SSC Selection Post 2026 Graduate 18 Sep Shift 1st

Solution: Using the identity \((a+b)(a-b)=a^2-b^2\), we get \((\sqrt{5})^2-(\sqrt{2})^2=5-2=3\). Trick: Apply the difference of squares identity instead of multiplying the brackets separately.

Solution: \(14=8+4+2=2^3+2^2+2^1\). Therefore, the binary representation is 1110. Trick: Express the decimal number as a sum of powers of 2 and write the corresponding binary digits.

Solution: The slope of the given line is \(m_1=\tan30^\circ=\frac{1}{\sqrt{3}}\). For two perpendicular lines, \(m_1m_2=-1\). Hence, \(m_2=-\sqrt{3}=-\tan60^\circ\). The required line is \(y=-\tan(60^\circ)\times x\). Trick: The slope of a line perpendicular to a line of slope m is \(-\frac{1}{m}\).

Solution: \(\sqrt{x+8}=10\). Squaring both sides gives \(x+8=100\). Therefore, \(x=100-8=92\). Trick: Square both sides of a square-root equation and then isolate the variable.

Solution: Squaring both sides, \((x+\frac{1}{x})^2=3^2\). Therefore, \(x^2+2+\frac{1}{x^2}=9\). Hence, \(x^2+\frac{1}{x^2}=9-2=7\). Trick: Use \((x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}\).

Solution: Like terms have the same variables raised to the same powers. In 7x and 9x, the variable part is x in both terms. Therefore, they are like terms. Trick: Compare the variable parts; coefficients may differ, but variables and their powers must match.

Solution: Volume of a cylinder \(V=\pi r^2h\). Substituting the values, \(V=3.14\times(2.5)^2\times10=3.14\times6.25\times10=196.25\text{ cm}^3\). Trick: Square the radius first, then multiply by the height and \(\pi\).

Solution: Arc length \(s=r\theta\), where the angle is measured in radians. Therefore, \(r=\frac{s}{\theta}=\frac{40}{5\pi/2}=\frac{16}{\pi}\) meters. Trick: For an angle in radians, radius equals arc length divided by the angle. Source note: Options C and D are mathematically equivalent; the PDF marks option C in green.

Solution: Using \(a^2-b^2=(a-b)(a+b)\), we get \(\frac{8^2-6^2}{8-6}=\frac{(8-6)(8+6)}{8-6}=8+6=14\). Trick: Factor the difference of two squares and cancel the common factor.

Solution: For a common external tangent, \(L=\sqrt{d^2-(r_2-r_1)^2}\). Here, \(d=20\), \(r_1=6\), and \(r_2=10\). Therefore, \(L=\sqrt{20^2-(10-6)^2}=\sqrt{400-16}=\sqrt{384}\approx19.6\) cm. Trick: For an external tangent, use the difference of the radii in the Pythagorean formula.

Solution: Each interior angle of a regular n-sided polygon is \(\frac{(n-2)\times180^\circ}{n}\). For \(n=9\), the angle is \(\frac{(9-2)\times180^\circ}{9}=\frac{7\times180^\circ}{9}=140^\circ\). Trick: Use the interior-angle formula for a regular polygon.

Solution: For a regular polygon, \(\frac{(n-2)\times180^\circ}{n}=179^\circ\). Hence, \(180^\circ-\frac{360^\circ}{n}=179^\circ\), so \(\frac{360}{n}=1\) and \(n=360\). Trick: Use the equivalent formula for an interior angle, \(180^\circ-\frac{360^\circ}{n}\).

Solution: The wire length equals the circle's circumference, \(2\pi r=2\times\frac{22}{7}\times14=88\) cm. This becomes the square's perimeter. Therefore, side \(=88\div4=22\) cm. Trick: Equate the circle's circumference to the square's perimeter.

Solution: Volume \(=\text{length}\times\text{breadth}\times\text{height}\). Thus, \(288=6\times4\times h\). So, \(h=\frac{288}{24}=12\) cm. Trick: Divide the volume by the area of the rectangular base.

