14 November 2025 · Shift 1

Partnership profit share In a partnership the profit is divided in the ratio of (capital × time) of each partner. P keeps ₹80,000 for all 3 years: $80000\times 3=240000$. Q keeps ₹1,20,000 for 1 year and only ₹80,000 for the next 2 years: $120000\times 1+80000\times 2=120000+160000=280000$. R keeps ₹1,60,000 for all 3 years: $160000\times 3=480000$. Ratio =240000:280000:480000=6:7:12, so there are 6+7+12=25 equal parts. R's share $=360000\times \frac{12}{25}=172800$, that is ₹1,72,800. Hence the correct option is (d).

Trigonometric ratios $\cos A=\frac{4}{5}$ means base 4 and hypotenuse 5, so the perpendicular is $\sqrt{5^2-4^2}=3$ and $\sin A=\frac{3}{5}$. $\sin B=\frac{12}{13}$ means perpendicular 12 and hypotenuse 13, so the base is $\sqrt{{13}^2-{12}^2}=5$ and $\cos B=\frac{5}{13}$. Now $\sin A\cos B+\cos A\sin B=\frac{3}{5}\times \frac{5}{13}+\frac{4}{5}\times \frac{12}{13}$. $=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$. The expression is in fact the expansion of $\sin \left(A+B\right)$, so the same value can be read off directly. Hence the correct option is (d).

Profit and loss The difference between selling price and cost price is exactly the profit, so profit = ₹600. Profit percentage is always taken on the cost price, so $\frac{30}{100}\times CP=600$. $CP=600\times \frac{100}{30}=2000$, that is ₹2,000. Selling price =2000+600=2600, that is ₹2,600. Hence the correct option is (b).

Percentage and ratio The comparison is made with the number of trucks, so let the number of trucks be 100. Jeeps are 40% more than this, so jeeps =100+40=140. Trucks : jeeps =100:140, and dividing both by 20 gives 5:7. The order asked for is trucks first, so the answer is 5:7 and not 7:5. Hence the correct option is (c).

Compound interest doubling Under compound interest the amount is multiplied by the same factor in equal periods, so every doubling takes the same 6 years. Write 32 as a power of 2: $32=2^5$, so the money has to double 5 times over. Required time $=5\times 6=30$ years. Hence the correct option is (b).

Weighted average The combined average is (total marks of all students) ÷ (total number of students), not the average of the three averages. Total marks $=60\times 30+70\times 50+80\times 30=1800+3500+2400=7700$. Total number of students =30+50+30=110. Combined average $=\frac{7700}{110}=70$. Hence the correct option is (d).

Remainder problem From the first condition the number is of the form 35k+28, so the candidates in increasing order are 28, 63, 98, 133, … Now test them against the second condition: division by 28 must leave remainder 14. $28=28\times 1+0$ and $63=28\times 2+7$ both fail, while $98=28\times 3+14$ works. Check the first condition once more: $98=35\times 2+28$, so both conditions hold together. The smallest such number is therefore 98. Hence the correct option is (b).

Percentage and passing marks Let the maximum marks be x; the pass mark is the same number in both statements, so express it twice. Ravi gets $35\%$ of x and falls 20 marks short, so pass mark =0.35x+20. Sohan gets $45\%$ of x and is 30 marks above the pass mark, so pass mark =0.45x-30. Equate them: 0.35x+20=0.45x-30. $20+30=0.45x-0.35x\Rightarrow 50=0.10x\Rightarrow x=500$. Hence the correct option is (d).

Average speed Average speed uses the total distance and the total time taken, and the halt is part of that total time. Gurugram → Jaipur: time $=\frac{360}{90}=4$ hours. Halt at Jaipur: 1.5 hours. Jaipur → Ajmer: time $=\frac{150}{60}=2.5$ hours. Total distance =360+150=510 km and total time =4+1.5+2.5=8 hours. Average speed $=\frac{510}{8}=63.75$ km/h. Hence the correct option is (c).

