20 November 2025 · Shift 2

Product of conjugate surds The two brackets are conjugates, so use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$. Here $a=\sqrt{3}$ and $b=\sqrt{2}$. $\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)={\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2$ =3-2=1 Hence option (b) is correct.

Algebraic identity with surds Rather than squaring each surd, use $a^2+b^2={\left(a+b\right)}^2-2ab$. $a+b=\left(2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)=4$ $ab=\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)=2^2-{\left(\sqrt{5}\right)}^2=4-5=-1$ $a^2+b^2={\left(4\right)}^2-2\times \left(-1\right)$ =16+2=18 Hence option (d) is correct.

Partnership profit ratio In a partnership the profit is divided in the ratio of capital multiplied by the time for which it stayed in the business. P keeps ₹14,000 for the whole year: $14000\times 12=168000$ Q keeps ₹20,000 for only 3 months: $20000\times 3=60000$ Profit ratio =168000:60000=168:60 Dividing both parts by 12 gives 14:5. Hence option (b) is correct.

Ratio of ages Let the present ages of P and Q be 5x and 7x years. After 6 years the ages are 5x+6 and 7x+6, and their ratio is 7:9. $\frac{5x+6}{7x+6}=\frac{7}{9}$ 9(5x+6)=7(7x+6) 45x+54=49x+42 $49x-45x=54-42\Rightarrow 4x=12\Rightarrow x=3$ P's present age $=5x=5\times 3=15$ years. Hence option (b) is correct.

Average of numbers The average is the sum of the numbers divided by how many numbers there are. Average $=\frac{4+6+8}{3}=\frac{18}{3}=6$ This average is then increased by 4, so the new average =6+4=10. Hence option (c) is correct.

Percentage decrease A percentage change is always taken on the ORIGINAL value, which here is Rs. 300. Decrease =300-250=50 Percentage decrease $=\frac{50}{300}\times 100$ $=\frac{50}{3}=16.67\%$ Hence option (d) is correct.

Percentage more or less The two comparisons use different bases, so fix the larger income first. Let Q's income be 100 units; P's income is 30% less, so P =100-30=70 units. Q exceeds P by 100-70=30 units, and this excess must now be compared with P's 70 units. Required percentage $=\frac{30}{70}\times 100$ $=\frac{300}{7}=42.85\%$ Hence option (c) is correct.

Selling price at a loss When there is a loss, $SP=CP\times \frac{100-\mathrm{loss}\%}{100}$. $SP=1200\times \frac{100-15}{100}$ $=1200\times \frac{85}{100}=12\times 85$ =1020 So the shoes were sold for ₹1,020. Hence option (c) is correct.

Equal SP profit and loss When two articles are sold at the SAME price, one at $x\%$ profit and the other at $x\%$ loss, the deal always ends in a loss of $\frac{x^2}{100}\%$. Here x=25, so loss $=\frac{{25}^2}{100}=\frac{625}{100}=6.25\%$. Verify it: cost of the first $=800\times \frac{100}{125}=640$ and cost of the second $=800\times \frac{100}{75}=1066.67$. Total cost =1706.67 against total sale =1600, so the loss is 106.67. Loss$\%=\frac{106.67}{1706.67}\times 100=6.25\%$, a loss as expected. Hence option (a) is correct.

Discount percentage Discount is always calculated on the marked price, not on the selling price. Discount =1800-1260=540 Discount$\%=\frac{540}{1800}\times 100$ $=\frac{3}{10}\times 100=30\%$ Hence option (b) is correct.

Mixture and ratio In 64 L the juice : water ratio is 5:3, so juice $=\frac{5}{8}\times 64=40$ L and water $=\frac{3}{8}\times 64=24$ L. Only juice is poured in, so the quantity of water stays 24 L throughout. Let x L of juice be added. Then $\frac{40+x}{24}=\frac{7}{1}$. $40+x=7\times 24=168$ x=168-40=128 Hence option (d) is correct.

Cylinder and cone volume Volume of a cylinder $=\pi r^{2}h$. Volume of a cone with the same base radius and the same height $=\frac{1}{3}\pi r^{2}h$. Required ratio $=\pi r^{2}h:\frac{1}{3}\pi r^{2}h$ $=1:\frac{1}{3}=3:1$ A cone is exactly one-third of the cylinder that just contains it, so the ratio never depends on the actual values of r and h. Hence option (b) is correct.

Interior angles of polygon For any polygon of n sides, sum of the interior angles $=\left(n-2\right)\times {180}^{\circ }$. The reason: diagonals from one vertex cut the polygon into (n-2) triangles. A hexagon has n=6, so the sum $=\left(6-2\right)\times {180}^{\circ }$ $=4\times {180}^{\circ }={720}^{\circ }$ The circular junction around the hexagon does not affect its angles. Hence option (b) is correct.