Solution: Volume of a square pyramid \(V=\frac13\times\text{base area}\times h\). Base area \(=6^2=36\text{ cm}^2\). Hence, \(V=\frac13\times36\times9=108\text{ cm}^3\). Trick: Find the square base area first, multiply by height, and divide by 3.

Solution: Since the gears touch externally, the distance between their centres is \(d=3+6=9\) cm. The direct common external tangent has length \(L=\sqrt{d^2-(r_2-r_1)^2}\). Thus, \(L=\sqrt{9^2-(6-3)^2}=\sqrt{81-9}=\sqrt{72}=6\sqrt{2}\approx8.49\) cm. Trick: Add the radii to find the centre distance, then use the difference of the radii in the tangent formula.

Source-marked answer: 120°. The PDF highlights option D in green. However, the printed description gives \(\angle ASB=60^\circ\) but does not specify a sufficient geometric relationship between this angle and \(\angle APB\) to derive 120° uniquely. The marked option is preserved from the source. Trick: For an exact solution to an angle-in-a-circle problem, identify the intercepted arc and establish the relationship between the centre angle and the inscribed angle.

Solution: Total cost price = ₹6000. Total selling price \(=5\times1100\) = ₹5500. Loss \(=6000-5500\) = ₹500. Loss percentage \(=\frac{500}{6000}\times100=8.33\%\) approximately. Trick: Calculate the total cost price and total selling price before finding the loss percentage.

Solution: Cost price of each part \(=\frac{100000}{2}\) = ₹50,000. Selling price of sarees \(=50000\times1.20\) = ₹60,000. Selling price of casual wear \(=50000\times0.90\) = ₹45,000. Total selling price = ₹1,05,000, so profit = ₹5,000. Profit percentage \(=\frac{5000}{100000}\times100=5\%\). Trick: When equal amounts are invested, the overall percentage is the average of the profit and loss percentages: \(\frac{20\%-10\%}{2}=5\%\) profit.

Solution: The mixture price ₹750/L is exactly midway between ₹900/L and ₹600/L. By alligation, the quantities are equal. Therefore, the ₹900/L perfume used \(=\frac{240}{2}=120\) L. The 15 L leakage occurred after blending and does not change the amount originally used. Trick: When the mean price is the average of the two prices, equal quantities are mixed.

Solution: The amount reinvested in India is ₹3,78,000. For 2 years at 10% compound interest, \(A=P(1+\frac{r}{100})^n\). Thus, \(A=378000(1+\frac{10}{100})^2=378000(1.1)^2\) = ₹4,57,380. Trick: Use the stated maturity amount as the principal for the second investment period.

Solution: For a square, area \(=s^2\). Thus, \(s=\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\) cm. Trick: Take the square root of the area and extract the largest perfect-square factor. Source note: Option B, \(\sqrt{72}\) cm, is mathematically equivalent to option A; the PDF marks option A in green.

Solution: Two separate non-touching circles have four common tangents. When the circles touch externally, the internal common tangent at the point of contact becomes a single shared tangent, so the total number is 3. Trick: Two externally touching circles have 3 common tangents.

Solution: Let the horizontal distance to the cliff be x. From the angle of depression, \(\tan30^\circ=\frac{25}{x}\), so \(x=25\sqrt{3}\) m. The height of the top above the observer is \(x\tan60^\circ=25\sqrt{3}\times\sqrt{3}=75\) m. Total cliff height \(=25+75=100\) m. Trick: Calculate the horizontal distance using the depression angle, then use the elevation angle to find the height above the observer.

Solution: Initially, milk \(=\frac{7}{10}\times60=42\) L and water \(=\frac{3}{10}\times60=18\) L. Let x litres of mixture be removed. Milk remaining \(=42-\frac{7x}{10}\). After removing the mixture and adding x litres of pure water, water becomes \(18-\frac{3x}{10}+x=18+\frac{7x}{10}\). Using the required ratio, \(\frac{42-\frac{7x}{10}}{18+\frac{7x}{10}}=\frac75\). Solving gives \(x=10\) litres. Trick: In replacement questions, the removed mixture contains milk and water in the original ratio; the replacement adds only water.