Time and work P finishes the whole work in 40 days, so in one day he finishes $\frac{1}{40}$ of it. Work done in 15 days $=15\times \frac{1}{40}=\frac{15}{40}$. Reducing the fraction, $\frac{15}{40}=\frac{3}{8}$ of the work. Hence the correct option is (a).

Profit after discount on marked price Let the cost price of the item be ₹100. It is marked $40\%$ above the cost price, so the marked price =100+40=140. A discount of $20\%$ is allowed on the marked price, so the selling price $=140\times \frac{80}{100}=112$. Profit =112-100=12, hence profit percentage $=\frac{12}{100}\times 100=12\%$. Shortcut: net effect of a $+40\%$ mark-up and a $-20\%$ discount $=40-20-\frac{40\times 20}{100}=12\%$. Hence option (c) is correct.

Simple interest and amount Formula: $SI=\frac{P\times R\times T}{100}$, where P is the principal, R the rate and T the time. For the first 2 years: $4800=\frac{P\times 12\times 2}{100}$. $P=\frac{4800\times 100}{24}=20000$, so the sum lent is ₹20,000. Interest for 6 years $=\frac{20000\times 12\times 6}{100}=14400$. Amount =P+SI=20000+14400=34400, i.e. ₹34,400. Hence option (c) is correct.

Algebraic identity, square of a sum Identity: ${\left(a+b\right)}^2=a^2+b^2+2ab$. Substituting a+b=9 and $a^2+b^2=47$: $9^2=47+2ab$. 81-47=2ab, so 2ab=34. $ab=\frac{34}{2}=17$. Hence option (a) is correct.

Square root by factorisation Break 9604 into prime factors: $9604=4\times 2401=2^2\times 7^4$. $\sqrt{9604}=\sqrt{2^2\times 7^4}=2\times 7^2=2\times 49=98$. Quick check: 9604 lies between ${90}^2=8100$ and ${100}^2=10000$, and it ends in 4, so its root must end in 2 or 8 — only 98 fits. Hence option (b) is correct.

Mixture, change of ratio Milk : water =5:6, so the mixture has 5+6=11 equal parts in 99 litres, i.e. 1 part $=\frac{99}{11}=9$ litres. Milk $=5\times 9=45$ litres and water $=6\times 9=54$ litres. Only water is added, so the milk stays 45 litres. Let x litres of water be added. $\frac{45}{54+x}=\frac{3}{4}$ $45\times 4=3\left(54+x\right)\Rightarrow 180=162+3x\Rightarrow 3x=18\Rightarrow x=6$. Hence 6 litres of water must be added — option (c).

Sphere melted and recast When a solid is melted and recast, the total volume does not change. Volume of a sphere $=\frac{4}{3}\pi r^3$. Number of small spheres $=\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3}={\left(\frac{R}{r}\right)}^3$. $={\left(\frac{15}{3}\right)}^3=5^3=125$. Hence option (c) is correct.

Volume of a hemispherical tank Volume of a hemisphere $=\frac{2}{3}\pi r^3$. $=\frac{2}{3}\times \pi \times 6^3=\frac{2}{3}\times \pi \times 216=144\pi$ cubic metres. Since 1 cubic metre =1000 litres, the capacity $=144\pi \times 1000=144000\pi$ litres. Time $=\frac{\mathrm{capacity}}{\mathrm{rate}}=\frac{144000\pi }{1440\pi }=100$ minutes. Hence option (b) is correct.

Recurring decimal to fraction Let $x=0.565656\dots$ The block '56' has 2 digits, so multiply by 100: $100x=56.565656\dots$ Subtracting the first equation from the second: 100x-x=56, i.e. 99x=56. $x=\frac{56}{99}$; since $56=2^3\times 7$ and $99=3^2\times 11$ share no common factor, this is already the simplest form. Rule: a purely recurring decimal equals the repeating block divided by as many 9s as there are repeating digits. Hence option (a) is correct.

