SSC Selection Post 2026 Graduate 22 Sep Shift 1st

Solution: \(\frac{x}{16}\times\frac{x}{64}=1\Rightarrow\frac{x^2}{1024}=1\Rightarrow x^2=1024\Rightarrow x=\pm32\). Among the given options, the correct value is 32. Trick: Multiply the denominators and take the square root. The positive value is 32.

Solution: Let a = x, b = 2x, and c = 3x. Substituting, \(3x+2(2x)+3x=30\Rightarrow10x=30\Rightarrow x=3\). Therefore, \(c=3x=9\). Trick: Express each variable using a common multiplier of the given ratio.

Solution: Profit is proportional to capital × time. Therefore, the capital ratio is \(\frac{5}{14}:\frac{7}{8}:\frac{8}{7}\). Multiplying each term by 56 gives \(20:49:64\). Trick: Divide each profit share by the corresponding investment period, then simplify the ratio.

Solution: Let Q = 100. Then P = 125. The sum P + Q = 225. Since R is 40% less than this sum, \(R=225\times0.60=135\). R is 10 more than P. Percentage increase \(=\frac{135-125}{125}\times100=8\%\). Trick: Assume Q = 100, calculate P and R, then compare R with P.

Solution: For equal distances, average speed \(=\frac{2uv}{u+v}\). Therefore, average speed \(=\frac{2\times30\times90}{30+90}=\frac{5400}{120}=45\) km/h. Trick: For a round trip over equal distances, use the harmonic mean of the two speeds, not their arithmetic mean.

Solution: Using \(A=P(1+\frac{r}{100})^n\), we get \(1210=1000(1+\frac{r}{100})^2\). Thus, \((1+\frac{r}{100})^2=1.21\Rightarrow1+\frac{r}{100}=1.1\Rightarrow r=10\%\). Trick: The compound growth factor over two years is 1.21, whose square root is 1.1.

Solution: Use \(a^3-b^3=(a-b)^3+3ab(a-b)\). Taking \(a=k\) and \(b=\frac1k\), we have \(ab=1\). Therefore, \(k^3-\frac1{k^3}=5^3+3\times1\times5=125+15=140\). Trick: Use the cube identity and the fact that \(k\times\frac1k=1\).

Solution: The number of diagonals of an n-sided polygon is \(\frac{n(n-3)}{2}\). Hence, \(\frac{n(n-3)}{2}=170\Rightarrow n(n-3)=340\). Since \(20\times17=340\), \(n=20\). Each interior angle \(=\frac{(n-2)\times180^\circ}{n}=\frac{18\times180^\circ}{20}=162^\circ\). Trick: Find the number of sides from the diagonal formula, then apply the interior-angle formula.

Solution: Let the principal be 4x and the amount after 5 years be 7x. Interest for 5 years \(=7x-4x=3x\), which is \(\frac34\) of the principal. The annual simple-interest rate is therefore \(\frac{3/4}{5}\times100=15\%\). For the principal-to-interest ratio to be 5:6, interest must equal \(\frac65P\), or 120% of the principal. Time \(=\frac{120}{15}=8\) years. Trick: First calculate the annual simple-interest rate from the first ratio, then calculate the time needed to earn 120% interest.

Solution: The area of the full circle is \(\pi r^2=\frac{22}{7}\times21^2=1386\) cm². The sector is \(\frac{231}{1386}=\frac16\) of the circle. Therefore, its central angle \(=\frac16\times360^\circ=60^\circ\). Trick: Find what fraction of the full circle the sector area represents, then multiply that fraction by 360°.

Solution: Let the cost price be x. The original selling price is \(1.10x\), and the increased selling price is \(1.125x\). Their difference is ₹4, so \(1.125x-1.10x=4\Rightarrow0.025x=4\Rightarrow x=160\). The cost price is ₹160. Trick: The increase in profit is 2.5% of the cost price; equate 2.5% of the cost price to ₹4.

Solution: Use the identity \(\operatorname{cosec}^2\theta-\cot^2\theta=1\). Therefore, \((\operatorname{cosec}\theta-\cot\theta)(\operatorname{cosec}\theta+\cot\theta)=1\). Hence, \(\operatorname{cosec}\theta+\cot\theta=\frac{1}{\sqrt2-1}=\sqrt2+1\). Trick: Rationalize \(\frac{1}{\sqrt2-1}\) by multiplying numerator and denominator by \(\sqrt2+1\).

Solution: Use \((\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\). Thus, \((\sin\theta+\cos\theta)^2=2-\frac{9}{25}=\frac{41}{25}\). Therefore, the magnitude of the sum is \(\frac{\sqrt{41}}5\). The green-marked option is C. Note: Since the question does not specify the quadrant of θ, the sign of the sum is not fully determined; the marked positive value is retained.

Solution: Using \(\sin A\cos B+\cos A\sin B=\sin(A+B)\), the bracket becomes \(\sin90^\circ=1\), so the first part is 2. Also, \(\sec54^\circ\sin36^\circ=1\), because \(\cos54^\circ=\sin36^\circ\). Thus, the numerator is \(2+3=5\). For the denominator, \(\tan75^\circ=\cot15^\circ\), so \(\operatorname{cosec}^2 15^\circ-\tan^2 75^\circ=\operatorname{cosec}^2 15^\circ-\cot^2 15^\circ=1\). The final value is 5. Trick: Use complementary-angle identities and \(\operatorname{cosec}^2 A-\cot^2 A=1\).

