SSC Selection Post 2026 Graduate 21 Sep Shift 1st

Solution: The hexadecimal number system uses sixteen symbols: digits 0-9 and letters A-F. Therefore, the correct answer is Hexadecimal. Trick: Remember that binary uses 0-1, octal uses 0-7, decimal uses 0-9, and hexadecimal uses 0-9 and A-F.

Solution: \(\sqrt{50}-\sqrt{8}=\sqrt{25\times2}-\sqrt{4\times2}=5\sqrt{2}-2\sqrt{2}=3\sqrt{2}\). Trick: Take out perfect-square factors from under each square root, then subtract like surds.

Solution: Since a, b, c are in A.P., \(a+c=2b\). Since a², b², c² are in G.P., \((b^2)^2=a^2c^2\), giving \(b^2=ac\) under the usual positive-term assumption. Squaring \(a+c=2b\), \(a^2+2ac+c^2=4b^2\). Substituting \(ac=b^2\), we get \(a^2+c^2=2b^2\). Trick: Combine the A.P. condition with the G.P. condition, then square the A.P. relation.

Solution: \(3x+5=20\Rightarrow3x=20-5=15\Rightarrow x=15\div3=5\). Trick: Move the constant term to the other side first, then divide by the coefficient of x.

Solution: Use the identity \(a^2+2ab+b^2=(a+b)^2\). Therefore, \(7^2+2\times7\times3+3^2=(7+3)^2=10^2=100\). Trick: Recognize the perfect-square identity to avoid lengthy multiplication.

Solution: The unit digit of powers of 7 repeats in the cycle 7, 9, 3, 1. Since \(83=4\times20+3\), the remainder on dividing 83 by 4 is 3. The third digit in the cycle is 3. Therefore, the unit digit of \(17^{83}\) is 3. Trick: For powers ending in 7, divide the exponent by 4 and use the remainder to select the digit in the cycle.

Solution: Each interior angle of a regular polygon with n sides is \(\frac{(n-2)\times180^\circ}{n}\). For a hexagon, \(n=6\), so the angle is \(\frac{(6-2)\times180^\circ}{6}=120^\circ\). Trick: The sum of the interior angles of a hexagon is 720°, and dividing by 6 gives 120°.

Solution: For a regular hexagon inscribed in a circle, the side of the hexagon equals the radius r. Its area is \(\frac{3\sqrt{3}}{2}r^2\). Therefore, \(\frac{3\sqrt{3}}{2}r^2=200\sqrt{3}\Rightarrow r^2=\frac{400}{3}\Rightarrow r=\frac{20}{\sqrt{3}}\approx11.55\) cm. The source green-marks option B, 11.54. Note: The calculation rounds to about 11.55 cm, so the printed option's decimal appears slightly inconsistent.

Solution: Distance in one round equals the circumference, \(2\pi r\). For two rounds, distance \(=2\times2\pi r=4\pi r\). Using \(\pi=\frac{22}{7}\) and \(r=28\), distance \(=4\times\frac{22}{7}\times28=352\) m. Trick: For two full rounds, multiply the circumference by 2.

Solution: The perpendicular from the centre bisects the chord, so half the chord is \(16\div2=8\) cm. Using Pythagoras, \(r^2=8^2+6^2=64+36=100\). Therefore, \(r=10\) cm. Trick: Use the half-chord, the distance from the centre, and the radius as a right triangle.

Solution: Volume of a prism = base area × height. Hence, \(V=35\times6.5=227.5\) cm³. Trick: Multiply the cross-sectional base area by the perpendicular height.

Solution: Base side \(a=60\div4=15\) m. Lateral surface area of a square pyramid is \(\frac12\times\text{perimeter}\times l\), where l is the slant height. Thus, \(500=\frac12\times60\times l\Rightarrow l=\frac{500}{30}=16.67\) m approximately. The vertical height is \(h=\sqrt{l^2-(a/2)^2}=\sqrt{16.67^2-7.5^2}\approx14.88\) m. Trick: Find the slant height from the lateral area first, then use Pythagoras to calculate the vertical height.

Solution: Since the selling price of each item is ₹500, the cost price of the item sold at 25% gain is \(500\div1.25=400\). The cost price of the item sold at 25% loss is \(500\div0.75=666.67\) approximately. Total cost price ≈ ₹1,066.67 and total selling price = ₹1,000. Loss percentage \(=\frac{66.67}{1066.67}\times100=6.25\%\). Trick: For equal selling prices with equal gain and loss percentages, the overall result is a loss of \(\frac{25^2}{100}=6.25\%\).

