13 November 2025 · Shift 1

Profit and loss Let the cost price be x. At a profit of 8% the selling price is $\frac{108}{100}x=1.08x$. In the second case the cost price is 12% less, so the new cost price is 0.88x, and the selling price is ₹36 less, so the new selling price is 1.08x-36. A gain of 20% on the new cost price gives $1.08x-36=1.2\times 0.88x=1.056x$. 1.08x-1.056x=36, that is 0.024x=36. $x=\frac{36}{0.024}=1500$. So the cost price is ₹1500 and option (d) is correct.

Partnership ratio Let 4A=6B=5C=k, where A, B and C stand for the three capitals. Then $A=\frac{k}{4}$, $B=\frac{k}{6}$ and $C=\frac{k}{5}$. So $A:B:C=\frac{1}{4}:\frac{1}{6}:\frac{1}{5}$. Multiplying every part by the LCM of 4, 6 and 5, that is 60, gives A:B:C=15:10:12. Total number of parts =15+10+12=37. C's share $=1950\times \frac{12}{37}=\frac{23400}{37}=632.4$ approximately. Hence option (a) is correct.

Trigonometric identity The expression $\sin A\cos B-\cos A\sin B$ is the expansion of $\sin \left(A-B\right)$. From $\cos A=\frac{24}{25}$, $\sin A=\sqrt{1-{\left(\frac{24}{25}\right)}^2}=\sqrt{\frac{49}{625}}=\frac{7}{25}$. From $\sin B=\frac{7}{25}$, $\cos B=\sqrt{1-{\left(\frac{7}{25}\right)}^2}=\sqrt{\frac{576}{625}}=\frac{24}{25}$. So the value is $\frac{7}{25}\times \frac{24}{25}-\frac{24}{25}\times \frac{7}{25}=\frac{168}{625}-\frac{168}{625}=0$. The two angles have the same sine and the same cosine, so A=B and $\sin \left(A-B\right)=\sin 0^{\circ }=0$. Hence option (c) is correct.

Percentage to ratio Note that $12.5\%=\frac{12.5}{100}=\frac{1}{8}$, so take the number of bananas as 8. Guavas are $\frac{1}{8}$ more than bananas, so guavas $=8+\frac{1}{8}\times 8=8+1=9$. Therefore guavas : bananas =9:8. Hence option (a) is correct.

Compound interest Under compound interest a sum is multiplied by the same factor in every equal interval of time. Here the factor is 5 for a block of 8 years, so after another 8 years the sum becomes $5\times 5=25$ times. Hence 25 times the original amount is reached in 8+8=16 years. So option (a) is correct.

LCM and remainders Compare each divisor with its remainder: 16-8=8, 20-12=8 and 24-16=8. The shortfall is the same 8 in every case, so the number is 8 less than a common multiple of 16, 20 and 24. LCM of 16, 20 and 24 =240. So the least such number is 240-8=232. Hence option (b) is correct.

Average of a series Let the smallest even number be a; the sixteen consecutive even numbers are $a,a+2,a+4,\dots ,a+30$. For numbers in arithmetic progression the average equals the mean of the first and the last term. Average $=\frac{a+\left(a+30\right)}{2}=a+15$. So a+15=111, which gives a=111-15=96. Hence the smallest number is 96 and option (b) is correct.

Successive percentage change A rise of 30% multiplies the population by $\frac{130}{100}$ and the fall of 25% in the next year multiplies it by $\frac{75}{100}$. Take the starting population as 100. After two years it is $100\times \frac{130}{100}\times \frac{75}{100}=100\times 0.975=97.5$. The net change is 97.5-100=-2.5, that is a decrease of 2.5%. Hence option (b) is correct.

Trains and platforms To cross a man the train covers only its own length; to cross a platform it covers its own length plus the length of the platform. Let the length of the train be L metres and its speed v m/s. Then L=18v and L+200=24v. Subtracting the first equation from the second, 200=6v, so $v=\frac{200}{6}=\frac{100}{3}$ m/s. Therefore $L=18\times \frac{100}{3}=600$. Hence the train is 600 m long and option (a) is correct.

