25 November 2025 · Shift 1

Discount Discount is always reckoned on the marked price. Discount $=8400\times \frac{15}{100}=1260$. Selling price = marked price - discount =8400-1260=7140. A quicker route is to use the surviving 85% at once: $8400\times \frac{85}{100}=7140$. So the selling price is Rs. 7,140 and option (b) is correct.

Average speed The two halves of the journey cover the same distance, so the plain average of the speeds is wrong; the harmonic mean is needed. Formula: average speed $=\frac{2xy}{x+y}$, where x and y are the two speeds. Substituting x=20 and y=30 gives $\frac{2\times 20\times 30}{20+30}$. $=\frac{1200}{50}=24$. So the average speed for the whole journey is 24 km/h and option (a) is correct.

Surface area of a cone A tent has no floor, so only the curved surface of the cone is made of canvas. Formula: curved surface area $=\pi rl$, with r=7 and l=10. For one tent, $\frac{22}{7}\times 7\times 10=220$ square metres. For 20 tents, $220\times 20=4400$ square metres. Hence option (b) is correct.

Profit and loss Purchase price of the 40 cheaper shirts $=40\times 500=20000$ and of the 60 costlier ones $=60\times 800=48000$. Total purchase price =20000+48000=68000, and transport adds $68000\times \frac{5}{100}=3400$. Total cost =68000+3400=71400; the transport charge is part of the cost price. Sales: 30 cheaper shirts at 20% profit fetch $30\times 500\times 1.2=18000$. The remaining 10 cheaper shirts at 10% loss fetch $10\times 500\times 0.9=4500$, and the 60 costlier shirts at 10% profit fetch $60\times 800\times 1.1=52800$. Total sale =18000+4500+52800=75300, so profit =75300-71400=3900. Profit per cent $=\frac{3900}{71400}\times 100=5.46$, which is about 5.5%. Hence option (c) is correct.

Percentage change in volume Volume depends on the cube of the edge, so a change in the edge is felt three times over. Take the edge as 10 units; the volume is ${10}^3=1000$ cubic units. A fall of 20% leaves an edge of $10\times \frac{80}{100}=8$ units, so the new volume is $8^3=512$ cubic units. Decrease =1000-512=488 cubic units. Percentage decrease $=\frac{488}{1000}\times 100=48.8$, that is 48.8%. Hence option (c) is correct.

Tangent to a circle A tangent is perpendicular to the radius drawn to the point of contact, so the radius, the tangent and the line joining the point to the centre make a right-angled triangle. With A as the point of contact and P the outside point, OA=7, AP=24 and the right angle is at A. By Pythagoras, $O{P}^2=O{A}^2+A{P}^2=7^2+{24}^2$. $O{P}^2=49+576=625$, so $OP=\sqrt{625}=25$. The point is 25 m from the centre, so option (a) is correct.

Time and work One day's work of X and Y together $=\frac{1}{12}$. One day's work of X alone $=\frac{1}{20}$. So one day's work of Y $=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30}$. Y does $\frac{1}{30}$ of the work in a day, so Y alone needs 30 days. Hence option (b) is correct.

Trigonometric ratios $\sin x=\frac{7}{25}$ means the side opposite x is 7 and the hypotenuse is 25. The third side is $\sqrt{{25}^2-7^2}=\sqrt{625-49}=\sqrt{576}=24$. Since x lies between 0° and 90°, the cosine is positive, so $\cos x=\frac{24}{25}$. $\frac{1+\cos x}{1-\cos x}=\frac{1+\frac{24}{25}}{1-\frac{24}{25}}=\frac{\frac{49}{25}}{\frac{1}{25}}$. =49, so option (d) is correct.

Height and distance The height of the kite, the ground and the taut string form a right-angled triangle in which the string is the hypotenuse. The angle of 30° is between the string and the ground, and the height 80 m is the side opposite to it. So $\sin 30°=\frac{80}{l}$, where l is the length of the string. $\frac{1}{2}=\frac{80}{l}$, hence $l=80\times 2=160$. The string is 160 metres long, so option (c) is correct.

Alligation In alligation the two quantities are in the ratio of the distances of the mean price from the two prices, taken crosswise. Cheaper : dearer =( dearer price - mean price ):( mean price - cheaper price ). =(300-280):(280-240)=20:40. 20:40=1:2. So the teas must be mixed in the ratio 1:2 and option (c) is correct.

Trigonometric identity The expression asked for is ${\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta$, and ${\sin }^2\theta +{\cos }^2\theta =1$, so it is simply $1-2\sin \theta \cos \theta$. So only the product $2\sin \theta \cos \theta$ has to be found; square the given relation to get it. ${\left(\sin \theta +\cos \theta \right)}^2={\left(\frac{6}{5}\right)}^2$ ${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{36}{25}$, i.e. $1+2\sin \theta \cos \theta =\frac{36}{25}$ $2\sin \theta \cos \theta =\frac{36}{25}-1=\frac{11}{25}$ Required value $=1-\frac{11}{25}=\frac{14}{25}$ Hence option (b) is correct.

Angle in a semicircle AB is a diameter and C lies on the circle, so $\angle ACB$ is the angle in a semicircle, which is always a right angle. Therefore $\angle ACB={90}^{\circ }$. In $△ABC$ the three angles add up to ${180}^{\circ }$. $\angle CAB+\angle ACB+\angle CBA={180}^{\circ }$ ${35}^{\circ }+{90}^{\circ }+\angle CBA={180}^{\circ }$ $\angle CBA={180}^{\circ }-{125}^{\circ }={55}^{\circ }$ Hence option (d) is correct.

