15 November 2025 · Shift 2

Natural numbers Natural numbers are the counting numbers $1,2,3,4,\dots$; the set begins at 1. 7, 6 and 4 are all counting numbers, so each of them is natural. 0 is not used for counting; it is the extra member of the set of whole numbers $0,1,2,3,\dots$. Hence the number that is not a natural number is 0, option (c).

Number systems Natural numbers are $1,2,3,\dots$ while integers are $\dots ,-2,-1,0,1,2,\dots$. 0 and -8 are integers, but neither is a counting number, so they are not natural numbers. 8 is a counting number, hence natural, and every natural number is also an integer. Hence 8 belongs to both sets, option (b).

Profit and loss Cost of the boxes $=220\times 160=35200$, i.e. ₹35,200. Overheads of ₹1,600 are part of the cost, so total cost price =35200+1600=36800, i.e. ₹36,800. Money from the first lot $=120\times 180=21600$, i.e. ₹21,600. Boxes left =220-120=100, and they fetch $100\times 170=17000$, i.e. ₹17,000. Total selling price =21600+17000=38600, i.e. ₹38,600. Profit =38600-36800=1800, i.e. ₹1,800. Profit $\%=\frac{1800}{36800}\times 100=\frac{45}{9.2}\approx 4.89\%$, i.e. about $4.8\%$. Hence option (d).

Incentre of a triangle The incentre O is the meeting point of the internal bisectors, so $\angle OQR=\frac{\angle Q}{2}$ and $\angle ORQ=\frac{\angle R}{2}$. In $△OQR$: $\angle QOR={180}^{\circ }-\left(\frac{\angle Q}{2}+\frac{\angle R}{2}\right)$. Since $\angle Q+\angle R={180}^{\circ }-\angle P$, we get $\angle QOR={180}^{\circ }-\frac{{180}^{\circ }-\angle P}{2}={90}^{\circ }+\frac{\angle P}{2}$. Putting the given value: ${130}^{\circ }={90}^{\circ }+\frac{\angle QPR}{2}$. $\frac{\angle QPR}{2}={40}^{\circ }$, so $\angle QPR={80}^{\circ }$. Hence option (c).

Trigonometric identities Use complementary angles: $\mathrm{cosec}{50}^{\circ }=\sec {40}^{\circ }$ and $\sec {50}^{\circ }=\mathrm{cosec}{40}^{\circ }$. First term $=\frac{{\tan }^2{40}^{\circ }}{{\sec }^2{40}^{\circ }}={\tan }^2{40}^{\circ }{\cos }^2{40}^{\circ }={\sin }^2{40}^{\circ }$. Second term $=\frac{{\cot }^2{40}^{\circ }}{{\mathrm{cosec}}^2{40}^{\circ }}={\cot }^2{40}^{\circ }{\sin }^2{40}^{\circ }={\cos }^2{40}^{\circ }$. So the first two terms add up to ${\sin }^2{40}^{\circ }+{\cos }^2{40}^{\circ }=1$. In the last part $\tan {55}^{\circ }=\cot {35}^{\circ }$ and $\tan {45}^{\circ }=1$, so $2\tan {35}^{\circ }\tan {45}^{\circ }\tan {55}^{\circ }=2\tan {35}^{\circ }\cot {35}^{\circ }=2$. Required value =1+2=3, i.e. option (b).

Trigonometric equation $\cos \left({90}^{\circ }-p\right)=\sin p$, so the equation becomes $2\sin p\cdot \sin p=1$. $2{\sin }^{2}p=1\Rightarrow {\sin }^{2}p=\frac{1}{2}$. As $0^{\circ }<p<{90}^{\circ }$, $\sin p=\frac{1}{\sqrt{2}}$, which gives $p={45}^{\circ }$. Then $2p={90}^{\circ }$. $\sin \left(2p\right)+\cos \left(2p\right)=\sin {90}^{\circ }+\cos {90}^{\circ }=1+0=1$. Hence option (c).

Simple interest Under simple interest the interest of every year is equal, so the growth from year 4 to year 8 is pure interest of 4 years. Interest of those 4 years =13384-10504=2880, i.e. ₹2,880. Interest of 1 year $=\frac{2880}{4}=720$, i.e. ₹720. Principal = amount after 4 years - interest of 4 years =10504-2880=7624, i.e. ₹7,624. $R=\frac{\mathrm{interest}\mathrm{of}1\mathrm{year}\times 100}{P}=\frac{720\times 100}{7624}$. $R=\frac{72000}{7624}\approx 9.44\%$. Hence option (c).

