SSC Selection Post 2026 Graduate 23 Sep Shift 1st

Solution: The hexadecimal number system uses the digits 0–9 and letters A–F. The number 4 is represented as 4 in hexadecimal as well. Trick: Numbers from 0 to 9 have the same representation in decimal and hexadecimal.

Solution: Rotating the line y = x by 90° about the origin maps it onto the line y = -x, which is perpendicular to the original line and passes through the origin. Trick: A 90° rotation changes the direction of a line to a perpendicular direction. Source note: Reflecting y = x about the x-axis also gives y = -x, so option B has the same geometric result; the green-marked option A is retained.

Solution: \(x^2-10x+25=x^2-2\times5\times x+5^2=(x-5)^2\). Trick: Recognize the identity \(a^2-2ab+b^2=(a-b)^2\).

Solution: \(4x-7=5x-3\). Subtract 4x from both sides: \(-7=x-3\). Add 3 to both sides: \(x=-4\). Trick: Collect the x-terms on one side and the constants on the other.

Solution: First, \(1.6-0.4=1.2\). Then \(2.5\times1.2\div0.2=3\div0.2=15\). Finally, \(1.25+15=16.25\). Trick: Follow BODMAS: brackets first, multiplication and division next, and addition last.

Solution: Expanding the square, \((x+y)^2=(x+y)(x+y)=x^2+xy+xy+y^2=x^2+2xy+y^2\). Trick: Remember the identity \((a+b)^2=a^2+2ab+b^2\).

Solution: The unit digit of 999^3 is the unit digit of 9^3, which is 9. The unit digit of 888^2 is the unit digit of 8^2, which is 4. The unit digit of 777^1 is 7. Adding the unit digits gives \(9+4+7=20\), whose unit digit is 0. Trick: For addition, calculate only the unit digits of the individual terms.

Solution: The number of diagonals of an n-sided polygon is \(\frac{n(n-3)}{2}\). For a pentagon, n = 5. Therefore, \(\frac{5(5-3)}{2}=\frac{5\times2}{2}=5\). Trick: Use the diagonal formula \(n(n-3)/2\).

Solution: The diagonal formula is \(\frac{n(n-3)}{2}=20\). Thus, \(n(n-3)=40\), or \(n^2-3n-40=0\). Factoring gives \((n-8)(n+5)=0\). Since the number of sides must be positive, n = 8. Trick: Set \(n(n-3)/2\) equal to the given number of diagonals and solve the quadratic equation.

Solution: Area of a quarter-circle \(=\frac14\pi r^2\). Taking \(\pi=\frac{22}{7}\) and r = 7 cm, the area is \(\frac14\times\frac{22}{7}\times7^2=38.5\text{ cm}^2\). Trick: Calculate the full circle's area and divide it by 4.

Solution: Lateral surface area of a square pyramid \(=\frac12\times\text{perimeter of base}\times\text{slant height}\). The base perimeter is \(4\times6=24\) cm. Therefore, LSA \(=\frac12\times24\times10=120\text{ cm}^2\). Trick: For a square pyramid, multiply the base perimeter by the slant height and divide by 2.

Solution: Sector area \(A=\frac12r^2\theta\), with θ in radians. Given \(32\pi=\frac12\times8^2\times\theta=32\theta\), so \(\theta=\pi\). Arc length \(s=r\theta=8\pi\) cm. Trick: For a sector, divide twice its area by \(r^2\) to obtain the angle in radians, then multiply the angle by the radius.

Solution: The longest diagonal of a rectangular prism is \(d=\sqrt{l^2+b^2+h^2}\). Therefore, \(d=\sqrt{9^2+12^2+16^2}=\sqrt{81+144+256}=\sqrt{481}\approx21.93\) cm. Trick: Square all three dimensions, add them, and take the square root.

Solution: The additional purchase needed to reach ₹500 is ₹500 − ₹480 = ₹20. The required percentage increase on ₹480 is \(\frac{20}{480}\times100\approx4.17\%\). Trick: Divide the additional amount required by the original purchase amount. Note: The source wording says 'over ₹500'; the marked answer treats ₹500 as the intended threshold.

Solution: Total cost price = ₹4,000 and overall gain = ₹200. Therefore, total selling price \(=4000+200\) = ₹4,200. Trick: Selling price = cost price + profit. Use the overall gain directly.

