30 November 2025 · Shift 3

Fraction of income Monthly income $=90,000$. Fraction saved $=1-\frac{4}{5}-\frac{1}{9}$. Taking 45 as the common denominator, $=\frac{45-36-5}{45}=\frac{4}{45}$. Saving $=90000\times \frac{4}{45}=2000\times 4=8000$.

Alligation of prices Selling the mixture at ₹50 per kg with neither gain nor loss means the cost price of the mixture is also ₹50 per kg. By the rule of alligation, quantity of cheaper : quantity of dearer =(dearer price - mean price$):($mean price - cheaper price). Ratio =(60-50):(50-40)=10:10=1:1. So the two varieties must be mixed in the ratio 1:1.

Compound interest For a period of one year the interest is added only once, so the compound interest is exactly the same as the simple interest. Amount $=3000{\left(1+\frac{10}{100}\right)}^1=3000\times \frac{11}{10}=3300$. CI=3300-3000=300. Hence the compound interest is ₹300.

Diagonal of a cuboid The shortest path inside a box from one bottom corner to the opposite top corner is its space diagonal. Formula: $d=\sqrt{l^2+b^2+h^2}$. $d=\sqrt{5^2+{12}^2+{13}^2}=\sqrt{25+144+169}=\sqrt{338}$. $\sqrt{338}=\sqrt{169\times 2}=13\sqrt{2}=13\times 1.414=18.38$ m.

Rational and irrational numbers A non-zero rational number times an irrational number is always irrational, for example $3\times \sqrt{2}=3\sqrt{2}$. But zero is itself a rational number, and $0\times \sqrt{2}=0$, which is rational. Since both outcomes are possible, the product is not always of one kind; it may be rational or irrational.

Profit and loss Let the total cost price be 100 units, so goods worth 40 units are sold at a profit and goods worth 60 units at a loss. Profit on the first part $=40\times \frac{25}{100}=10$ units. Loss on the second part $=60\times \frac{10}{100}=6$ units. Net gain =10-6=4 units, and this net gain is given as ₹150. So 1 unit $=\frac{150}{4}=37.5$ and total cost price $=100\times 37.5=3750$, that is ₹3750.

Centroid of a triangle Formula: centroid $G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$. $G=\left(\frac{1+5+3}{3},\frac{1+1+7}{3}\right)=\left(\frac{9}{3},\frac{9}{3}\right)=\left(3,3\right)$. $RG=\sqrt{{\left(3-3\right)}^2+{\left(7-3\right)}^2}=\sqrt{0+16}=4$. So the distance from vertex R to the centroid is 4 units.

Average with correction Sum of the top 4 scores $=4\times 88.5=354$. Sum of the bottom 4 scores as recorded $=4\times 74=296$. One score was written as 71.2 instead of 91.2, so the sum has to be raised by 91.2-71.2=20. Corrected sum of the bottom 4=296+20=316. Corrected total of all 8 scores =354+316=670. Corrected average $=\frac{670}{8}=83.75$.

Volume and mass of a prism The base is a right-angled triangle, so its area $=\frac{1}{2}\times 0.5\times 1.2=0.3$ square metre. Volume of the prism = base area $\times$ length $=0.3\times 3.6=1.08$ cubic metre. Since 1 cubic metre $={10}^6$ cubic centimetre, the volume $=1.08\times {10}^6$ cubic centimetre. Mass = density $\times$ volume $=0.80\times 1.08\times {10}^6=8.64\times {10}^5$ gram. $8.64\times {10}^5$ gram $=\frac{864000}{1000}=864.00$ kg.

Trigonometric ratios $\cot \theta =\frac{\mathrm{base}}{\mathrm{perpendicular}}=\frac{7}{24}$, so take base =7 and perpendicular =24. By Pythagoras, hypotenuse $=\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}=25$. $cosec\theta =\frac{\mathrm{hypotenuse}}{\mathrm{perpendicular}}=\frac{25}{24}$. Hence $cosec\theta =\frac{25}{24}$.