Solution: Volume of a hemisphere \(=\frac23\pi r^3\). Thus, \(V=\frac23\times\frac{22}{7}\times21^3=19404\text{ m}^3\). Converting to litres gives \(19,404,000\) litres. Time \(=\frac{19,404,000}{100}=194,040\) seconds. Converting to hours and minutes gives 53 hours and 54 minutes. Trick: Calculate the hemispherical volume in cubic metres, convert it to litres, and divide by the pump rate.

Solution: Substitute \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and \(\sec\theta=\frac1{\cos\theta}\). Then the expression becomes \(\frac{\sin\theta-1+\cos\theta}{\sin\theta+1-\cos\theta}\). Simplifying using standard trigonometric identities gives \(\frac{1-\sin\theta}{\cos\theta}\). Trick: Replace tan and sec with sine and cosine, then simplify the fraction.

Solution: Interior angle \(A=180^\circ-130^\circ=50^\circ\). Since O is the incenter, \(\angle BOC=90^\circ+\frac{A}{2}\). Therefore, \(\angle BOC=90^\circ+25^\circ=115^\circ\). Trick: The angle between the internal bisectors of B and C is \(90^\circ+\frac12\angle A\).

Solution: Radius \(=30\div2=15\) cm. Half of the 24 cm chord is 12 cm, so its distance from the centre is \(\sqrt{15^2-12^2}=\sqrt{81}=9\) cm. The other chord lies on the opposite side, so its distance from the centre is \(21-9=12\) cm. Its half-length is \(\sqrt{15^2-12^2}=9\) cm, giving a full length of 18 cm. Trick: Use Pythagoras to find the first chord's distance from the centre, then use the distance between the chords.

Solution: Let the initial amounts be P = 5x and Q = 2x. Removing 14 litres in the ratio 5:2 removes 10 litres of P and 4 litres of Q. After removal, P = \(5x-10\) and Q = \(2x-4\). Adding 7 litres of Q gives Q = \(2x+3\). Using the new ratio, \(\frac{5x-10}{2x+3}=2\Rightarrow5x-10=4x+6\Rightarrow x=16\). Initial quantity of P \(=5x=80\) litres. Trick: Remove each component according to the original ratio before adding the extra liquid.

Solution: The smaller pyramid's height is half the original height, so its linear scale factor is \(\frac12\). Volumes of similar solids are proportional to the cubes of corresponding dimensions. The smaller pyramid's volume is therefore \((\frac12)^3=\frac18\) of the original. The frustum volume is \(1-\frac18=\frac78\). Required ratio \(=\frac18:\frac78=1:7\). Trick: Halving every linear dimension makes the volume one-eighth as large.

Solution: Let the marked price per item be 100. The total cost paid is \(198\times100=19800\). The selling price of 220 items after a 5% discount is \(220\times95=20900\). Profit \(=20900-19800=1100\). Gain percentage \(=\frac{1100}{19800}\times100\approx5.56\%\), approximately 6%. Trick: Assume a marked price of 100 per item to simplify the percentage calculations.

Solution: The first saree's marked price is \(644\div0.92=700\). The second saree's marked price is \(594\div0.99=600\). Total marked price = ₹1,300 and total selling price = ₹1,238. Total discount = ₹62. Overall discount percentage \(=\frac{62}{1300}\times100\approx4.77\%\). Trick: Recover each marked price, then calculate the total discount as a percentage of the combined marked price.

Solution: Let the radii be r₁ and r₂ and the distance between centres be d. For the direct tangent, \(d^2-(r_1-r_2)^2=25^2\). For the transverse tangent, \(d^2-(r_1+r_2)^2=15^2\). Subtracting the two equations gives \((r_1+r_2)^2-(r_1-r_2)^2=25^2-15^2=400\). Since the left side is \(4r_1r_2\), \(4r_1r_2=400\Rightarrow r_1r_2=100\). Trick: The difference between the squares of direct and transverse tangent lengths is four times the product of the radii.

Solution: Use \((x^2+\frac1{x^2})^2=x^4+\frac1{x^4}+2\). Therefore, \((x^2+\frac1{x^2})^2=322+2=324\), giving \(x^2+\frac1{x^2}=18\). Now, \((x-\frac1x)^2=x^2+\frac1{x^2}-2=18-2=16\). Since x > 1, \(x-\frac1x\) is positive, so the value is 4. Trick: Convert the fourth-power expression into a square, then use the identity for the square of a difference.

Solution: Since OA and OB are perpendicular to the tangents at A and B, the angle between the radii is \(\angle AOB=180^\circ-\theta\). Since M lies on the minor arc AB, the tangent at M forms the corresponding angles with the tangents PA and PB. Using the tangent-angle relationships, \(\angle XOY=90^\circ-\frac{\theta}{2}\). Trick: First use \(\angle AOB=180^\circ-\angle APB\), then apply the tangent and angle-bisector relationships for the tangent at M.