12 November 2025 · Shift 2

Algebraic identity Use the identity ${\left(A+B+C\right)}^2=A^2+B^2+C^2+2\left(AB+BC+CA\right)$. Substituting A+B+C=0 and $A^2+B^2+C^2=64$: $0^2=64+2\left(AB+BC+CA\right)$ 2(AB+BC+CA)=-64 AB+BC+CA=-32 Hence option (a) is correct.

Partnership profit share Take the capitals as 4, 5 and 6 units; the whole period is 2 years, i.e. 24 months. P keeps 4 units for the first 8 months and then $4\times 1.5=6$ units for the remaining 16 months. P's capital-months $=4\times 8+6\times 16=32+96=128$. Q's capital-months $=5\times 24=120$ and R's $=6\times 24=144$. So the profit is shared in the ratio 128:120:144, whose parts total 128+120+144=392. P's share $=\frac{128}{392}\times 140000=45714.28$, i.e. Rs. 45,714.28. Hence option (d) is correct.

Trigonometric identity Since $\sin A=\frac{15}{17}$, we get $\cos A=\sqrt{1-\frac{225}{289}}=\frac{8}{17}$. Since $\cos B=\frac{8}{17}$, we get $\sin B=\sqrt{1-\frac{64}{289}}=\frac{15}{17}$. The expression $\cos A\cos B+\sin A\sin B$ is the expansion of $\cos \left(A-B\right)$. $=\frac{8}{17}\times \frac{8}{17}+\frac{15}{17}\times \frac{15}{17}$ $=\frac{64+225}{289}=\frac{289}{289}=1$ Hence option (c) is correct.

Profit percentage Cost price of one apple $=\frac{80}{6}=\frac{40}{3}$. Selling price of one apple $=\frac{90}{4}=\frac{45}{2}$. Profit on one apple $=\frac{45}{2}-\frac{40}{3}=\frac{135-80}{6}=\frac{55}{6}$. Profit % $=\frac{55}{6}\times \frac{3}{40}\times 100=\frac{55}{80}\times 100=68.75$. The selling price is above the cost price, so it is a profit of 68.75%. Hence option (d) is correct.

Ratio and division of money Let P have P and Q have Q, so that P+Q=12000. It is given that $\frac{5}{11}P=\frac{4}{7}Q$. $\frac{P}{Q}=\frac{4}{7}\times \frac{11}{5}=\frac{44}{35}$ So P : Q = 44 : 35 and the total number of parts is 44+35=79. P's amount $=\frac{44}{79}\times 12000=\frac{528000}{79}=6683.5$ (approx.), i.e. Rs. 6683.5. Hence option (a) is correct.

Compound interest doubling Under compound interest the amount is multiplied by the same factor in every equal interval of time. The amount becomes 2 times in 8 years, so in 8n years it becomes $2^n$ times. We need 16 times, and $16=2^4$, so n=4. Required time $=4\times 8=32$ years. Hence option (c) is correct.

Ratio of speeds Speed of the bike $=\frac{180}{6}=30$ km/h. Let the speeds be 5x and 7x; then 5x=30, so x=6. Speed of the car $=7\times 6=42$ km/h. Distance covered by the car in 8 hours $=42\times 8=336$ km. Hence option (b) is correct.

Average after replacement Total marks of the 40 students $=40\times 66=2640$. Replacing the student who scored 50 by one who scored 90 changes the total by 90-50=40. New total =2640+40=2680. New average $=\frac{2680}{40}=67$. Hence option (a) is correct.

LCM with a remainder A number that leaves the same remainder 5 with each of the divisors must be a common multiple of them increased by 5. 11=11, $15=3\times 5$, $24=2^3\times 3$, so LCM $=2^3\times 3\times 5\times 11=1320$. The smallest such number =1320+5=1325. Check: $1325=11\times 120+5=15\times 88+5=24\times 55+5$. Hence option (b) is correct.