Solution: Using alligation, the ratio of investment at 12% to investment at 18% is \((18-15):(15-12)=3:3=1:1\). Thus, each scheme receives half the total investment. Investment in the lower-return scheme \(=\frac{600000}{2}\) = ₹3,00,000. Trick: When the average rate is exactly midway between two rates, the investments are equal.

Solution: To convert degrees to radians, multiply by \(\frac{\pi}{180^\circ}\). Thus, \(-390^\circ\times\frac{\pi}{180^\circ}=-\frac{390\pi}{180}=-\frac{13\pi}{6}\). Trick: Multiply the degree measure by \(\pi/180\) and simplify the fraction.

Solution: After 1 year, the amount is \(2500\times1.20=3000\). For the remaining 9 months, calculate simple interest on ₹3,000: \(SI=\frac{3000\times20\times9}{100\times12}=450\). Total amount \(=3000+450=3450\). Compound interest \(=3450-2500\) = ₹950. Trick: With annual compounding and a fractional final year, compound for the completed year and calculate simple interest for the remaining months.

Solution: Squaring both sides, \((\sqrt{x}+\frac{1}{\sqrt{x}})^2=7\). Thus, \(x+2+\frac1x=7\), so \(x+\frac1x=5\). Trick: Use \((a+b)^2=a^2+2ab+b^2\); here the middle term is 2.

Solution: Expand the square: \((x+\frac1x)^2=x^2+2+\frac1{x^2}\). Substituting the given value gives \(14+2=16\). Trick: Add 2 to \(x^2+\frac1{x^2}\) to obtain \((x+\frac1x)^2\).

Solution: The central angle subtended by the chord is twice the angle at the circumference, so it is \(2\times30^\circ=60^\circ\). The chord length is 6 cm, giving radius \(r=6\) cm because \(6=2r\sin30^\circ\). The perpendicular from the centre bisects the chord, so half the chord is 3 cm. The distance from the centre is \(\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt3\) cm. Trick: Use the central-angle theorem, then use Pythagoras with the radius and half-chord.

Solution: Since OP is perpendicular to the tangent PQ, \(\angle OPR=90^\circ-30^\circ=60^\circ\). In triangle OPR, OP and OR are radii, so OP = OR and the base angles are equal. Thus, \(\angle ORP=60^\circ\), and \(\angle POR=180^\circ-60^\circ-60^\circ=60^\circ\). Trick: Use the perpendicular radius-tangent relationship and the isosceles triangle formed by two radii.

Solution: Let the radii be r₁ and r₂, and the distance between the centres be d. For the external tangent, \(17^2=d^2-(r_1-r_2)^2\). For the internal tangent, \(9^2=d^2-(r_1+r_2)^2\). Subtracting the equations gives \(17^2-9^2=(r_1+r_2)^2-(r_1-r_2)^2=4r_1r_2\). Thus, \(289-81=4r_1r_2\Rightarrow208=4r_1r_2\Rightarrow r_1r_2=52\). Trick: The difference between the squares of the external and internal tangent lengths equals four times the product of the radii.

Solution: Since the point is midway between the towers, the horizontal distance to each tower is d/2. Therefore, \(h_1=\frac d2\tan30^\circ=\frac{d}{2\sqrt3}\), and \(h_2=\frac d2\tan60^\circ=\frac{\sqrt3d}{2}\). Hence, \(h_1:h_2=\frac1{\sqrt3}:\sqrt3=1:3\). Trick: At the midpoint, the horizontal distances are equal, so the height ratio is the ratio of the tangents of the angles.

Solution: Let the quantity of each of the first two solutions be x, so the third quantity is 2x. The acid fractions are 1/3, 2/5, and 3/5. Total acid \(=\frac{x}{3}+\frac{2x}{5}+\frac{6x}{5}=\frac{29x}{15}\). The water fractions are 2/3, 3/5, and 2/5. Total water \(=\frac{2x}{3}+\frac{3x}{5}+\frac{4x}{5}=\frac{31x}{15}\). Therefore, the ratio of acid to water is 29:31. Trick: Multiply each solution's acid and water fractions by its relative quantity, then add each component separately.