Source-marked answer: ₹256 (B). If the intended meaning is a 25% profit on cost price after the discount, the selling price is \(400\times0.80=320\), and \(CP=320\div1.25=256\). Note: The printed wording says '25% profit on marked price'; interpreted literally, that would not give ₹256. The green-marked option is retained, but the wording and answer are inconsistent.

Source-marked answer: 70 (C). However, under the standard interpretation, the initial salt is 40 liters. If x liters of water are added, the concentration becomes \(40/(100+x)\). Setting this to 25% gives \(40/(100+x)=0.25\Rightarrow100+x=160\Rightarrow x=60\) liters. If 10 liters of a homogeneous solution subsequently overflow, its concentration remains unchanged. Therefore, the printed conditions lead to 60 liters, not the green-marked 70 liters. The source-marked answer is preserved, but the item appears inconsistent.

Source-marked answer: \(\frac{327}{99}\) (A). Direct conversion of the printed repeating decimal gives \(x=3.272727...\). Then \(100x=327.272727...\). Subtracting, \(99x=324\), so \(x=\frac{324}{99}=\frac{36}{11}\), which is option B. The green-marked option A conflicts with the printed decimal, so the source-marked answer is retained with this note.

Solution: \(\sqrt{75}-\sqrt{27}=\sqrt{25\times3}-\sqrt{9\times3}=5\sqrt3-3\sqrt3=2\sqrt3\). Trick: Extract the perfect-square factors before subtracting the surds.

Solution: The equation \(3x-2y=6\) is a linear equation in two variables. For every chosen value of x, a corresponding value of y can be found using \(y=(3x-6)/2\). Therefore, it has infinitely many real solutions. Trick: A single linear equation in two variables generally has infinitely many real solutions.

Solution: Convert radians to degrees by multiplying by \(180^\circ/\pi\). Thus, \(\frac{5\pi}{3}\times\frac{180^\circ}{\pi}=5\times60^\circ=300^\circ\). Trick: Remember that \(\pi\) radians equals 180°.

Solution: Let each annual installment be x. Taking the present value of both installments, \(\frac{x}{1.1}+\frac{x}{1.1^2}=20000\). Thus, \(x(0.90909+0.82645)=20000\), giving \(x\approx11523.81\). Therefore, each installment is approximately ₹11,524. Trick: Discount each future installment to the present using the compound-interest factor.

Solution: When two chords intersect inside a circle, \(\angle APC=\frac12(\text{arc AC}+\text{arc BD})\). Substituting, \(50^\circ=\frac12(60^\circ+\text{arc BD})\). Therefore, \(100^\circ=60^\circ+\text{arc BD}\), giving the central angle for minor arc BD as 40°. Trick: The angle formed by two intersecting chords inside a circle is half the sum of the intercepted arcs.

Solution: The radii OA and OB are perpendicular to the tangents at A and B. In quadrilateral AOBT, the angles at A and B are 90° each. Therefore, \(\angle ATB=360^\circ-90^\circ-90^\circ-120^\circ=60^\circ\). Trick: The angle between two tangents equals \(180^\circ\) minus the central angle between the radii to the points of contact.

Solution: Using the geometric relation for this configuration, \(PQ^2=AB^2+MN^2\). Therefore, \(AB^2=PQ^2-MN^2=5^2-3^2=25-9=16\). Hence, \(AB=\sqrt{16}=4\) cm. Trick: Square the given lengths, subtract the square of MN from the square of PQ, and take the square root.

Solution: Let the balloon's height be h, and let its vertical projection be between A and B. The horizontal distances from A and B are \(h/\tan60^\circ=h/\sqrt3\) and \(h/\tan30^\circ=\sqrt3h\), respectively. Their sum is 1200 m: \(\frac{h}{\sqrt3}+\sqrt3h=1200\). Hence, \(\frac{4h}{\sqrt3}=1200\Rightarrow h=300\sqrt3\) m. Trick: When the object lies between two observation points, add the two horizontal distances and equate them to the distance between the points.

Solution: Initially, milk = 80 L and water = 20 L. Removing 20 L in the ratio 4:1 removes 16 L milk and 4 L water, leaving 64 L milk and 16 L water. Add 20 L pure water: milk = 64 L and water = 36 L. Now remove 25 L of the new mixture. Since the ratio is 64:36 = 16:9, the removed mixture contains 16 L milk and 9 L water. Remaining quantities are 48 L milk and 27 L water. Add 25 L pure milk: milk = 73 L and water = 27 L. Final ratio = 73:27. Trick: In each replacement, remove the components in their current ratio, then add the new liquid.