SSC CHSL Tier 1 12 Nov 2025 Shift 2

Using the identity \((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\), substitute the given values: \(0=32+2(xy+yz+zx)\). Therefore, \(2(xy+yz+zx)=-32\), so \(xy+yz+zx=-16\). Trick: \(xy+yz+zx=-\frac{x^2+y^2+z^2}{2}=-\frac{32}{2}=-16\).

Let the initial investments be \(2x,3x,4x\). A's capital-time investment is \(2x\times6+3x\times18=66x\). B's investment is \(3x\times24=72x\), and C's investment is \(4x\times24=96x\). The profit-sharing ratio is \(66:72:96=11:12:16\). A's profit is \(\frac{66}{234}\times90000=25,384.62\) approximately. Trick: Profit is divided according to capital multiplied by time.

Using \(\sin A\cos B+\cos A\sin B=\sin(A+B)\), find the missing ratios. Since \(\cos A=\frac{12}{13}\), \(\sin A=\frac{5}{13}\). Since \(\sin B=\frac{5}{13}\), \(\cos B=\frac{12}{13}\). Therefore, \(\frac{5}{13}\\times\frac{12}{13}+\frac{12}{13}\\times\frac{5}{13}=\frac{120}{169}\). Trick: Use the \(5\text{-}12\text{-}13\) right-triangle ratios.

Cost price per apple is \(120/10=12\)₹. Selling price per apple is \(100/8=12.50\)₹. Profit per apple is \(12.50-12=0.50\)₹. Profit percentage is \(\frac{0.50}{12}\times100=4.17\%\) approximately. Trick: Convert both prices to a per-apple price before calculating the percentage.

Given \(\frac{4}{11}A=\frac{3}{5}B\). Cross-multiplying gives \(20A=33B\), so \(A:B=33:20\). The total ratio is \(53\). Therefore, \(A=\frac{33}{53}\times9000=5603.7735\ldots\)₹, approximately Rs. 5603.8. Trick: Convert the fractional equality into a ratio, then divide the total amount in that ratio.

The amount doubles every 5 years. Since \(8=2^3\), it must double three times to become eight times its original value. Required time \(=3\times5=15\) years. Trick: Express the required multiple as a power of 2.

Bike's speed \(=160/4=40\) km/h. The speed ratio is \(4:5\), so the car's speed is \(40\times\frac54=50\) km/h. Distance travelled by the car in 5 hours \(=50\times5=250\) km. Trick: Calculate the bike's speed first, use the speed ratio to find the car's speed, and then multiply by time.

Original total marks \(=30\times55=1650\). The increase in total marks is \(70-40=30\). New total marks \(=1650+30=1680\). New average \(=1680/30=56\). Trick: New average \(=\) old average \(+\frac{\text{difference in marks}}{\text{number of students}}=55+\frac{30}{30}=56\).

Find the LCM of 9, 12, and 15. \(9=3^2\), \(12=2^2\times3\), and \(15=3\times5\). Therefore, \(\operatorname{LCM}=2^2\times3^2\times5=180\). Adding the required remainder gives \(180+4=184\). Trick: Add the common remainder to the LCM.

Let the original price be ₹100. After a 20% increase, the price becomes ₹120. After another 10% increase, it becomes \(120\times1.10=132\)₹. The net increase is ₹32 on ₹100, so the net percentage increase is 32%. Trick: Successive increase \(=a+b+\frac{ab}{100}=20+10+\frac{20\times10}{100}=32\%\).

A's one-day work is \(\frac{1}{10}\). In 4 days, A completes \(\frac{4}{10}=\frac25\) of the work, leaving \(\frac35\). Their combined one-day work is \(\frac1{10}+\frac1{15}=\frac16\). Required additional time \(=\frac35\div\frac16=\frac35\times6=3.6\) days. Trick: Divide the remaining work by their combined daily work.

Marked price \(=5000\)₹. Selling price after a 20% discount \(=5000\times\frac{80}{100}=4000\)₹. A 25% profit means the selling price is 125% of the cost price. Therefore, \(CP=\frac{4000\times100}{125}=3200\)₹. Trick: Apply the discount first, then divide the selling price by 1.25.

The difference between the amounts is \(7680-6720=960\)₹, which is the simple interest for 2 years. Annual interest \(=960/2=480\)₹. Principal \(=6720-(4\times480)=4800\)₹. Rate \(=\frac{480\times100}{4800}=10\%\). Trick: Subtract the amounts to find the interest for the difference in years.

