25 November 2025 · Shift 3

Discount and marked price A discount of $20\%$ leaves $80\%$ of the marked price as the selling price. Formula: $SP=MP\times \frac{100-d}{100}$ $960=MP\times \frac{80}{100}$ $MP=960\times \frac{100}{80}=1200$ So the marked price is ₹1,200 and option (c) is correct.

Average on replacement Total weight of the 9 crates $=9\times 94.5=850.5$ kg. After the exchange the total becomes $9\times 97.1=873.9$ kg. The number of crates has not changed, so the whole rise comes from the exchange: 873.9 - 850.5 = 23.4 kg. Weight of the new crate = 89.2 + 23.4 = 112.6 kg. The five crates added later come after this and do not affect the answer, so option (c) is correct.

Cone: slant height and CSA Formula: slant height $l=\sqrt{h^2+r^2}$ $l=\sqrt{{15}^2+8^2}=\sqrt{225+64}=\sqrt{289}=17$ cm Formula: curved surface area $=\pi rl$ $=\frac{22}{7}\times 8\times 17=\frac{2992}{7}=427.42$ sq cm So the slant height is 17 cm and the curved surface area is 427.42 sq cm, and option (a) is correct.

Trigonometric ratios $\sin A=\frac{8}{17}$ means the side opposite A is 8 and the hypotenuse is 17. Third side $=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ So $\cos A=\frac{15}{17}$ and $\cot A=\frac{15}{8}$. $\cot A-\cos A=\frac{15}{8}-\frac{15}{17}=\frac{15\times 17-15\times 8}{136}$ $=\frac{255-120}{136}=\frac{135}{136}$ Hence the value is $\frac{135}{136}$ and option (d) is correct.

Common internal tangent Formula: length of a common internal tangent $=\sqrt{d^2-{\left(r_1+r_2\right)}^2}$ $=\sqrt{{13}^2-{\left(7+5\right)}^2}$ $=\sqrt{169-144}=\sqrt{25}=5$ So $A_1{A}_2=5$ cm and option (c) is correct.

Height and distance The cable is the hypotenuse of a right triangle and the distance asked for is the side lying along the ground. Formula: $\cos {60}^{\circ }=\frac{\mathrm{base}}{\mathrm{hypotenuse}}$ $\frac{1}{2}=\frac{\mathrm{base}}{540\sqrt{2}}$ base $=\frac{540\sqrt{2}}{2}=270\sqrt{2}$ So the distance from A to the foot of the cliff is $270\sqrt{2}$ metres and option (c) is correct.

Ratio and cost Let the bracelet hold 5x units of gold and 8x units of copper. Let copper cost p per unit; then gold costs 12p per unit. Cost of the copper $=8x\times p=32000$, so xp = 4000. Cost of the gold $=5x\times 12p=60xp=60\times 4000=240000$ Total cost = 240000 + 32000 = 272000 So the bracelet costs ₹2,72,000 and option (c) is correct.

Mixture and alligation Take the quantities as 2 kg, 2 kg and 3 kg, so the mixture weighs 7 kg. The money spent on the parts must equal the value of the mixture. 2(140) + 2(150) + 3x = 7(160) 280 + 300 + 3x = 1120 3x = 540, so x = 180. So the third variety costs ₹180 per kg and option (a) is correct.

Time and work Ajay works three times as fast as Vipin, so efficiency Ajay : Vipin = 3 : 1. Time is inversely proportional to efficiency, so time Ajay : Vipin = 1 : 3. The gap of 2 parts stands for 40 days, so 1 part = 20 days. Ajay therefore takes 20 days and Vipin 60 days. Take the whole job as 60 units; Ajay does 3 units a day and Vipin 1 unit a day. Together they do 4 units a day, so the time $=\frac{60}{4}=15$ days. Hence option (c) is correct.

Trigonometric identity Given $\cot \theta +\tan \theta =8$. Square both sides: ${\left(\cot \theta +\tan \theta \right)}^2=8^2$ ${\cot }^2\theta +{\tan }^2\theta +2\cot \theta \tan \theta =64$ But $\cot \theta \times \tan \theta =1$, so the middle term is just 2. ${\cot }^2\theta +{\tan }^2\theta =64-2=62$ So the value is 62 and option (c) is correct.

Speed, time and distance Formula: Distance = Speed × Time, applied separately to each part of the journey. Distance covered by car $=65\times 6.6=429$ km. Distance covered by bus $=57.5\times 6.2=356.5$ km. Total distance = 429 + 356.5 = 785.5 km. Hence option (d) is correct.

Chord and central angle OP and OQ are radii of the same circle, so OP = OQ = 12 cm and triangle POQ is isosceles. Its base angles are equal and together they make ${180}^{\circ }-{60}^{\circ }={120}^{\circ }$, so each of them is ${60}^{\circ }$. All three angles are ${60}^{\circ }$, so triangle POQ is equilateral and PQ = OP = 12 cm. The perpendicular OM gives the same result: $PM=OP\sin {30}^{\circ }=12\times \frac{1}{2}=6$ cm, so $PQ=2\times 6=12$ cm. Hence option (c) is correct.