Simple interest Let the sum be P. After 8 years the amount is $340\%$ of $P=\frac{340P}{100}=3.4P$. Interest of 8 years =3.4P-P=2.4P, i.e. $240\%$ of P. Rate $=\frac{240\%}{8}=30\%$ per annum. For ₹4,000 and 3 years: $SI=\frac{P\times R\times T}{100}=\frac{4000\times 30\times 3}{100}$. $SI=\frac{360000}{100}=3600$. So the simple interest is ₹3,600, option (a).

Simple interest and ratio Let the total money be x. The part distributed $=80\%$ of $x=\frac{4x}{5}$. Saroj's share $=\frac{4x}{5}\times \frac{7}{16}=\frac{7x}{20}$ and Sita's share $=\frac{4x}{5}\times \frac{9}{16}=\frac{9x}{20}$. Saroj's interest $=\frac{7x}{20}\times \frac{7\times 5}{100}=\frac{49x}{400}$. Sita's interest $=\frac{9x}{20}\times \frac{11\times 5}{100}=\frac{99x}{400}$. Difference $=\frac{99x-49x}{400}=\frac{50x}{400}=\frac{x}{8}$. $\frac{x}{8}=3304\Rightarrow x=3304\times 8=26432$. So Mohit had ₹26,432 in all, option (b).

Heights and distances The wall, the floor and the stick form a right triangle, the right angle being at C where the wall meets the floor. The stick AB is the hypotenuse, AB=38 cm, and its inclination at the foot B is ${30}^{\circ }$. The required length is the height AC of the touching point above the floor. $\sin {30}^{\circ }=\frac{AC}{AB}\Rightarrow AC=AB\sin {30}^{\circ }$. $AC=38\times \frac{1}{2}=19$ cm. Hence option (c).

Heights and distances The rope, the wall of the house and the ground form a right triangle in which the rope is the hypotenuse. Let the height of the house be h metres. Formula: $\sin \theta =\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$ $\sin {60}^{\circ }=\frac{h}{32}$ $\frac{\sqrt{3}}{2}=\frac{h}{32}$ $h=32\times \frac{\sqrt{3}}{2}=16\sqrt{3}$ metres, so option (a) is correct.

Pipes and cisterns Take the capacity of the tank as $LCM\left(16,10\right)=80$ units. Filling pipe in 1 hour $=\frac{80}{16}=+5$ units. Emptying pipe in 1 hour $=\frac{80}{10}=-8$ units. With both pipes open, the net work in 1 hour =8-5=3 units of emptying. Water already present $=\frac{1}{5}\times 80=16$ units. Time to empty $=\frac{16}{3}=5\frac{1}{3}$ hours. $\frac{1}{3}$ hour =20 minutes, so the tank empties in 5 hours 20 minutes and option (b) is correct.

Time and work Let one day's work of a woman be w and that of a girl be g. The same work is completed both times, so 16(5w+8g)=4(14w+36g). 80w+128g=56w+144g $24w=16g\Rightarrow \frac{w}{g}=\frac{2}{3}$ Take w=2 and g=3 units per day. Total work $=16\times \left(5\times 2+8\times 3\right)=16\times 34=544$ units. One day's work of 4 women and 3 girls $=4\times 2+3\times 3=17$ units. Required time $=\frac{544}{17}=32$ days, so option (c) is correct.

Heights and distances Draw a horizontal line from the top of the smaller pole; it meets the taller pole at a point 20 m above the ground. This gives a right triangle whose base is the distance between the poles, 40 m, and whose angle of elevation is ${60}^{\circ }$. Let x be the part of the taller pole lying above that horizontal line. Formula: $\tan \theta =\frac{\mathrm{perpendicular}}{\mathrm{base}}$ $\tan {60}^{\circ }=\frac{x}{40}$ $\sqrt{3}=\frac{x}{40}\Rightarrow x=40\sqrt{3}$ m Height of the taller pole $=x+20=40\sqrt{3}+20$ m, so option (a) is correct.

Chord of a circle The perpendicular drawn from the centre to a chord bisects the chord, so $AM=MB=\frac{16}{2}=8$ cm. OM=15 cm is the distance of the chord from the centre and $\angle OMB={90}^{\circ }$. By Pythagoras' theorem in $△OMB$: $O{B}^2=O{M}^2+M{B}^2$ $O{B}^2={15}^2+8^2=225+64=289$ $OB=\sqrt{289}=17$ cm, so the radius is 17 cm and option (b) is correct.

Parts of a circle A chord is a line segment whose two end points lie on the circle. As a chord is shifted nearer to the centre its length increases, and it is longest when it passes through the centre itself. A chord passing through the centre is the diameter, whose length is 2r; no other chord can exceed this. The radius r is only half of it, a tangent meets the circle at just one point and an arc is a curved part of the circumference, so none of them is a chord at all. Hence the diameter is the largest chord and option (b) is correct.