Let \(x=\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}\). Since the expression repeats indefinitely, \(x=\sqrt{12+x}\). Squaring gives \(x^2=12+x\), so \(x^2-x-12=0\). Factoring gives \((x-4)(x+3)=0\). Since a square root is non-negative, \(x=4\). Trick: Let the repeating radical equal \(x\) and solve the resulting quadratic equation.

Initial acid \(=80\times\frac{90}{100}=72\) litres. Acid removed \(=8\times\frac{90}{100}=7.2\) litres. Remaining acid \(=72-7.2=64.8\) litres. Adding pure water does not change the quantity of acid, and the total volume remains 80 litres. New acid percentage \(=\frac{64.8}{80}\times100=81\%\). Trick: The removed solution contains the same 90% acid concentration as the original mixture.

The perpendicular from the centre of a circle bisects the chord. Half the chord is \(16/2=8\) cm. Using the Pythagorean theorem, \(r^2=8^2+6^2=64+36=100\). Therefore, \(r=10\) cm. Trick: Use the right triangle with sides 6 cm, 8 cm, and the radius as the hypotenuse.

The total volume remains unchanged when the metal is melted and reshaped. Since the volume of a sphere is \(V=\frac43\pi r^3\), the number of smaller spheres is \(\frac{\frac43\pi(3)^3}{\frac43\pi(1)^3}=\frac{27}{1}=27\). Trick: Number of smaller spheres \(=(\frac{\text{original radius}}{\text{smaller radius}})^3=(3/1)^3=27\).

Volume of a hemisphere \(=\frac23\pi r^3\). Volume of the cylinder \(=\pi r^2h\). Equating the volumes gives \(\frac23\pi r^3=\pi r^2h\). Cancelling \(\pi r^2\), \(h=\frac{2r}{3}\). Therefore, the height as a fraction of the radius is \(\frac hr=\frac23\). Trick: Equate the volumes and cancel the common radius terms.

Calculate each part: \((3.5)^2=12.25\), \(\sqrt{1.44}=1.2\), and \(2.5\div0.5=5\). Therefore, \(12.25+1.2-5=13.45-5=8.45\). Trick: Follow BODMAS and evaluate the square, square root, and division before addition and subtraction.

Taking the LCM, \(\frac{1}{\sqrt5-2}+\frac{1}{\sqrt5+2}=\frac{(\sqrt5+2)+(\sqrt5-2)}{(\sqrt5-2)(\sqrt5+2)}\). Using \((a-b)(a+b)=a^2-b^2\), this becomes \(\frac{2\sqrt5}{5-4}=2\sqrt5\). Trick: Use the difference of squares for conjugate denominators.

Use the point-slope form \(y-y_1=m(x-x_1)\). Given \((x_1,y_1)=(5,7)\) and \(m=-1\), \(y-7=-(x-5)\). Expanding gives \(y-7=-x+5\), so \(y=-x+12\). Trick: Substitute the point and slope directly into the point-slope formula.

For real and distinct roots, the discriminant must be positive: \(D=b^2-4ac>0\). For \(x^2+kx+16=0\), \(k^2-4(1)(16)>0\), so \(k^2-64>0\). Thus \(k^2>64\), giving \(k>8\) or \(k8\), as correct, but the complete mathematical condition is \(k>8\) or \(k<-8\).

Using \(1+\tan^2x=\sec^2x\) and \(1+\cot^2x=\csc^2x\), the first fraction simplifies to \(\frac{\sec^2x}{\csc^2x}=\tan^2x\). Multiplying the numerator and denominator of the second fraction by \(\cos x\) gives \(\frac{1-\sin x}{1+\sin x}\). Therefore, the expression becomes \(\frac{\tan^2x(1-\sin x)}{1+\sin x}\). Trick: Apply the Pythagorean identities first, then multiply the second fraction by \(\frac{\cos x}{\cos x}\).

Using \(\sin^2\theta+\cos^2\theta=1\), we get \(\cos^2\theta=1-\sin^2\theta\). Substitute \(\sin\theta=\frac7{25}\): \(\cos^2\theta=1-\left(\frac7{25}\right)^2=1-\frac{49}{625}=\frac{576}{625}\). Trick: Use \(\cos^2\theta=1-\sin^2\theta\) directly.

Given \(\tan A=1\), so \(A=45^\circ\) because A is an acute angle in a right triangle. Thus, \(\sin A=\frac1{\sqrt2}\) and \(\cos A=\frac1{\sqrt2}\). Therefore, \(2\sin A\cos A=2\times\frac1{\sqrt2}\times\frac1{\sqrt2}=1\). Trick: Use \(2\sin A\cos A=\sin2A=\sin90^\circ=1\).