17 November 2025 · Shift 1

Overall profit percentage All three fans are identical, so the actual figure ₹5,320 is not needed — take the cost price of one fan as ₹100, making the total cost price ₹300. Two fans sold at a gain of $16\%$: their selling price $=2\times 116=232$. The third fan sold at a loss of $12\%$: its selling price =88. Total selling price =232+88=320, so profit =320-300=20 on a cost of 300. Profit $\%=\frac{20}{300}\times 100=\frac{20}{3}=6\frac{2}{3}\%\approx 6.67\%$ profit. Hence, option (c) is the most appropriate answer.

Successive discount Two successive discounts are applied one after the other, so they are multiplied — they are never simply added. Price after the first discount $=12000\times \frac{80}{100}=9600$. Price after the extra $4\%$ cash discount $=9600\times \frac{96}{100}=9216$. So Gita pays ₹9,216. Hence, option (b) is the most appropriate answer.

Marked price from discount Discount is always reckoned on the marked price, so the selling price is $\left(100-8\right)\%=92\%$ of the marked price. SP = MP $\times \frac{92}{100}$, i.e. 2668= MP $\times \frac{92}{100}$. MP $=\frac{2668\times 100}{92}=\frac{266800}{92}=2900$. Check: $8\%$ of 2900=232 and 2900-232=2668, which matches the given selling price. Hence, option (a) is the most appropriate answer.

Average of two groups Total age of the 105 old students $=105\times 30=3150$ years. Let the new batch have x students; their total age =28x years. The average of all of them together is 29.5, so $\frac{3150+28x}{105+x}=29.5$. $3150+28x=29.5\times 105+29.5x=3097.5+29.5x$ $3150-3097.5=29.5x-28x\Rightarrow 52.5=1.5x$ $x=\frac{52.5}{1.5}=35$ students. Hence, option (b) is the most appropriate answer.

Work and daily output Days on which work is actually possible $=540\times \frac{70}{100}=378$ days. The whole project must still be finished on time, so the work planned for 540 days has to be done in 378 days. (daily output) $\times$ (number of days) is constant, so new output : old output =540:378=10:7. Required increase $=\frac{10-7}{7}\times 100=\frac{300}{7}=42\frac{6}{7}\%$. Hence, option (a) is the most appropriate answer.

Percentage and sets Students who like Mathematics or Physics $=60\%+50\%-20\%=90\%$ — the common $20\%$ is subtracted so that it is not counted twice. Remaining students $=100\%-90\%=10\%$ of $500=\frac{10}{100}\times 500=50$. Of these 50, those preferring Computer Science $=\frac{30}{100}\times 50=15$. Students with no preference =50-15=35. Hence, option (a) is the most appropriate answer.

Perfect squares A number is a perfect square only if its square root is a whole number. $1600=40\times 40={40}^2$, $625=25\times 25={25}^2$ and $1225=35\times 35={35}^2$. For 840, ${28}^2=784$ and ${29}^2=841$, so 840 lies between two consecutive squares and cannot be a perfect square. Hence, option (d) is the most appropriate answer.

Partnership profit share Both partners invest for the same period, so the profit is divided in the ratio of their investments, i.e. 5:4. Q's share is 4 parts of the profit, and these 4 parts =₹1200. One part $=\frac{1200}{4}=₹300$. Total profit $=\left(5+4\right)\times 300=9\times 300=₹2700$. Hence, option (a) is the most appropriate answer.

Share of profit in percent All three invest for the same period of one year, so the profit is shared in the ratio of the capitals. 16000:20000:24000=16:20:24=4:5:6, so the profit is cut into 4+5+6=15 equal parts. R gets 6 of these 15 parts $=\frac{6}{15}\times 100=40\%$ of the total profit. Hence, option (c) is the most appropriate answer.

Partnership profit difference The time is the same for both, so the profit is shared in the ratio of the capitals: 25000:33000=25:33. Q gets 33-25=8 parts more than P, and this extra amount is given as ₹1000. One part $=\frac{1000}{8}=₹125$. Total profit $=\left(25+33\right)\times 125=58\times 125=₹7250$. Hence, option (a) is the most appropriate answer.

Area of circle and path The area of a circle is $\pi r^2$, so each radius comes from the area that belongs to it. Inner circle (field only): $\pi r^2=314\Rightarrow r^2=\frac{314}{3.1416}=99.95$ $r=\sqrt{99.95}\approx 9.99$ m Outer circle (field with the path): $\pi R^2=452.16\Rightarrow R^2=\frac{452.16}{3.1416}=143.93$ $R=\sqrt{143.93}\approx 11.99$ m Check: $R-r\approx 11.99-9.99=2$ m, which is exactly the given width of the path. Hence the correct option is (b).

Algebraic identity, cubes Identity: $a^3-b^3={\left(a-b\right)}^3+3ab\left(a-b\right)$. Substituting a-b=6 and $a^3-b^3=540$: $540=6^3+3ab\times 6$ 540=216+18ab 18ab=324 ab=18 Hence the correct option is (c).

Algebraic identities Square the first condition: ${\left(x+y\right)}^2={10}^2\Rightarrow x^2+y^2+2xy=100$ Using $x^2+y^2=58$: $2xy=100-58=42\Rightarrow xy=21$ For the required expression use $x^4+x^2{y}^2+y^4={\left(x^2+y^2\right)}^2-x^2{y}^2$. $={\left(58\right)}^2-{\left(21\right)}^2$ =3364-441=2923 Hence the correct option is (a).

