SSC Selection Post 2026 Graduate 19 Sep Shift 1st

Solution: \(\sqrt{282429536481}=531441\). The number 531441 contains 6 digits. Therefore, the answer is 6. Trick: A 12-digit perfect square has a 6-digit square root. You can also verify that \(531441^2=282429536481\).

Solution: Let p = x, q = 3x, and r = 4x. Substituting, \(2x+3x+3(4x)=51\Rightarrow17x=51\Rightarrow x=3\). Therefore, \(r=4x=4\times3=12\). Trick: Express all three variables using a common multiplier of their ratio.

Solution: C's share of profit \(=1-\frac17-\frac15=\frac{23}{35}\). Therefore, the profit-sharing ratio is \(\frac17:\frac15:\frac{23}{35}=5:7:23\). Profit is proportional to capital × time. C's capital-time \(=4600\times9=41400\), corresponding to 23 parts. One part \(=41400\div23=1800\). A's capital for 5 months corresponds to 5 parts, so A's capital \(=(1800\times5)\div5=1800\). Therefore, A invested ₹1,800. Trick: Convert profit fractions into a common integer ratio, then compare capital × time.

Solution: Initial savings \(=24000-16000=8000\). New income \(=24000\times1.25=30000\). New expenditure \(=16000\times1.10=17600\). New savings \(=30000-17600=12400\). Increase in savings \(=12400-8000=4400\). Percentage increase \(=\frac{4400}{8000}\times100=55\%\). Trick: Calculate the increase using the original savings as the base, not the original income.

Solution: Use the identity \(a^3+b^3=(a+b)^3-3ab(a+b)\). Substituting the given values, \(a^3+b^3=3^3-3\times2\times3=27-18=9\). Trick: When a + b and ab are given, use the identity directly instead of finding a and b separately.

Solution: The area of a regular hexagon is proportional to the square of its side. If the side changes from s to 2s, the area changes by a factor of \(\frac{(2s)^2}{s^2}=4\). Therefore, its area becomes 4 times the original area. Trick: If every linear dimension is multiplied by k, the area is multiplied by \(k^2\).

Solution: Interest for 3 years \(=6500-5000=1500\). Annual simple interest \(=1500\div3=500\), so the original rate is \(\frac{500}{5000}\times100=10\%\). The doubled rate is 20%. Interest for 5 years \(=\frac{5000\times20\times5}{100}=5000\). Amount \(=5000+5000\) = ₹10,000. Trick: First find the original rate from the given interest, then double the rate and calculate the new simple interest.

Solution: For five consecutive odd numbers, the middle number equals their average. The numbers are 23, 25, 27, 29, and 31. Product of the first and last numbers \(=23\times31=713\). Trick: For an odd number of consecutive numbers, the average is the middle number.

Solution: Let the manufacturer's cost price be CP. The retailer's price is \(CP\times1.10\times1.20=1320\). Thus, \(CP=\frac{1320}{1.10\times1.20}=\frac{1320}{1.32}=1000\). Therefore, the manufacturer's cost price is ₹1,000. Trick: Successive profits are multiplied, not added.

Solution: Sector area \(=\frac{\theta}{360^\circ}\times\pi r^2\). Substituting, \(308=\frac{\theta}{360^\circ}\times\frac{22}{7}\times28^2\). The full circle's area is 2464 cm², so \(\theta=\frac{308}{2464}\times360^\circ=45^\circ\). Trick: The sector is \(\frac{308}{2464}=\frac18\) of the circle, so its central angle is \(\frac18\times360^\circ=45^\circ\).

Solution: Let each annual installment be x. Discounting both installments to the present, \(\frac{x}{1.1}+\frac{x}{1.1^2}=12400\). Therefore, \(x\left(\frac{1}{1.1}+\frac{1}{1.21}\right)=12400\), which gives \(x\approx7145\). The approximate installment is ₹7,145. Trick: Bring each future installment to present value using the compound-interest discount factor.

Solution: After a 20% discount, the selling price is 80% of the marked price. Thus, \(0.80y=8800\Rightarrow y=\frac{8800}{0.80}=11000\). Therefore, the marked price is ₹11,000. Trick: Divide the selling price by the remaining percentage, 80%.

Solution: Square and add the two expressions: \((3\sin\theta+5\cos\theta)^2+(5\sin\theta-3\cos\theta)^2=34(\sin^2\theta+\cos^2\theta)=34\). The first squared expression equals \(5^2=25\). Therefore, \((5\sin\theta-3\cos\theta)^2=34-25=9\), so \(5\sin\theta-3\cos\theta=\pm3\). Trick: Choose coefficients so that the cross terms cancel when the expressions are squared and added.

Solution: Using \(\sin A\cos B+\cos A\sin B=\sin(A+B)\), the first and third terms sum to \(\sin90^\circ=1\). Also, \(\sec61^\circ\sin29^\circ=1\), because \(\cos61^\circ=\sin29^\circ\), and \(\operatorname{cosec}61^\circ\cos29^\circ=1\), because \(\sin61^\circ=\cos29^\circ\). The numerator is 3. The denominator is \(\sec^2 48^\circ-\tan^2 48^\circ=1\), since \(\cot42^\circ=\tan48^\circ\). Therefore, the value is 3. Trick: Use complementary-angle identities and \(\sec^2 A-\tan^2 A=1\).