Regular octagon angle Sum of the interior angles of an n-sided polygon $=\left(n-2\right)\times {180}^{\circ }$. In a regular polygon all the angles are equal, so each angle $=\frac{\left(n-2\right)\times {180}^{\circ }}{n}$. For an octagon n=8: each angle $=\frac{\left(8-2\right)\times {180}^{\circ }}{8}$ $=\frac{{1080}^{\circ }}{8}={135}^{\circ }$ Hence option (a) is correct.

Melting and recasting Volume of the cylinder $=\pi r^{2}h=\pi \times 4^2\times 21=336\pi$ cm³. Volume of one hemispherical ball $=\frac{2}{3}\pi r^3=\frac{2}{3}\pi \times 2^3=\frac{16\pi }{3}$ cm³. Melting only reshapes the metal, so the total volume is unchanged. Number of balls $=\frac{336\pi }{\frac{16\pi }{3}}=\frac{336\times 3}{16}$ =63 Hence option (c) is correct.

Curved surface of cylinder Only the curved (lateral) face is wrapped, so the two circular ends are not counted. Cut the curved face along a vertical line and unroll it: it becomes a rectangle of length $2\pi r$ and breadth h. Lateral surface area $=2\pi rh=2\times \pi \times 10\times 18$ $=360\pi$ cm² Hence option (b) is correct.

Surface area of prism Check the base: $5^2+{12}^2=25+144=169={13}^2$, so the triangle is right-angled with legs 5 cm and 12 cm. Area of the top triangular face $=\frac{1}{2}\times 5\times 12=30$ cm² (the bottom face is not painted). Perimeter of the triangular base =5+12+13=30 cm. Area of the three rectangular faces = perimeter $\times$ length $=30\times 10=300$ cm². Total area to be painted =30+300=330 cm² Hence option (d) is correct.

Volume of hemisphere Volume of a hemisphere $=\frac{2}{3}\pi r^3$. $\frac{2}{3}\pi r^3=144\pi$ $r^3=\frac{144\times 3}{2}=216$ $r=\sqrt[3]{216}=6$ So the radius of the tank is 6 m. Hence option (a) is correct.

Recurring decimal to fraction Let $x=7.181818\dots$ Two digits repeat, so multiply by 100: $100x=718.181818\dots$ Subtracting removes the endless tail: $100x-x=718.1818\dots -7.1818\dots$ 99x=711 $x=\frac{711}{99}=\frac{79}{11}$ (dividing numerator and denominator by 9) Quick check: $\frac{79}{11}=7.1818\dots$, which matches the given score. Hence option (c) is correct.

Complementary angle identity Use the complementary-angle rule $\cos \left({90}^{\circ }-\theta \right)=\sin \theta$. So the required product $=\sin \theta \times \sin \theta ={\sin }^2\theta$. $\tan \theta =\frac{3}{4}$ means perpendicular : base =3:4, so hypotenuse $=\sqrt{3^2+4^2}=\sqrt{25}=5$. $\sin \theta =\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}=\frac{3}{5}$ ${\sin }^2\theta ={\left(\frac{3}{5}\right)}^2=\frac{9}{25}$ Hence option (b) is correct.

Counting perfect squares A perfect square lying between 30 and 900 is $n^2$ with $30<n^2<900$. Smallest one: $5^2=25$ is below 30, while $6^2=36$ is above 30, so n starts at 6. Largest one: ${30}^2=900$ is not less than 900, while ${29}^2=841$ is, so n ends at 29. Count of values of n =29-6+1=24 Hence option (c) is correct.

Simple interest Simple interest $=\frac{P\times R\times T}{100}$, where P=8000, R=5 and T=4. $SI=\frac{8000\times 5\times 4}{100}$ $=\frac{160000}{100}=1600$ Amount = principal + interest = ₹8,000 + ₹1,600 = ₹9,600 Hence option (a) is correct.

Sum of cubes identity Write a=545 and b=315. The expression is $\frac{a^3+b^3}{a^2-ab+b^2}$. Identity: $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$. $\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}=a+b$ =545+315=860 Whenever the numerator is a sum of cubes and the denominator is exactly $a^2-ab+b^2$, the answer is simply a+b. Hence option (a) is correct.

Up-and-down series sum The series climbs from 1 up to 31 and then comes back down to 1. For such a series, $1+2+\cdots +n+\cdots +2+1=n^2$, where n is the peak term. Here the peak is n=31, so the sum $={31}^2=961$. Check: $2\left(1+2+\cdots +30\right)+31=2\times \frac{30\times 31}{2}+31=930+31=961$ Hence option (b) is correct.

Trigonometric simplification Split the single fraction over the common denominator: $\frac{1+\sin x}{\cos x}=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$. Now use the basic ratios: $\frac{1}{\cos x}=\sec x$ and $\frac{\sin x}{\cos x}=\tan x$. So the expression $=\sec x+\tan x$. Option (a) fails because $\cot x=\frac{\cos x}{\sin x}$, not $\frac{\sin x}{\cos x}$. Hence option (b) is correct.