15 November 2025 · Shift 3

Rational and irrational numbers By definition the set of real numbers is made up of the rational numbers together with the irrational numbers, and of nothing else. So every real number is either rational or irrational, which makes option (a) true. Option (b) is false: a rational number can be written as $\frac{p}{q}$ with $q\ne 0$, which is exactly what an irrational number cannot be. Option (c) is false: $\frac{3}{4}$ and 2 are real numbers and both are rational. Option (d) is false: every integer n can be written as $\frac{n}{1}$, so integers are rational. Hence option (a) is the answer.

Volume of a hexagonal prism Area of a regular hexagon of side a is $\frac{3\sqrt{3}}{2}{a}^2$. Here a=4, so the base area $=\frac{3\times 1.732}{2}\times 4^2=2.598\times 16=41.568{cm}^2$. Volume of the prism = base area $\times$ height $=41.568\times 15=623.52{cm}^3$. The cavity is a cuboid of square cross-section running through the whole height: $2\times 2\times 15=60{cm}^3$. Material remaining $=623.52-60=563.52\approx 563.5{cm}^3$. Hence option (a) is the answer.

Volume and profit The base is a square, so the volume of one prism = base area $\times$ height $=12\times 12\times 30=4320{cm}^3$. Profit on each cubic centimetre = ₹0.80 − ₹0.50 = ₹0.30. Profit on one prism $=4320\times 0.30=1296$, that is ₹1,296. Profit on 20 prisms $=1296\times 20=25920$. So the total profit is ₹25,920, and option (d) is the answer.

Partnership ratio In a partnership the profit is divided in the ratio of capital $\times$ time. A's capital worked for only 8 months: $40000\times 8=320000$. B never withdrew, so his capital worked for all 12 months: $30000\times 12=360000$. Ratio =320000:360000=32:36=8:9. Hence option (d) is the answer.

Partnership ratio Profit is shared in the ratio of capital $\times$ time, so count the months each sum actually worked. A started the business, so his money worked for all 12 months: $50000\times 12=600000$. B joined after 6 months, so his money worked for the remaining 6 months: $20000\times 6=120000$. Ratio =600000:120000=5:1. Hence option (a) is the answer.

Sphere and cone volumes The height of a sphere of radius r is its diameter, so the cone also has h=2r. Volume of the sphere $=\frac{4}{3}\pi r^3$. Volume of the cone $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2\left(2r\right)=\frac{2}{3}\pi r^3$. Ratio $=\frac{4}{3}\pi r^3:\frac{2}{3}\pi r^3=4:2=2:1$. Hence option (c) is the answer.

Centres of a triangle Every point on the perpendicular bisector of a side is equally far from the two end points of that side. So the point where all three perpendicular bisectors meet is equally far from A, B and C, and a circle drawn with that point as centre passes through all three vertices. That point is therefore the circumcentre, the centre of the circumcircle. For contrast: the incentre is where the angle bisectors meet, the centroid where the medians meet and the orthocentre where the altitudes meet. Hence option (b) is the answer.

Sphere in a cylinder A sphere of radius r that just fits inside a cylinder touches the curved surface and both flat faces, so the cylinder has radius r and height h=2r. Volume of the sphere $=\frac{4}{3}\pi r^3$. Volume of the cylinder $=\pi r^{2}h=\pi r^2\left(2r\right)=2\pi r^3$. Ratio $=\frac{4}{3}\pi r^3:2\pi r^3=\frac{4}{3}:2=4:6=2:3$. Hence option (a) is the answer.

Trigonometric identity Square both expressions and add them; the cross terms will cancel. $x^2={\left(7\cos A+24\sin A\right)}^2=49{\cos }^{2}A+336\sin A\cos A+576{\sin }^{2}A$. $y^2={\left(7\sin A-24\cos A\right)}^2=49{\sin }^{2}A-336\sin A\cos A+576{\cos }^{2}A$. On adding, $336\sin A\cos A$ cancels: $x^2+y^2=49\left({\sin }^{2}A+{\cos }^{2}A\right)+576\left({\sin }^{2}A+{\cos }^{2}A\right)$. Since ${\sin }^{2}A+{\cos }^{2}A=1$, this gives $x^2+y^2=49+576=625$. Worth remembering: ${\left(a\cos A+b\sin A\right)}^2+{\left(a\sin A-b\cos A\right)}^2=a^2+b^2$. Therefore $2\left(x^2+y^2\right)=2\times 625=1250$. Hence option (a) is the answer.