Largest circles cut from a circle Let the radius of the sheet be R and the radius of each plate be r. Two equal circles of the largest possible size fit inside a circle when they lie along a diameter, touching each other at the centre and touching the sheet from inside. Then the two diameters together make up the diameter of the sheet: $2\times 2r=2R$, so 2r=R. Circumference of a plate $=2\pi r=\pi R=\frac{2\pi R}{2}$, which is exactly half the circumference of the sheet. $=\frac{424}{2}=212$ cm. Hence option (d) is correct.

Laws of indices Write every term as a power of 2: $512=2^9$, $\sqrt{8}=8^{\frac{1}{2}}={\left(2^3\right)}^{\frac{1}{2}}=2^{\frac{3}{2}}$ and ${64}^4={\left(2^6\right)}^4=2^{24}$. So $2^9\times {\left(2^{\frac{3}{2}}\right)}^n=2^{24}$. $2^{9+\frac{3n}{2}}=2^{24}$. The bases are equal, so the powers must be equal: $9+\frac{3n}{2}=24$. $\frac{3n}{2}=15\Rightarrow n=10$. Hence option (b) is correct.

Equation of a line, point-slope form Point-slope form of a straight line: $y-y_1=m\left(x-x_1\right)$. Here the slope m=-6 and the point is $\left(x_1,y_1\right)=\left(4,5\right)$. y-5=-6(x-4) y-5=-6x+24 y=-6x+29 Check: putting x=4 gives y=-24+29=5, which is the given point. Hence option (d) is correct.

Algebraic identities, sum and difference Squaring $x+y=\sqrt{19}$: $x^2+y^2+2xy=19$ … (i) Note $\sqrt{1}=1$, so squaring x-y=1: $x^2+y^2-2xy=1$ … (ii) Adding (i) and (ii): $2\left(x^2+y^2\right)=20$, so $x^2+y^2=10$. Subtracting (ii) from (i): 4xy=18, so $xy=\frac{9}{2}$. $\frac{x^2+y^2}{xy}=\frac{10}{\frac{9}{2}}=10\times \frac{2}{9}=\frac{20}{9}$. Hence option (c) is correct.

Trigonometric simplification Take the common denominator: $\left(1-\sin x\right)\left(1+\sin x\right)=1-{\sin }^{2}x={\cos }^{2}x$. So the expression $=\frac{\left(1+\cos x\right)\left(1+\sin x\right)+\left(1-\cos x\right)\left(1-\sin x\right)}{{\cos }^{2}x}$. Numerator $=\left(1+\sin x+\cos x+\sin x\cos x\right)+\left(1-\sin x-\cos x+\sin x\cos x\right)$. The $\sin x$ and $\cos x$ terms cancel, leaving $2+2\sin x\cos x=2\left(1+\cos x\sin x\right)$. Therefore the expression $=\frac{2\left(1+\cos x\sin x\right)}{{\cos }^{2}x}$. Hence option (b) is correct.

Trigonometric ratio to sides $\cot \theta =\frac{\mathrm{base}}{\mathrm{perpendicular}}=\frac{15}{8}$, so base =15k and perpendicular =8k. By the Pythagoras theorem, hypotenuse $=\sqrt{{\left(15k\right)}^2+{\left(8k\right)}^2}=\sqrt{225k^2+64k^2}=\sqrt{289k^2}=17k$. Taking the smallest whole-number case k=1, the hypotenuse is 17 units. Hence option (a) is correct.

Finding cos and tan from sin $\sin A=\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}=\frac{5}{13}$, so perpendicular =5k and hypotenuse =13k. By the Pythagoras theorem, base $=\sqrt{{\left(13k\right)}^2-{\left(5k\right)}^2}=\sqrt{169k^2-25k^2}=\sqrt{144k^2}=12k$. $\cos A=\frac{\mathrm{base}}{\mathrm{hypotenuse}}=\frac{12k}{13k}=\frac{12}{13}$. $\tan A=\frac{\mathrm{perpendicular}}{\mathrm{base}}=\frac{5k}{12k}=\frac{5}{12}$. Hence option (a) is correct.