SSC CHSL Tier 1 2025 14 Nov Shift 1st

Profit is divided in the ratio of capital multiplied by time. A's investment = \(75000\times3=225000\). B's investment = \(100000\times1+80000\times2=260000\). C's investment = \(150000\times3=450000\). Investment ratio = \(225:260:450=45:52:90\). C's share = \(\frac{90}{187}\times450000\) = ₹216,577.54 approximately. Trick: Calculate capital multiplied by time for each partner before dividing the profit.

Using the identity \(\sin A\cos B+\cos A\sin B=\sin(A+B)\). Assuming acute angles, \(\cos A=\sqrt{1-\sin^2 A}=\frac{4}{5}\) and \(\sin B=\sqrt{1-\cos^2 B}=\frac{12}{13}\). Therefore, \(\frac{3}{5}\times\frac{5}{13}+\frac{4}{5}\times\frac{12}{13}=\frac{15+48}{65}=\frac{63}{65}\). Trick: Recognise the sine addition identity.

Profit = ₹500. Since profit is 25% of cost price, \(CP=\frac{500\times100}{25}=2000\)₹. Selling price = Cost price + Profit = ₹2000 + ₹500 = ₹2500. Trick: A profit of 25% means the profit is one-fourth of the cost price.

Assume the number of trucks is 100. Cars = \(100+20=120\). Required ratio = \(100:120=5:6\). Trick: If one quantity is 20% more than another, the ratio of the smaller to the larger is \(100:120=5:6\).

The amount doubles every 7 years. Since \(16=2^4\), the amount must double 4 times. Time = \(4\times7=28\) years. Trick: Express the required multiple as a power of 2.

Combined average = \(\frac{50\times20+60\times30+70\times50}{20+30+50}\) = \(\frac{1000+1800+3500}{100}\) = \(\frac{6300}{100}=63\). Trick: Divide the total marks of all groups by the total number of students.

Check 117: \(117=45\times2+27\), so the remainder is 27 when divided by 45. Also, \(117=27\times4+9\), so the remainder is 9 when divided by 27. Therefore, the smallest such number is 117. Trick: Check the given options against both remainder conditions.

Let maximum marks be x and passing marks be p. \(0.4x=p-30\) and \(0.5x=p+20\). Subtracting the equations gives \(0.1x=50\). Therefore, \(x=\frac{50}{0.1}=500\). Trick: The difference in percentages is 10%, and the difference in marks is \(30+20=50\). Hence, maximum marks = \(\frac{50\times100}{10}=500\).

Total distance = \(240+130=370\) km. First journey time = \(\frac{240}{60}=4\) hours. Second journey time = \(\frac{130}{50}=2.6\) hours. Including the break, total time = \(4+1+2.6=7.6\) hours. Average speed = \(\frac{370}{7.6}=48.68\) km/h approximately. Trick: Average speed = total distance divided by total time, including the break.

Work completed in one day = \(\frac{1}{30}\). Work completed in 6 days = \(6\times\frac{1}{30}=\frac{1}{5}\). Trick: Work completed = days worked divided by total days required.

Assume CP = ₹100. Marked price = \(100\times\frac{120}{100}=120\)₹. Selling price after 10% discount = \(120\times\frac{90}{100}=108\)₹. Profit = ₹108 - ₹100 = ₹8. Profit percentage = \(\frac{8}{100}\times100=8\%\). Trick: Net profit percentage = \(20-10-\frac{20\times10}{100}=8\%\).

Using \(SI=\frac{P\times R\times T}{100}\), \(2400=\frac{P\times8\times3}{100}\). Therefore, \(P=10000\)₹. Interest for 5 years = \(\frac{10000\times8\times5}{100}=4000\)₹. Amount = ₹10000 + ₹4000 = ₹14000. Trick: Interest for 5 years = \(2400\times\frac{5}{3}=4000\)₹.

Using \((x+y)^2=x^2+y^2+2xy\), \(10^2=58+2xy\). Therefore, \(100-58=2xy\), so \(42=2xy\) and \(xy=21\). Trick: \(2xy=(x+y)^2-(x^2+y^2)\).