Angles of a triangle Let the two given angles be x and 3x, so their sum is 4x. The third angle is 50% more than this sum: $4x+\frac{50}{100}\times 4x=4x+2x=6x$. The three angles of a triangle add up to ${180}^{\circ }$, so $x+3x+6x={180}^{\circ }$. $10x={180}^{\circ }$, which gives $x={18}^{\circ }$. The angles are ${18}^{\circ }$, ${54}^{\circ }$ and ${108}^{\circ }$, so the largest angle is ${108}^{\circ }$. Hence option (a) is the answer.

Apollonius theorem — median Since QS=SR, the point S is the mid-point of QR, so PS is the median drawn from P. Perimeter =PQ+QR+RP, so QR=54-10-20=24 cm and hence QS=SR=12 cm. Apollonius theorem for the median PS: $P{Q}^2+P{R}^2=2\left(P{S}^2+Q{S}^2\right)$. ${10}^2+{20}^2=2\left(P{S}^2+{12}^2\right)$ $100+400=2P{S}^2+288$ $2P{S}^2=500-288=212\Rightarrow P{S}^2=106$ $PS=\sqrt{106}$ cm. Hence the correct option is (d).

Basic proportionality theorem $MN\parallel QR$, so by the Basic Proportionality (Thales) theorem the two sides are cut in the same ratio: $\frac{PM}{MQ}=\frac{PN}{NR}$. $\frac{4}{8}=\frac{5}{NR}$ $NR=\frac{5\times 8}{4}=10$ cm. Hence the correct option is (a).

Alloys and ratio Take 1 kg of each alloy, so the new alloy weighs 1+1=2 kg. The first alloy has gold : silver : brass =2:3:3, i.e. 8 parts, and it contains no tin at all. The second alloy has silver : brass : tin =3:5:2, i.e. 10 parts, so tin in it $=\frac{2}{10}\times 1=0.2$ kg. Total tin in the 2 kg of new alloy =0+0.2=0.2 kg. Tin per kg $=\frac{0.2}{2}=0.1$ kg. Hence the correct option is (c).

Mixture — weighted average The price of a mixture is the weighted average of the two prices, the weights being the quantities mixed. Cost per kg $=\frac{32\times 2+48\times 3}{2+3}$ $=\frac{64+144}{5}=\frac{208}{5}=41.6$ By alligation the same value satisfies (48-M):(M-32)=2:3, which again gives M=41.6. So the mixed rice costs ₹41.6 per kg. Hence the correct option is (c).

Mixture — removal and addition Initially water $=25\%$ of 120=30 L, so sugar syrup =120-30=90 L. The 24 L that is sold has the same composition, so water sold $=25\%$ of 24=6 L and syrup sold =24-6=18 L. Water left =30-6=24 L and syrup left =90-18=72 L. Let x litres of sugar syrup and x litres of water be added. $\frac{24+x}{72+x}=\frac{1}{2}$ $2\left(24+x\right)=72+x\Rightarrow 48+2x=72+x$ x=24 So 24 litres of water was added later on. Hence the correct option is (c).

Percentage in a mixture The mixture is syrup + water =(8+x) litres. Syrup is $25\%$ of the mixture: $\frac{8}{8+x}=\frac{25}{100}=\frac{1}{4}$ $8+x=8\times 4=32$ x=32-8=24 litres. Check: syrup $=\frac{8}{32}\times 100=25\%$, as required. Hence the correct option is (b).

Time and work — efficiency Let the total work $=\mathrm{LCM}\left(12,18\right)=36$ units. Ravi's efficiency $=\frac{36}{12}=3$ units/day and Sohan's $=\frac{36}{18}=2$ units/day. Work done by both in 4 days $=\left(3+2\right)\times 4=20$ units, so work left =36-20=16 units. Ravi and Kiran finish these 16 units in 4 days, so their combined efficiency $=\frac{16}{4}=4$ units/day. Kiran's efficiency =4-3=1 unit/day. Time taken by Kiran alone $=\frac{36}{1}=36$ days. Hence the correct option is (c).