Trigonometric identity Given $\cos x+{\cos }^{2}x=1$, so $\cos x=1-{\cos }^{2}x$. But $1-{\cos }^{2}x={\sin }^{2}x$, hence ${\sin }^{2}x=\cos x$. Squaring both sides: ${\sin }^{4}x={\cos }^{2}x$ Adding the two results: ${\sin }^{2}x+{\sin }^{4}x=\cos x+{\cos }^{2}x$ The right-hand side is exactly the given expression, whose value is 1. $\therefore {\sin }^{2}x+{\sin }^{4}x=1$ Hence the correct option is (c).

Trigonometric ratios, angles Complementary angles: $\sin \left({90}^{\circ }-q\right)=\cos q$, so the equation becomes $2\cos q\times \cos q=\frac{1}{2}$. $2{\cos }^{2}q=\frac{1}{2}\Rightarrow {\cos }^{2}q=\frac{1}{4}$ $\cos q=\frac{1}{2}$ (positive, since $0^{\circ }<q<{90}^{\circ }$), so $q={60}^{\circ }$. ${\tan }^{2}2q={\tan }^2{120}^{\circ }={\left(-\sqrt{3}\right)}^2=3$ ${\sin }^{2}3q={\sin }^2{180}^{\circ }=0^2=0$ ${\tan }^{2}2q+{\sin }^{2}3q=3+0=3$ Hence the correct option is (b).

Height and distance The ladder is the hypotenuse of a right triangle, and $\sin x=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$. Since $\sin x=\frac{28}{53}$ and the hypotenuse is 53 m, the height reached on the wall is 28 m. By Pythagoras' theorem, base $=\sqrt{{53}^2-{28}^2}$ $=\sqrt{2809-784}=\sqrt{2025}$ =45 So the foot of the ladder is 45 metres from the wall. Hence the correct option is (a).

Simple and compound interest Scheme P is simple interest: $SI=\frac{P\times R\times T}{100}=\frac{14000\times 12\times 5}{100}=8400$ Amount received from P = 14000 + 8400 = ₹22,400. This ₹22,400 becomes the principal of scheme Q, at 10% p.a. compounded annually for 2 years. $A=P{\left(1+\frac{R}{100}\right)}^n=22400\times {\left(\frac{11}{10}\right)}^2$ $A=22400\times \frac{121}{100}=27104$ So the amount received from scheme Q is ₹27,104. Hence the correct option is (a).

CI minus SI, two years For 2 years at the same rate, the compound interest exceeds the simple interest by $P{\left(\frac{R}{100}\right)}^2$. $=22000\times {\left(\frac{25}{100}\right)}^2=22000\times \frac{1}{16}$ =1375 So the difference between the two interests is ₹1,375. Hence the correct option is (a).

Half-yearly compound interest Compounding every 6 months halves the rate and doubles the count of periods: rate per period $=\frac{20}{2}=10$ per cent, periods $=\frac{18}{6}=3$. $A=P{\left(1+\frac{10}{100}\right)}^3=70000\times {\left(\frac{11}{10}\right)}^3$ $A=70000\times \frac{1331}{1000}=93170$ Interest =93170-70000=23170 So the interest earned is ₹23,170. Hence the correct option is (a).

Ratio of coins The ratio counts the coins, not the money, so let the numbers be 8x, 4x and 3x. Total value =5(8x)+2(4x)+1(3x) =40x+8x+3x=51x $51x=1122\Rightarrow x=22$ Number of ₹2 coins $=4x=4\times 22=88$ Hence the correct option is (c).

Mixture replacement, ratio Let the initial quantities be A=16k and B=14k litres, so the mixture totals 30k litres. The 60 litres drawn out keep the same ratio: A removed $=60\times \frac{16}{30}=32$, B removed $=60\times \frac{14}{30}=28$. After the removal and the addition of 60 litres of B: A=16k-32 and B=14k-28+60=14k+32. $\frac{16k-32}{14k+32}=\frac{9}{11}$ 11(16k-32)=9(14k+32) 176k-352=126k+288 $50k=640\Rightarrow k=12.8$ Initial quantity of B $=14k=14\times 12.8=179.2$ litres. Hence the correct option is (a).

Alligation, mixtures Take the two mixtures as 4 parts and 5 parts, so the final mixture is 9 parts. The methanol must balance: 4(24)+5(b)=9(20) 96+5b=180 5b=84 b=16.8 The alligation cross gives the same relation, since 20 lies between b and 24. Hence the correct option is (b).

Average speed Distance by car $=\frac{60}{100}\times 300=180$ km, so distance by bike =300-180=120 km. Time by car $=\frac{180}{60}=3$ hours Time by bike $=\frac{120}{40}=3$ hours Average speed = (total distance) ÷ (total time) $=\frac{300}{3+3}$ $=\frac{300}{6}=50$ So the average speed for the whole journey is 50 km/h. Hence the correct option is (b).

Counting perfect squares We need every integer n with $15000\le n^2\le 25000$. ${122}^2=14884$ is below the range and ${123}^2=15129$ is inside, so the smallest such n is 123. ${158}^2=24964$ is inside and ${159}^2=25281$ is above the range, so the largest such n is 158. Count of integers from 123 to 158 =158-123+1 =36 Hence the correct option is (c).

Average speed, unknown leg Total time = (total distance) ÷ (average speed) $=\frac{240}{48}=5$ hours. Time for the first 80 km $=\frac{80}{40}=2$ hours Time for the next 80 km $=\frac{80}{80}=1$ hour Remaining distance =240-160=80 km and remaining time =5-3=2 hours. $x=\frac{80}{2}=40$ So the speed over the last stretch is 40 km/h. Hence the correct option is (c).