Trigonometric identities Use the two Pythagorean identities: $1+{\tan }^{2}A={\sec }^{2}A$ and $1+{\cot }^{2}A={cosec}^{2}A$. So the expression $=\frac{{\sec }^{2}A}{{cosec}^{2}A}=\frac{1/{\cos }^{2}A}{1/{\sin }^{2}A}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$. Given $\cos A=\frac{12}{13}$ with A acute, take the base =12 and the hypotenuse =13. Perpendicular $=\sqrt{{13}^2-{12}^2}=\sqrt{169-144}=\sqrt{25}=5$. $\tan A=\frac{5}{12}$, so ${\tan }^{2}A=\frac{25}{144}$. Hence the required value is $\frac{25}{144}$, and option (c) is correct.

Congruent triangles Congruent triangles are identical in both shape and size - one can be placed exactly on the other. So every pair of corresponding sides is equal and every pair of corresponding angles is equal. Equal sides give equal perimeters, since the perimeter is just the sum of the three sides. A median joins a vertex to the mid-point of the opposite side; the corresponding halves are equal, so the corresponding medians are equal too. The circumradius is $R=\frac{abc}{4\Delta }$, and both the sides and the area are the same in the two triangles, so the circumradii are equal. All three statements hold, so option (d) is correct.

Chord and central angle Let AB be the chord and O the centre; join OA and OB. OA=OB=5 cm because both are radii, and AB=5 cm is given. All three sides of $△OAB$ are equal, so it is an equilateral triangle. Each angle of an equilateral triangle is ${60}^{\circ }$. Hence the angle subtended by the chord at the centre is $\angle AOB={60}^{\circ }$, so option (b) is correct.

Speed, time and distance Speed of the train $=\frac{240}{6}=40$ km/h. Speed of the car $=40\times \frac{120}{100}=48$ km/h. Speed of the bike $=48\times \frac{50}{100}=24$ km/h. Time taken by the car for 288 km $=\frac{288}{48}=6$ hours. Time taken by the bike for 288 km $=\frac{288}{24}=12$ hours. Difference =12-6=6 hours $=6\times 60=360$ minutes, so option (d) is correct.

Tangents and cyclic quadrilateral C lies on the minor arc, so $\angle ACB={102}^{\circ }$ is the angle in the minor segment. Take any point D on the major arc; ACBD is a cyclic quadrilateral, so $\angle ACB+\angle ADB={180}^{\circ }$. $\angle ADB={180}^{\circ }-{102}^{\circ }={78}^{\circ }$. The angle at the centre is twice the angle at the remaining part of the circle, so $\angle AOB=2\times {78}^{\circ }={156}^{\circ }$. A tangent is perpendicular to the radius at the point of contact, so $\angle OAP=\angle OBP={90}^{\circ }$. The four angles of quadrilateral AOBP add up to ${360}^{\circ }$: $\angle APB={360}^{\circ }-\left({90}^{\circ }+{90}^{\circ }+{156}^{\circ }\right)={360}^{\circ }-{336}^{\circ }={24}^{\circ }$. Hence option (b) is correct.

Surface area of hemispheres The curved surface area of one hemisphere of radius r is $2\pi r^2$ (its flat circular face is not part of the curved surface). For the two hemispheres together, curved surface $=2\times 2\pi r^2=4\pi r^2$. When the two flat faces are pressed together they vanish inside the join, so the outside of the ball is made up only of the two curved surfaces. Total surface area of the sphere formed $=4\pi r^2$. Required percentage $=\frac{4\pi r^2}{4\pi r^2}\times 100=100\%$, so option (b) is correct.

Lateral surface of a pyramid The lateral surface depends only on the base side and the height, so the ant's path is extra information here. Half of the base side $=\frac{16}{2}=8$ cm; this is the distance from the centre of the base to the mid-point of a base edge. Slant height $l=\sqrt{h^2+{\left(\frac{a}{2}\right)}^2}=\sqrt{{12}^2+8^2}=\sqrt{144+64}=\sqrt{208}\approx 14.42$ cm. Lateral surface area $=\frac{1}{2}\times \left(\mathrm{perimeter}\mathrm{of}\mathrm{base}\right)\times l=\frac{1}{2}\times \left(4\times 16\right)\times 14.42$. $=32\times 14.42=461.44$ cm${}^2$, so option (b) is correct.