Using the identity \((a+b)^2=a^2+2ab+b^2\), \(88^2=(80+8)^2=6400+1280+64=7744\). Therefore, \(\sqrt{7744}=88\). Trick: Remember common perfect squares.

Milk = \(75\times\frac{2}{3}=50\) litres. Water = \(75\times\frac{1}{3}=25\) litres. Let x litres of water be added. Then \(\frac{50}{25+x}=\frac{1}{2}\). Therefore, \(100=25+x\), so \(x=75\) litres. Trick: Required water = \(2\times50=100\) litres. Additional water = \(100-25=75\) litres.

The total volume remains unchanged. Number of smaller spheres = \(\frac{R^3}{r^3}=\left(\frac{12}{2}\right)^3=6^3=216\). Trick: Number of smaller spheres = \((R/r)^3\).

Volume of a hemisphere = \(\frac{2}{3}\pi r^3\). For \(r=3\), volume = \(\frac{2}{3}\pi\times3^3=18\pi\) m^3. In litres, volume = \(18\pi\times1000=18000\pi\) litres. Time = \(\frac{18000\pi}{1800\pi}=10\) minutes. Trick: Divide the total volume by the filling rate.

Let \(x=0.454545\ldots\). Since two digits repeat, \(100x=45.454545\ldots\). Subtracting, \(100x-x=45\), so \(99x=45\). Therefore, \(x=\frac{45}{99}=\frac{5}{11}\). Trick: Write the repeating two-digit block over 99 and simplify.

Let the radius of the original circular sheet be R. Each of the two identical maximum-sized circles has radius R/2. Original circumference = \(2\pi R=352\) cm. Circumference of each smaller circle = \(2\pi\times\frac{R}{2}=\pi R=\frac{352}{2}=176\) cm. Trick: Each smaller circle has half the original diameter, so its circumference is half the original circumference.

Write all terms with base 7. \(343=7^3\), \(\sqrt7=7^{1/2}\), and \(49^4=(7^2)^4=7^8\). Thus, \(7^3\times7^{n/2}=7^8\). Therefore, \(3+\frac{n}{2}=8\), so \(\frac{n}{2}=5\) and \(n=10\). Trick: Convert all terms to the same base and compare exponents.

Using the point-slope formula \(y-y_1=m(x-x_1)\), where \(m=-5\) and \((x_1,y_1)=(2,3)\), we get \(y-3=-5(x-2)\). Therefore, \(y-3=-5x+10\), so \(y=-5x+13\). Trick: Substitute the given point into the options and check the slope.

Given \(a+b=\sqrt{17}\) and \(a-b=1\). Squaring gives \((a+b)^2=17\) and \((a-b)^2=1\). Adding, \(2(a^2+b^2)=18\), so \(a^2+b^2=9\). Subtracting, \(4ab=16\), so \(ab=4\). Therefore, \(\frac{a^2+b^2}{ab}=\frac{9}{4}\). Trick: Use the identities for \((a+b)^2\) and \((a-b)^2\).

Taking the LCM, the numerator becomes \((1+\sin x)(1+\cos x)+(1-\sin x)(1-\cos x)\), and the denominator becomes \((1-\cos x)(1+\cos x)\). Expanding the numerator gives \(2+2\sin x\cos x\). The denominator is \(1-\cos^2x=\sin^2x\). Therefore, the expression simplifies to \(\frac{2(1+\sin x\cos x)}{\sin^2x}\). Trick: Use the identity \(1-\cos^2x=\sin^2x\).

Since \(\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{12}{5}\), take the perpendicular as 12 and the base as 5. By the Pythagorean theorem, \(H^2=12^2+5^2=144+25=169\). Therefore, \(H=\sqrt{169}=13\). Trick: Recognise the Pythagorean triplet 5, 12, 13.

Given \(\cos A=\frac{12}{13}\). Using \(\sin^2A+\cos^2A=1\), \(\sin^2A=1-\left(\frac{12}{13}\right)^2=\frac{25}{169}\). Therefore, \(\sin A=\frac{5}{13}\). Now, \(\cot A=\frac{\cos A}{\sin A}=\frac{12/13}{5/13}=\frac{12}{5}\). Trick: Recognise the 5-12-13 right triangle.