Time and work Take the total work as the LCM of 24 and 60, that is 120 units. Anuj does $\frac{120}{24}=5$ units a day and Bhagat does $\frac{120}{60}=2$ units a day. Working together they finish 5+2=7 units in one day. Time required $=\frac{120}{7}=17\frac{1}{7}$ days. Hence option (c) is correct.

Successive discounts Marked price = ₹20,000 and the first discount is 15%, so the price after it is $20000\times \frac{85}{100}=17000$. The second discount is applied on ₹17,000 (not on the marked price), and it brings the price down to ₹12,750. Discount given in the second step =17000-12750=4250 Second discount $=\frac{4250}{17000}\times 100=25\%$ Hence option (d) is correct.

Simple interest — rate Let the principal be P; after 13 years the amount is $\frac{19}{6}P$. Simple interest earned $=\frac{19}{6}P-P=\frac{13}{6}P$ Formula: $SI=\frac{P\times R\times T}{100}$ $\frac{13}{6}P=\frac{P\times R\times 13}{100}$ $R=\frac{100}{6}=16.67\%$ per annum Hence option (d) is correct.

Algebraic identity Identity: $A^2-B^2=\left(A+B\right)\left(A-B\right)$ Substituting the given values: $96=16\times \left(A-B\right)$ $A-B=\frac{96}{16}=6$ (Solving further, A=11 and B=5, which indeed satisfy both conditions.) Hence option (b) is correct.

Square root 7569 lies between ${80}^2=6400$ and ${90}^2=8100$, so its square root is a two-digit number starting with 8. The number ends in 9, and only $3^2=9$ and $7^2=49$ end in 9, so the root must end in 3 or 7. Checking the two candidates: ${83}^2=6889$ and ${87}^2=7569$. So $\sqrt{7569}=87$, and option (d) is correct.

Mixture — replacement Total ratio parts =11+9=20, so milk $=120\times \frac{11}{20}=66$ litres and water $=120\times \frac{9}{20}=54$ litres. The 20 litres taken out is mixture, so it carries milk and water in the same 11 : 9 ratio: milk $=20\times \frac{11}{20}=11$ litres, water $=20\times \frac{9}{20}=9$ litres. Left in the vessel: milk =66-11=55 litres, water =54-9=45 litres. Now 20 litres of pure milk is added, so milk =55+20=75 litres while water stays 45 litres. New ratio =75:45=5:3 Hence option (a) is correct.

Volume — melting and recasting Radius of the cannonball $=\frac{36}{2}=18$ cm. Melting only changes the shape, not the amount of metal, so volume of sphere = volume of the block. $\frac{4}{3}\pi r^3=l\times b\times h$ $\frac{4}{3}\times \frac{22}{7}\times {18}^3=66\times 18\times h$ $\frac{4}{3}\times \frac{22}{7}\times 5832=1188h$ $h=\frac{171072}{7\times 1188}=\frac{171072}{8316}=20.57$ Hence the height of the block is 20.57 cm, so option (a) is correct.

Volume — hemisphere and cylinder Volume of the hemispherical bowl $=\frac{2}{3}\pi r^3$ with r=42 cm. $=\frac{2}{3}\times \frac{22}{7}\times 42\times 42\times 42=155232$ cubic cm For one bottle, radius $=\frac{14}{2}=7$ cm and height =16 cm. Volume of one bottle $=\pi r^{2}h=\frac{22}{7}\times 7\times 7\times 16=2464$ cubic cm Number of bottles $=\frac{155232}{2464}=63$ Hence 63 bottles are needed to empty the bowl, so option (b) is correct.