Successive percentage increase Take the original price as ₹100. After the first rise the price is $100\times \frac{130}{100}=130$. After the second rise it is $130\times \frac{120}{100}=156$. Net increase =156-100=56 on 100, that is 56%. Shortcut: $30+20+\frac{30\times 20}{100}=56$. Hence option (b) is correct.

Time and work Take the total work as the LCM of 20 and 30, i.e. 60 units. One day's work of $P=\frac{60}{20}=3$ units and of $Q=\frac{60}{30}=2$ units. Work done by P alone in 8 days $=8\times 3=24$ units. Remaining work =60-24=36 units. After Q joins, they together do 3+2=5 units in a day. Required time $=\frac{36}{5}=7.2$ days. Hence the correct option is (a).

Discount and profit Selling price = marked price $\times \left(1-\frac{25}{100}\right)$ $=8000\times \frac{3}{4}=6000$, i.e. the fan is sold for ₹6,000. A profit of 20% means selling price = cost price $\times \frac{120}{100}$. $6000=CP\times \frac{6}{5}$ $CP=6000\times \frac{5}{6}=5000$ So the fan cost the retailer ₹5,000. Hence the correct option is (d).

Simple interest In simple interest the same amount of interest is added every year, so the rise in the amount is proportional to the time. Interest of 9-7=2 years =9880-8840=1040. Interest of 1 year $=\frac{1040}{2}=520$, i.e. ₹520. Interest of 7 years $=7\times 520=3640$, so principal =8840-3640=5200, i.e. ₹5,200. Rate $=\frac{SI\times 100}{P\times T}=\frac{520\times 100}{5200\times 1}=10\%$. Hence the correct option is (c).

Infinite nested radical Let $x=\sqrt{15+\sqrt{15+\sqrt{15+\dots }}}$. The expression sitting under the first root is the very same infinite radical, so it is again x. $x=\sqrt{15+x}$ Squaring: $x^2=15+x$, i.e. $x^2-x-15=0$. $x=\frac{1+\sqrt{1+60}}{2}=\frac{1+\sqrt{61}}{2}$ (the negative root is rejected because a square root is non-negative). $\sqrt{61}\approx 7.81$, so $x\approx \frac{1+7.81}{2}\approx 4.4$, a value lying between 4 and 5. Of the four choices only 5 is near this value (3 is too small, 8 and 15 far too large), so the value of the radical is taken as 5. Hence the correct option is (a).

Mixture - replacement Acid at the start $=\frac{80}{100}\times 120=96$ litres. The 16 litres taken out is ordinary mixture, so it carries away acid in the same ratio: $\frac{80}{100}\times 16=12.8$ litres. Acid left =96-12.8=83.2 litres. The 16 litres put back is pure water, so it adds no acid, and the drum is full again with 120 litres. New percentage of acid $=\frac{83.2}{120}\times 100=69.33\%$. Hence the correct option is (c).

Chord of a circle The perpendicular drawn from the centre of a circle to a chord bisects that chord. So $AM=MB=\frac{24}{2}=12$ cm and OM=9 cm. In the right triangle OMB, by Pythagoras theorem $O{B}^2=O{M}^2+M{B}^2$. $r^2=9^2+{12}^2=81+144=225$ $r=\sqrt{225}=15$ cm. Hence the correct option is (a).

Volume of spheres Melting and recasting does not change the volume of metal, so the big sphere's volume is shared equally by the small spheres. Number of small spheres $=\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}={\left(\frac{R}{r}\right)}^3$ $={\left(\frac{8}{2}\right)}^3=4^3=64$ Hence the correct option is (c).

Sphere recast into cylinder Let the radius of the sphere be r; the cylinder is given the same radius r and height h. Recasting keeps the volume unchanged: $\frac{4}{3}\pi r^3=\pi r^{2}h$ $h=\frac{\frac{4}{3}\pi r^3}{\pi r^2}=\frac{4r}{3}$ Required fraction $=\frac{h}{r}=\frac{4}{3}$. Hence the correct option is (a).

