SSC Selection Post 2026 Graduate 17 Sep Shift 1st

Solution: \(\frac{x^2}{27\times108}=1\Rightarrow x^2=2916\Rightarrow x=\pm54\). The green-marked option is 54 (D). Note: The question does not specify that x must be positive. Trick: Multiply 27 and 108, then take the square root. The positive value among the options is 54.

Solution: Let the numbers be \(5x\) and \(9x\). \(\frac{5x-4}{9x+6}=\frac12\Rightarrow10x-8=9x+6\Rightarrow x=14\). Sum \(=(5+9)\times14=196\). Trick: Represent the numbers using the ratio parts and solve the resulting linear equation.

Solution: Profit is proportional to investment × time. Let the investments be \(5x\) and \(4x\), and Y's investment duration be t months. \(\frac{5x\times8}{4x\times t}=\frac59\Rightarrow\frac{40}{4t}=\frac59\Rightarrow t=18\) months. Trick: Compare capital-months to the profit ratio. Source note: The green-marked answer is 18 months, but this exceeds the stated one-year period, so the original question appears inconsistent.

Solution: Initial savings \(=18000-12000=6000\). New income \(=18000\times1.20=21600\). New expenditure \(=12000\times1.15=13800\). New savings \(=21600-13800=7800\). Percentage increase \(=\frac{7800-6000}{6000}\times100=30\%\). Trick: Calculate the increase in savings over the original savings, not over the income.

Solution: For 4 years at 10% simple interest, \(SI=\frac{P\times10\times4}{100}=0.4P\). Given \(P-SI=3600\), so \(P-0.4P=3600\Rightarrow0.6P=3600\Rightarrow P=6000\). The invested sum is ₹6000. Trick: The interest is 40% of the principal, so the difference is 60% of the principal.

Solution: \(A=P(1+\frac{r}{100})^n\). Thus, \(A=6400(1+\frac{5}{100})^3=6400(1.05)^3\) = ₹7408.80. Trick: Apply the 5% increase successively for all three years; compound interest is calculated on the updated amount each year.

Solution: \(a+\frac1a=2\Rightarrow a^2+1=2a\Rightarrow(a-1)^2=0\), so \(a=1\). Therefore, \(a^{2021}+\frac1{a^{2027}}=1+1=2\). Trick: When \(a+\frac1a=2\), equality occurs at \(a=1\); all powers of 1 remain 1.

Solution: Square side \(=24\div4=6\) cm. The regular hexagon has six equal sides, each 6 cm. Its perimeter \(=6\times6=36\) cm. Trick: Find the square's side first, then multiply it by 6 for the regular hexagon's perimeter.

Solution: Use \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Thus, \(8x^3-27y^3=(2x-3y)(4x^2+6xy+9y^2)\). Comparing coefficients gives \(M=4\), \(N=9\), and \(O=6\). Therefore, \(M^2+N^2+O^2=16+81+36=133\). Trick: Apply the difference-of-cubes identity and compare coefficients.

Solution: Original total age \(=12\times40=480\) years. Replacing the 55-year-old with the 31-year-old decreases the total by \(55-31=24\) years. New total \(=480-24=456\). New average \(=456\div12=38\) years. Trick: New average \(=40+\frac{31-55}{12}=38\) years.

Solution: Arc length \(=\frac{\theta}{360}\times2\pi r\). Therefore, \(\frac{120}{360}\times2\times\frac{22}{7}\times21=44\) cm. Trick: An angle of 120 degrees is one-third of 360 degrees, so the arc is one-third of the circle's circumference.

Solution: Assume the cost price of one item is 100. At 25% profit, its selling price is 125. The cost price of 20 items is 2000, so \(125x=2000\Rightarrow x=16\). Trick: \(x=\frac{20\times100}{125}=16\).

Solution: After a 12% discount, the selling price is 88% of the marked price. Hence, \(0.88x=4400\Rightarrow x=\frac{4400\times100}{88}\) = ₹5000. Trick: Divide the selling price by 0.88 to recover the marked price.

Solution: Expand the squares and use \(\sin^2 A+\cos^2 A=1\), \(\sec^2 A-\tan^2 A=1\), and \(\operatorname{cosec}^2 A-\cot^2 A=1\). The expression simplifies to \(1+4+1+1=7\). Trick: Pair cosec² with cot² and sec² with tan² to use the standard identities.