Simplification of fractions By BODMAS the bracket is taken first: $\frac{5}{8}+\frac{3}{8}=\frac{8}{8}=1$ The expression becomes $\frac{1}{4}\times 1÷\frac{1}{4}$ Dividing by a fraction means multiplying by its reciprocal: $\frac{1}{4}\times \frac{4}{1}=1$ Hence the value is 1, so option (d) is correct.

Laws of exponents Write 49 as a power of 7: $49=7^2$ Law of exponents: $a^m\times a^n=a^{m+n}$ $7^2\times 7^3=7^{2+3}=7^5$ Comparing $7^5=7^x$, the bases are equal, so x=5. Hence option (b) is correct.

Circumference and revolutions In one complete revolution a wheel covers a distance equal to its circumference. Circumference $=\pi d=\frac{22}{7}\times 49=154$ cm Distance in 1500 revolutions $=154\times 1500=231000$ cm Converting: 231000 cm =2310 m =2.31 km Hence option (a) is correct.

Equation of perpendicular line Comparing y=5x+6 with y=mx+c, the slope of the given line is $m_1=5$. For two perpendicular lines $m_1\times m_2=-1$, so $m_2=-\frac{1}{5}$. Point-slope form through (3,4): $y-4=-\frac{1}{5}\left(x-3\right)$ $y=-\frac{1}{5}x+\frac{3}{5}+4$ $y=-\frac{1}{5}x+\frac{23}{5}$ Hence option (d) is correct.

Perfect-square identity Identity: ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca$ Take a=x, b=3y, c=-4z; then $a^2+b^2+c^2=x^2+9y^2+16z^2$ and 2ab+2bc+2ca=6xy-24yz-8zx. So the whole expression is the perfect square ${\left(x+3y-4z\right)}^2$. Substituting x=6, y=4, z=3: x+3y-4z=6+12-12=6 Required value $=6^2=36$ Hence option (a) is correct.

Trigonometric simplification Write both functions in terms of sine and cosine: $cosecx=\frac{1}{\sin x}$ and $\cot x=\frac{\cos x}{\sin x}$. Numerator $=\frac{1}{\sin x}-\frac{\cos x}{\sin x}=\frac{1-\cos x}{\sin x}$ Denominator $=\frac{1}{\sin x}+\frac{\cos x}{\sin x}=\frac{1+\cos x}{\sin x}$ On dividing, the common $\sin x$ cancels: $\frac{1-\cos x}{\sin x}\times \frac{\sin x}{1+\cos x}=\frac{1-\cos x}{1+\cos x}$ Hence option (b) is correct.

Trigonometric ratios Since $\cos 3A=\frac{\mathrm{base}}{\mathrm{hypotenuse}}=\frac{15}{17}$, take base =15 and hypotenuse =17. By Pythagoras, perpendicular $=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8$ $\sin 3A=\frac{8}{17}$ The same result follows from ${\sin }^{2}3A=1-{\left(\frac{15}{17}\right)}^2=\frac{64}{289}$. Hence option (c) is correct.

Right triangle trigonometry The right angle is at R, so for $\angle P$ the base is PR, the perpendicular is QR and the hypotenuse is PQ. $\cot P=\frac{\mathrm{base}}{\mathrm{perpendicular}}=\frac{PR}{QR}=2$, so take PR=2k and QR=k. By Pythagoras, $PQ=\sqrt{{\left(2k\right)}^2+k^2}=\sqrt{5k^2}=k\sqrt{5}$ $\sin P=\frac{QR}{PQ}=\frac{k}{k\sqrt{5}}=\frac{1}{\sqrt{5}}$ and $\cos P=\frac{PR}{PQ}=\frac{2k}{k\sqrt{5}}=\frac{2}{\sqrt{5}}$ $2\sin P\cos P=2\times \frac{1}{\sqrt{5}}\times \frac{2}{\sqrt{5}}=\frac{4}{5}$ The value of k cancels, so the ratio alone fixes the answer. Hence option (a) is correct.