Decimal simplification Work out each part separately and then apply BODMAS. ${\left(4.5\right)}^2=4.5\times 4.5=20.25$ $\sqrt{2.25}=\sqrt{\frac{225}{100}}=\frac{15}{10}=1.5$ $6.5÷1.3=\frac{65}{13}=5$ So the expression =20.25+1.5-5=16.75. Hence the correct option is (d).

Rationalising surds Rationalise each term by multiplying with the conjugate of its denominator. $\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{6}\right)}^2}=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}$ $\frac{1}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}-\sqrt{6}}{7-6}=\sqrt{7}-\sqrt{6}$ Adding: $\left(\sqrt{7}+\sqrt{6}\right)+\left(\sqrt{7}-\sqrt{6}\right)=2\sqrt{7}$. Hence the correct option is (a).

Equation of a line Point-slope form: $y-y_1=m\left(x-x_1\right)$. Here m=-3, $x_1=6$ and $y_1=8$. y-8=-3(x-6) y-8=-3x+18 y=-3x+26 Check: putting x=6 gives y=-18+26=8, so the point (6,8) does lie on it. Hence the correct option is (a).

Discriminant of a quadratic Roots are real and distinct only when the discriminant $D=b^2-4ac>0$. Comparing $y^2+ky+36=0$ with $a{y}^2+by+c=0$: a=1, b=k, c=36. $k^2-4\times 1\times 36>0$ $k^2>144$ k>12 or k<-12. Among the options only k>12 satisfies this (k<-11 fails, for example k=-11.5 gives D<0). Hence the correct option is (a).

Trigonometric identities Use $1+{\cot }^{2}x={\mathrm{cosec}}^{2}x$ and $1+{\tan }^{2}x={\sec }^{2}x$. First factor $=\frac{{\mathrm{cosec}}^{2}x}{{\sec }^{2}x}=\frac{1/{\sin }^{2}x}{1/{\cos }^{2}x}=\frac{{\cos }^{2}x}{{\sin }^{2}x}={\cot }^{2}x$. For the second factor multiply numerator and denominator by $\left(\mathrm{cosec}x-\cot x\right)$: $\frac{{\left(\mathrm{cosec}x-\cot x\right)}^2}{{\mathrm{cosec}}^{2}x-{\cot }^{2}x}={\left(\mathrm{cosec}x-\cot x\right)}^2$, since ${\mathrm{cosec}}^{2}x-{\cot }^{2}x=1$. ${\left(\mathrm{cosec}x-\cot x\right)}^2={\left(\frac{1-\cos x}{\sin x}\right)}^2=\frac{{\left(1-\cos x\right)}^2}{1-{\cos }^{2}x}=\frac{1-\cos x}{1+\cos x}$ Multiplying the two factors: $\frac{{\cot }^{2}x\left(1-\cos x\right)}{1+\cos x}$. Hence the correct option is (d).

Fundamental trigonometric identity Identity: ${\sin }^2\theta +{\cos }^2\theta =1$, so ${\sin }^2\theta =1-{\cos }^2\theta$. $\cos \theta =\frac{8}{27}\Rightarrow {\cos }^2\theta =\frac{64}{729}$ ${\sin }^2\theta =1-\frac{64}{729}=\frac{729-64}{729}=\frac{665}{729}$ (Option (b) is impossible anyway, since ${\sin }^2\theta$ can never exceed 1.) Hence the correct option is (a).

Trigonometric ratios of 45 $\cot B=1\Rightarrow \tan B=1\Rightarrow B={45}^{\circ }$. $\sin {45}^{\circ }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$ $2\sin B\cos B=2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=2\times \frac{1}{2}=1$ The same result follows from the double-angle form $2\sin B\cos B=\sin 2B=\sin {90}^{\circ }=1$. Hence the correct option is (c).