Time and work — efficiency Let the total work =36 units. X and Y together take 12 days, so their combined efficiency $=\frac{36}{12}=3$ units/day. Z alone takes 36 days, so Z's efficiency $=\frac{36}{36}=1$ unit/day. Given that X works as fast as Y and Z together: X=Y+Z, so Y=X-Z=X-1. $X+Y=3\Rightarrow X+\left(X-1\right)=3\Rightarrow 2X=4\Rightarrow X=2$ units/day. Time taken by X alone $=\frac{36}{2}=18$ days. Hence the correct option is (a).

Time and work — remaining work Let the total work $=\mathrm{LCM}\left(20,15\right)=60$ units. P's efficiency $=\frac{60}{20}=3$ units/day and Q's efficiency $=\frac{60}{15}=4$ units/day. Work done by Q in 6 days $=4\times 6=24$ units. Work left =60-24=36 units. Time taken by P alone $=\frac{36}{3}=12$ days. Hence the correct option is (b).

Area — tiles on a floor Perimeter of a tile =2(l+b)=200 cm, so l+b=100 cm. Since l:b=3:2 (5 parts), $l=\frac{3}{5}\times 100=60$ cm and $b=\frac{2}{5}\times 100=40$ cm. Area of one tile $=60\times 40=2400$ cm${}^2$. Floor: 32.4 m =3240 cm and 19.6 m =1960 cm, so area of floor $=3240\times 1960=6350400$ cm${}^2$. Number of tiles $=\frac{6350400}{2400}=2646$. Hence the correct option is (d).

Cylinder — CSA and volume For a cylinder, curved surface area $=2\pi rh$ and volume $=\pi r^{2}h$. Dividing the volume by the curved surface area removes $\pi$ and h: $\frac{\pi r^{2}h}{2\pi rh}=\frac{r}{2}$. $\frac{r}{2}=\frac{1848}{528}=3.5\Rightarrow r=7$ m. Putting r=7 in $2\pi rh=528$: $2\times \frac{22}{7}\times 7\times h=528$ $44h=528\Rightarrow h=12$ m. Required ratio r:h=7:12. Hence the correct option is (b).

Circles — common tangents Here $r_1=4$, $r_2=5$ and the distance between the centres is d=15. $r_1+r_2=4+5=9$, and 15>9, so $d>r_1+r_2$. When the distance between the centres exceeds the sum of the radii, the two circles lie completely outside each other without touching or cutting. In that position both direct (external) tangents and both transverse (internal) tangents exist, so the number of common tangents =2+2=4. Hence the correct option is (a).

Angles of a quadrilateral The four angles of a quadrilateral add up to ${360}^{\circ }$. Sum of the remaining three angles $={360}^{\circ }-{94}^{\circ }={266}^{\circ }$. Let them be 6x, 5x and 3x, so $6x+5x+3x={266}^{\circ }$. $14x={266}^{\circ }\Rightarrow x={19}^{\circ }$ So the three angles are $6x={114}^{\circ }$, $5x={95}^{\circ }$ and $3x={57}^{\circ }$. The four angles are ${94}^{\circ },{114}^{\circ },{95}^{\circ },{57}^{\circ }$; the largest is ${114}^{\circ }$ and the smallest is ${57}^{\circ }$. Required difference $={114}^{\circ }-{57}^{\circ }={57}^{\circ }$. Hence the correct option is (c).

Equation of a line Point-slope form of a straight line: $y-y_1=m\left(x-x_1\right)$. Here the slope m=5 and the point is $\left(x_1,y_1\right)=\left(2,3\right)$. y-3=5(x-2) y-3=5x-10 y=5x-7 Check: at x=2, $y=5\times 2-7=3$, so the line does pass through (2,3). Hence the correct option is (a).

Work done in cycles Let the total work $=\mathrm{LCM}\left(10,12,15\right)=60$ units. $P=\frac{60}{10}=6$, $Q=\frac{60}{12}=5$ and $R=\frac{60}{15}=4$ units per day. One cycle of 3 days gives 6+5+4=15 units. $\frac{60}{15}=4$ complete cycles are needed, and no part-cycle is left over. Total time $=4\times 3=12$ days. Hence the correct option is (d).