Product $=\frac{2}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{2}{3}$. Hence, option (d) is the most appropriate answer.

Trigonometric identities Square the first: $a^2=25{\cos }^{2}A+120\sin A\cos A+144{\sin }^{2}A$. Square the second: $b^2=25{\sin }^{2}A-120\sin A\cos A+144{\cos }^{2}A$. Add them; the $120\sin A\cos A$ terms cancel: $a^2+b^2=25\left({\cos }^{2}A+{\sin }^{2}A\right)+144\left({\sin }^{2}A+{\cos }^{2}A\right)$. Since ${\sin }^{2}A+{\cos }^{2}A=1$, $a^2+b^2=25+144=169$. So $6\left(a^2+b^2\right)=6\times 169=1014$. Note the pattern: for $a=p\cos A+q\sin A$ and $b=p\sin A-q\cos A$, always $a^2+b^2=p^2+q^2$. Hence, option (d) is the most appropriate answer.

Heights and distances The rope, the flagpole and the ground form a right triangle: the rope is the hypotenuse (30 m) and the angle at Ramesh is ${30}^{\circ }$. Height of the tie point $=30\times \sin {30}^{\circ }=30\times \frac{1}{2}=15$ metres. Horizontal distance $=30\times \cos {30}^{\circ }=30\times \frac{\sqrt{3}}{2}=15\sqrt{3}$. $15\sqrt{3}=15\times 1.732=25.98\approx 26$ metres. So the rope is tied 15 m up the pole and Ramesh stands about 26 m from its base. Hence, option (c) is the most appropriate answer.

Simple and compound interest Simple interest for 5 years at $11\%$ is $5\times 11=55\%$ of the sum, so that amount $=155\%$ of Y=1.55Y. Compound interest at $20\%$ for 2 years multiplies the sum by ${\left(1+\frac{20}{100}\right)}^2={\left(1.2\right)}^2=1.44$, so that amount =1.44Y. The simple-interest amount exceeds the compound-interest amount by 1.55Y-1.44Y=0.11Y. $0.11Y=88\Rightarrow Y=\frac{88}{0.11}=800$. Check: $1.55\times 800=1240$ and $1.44\times 800=1152$, and 1240-1152=88. Hence, option (b) is the most appropriate answer.

CI and SI difference For 2 years the whole gap between compound and simple interest is just the interest earned on the first year's interest. Formula: $\mathrm{Difference}=P{\left(\frac{x}{100}\right)}^2$. $5000\times {\left(\frac{x}{100}\right)}^2=800$. ${\left(\frac{x}{100}\right)}^2=\frac{800}{5000}=0.16$, so $\frac{x}{100}=0.4$ and x=40. Check: CI $=5000\left({\left(1.4\right)}^2-1\right)=5000\times 0.96=4800$ and SI $=5000\times 0.4\times 2=4000$; the difference is ₹800. Hence, option (c) is the most appropriate answer.

Interest and equations Nikhil's interest is simple: $\frac{20000\times 20\times 2}{100}=8000$, i.e. ₹8,000. This is given as 3y-700, so $3y-700=8000\Rightarrow 3y=8700\Rightarrow y=2900$. So Neeraj invested ₹2,900 at $11\%$ compound interest for 3 years. Growth factor $={\left(1.11\right)}^3=1.367631$, so the effective interest for 3 years is $36.7631\%$. Interest $=2900\times 0.367631=1066.13\approx 1066$. Hence, option (d) is the most appropriate answer.

Ratio and division Let the numbers be 3x, 6x and 4x. Their sum: 3x+6x+4x=13x=169. $x=\frac{169}{13}=13$. The largest ratio term is 6, so the largest number $=6x=6\times 13=78$. Hence, option (b) is the most appropriate answer.

Mixture and alligation Take the ratio as actual quantities: 3 kg from box X and 7 kg from box Y, a total of 10 kg. Total cost $=3\times 400+7\times 500=1200+3500=4700$. Price per kg of the mixture $=\frac{4700}{10}=470$. By alligation the mean lies nearer the costlier coffee, because more of it (7 parts) is used. Hence, option (c) is the most appropriate answer.

Proportion in recipes The recipe fixes sugar : water =3:4, whatever quantity of juice is made. Water used is 16 cups, and water is 4 parts, so 1 part $=\frac{16}{4}=4$ cups. Sugar is 3 parts $=3\times 4=12$ cups. In proportion form: $\frac{3}{4}=\frac{s}{16}\Rightarrow s=\frac{3\times 16}{4}=12$. Hence, option (c) is the most appropriate answer.

Trains crossing each other Speed of train A =72 km/h $=72\times \frac{5}{18}=20$ m/s. Crossing a pole means covering only its own length: length of A $=20\times 10=200$ m. Lengths are in the ratio 4:3, so length of B $=200\times \frac{3}{4}=150$ m. In opposite directions the relative speed is the sum of the speeds, and the trains together cover 200+150=350 m in 10 seconds. $20+v=\frac{350}{10}=35$, so v=35-20=15 m/s. Hence, option (b) is the most appropriate answer.

Relative speed and chase Both run in the same direction, so the gap closes at the relative speed =32-24=8 m/s. The gap to be closed is 1,000 m. Time $=\frac{\mathrm{gap}}{\mathrm{relative}\mathrm{speed}}=\frac{1000}{8}=125$ seconds. Hence, option (c) is the most appropriate answer.

Successive discount and tax Let the original price be P. After a 10% discount the price is 0.9P; after a further 12.5% discount it is $0.9P\times 0.875=0.7875P$. Adding 12% tax multiplies it by 1.12: $0.7875P\times 1.12=0.882P$. $0.882P=748\Rightarrow P=\frac{748}{0.882}=848.07$ (to two decimal places). Rounded to the nearest rupee the original price is ₹848. Hence, option (d) is the most appropriate answer.