Solution: Given \(p=\operatorname{cosec}\theta+\cot\theta=\frac{1+\cos\theta}{\sin\theta}\). Squaring, \(p^2=\frac{1+\cos\theta}{1-\cos\theta}\). Therefore, \(\frac{p^2-1}{p^2+1}=\cos\theta\). Trick: Express cosec θ + cot θ in terms of sine and cosine, then simplify the given ratio.

Solution: The volume of a hemisphere is proportional to \(r^3\). Thus, the ratio of radii is \(\sqrt[3]{27}:\sqrt[3]{64}=3:4\). Total surface area of a hemisphere \(=3\pi r^2\), so the ratio of total surface areas is \(3^2:4^2=9:16\). Trick: Convert the volume ratio to a radius ratio by taking cube roots, then square the radius ratio.

Solution: Let \(x=\sin\theta\cos\theta\). Since \(\sin^2\theta+\cos^2\theta=1\), we have \((\sin\theta-\cos\theta)^2=1-2x\), so the fourth power is \((1-2x)^2\). Also, \((\sin\theta+\cos\theta)^2=1+2x\), and \(\sin^6\theta+\cos^6\theta=1-3x^2\). The expression becomes \(3(1-2x)^2+6(1+2x)+4(1-3x^2)=13\). Trick: Substitute \(x=\sin\theta\cos\theta\) and use \(\sin^2\theta+\cos^2\theta=1\) to simplify the higher powers.

Solution: Since DE is parallel to BC, by the Basic Proportionality Theorem, \(\frac{AD}{DB}=\frac{AE}{EC}\). Substituting, \(\frac{x}{x-2}=\frac{x+2}{x-1}\). Cross-multiplying gives \(x(x-1)=(x-2)(x+2)=x^2-4\). Thus, \(x^2-x=x^2-4\Rightarrow x=4\). Trick: For a line parallel to one side of a triangle, the corresponding segments on the other two sides are proportional.

Solution: The perpendicular from the centre bisects each chord. For chord AB, half its length is 5 cm, so its distance from the centre is \(\sqrt{13^2-5^2}=\sqrt{144}=12\) cm. For chord CD, half its length is 12 cm, so its distance from the centre is \(\sqrt{13^2-12^2}=\sqrt{25}=5\) cm. Since the chords are on opposite sides of the centre, the distance between them \(=12+5=17\) cm. Trick: Find each chord's distance from the centre using the half-chord and Pythagoras, then add the distances.

Solution: Let the initial quantities be P = 4x and Q = x. The total mixture is 5x liters. Removing 10 liters in the ratio 4:1 removes 8 liters of P and 2 liters of Q. After adding 10 liters of Q, the quantities are P = \(4x-8\) and Q = \(x-2+10=x+8\). Using the new ratio, \(\frac{4x-8}{x+8}=\frac23\). Thus, \(12x-24=2x+16\Rightarrow10x=40\Rightarrow x=4\). Initial P \(=4x=16\) liters. Trick: Remove each component in its original ratio before adding the new liquid.

Solution: If the base side is a, then \(a\sqrt2=10\sqrt2\), so \(a=10\) cm. The lateral surface area of a square pyramid is \(2al\), where l is the slant height. Thus, \(2\times10\times l=260\Rightarrow l=13\) cm. The vertical height is \(h=\sqrt{13^2-(10/2)^2}=\sqrt{144}=12\) cm. Volume \(=\frac13\times10^2\times12=400\text{ cm}^3\). Trick: Find the base side from the diagonal, the slant height from the lateral area, and then the vertical height using Pythagoras.

Solution: Total marked price \(=3\times1200+2\times2000=7600\). Discount on 3 shirts \(=3\times1200\times\frac{15}{100}=540\). Discount on 2 pairs of trousers \(=2\times2000\times\frac{25}{100}=1000\). Total discount \(=540+1000=1540\). Overall discount percentage \(=\frac{1540}{7600}\times100\approx20.26\%\). Trick: Use the total discount divided by the total marked price; do not take a simple average of the two discount rates.

Solution: Using the standard common-tangent relationship for this configuration, \(XY^2=AB^2+PQ^2\). Therefore, \(AB^2=17^2-8^2=289-64=225\), so \(AB=\sqrt{225}=15\) cm. Trick: Square the given lengths, subtract the square of PQ from the square of XY, and take the square root.

Solution: Given \(x+\frac1x=9\). First, \(x^2+\frac1{x^2}=9^2-2=79\). Next, \(x^3+\frac1{x^3}=(x+\frac1x)^3-3(x+\frac1x)=9^3-3\times9=702\). The required sum \(=79+702=781\). Trick: Use the square identity first, then the cube identity, and add the two results.

Solution: The radius to the point of tangency is perpendicular to the tangent. By Pythagoras, tangent length \(=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\) cm. Trick: Use tangent length \(=\sqrt{(\text{distance from centre})^2-(\text{radius})^2}\).