Proper fractions Statement I: take the proper fraction $\frac{3}{4}$ and the whole number 8. $\frac{3}{4}\times 8=6$, which is a whole number, so such a product can indeed be a whole number. Statement I is true. Statement II: a proper fraction lies between 0 and 1. If $0<\frac{a}{b}<1$ and $0<\frac{c}{d}<1$, then $\frac{a}{b}\times \frac{c}{d}<1\times 1=1$, and the product is still positive. For example $\frac{2}{3}\times \frac{3}{5}=\frac{2}{5}$, again a proper fraction. Statement II is true. Both statements are true, so option (b) is correct.

Complementary angles Use $\sin \theta =\cos \left({90}^{\circ }-\theta \right)$: if a sine equals a cosine, the two angles are complementary. So $\left(x+{10}^{\circ }\right)+\left(3x-{20}^{\circ }\right)={90}^{\circ }$. $4x-{10}^{\circ }={90}^{\circ }$ $4x={100}^{\circ }$, hence $x={25}^{\circ }$. Check: $\sin \left({25}^{\circ }+{10}^{\circ }\right)=\sin {35}^{\circ }$ and $\cos \left({75}^{\circ }-{20}^{\circ }\right)=\cos {55}^{\circ }$, and $\sin {35}^{\circ }=\cos {55}^{\circ }$. So option (b) is correct.

Successive percentage change Each change multiplies the previous population by a fixed factor: $+20\%\to \frac{6}{5}$, $-10\%\to \frac{9}{10}$, $+25\%\to \frac{5}{4}$. After three hours of $20\%$ growth: $10000\times {\left(\frac{6}{5}\right)}^3=10000\times \frac{216}{125}=17280$. Fall of $10\%$ in the 4th hour: $17280\times \frac{9}{10}=15552$. Rise of $25\%$ in the 5th hour: $15552\times \frac{5}{4}=19440$. So the population after 5 hours is 19,440, and option (c) is correct.

Slant height of a cone For a cone, $l^2=r^2+h^2$. ${10}^2=6^2+h^2$, so $h^2=100-36=64$ and h=8 cm. New height $=8\times \frac{150}{100}=12$ cm, while the radius stays 6 cm. New slant height $=\sqrt{6^2+{12}^2}=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}$. $6\sqrt{5}\approx 6\times 2.236=13.41$ cm, so option (b) is correct.

Consecutive odd numbers Let the two consecutive odd numbers be n and n+2. Their product $=n\left(n+2\right)=n^2+2n$. Adding 1 gives $n^2+2n+1={\left(n+1\right)}^2$, a perfect square - and n+1 is the even number lying between them. Since 1 is the smallest positive number there is, nothing smaller can work. Hence the required number is 1, and option (a) is correct.

Tangents from an external point A tangent is perpendicular to the radius at the point of contact, so $\angle OAP=\angle OBP={90}^{\circ }$. In right triangle OAP, $AP=\sqrt{O{P}^2-O{A}^2}=\sqrt{{61}^2-{11}^2}=\sqrt{3721-121}=\sqrt{3600}=60$ units. Area of $△OAP=\frac{1}{2}\times OA\times AP=\frac{1}{2}\times 11\times 60=330$ square units. The two tangents from an external point are equal, so $△OAP\cong △OBP$ and the quadrilateral is made of these two equal triangles. Area of PAOB $=2\times 330=660$ square units, so option (b) is correct.

Centroid of a triangle The centroid is the average of the coordinates of the three vertices: $G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$ x-coordinate $=\frac{1+3+5}{3}=\frac{9}{3}=3$. y-coordinate $=\frac{1+2+3}{3}=\frac{6}{3}=2$. So the centroid is (3,2), and option (b) is correct.

Pythagorean triple and sine The right angle is at B, so AC=41 is the hypotenuse and BC=9 is one leg. Third side $=\sqrt{{41}^2-9^2}=\sqrt{1681-81}=\sqrt{1600}=40$, so the sides form the 9-40-41 triple. In this triangle the two acute angles have sines $\frac{40}{41}$ and $\frac{9}{41}$. A sine can never exceed 1, which rules out $\frac{41}{40}$, and $\frac{7}{40}$ and $\frac{7}{41}$ do not arise from a 9-40-41 triangle at all. So $\sin A=\frac{40}{41}$, and option (c) is correct.