Cylinder: curved surface area Let the length and breadth of the rectangle be 16x and 14x. Perimeter: $2\left(16x+14x\right)=180\Rightarrow 60x=180\Rightarrow x=3$ So length =48 cm, breadth =42 cm and area $=48\times 42=2016$ cm${}^2$. The top surface of a solid cylinder is a circle, so $\pi r^2=2016$ cm${}^2$. The radius is $110\%$ of the height, i.e. r=1.1h, hence $h=\frac{r}{1.1}$. Curved surface area $=2\pi rh=2\pi r\times \frac{r}{1.1}=\frac{2\pi r^2}{1.1}$ $=\frac{2\times 2016}{1.1}=\frac{4032}{1.1}=3665.45$ cm${}^2$, so option (a) is correct.

Tangent to a circle The radius drawn to the point of contact is perpendicular to the tangent, so $\angle OCP={90}^{\circ }$. In $△OCP$: $\angle COP={180}^{\circ }-\angle OCP-\angle CPO={180}^{\circ }-{90}^{\circ }-{44}^{\circ }={46}^{\circ }$ P, A and O lie on one straight line, so $\angle AOC=\angle COP={46}^{\circ }$. In $△OAC$, OA=OC (both radii), so it is isosceles and $\angle OCA=\angle OAC$. $\angle OCA=\frac{{180}^{\circ }-{46}^{\circ }}{2}=\frac{{134}^{\circ }}{2}={67}^{\circ }$ $\angle ACP=\angle OCP-\angle OCA={90}^{\circ }-{67}^{\circ }={23}^{\circ }$, so option (b) is correct.

Simplification of surds Write each surd in its simplest form by taking out perfect squares. $\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$ $\sqrt{45}=\sqrt{9\times 5}=3\sqrt{5}$ $\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}$ $\frac{5\sqrt{5}}{\sqrt{5}}=5$, so option (c) is correct.

Algebraic identity The product is of the form $\left(a+b\right)\left(a-b\right)=a^2-b^2$, with a=6 and $b=2\sqrt{6}$. $\left(6+2\sqrt{6}\right)\left(6-2\sqrt{6}\right)=6^2-{\left(2\sqrt{6}\right)}^2$ ${\left(2\sqrt{6}\right)}^2=4\times 6=24$ =36-24=12, so option (b) is correct.

Nested surds Simplify from the innermost radical outwards; each bracket becomes a perfect square. $\sqrt{100}=10$, so $\sqrt{90+10}=\sqrt{100}=10$. $\sqrt{54+10}=\sqrt{64}=8$ $\sqrt{28+8}=\sqrt{36}=6$ $\sqrt{75+6}=\sqrt{81}=9$ $\sqrt{72+9}=\sqrt{81}=9$, so option (a) is correct.

Rationalising the denominator To rationalise, multiply the numerator and the denominator by the conjugate of the denominator, $\left(\sqrt{7}-\sqrt{2}\right)$. $\frac{1}{\sqrt{7}+\sqrt{2}}\times \frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}}=\frac{\sqrt{7}-\sqrt{2}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{2}\right)}^2}$ Denominator =7-2=5, which is rational. $=\frac{\sqrt{7}-\sqrt{2}}{5}$, so option (a) is correct.

Square of a binomial surd Use ${\left(a+b\right)}^2=a^2+2ab+b^2$ with $a=4\sqrt{3}$ and $b=\sqrt{2}$. $a^2={\left(4\sqrt{3}\right)}^2=16\times 3=48$ $b^2={\left(\sqrt{2}\right)}^2=2$ $2ab=2\times 4\sqrt{3}\times \sqrt{2}=8\sqrt{6}$ Adding: $48+2+8\sqrt{6}=50+8\sqrt{6}$, so option (a) is correct.

Algebraic identity The expression has the pattern $a^2+b^2+2ab$, which is ${\left(a+b\right)}^2$. Here $a=\sqrt{6}+\sqrt{3}$ and $b=\sqrt{6}-\sqrt{3}$. $a+b=\left(\sqrt{6}+\sqrt{3}\right)+\left(\sqrt{6}-\sqrt{3}\right)=2\sqrt{6}$ ${\left(a+b\right)}^2={\left(2\sqrt{6}\right)}^2=4\times 6=24$ So the value of the expression is 24 and option (d) is correct.

Average Total weight of the 28 students $=28\times 65=1820$ kg. Including the teacher there are 29 people, so their total weight $=29\times 67=1943$ kg. Weight of the teacher =1943-1820=123 kg. Shortcut: teacher's weight = new average + (number of students $\times$ rise in average) $=67+28\times 2=123$ kg, so option (b) is correct.