Percentage change in volume For a sphere $V=\frac{4}{3}\pi R^3$, so the volume varies as the cube of the radius. New radius $=R+20\%$ of R = 1.2R. New volume : old volume $={\left(1.2R\right)}^3:R^3=1.728:1$. Increase in volume = 1.728 - 1 = 0.728 of the original. Percentage increase $=0.728\times 100=72.8\%$. Hence option (b) is correct.

Parallelogram with perpendicular diagonals The diagonals of a parallelogram always bisect each other; here they also meet at ${90}^{\circ }$, so the figure is a rhombus and all four sides are equal. Half of each diagonal: $\frac{34}{2}=17$ cm and $\frac{16}{2}=8$ cm. Every side is the hypotenuse of a right triangle with legs 17 cm and 8 cm. Side $=\sqrt{{17}^2+8^2}=\sqrt{289+64}=\sqrt{353}$ cm. When the diagonals are perpendicular, area $=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 34\times 16=272$ cm². Hence option (a) is correct.

Successive discounts and profit Cost price = ₹520, and the mark-up is 20% of it. Marked price $=520\times 1.2=624$. Successive discounts of 10% and then 5% leave $0.9\times 0.95=0.855$ of the marked price. Selling price $=624\times 0.855=533.52$. Since the selling price exceeds the cost price, it is a profit: 533.52 - 520 = 13.52. Profit percent $=\frac{13.52}{520}\times 100=2.6\%$. Hence option (a) is correct.

Partnership profit sharing Both partners invest for the same period, so the profit is divided in the ratio of their capitals, 3 : 5. Total number of parts = 3 + 5 = 8. Shivan's share $=\frac{3}{8}\times 16000=6000$, i.e. Rs 6,000. Hence option (d) is correct.

Rectangle from ratio and perimeter Let the length be 8x and the width be 5x. Perimeter = 2(8x + 5x) = 26x. $26x=104\Rightarrow x=4$. So length $=8\times 4=32$ cm and width $=5\times 4=20$ cm. Area $=32\times 20=640$ cm². Hence option (a) is correct.

Average and missing value Total of all six days = average × number of days $=3\times 6=18$ L. Total of the first five days = 2.5 + 3 + 2 + 3.5 + 3 = 14 L. Intake on the 6th day = 18 - 14 = 4 L. Hence option (c) is correct.

Distance of a chord from centre The perpendicular drawn from the centre to a chord bisects it, so each half is $\frac{36}{2}=18$ cm. The radius, the half-chord and this perpendicular distance form a right-angled triangle. $d=\sqrt{{30}^2-{18}^2}=\sqrt{900-324}=\sqrt{576}=24$ cm. Hence option (b) is correct.

Intersecting chords theorem When two chords cut each other inside a circle, the products of the two parts of each chord are equal: $AP\times PB=CP\times PD$. $7\times 9=5\times PD$ $PD=\frac{63}{5}=12.6$ cm. CD = CP + PD = 5 + 12.6 = 17.6 cm. Hence option (b) is correct.

x-intercept of a line The x-intercept is the x-coordinate of the point where the line meets the x-axis, and every point on the x-axis has y = 0. Put y = 0 in 5x - 2y = 20: 5x - 0 = 20. $x=\frac{20}{5}=4$. Hence option (b) is correct.

Chain of percentage changes Take T = 100 and move down the chain, one statement at a time. S is 40% less than T: $S=0.6\times 100=60$. R is 10% more than S: $R=1.1\times 60=66$. Q is 20% less than R: $Q=0.8\times 66=52.8$. P is 25% more than Q: $P=1.25\times 52.8=66$. Comparing with T = 100: P is short by 100 - 66 = 34, i.e. $34\%$ less than T. Hence option (b) is correct.

Simple interest rate Let the sum be P. It becomes 3P, so the simple interest earned is 3P - P = 2P. Time = 7 years 6 months $=7\frac{1}{2}=\frac{15}{2}$ years. Formula: $SI=\frac{P\times R\times T}{100}$. $2P=\frac{P\times R\times 15}{100\times 2}$ $R=\frac{2\times 100\times 2}{15}=\frac{400}{15}=26.66$ percent. Hence option (c) is correct.

Simple interest difference Under simple interest the interest of every year is the same, so the difference between 12 years' interest and 10 years' interest is just the interest of 2 years. $\frac{P\times 15\times 2}{100}=120$ $\frac{30P}{100}=120$ $P=\frac{120\times 100}{30}=400$, i.e. Rs. 400. Hence option (b) is correct.

Pipes and cisterns Let the capacity of the tank be the LCM of 20, 24 and 28, i.e. 840 units. $P+Q=\frac{840}{20}=42$, $Q+R=\frac{840}{24}=35$, $R+P=\frac{840}{28}=30$ units per minute. Adding the three: 2(P + Q + R) = 42 + 35 + 30 = 107, so P + Q + R = 53.5 units per minute. R = (P + Q + R) - (P + Q) = 53.5 - 42 = 11.5 units per minute. R runs alone for M minutes and then all three run for 8 minutes, filling 840 units in all: $11.5M+53.5\times 8=840$ 11.5M = 840 - 428 = 412 $M=\frac{412}{11.5}=\frac{824}{23}$ minutes. Hence option (d) is correct.