Solution: The triangle with sides 9, 12, and 15 is right-angled because \(9^2+12^2=15^2\). Its area is \(\frac12\times9\times12=54\) cm². The area of the original triangle is \(\frac43\) times the area of the triangle whose sides are the three medians. Thus, \(\frac43\times54=72\) cm². Trick: Recognize the 9–12–15 right-triangle triple and multiply its area by \(\frac43\).

Solution: Expand the first two squares and use \(\sin^2\theta+\cos^2\theta=1\), \(\operatorname{cosec}^2\theta=1+\cot^2\theta\), and \(\sec^2\theta=1+\tan^2\theta\). After subtracting \((\tan\theta+\cot\theta)^2\), the terms cancel, leaving 5. Trick: Use the Pythagorean identities before combining the terms.

Solution: Use \((\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\). Thus, \((\sin\theta+\cos\theta)^2=2-\frac{16}{25}=\frac{34}{25}\). Taking the positive value given in the options, \(\sin\theta+\cos\theta=\frac{\sqrt{34}}5\). Trick: Square the sum and difference; their squares add up to 2.

Solution: \(\sec^4\theta-\tan^4\theta=(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)\). Since \(\sec^2\theta-\tan^2\theta=1\), the expression is \(1\times\frac{7}{12}=\frac{7}{12}\). Trick: Factor the difference of two squares. Source note: The printed condition \(\sec^2\theta+\tan^2\theta=\frac{7}{12}\) is not possible for a real angle, but the green-marked answer is retained.

Solution: Half of the 16 cm chord is 8 cm. Its distance from the centre is \(\sqrt{17^2-8^2}=\sqrt{225}=15\) cm. Since the chords lie on opposite sides, the other chord is \(23-15=8\) cm from the centre. Its length is \(2\sqrt{17^2-8^2}=30\) cm. Difference \(=30-16=14\) cm. Trick: Use the perpendicular from the centre to bisect a chord, then apply Pythagoras.

Solution: Let the initial mixture be V liters. Initially, A \(=\frac{3V}{4}\) and B \(=\frac V4\). Removing 16 liters removes 12 liters of A and 4 liters of B. After adding 9 liters of B, the amounts are A \(=\frac{3V}{4}-12\) and B \(=\frac V4+5\). Using the new ratio, \(\frac{\frac{3V}{4}-12}{\frac V4+5}=\frac34\), which gives \(V=28\). Initial A \(=\frac34\times28=21\) liters. Trick: Removing a mixture in the ratio 3:1 removes A and B in that same ratio.

Solution: Total surface area of a hemisphere \(=3\pi r^2=462\). Thus, \(r^2=\frac{462\times7}{3\times22}=49\). Curved surface area \(=2\pi r^2=2\times\frac{22}{7}\times49=308\) cm². Trick: Hemisphere TSA is \(3\pi r^2\), while its curved surface area is \(2\pi r^2\).

Solution: Tangents drawn from the same external point are equal. Let the tangent length from A be x. Then AB \(=x+8\), AC \(=x+6\), and BC \(=8+6=14\). The semiperimeter is \(s=x+14\). By Heron's formula, the area is \(\sqrt{48x(x+14)}\); by the inradius formula, area \(=4(x+14)\). Equating the two areas gives \(x=7\). Thus, AB \(=15\) cm and AC \(=13\) cm. Trick: Use equal tangent lengths from each vertex and the area formula \(\Delta=rs\).

Solution: Let the square base side be a and the slant height be l. Base area \(=a^2\); lateral face area \(=\frac12 al\). From the ratio, \(\frac{a^2}{\frac12 al}=2\Rightarrow l=a\). In the cross-section through the apex and the midpoint of a base side, \(\cos\theta=\frac{a/2}{l}=\frac12\). Hence, \(\theta=60^\circ\). Trick: Convert the area ratio into a relation between the base side and slant height, then use the right triangle.

Solution: Total marked price \(=5\times1050+4\times1750\) = ₹12,250. Total discount \(=0.11(5\times1050)+0.32(4\times1750)\) = ₹2,817.50. Effective discount \(=\frac{2817.50}{12250}\times100=23\%\). Trick: Calculate the total discount in rupees, then divide by the total marked price; do not simply average 11% and 32%.

Solution: Let the radii be r₁ and r₂ and the distance between centres be d. For the external tangent, \(d^2-(r_1-r_2)^2=9^2=81\). For the internal tangent, \(d^2-(r_1+r_2)^2=5^2=25\). Subtracting gives \((r_1+r_2)^2-(r_1-r_2)^2=56\Rightarrow4r_1r_2=56\Rightarrow r_1r_2=14\). Trick: Subtract the squared tangent lengths to eliminate the distance between centres.