Circle through three points Join the three points and draw the perpendicular bisectors of two of the joining segments; since the points are not collinear these bisectors are not parallel, so they meet at exactly one point O. O is equidistant from all three points, so it is the only possible centre and OA is the only possible radius. One centre with one radius gives exactly one circle, so exactly one circle passes through three non-collinear points. If the three points were collinear the perpendicular bisectors would be parallel, and no circle at all could be drawn through them. Hence option (a) is correct.

Chord and central angle Formula: a chord subtending an angle $\theta$ at the centre of a circle of radius R has length $2R\sin \frac{\theta }{2}$. Here R=15 cm and $\theta ={60}^{\circ }$. $AB=2\times 15\times \sin {30}^{\circ }$ $AB=30\times \frac{1}{2}=15$ cm Quick check: OA=OB=15 cm and the angle between them is ${60}^{\circ }$, so $△OAB$ is equilateral and AB must equal the radius. Hence option (a) is correct.

Rectangle area and perimeter Let the length be l and the breadth be b; perimeter =2(l+b). Given l:2(l+b)=1:3, so 3l=2(l+b). $3l=2l+2b\Rightarrow l=2b$ Area $=l\times b=2b\times b=2b^2$ $2b^2=200\Rightarrow b^2=100\Rightarrow b=10$ cm Hence option (a) is correct.

Area of a parallelogram Formula: for adjacent sides a and b with included angle $\theta$, area of a parallelogram $=ab\sin \theta$. Area $=24\times 16\times \sin {60}^{\circ }=384\times \frac{\sqrt{3}}{2}$ Area $=192\sqrt{3}{\mathrm{cm}}^2$ The longer side is 24 cm, and area = base $\times$ corresponding height. $192\sqrt{3}=24\times h$ $h=\frac{192\sqrt{3}}{24}=8\sqrt{3}$ cm Hence option (b) is correct.

Ratio of three numbers Let x+y=8k, y+z=9k and z+x=7k. Adding all three: 2(x+y+z)=8k+9k+7k=24k, so x+y+z=12k. x=(x+y+z)-(y+z)=12k-9k=3k y=(x+y+z)-(z+x)=12k-7k=5k z=(x+y+z)-(x+y)=12k-8k=4k Therefore x:y:z=3k:5k:4k=3:5:4. Hence option (b) is correct.

Average and overlapping groups Sum of all 11 numbers $=11\times 50=550$ Sum of the first 6 numbers $=6\times 48=288$ Sum of the last 6 numbers $=6\times 55=330$ The first six are the numbers in positions 1 to 6 and the last six are those in positions 6 to 11, so adding these two sums counts the 6th number twice. 6th number =(288+330)-550=618-550=68 Hence option (d) is correct.

Simple interest, change in rate Formula: $SI=\frac{P\times R\times T}{100}$, with T=4 years in both cases. At the original rate: $\frac{P\times r\times 4}{100}=2400$ At the higher rate: $\frac{P\times \left(r+3\right)\times 4}{100}=3120$ Subtracting, the whole r cancels and only the extra 3% is left: $\frac{P\times 3\times 4}{100}=3120-2400=720$ $\frac{12P}{100}=720\Rightarrow P=\frac{720\times 100}{12}=6000$ So the original principal was ₹6,000, and option (c) is correct.

Partnership profit sharing Both invested for the same period (one full year), so the profit is divided in the ratio of the investments alone. Arjun : Bhavna =50000:75000=2:3 Total parts =2+3=5 Bhavna's share $=\frac{3}{5}\times 90000=54000$ So Bhavna gets ₹54,000, and option (c) is correct.

Average of five days Total needed in 5 days = average $\times$ number of days $=5\times 150=750$ Produced in the first 4 days =140+160+135+145=580 Production on the 5th day =750-580=170 Hence option (a) is correct.

Parallel straight lines Two lines are parallel when their slopes are equal. For a line ax+by=c the slope is $-\frac{a}{b}$, so the slope of 5x+4y=20 is $-\frac{5}{4}$. Slopes of the options: (a) $\frac{5}{4}$, (b) $-\frac{4}{5}$, (c) $-\frac{5}{4}$, (d) $-\frac{5}{3}$. Only 5x+4y=35 has the same coefficients of x and y, hence the same slope $-\frac{5}{4}$. Its constant term is 35 instead of 20, so it is a different line — parallel to the given one, not identical with it. Hence option (c) is correct.

Percentage and ratio '$x\%$ more than' means multiply by $\left(1+\frac{x}{100}\right)$. D=200 and C is 50% more than D, so $C=200\times \frac{150}{100}=300$. C+D=300+200=500 A+B is 200% more than C+D, so $A+B=500\times \frac{300}{100}=1500$. A is 50% more than B, so A:B=150:100=3:2, i.e. 5 parts in all. $A=\frac{3}{5}\times 1500=900$ Hence option (b) is correct.

Simple interest, time Simple interest = Amount - Principal =18500-12500=6000 Formula: $SI=\frac{P\times R\times T}{100}$ $6000=\frac{12500\times 8\times T}{100}$ $6000=1000\times T$ T=6 years Hence option (c) is correct.

Work on alternate days Take the total work as LCM(12,24)=24 units. A's one day work $=\frac{24}{12}=2$ units and B's one day work $=\frac{24}{24}=1$ unit. They work alternately with A on day 1, so one 2-day cycle finishes 2+1=3 units. Number of complete cycles needed $=\frac{24}{3}=8$ Days taken $=8\times 2=16$ days, and the work ends exactly on B's turn, so no part-day is left over. Hence option (b) is correct.