30 November 2025 · Shift 1

Real number system The real numbers are exactly the rational numbers together with the irrational numbers. Every integer n can be written as $\frac{n}{1}$, so integers sit inside the rationals and hence inside the reals. An imaginary number such as 3i, where $i=\sqrt{-1}$, has no place on the number line. So the imaginary numbers are not a subset of the reals, and option (d) is correct.

Pure recurring decimal In a recurring decimal one block of digits repeats without end. It is called pure recurring only when that repeating block begins the instant the decimal point is crossed, with no digit standing in between. $0.6=0.666\dots$ is pure recurring, while $0.16=0.1666\dots$ carries the non-repeating digit 1 first and is mixed recurring. Hence option (c) is correct.

Profit and loss Let the cost price of 1 kg be ₹x. At a 10% loss the selling price is $90\%$ of x; at an 8% gain it would be $108\%$ of x. The two selling prices differ by exactly ₹0.40. $\left(108-90\right)\%\mathrm{of}x=0.40$ $\frac{18x}{100}=0.40$ $x=\frac{0.40\times 100}{18}=\frac{40}{18}=2.22$ (approximately). So the cost price is ₹2.22 per kg and option (c) is correct.

Average of a group Always turn averages into totals first. Total of the first 20 students $=20\times 68.5=1370$. Total of all 24 students $=24\times 70.2=1684.8$. So the 4 new students together score 1684.8-1370=314.8. Two of them score 72.5+75.5=148, so the other two together score 314.8-148=166.8. As their scores are equal, each of them scores $\frac{166.8}{2}=83.4$. Hence option (c) is correct.

Percentage change in costs Split the estimate of ₹6,00,000 into its three heads. Materials $=50\%$ of $6,00,000=3,00,000$ and labour $=30\%$ of $6,00,000=1,80,000$. Contingency $=6,00,000-\left(3,00,000+1,80,000\right)=1,20,000$. New material cost $=3,00,000\times 1.10=3,30,000$. New labour cost $=1,80,000\times 1.20=2,16,000$. New total $=3,30,000+2,16,000+1,20,000=6,66,000$. So the new cost is ₹6,66,000 and option (b) is correct.

Surface area of a prism For any right prism, total surface area = lateral surface area $+2\times$ (base area). Perimeter of the regular pentagon $=5\times 5.8=29$ cm. Lateral surface area = perimeter $\times$ height $=29\times 12=348{\mathrm{cm}}^2$. Total surface area $=348+2\times 58.0=348+116=464{\mathrm{cm}}^2$. Hence option (a) is correct.

Surface area of a prism For a right prism, total surface area = (perimeter of base $\times$ height) $+2\times$ (base area). Perimeter of the regular hexagon $=6\times 6.5=39$ cm. Lateral surface area $=39\times 20=780{\mathrm{cm}}^2$. Two bases contribute $2\times 109.60=219.20{\mathrm{cm}}^2$. Total surface area $=780+219.20=999.20{\mathrm{cm}}^2$. Hence option (a) is correct.

Centroid of a triangle The centroid is the point where the three medians of a triangle meet. It divides every median in the ratio 2:1, measured from the vertex towards the midpoint of the opposite side. So the vertex-to-centroid part is $\frac{2}{3}$ of the whole median. Required distance $=\frac{2}{3}\times 18=12$ cm. Hence option (c) is correct.

Volume of spheres Melting and recasting does not change the total volume, so (volume of the big sphere) $=n\times$ (volume of one small sphere). $\frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3$ $R^3=n\times \frac{R^3}{27}$ n=27 So 27 small spheres can be formed and option (b) is correct.

Simple interest Formula: $\mathrm{SI}=\frac{P\times R\times T}{100}$. $58410=\frac{P\times 6\times 20}{100}$ $58410=\frac{120P}{100}$ $P=\frac{58410\times 100}{120}=48675$ So the sum invested is ₹48,675 and option (b) is correct.

Height and distance The wall, the ground and the ladder form a right triangle in which the ladder is the hypotenuse and the required distance is the side lying along the ground. $\tan x=\frac{\mathrm{height}\mathrm{reached}\mathrm{on}\mathrm{the}\mathrm{wall}}{\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{foot}\mathrm{from}\mathrm{the}\mathrm{wall}}=\frac{117}{44}$, so the two legs are in the ratio 44:117. Hypotenuse $=\sqrt{{44}^2+{117}^2}=\sqrt{1936+13689}=\sqrt{15625}=125$, so the three sides are in the ratio 44:117:125. The ladder is the hypotenuse, so $125k=25\Rightarrow k=\frac{25}{125}=\frac{1}{5}$. Distance of the foot of the ladder from the wall $=44k=\frac{44}{5}=8.8$ metres. Hence option (a) is correct.

Similar triangles In two similar triangles the corresponding angles are equal, so $△XYZ$ has exactly the same three angles as $△PQR$, vertex for vertex. Two angles of $△PQR$ are given as ${30}^{\circ }$ and ${70}^{\circ }$. By the angle-sum property the remaining angle $={180}^{\circ }-{30}^{\circ }-{70}^{\circ }={80}^{\circ }$. Matching the vertices in the order $P\leftrightarrow X,Q\leftrightarrow Y,R\leftrightarrow Z$, the angle asked for measures ${80}^{\circ }$. Hence option (c) is correct.

Alligation and mixture The mixture is sold at ₹20 per kg with a gain of $10\%$, so its cost price per kg $=\frac{20\times 100}{110}=\frac{200}{11}$. So the ₹10 sugar and the ₹20 sugar have to be blended to give a mean cost of ₹$\frac{200}{11}$ per kg. By alligation, cheaper : dearer $=\left(20-\frac{200}{11}\right):\left(\frac{200}{11}-10\right)=\frac{20}{11}:\frac{90}{11}=2:9$. The cheaper sugar is 40 kg, so 2 parts =40 kg and 1 part =20 kg. Quantity of the ₹20 sugar $=9\times 20=180$ kg. Hence option (c) is correct.

Time and work Amit copies 30 pages in 6 hours, so his rate $=\frac{30}{6}=5$ pages per hour. Amit and Ravi together copy 250 pages in 25 hours, so their combined rate $=\frac{250}{25}=10$ pages per hour. Ravi's own rate =10-5=5 pages per hour. Time taken by Ravi for 40 pages $=\frac{40}{5}=8$ hours. Hence option (c) is correct.

Relative speed Both run in the same direction, so the gap between them closes at the relative speed =15-5=10 m/s. The gap to be covered is 500 m, so the time taken to catch him $=\frac{500}{10}=50$ seconds. In these 50 seconds the chain snatcher runs $5\times 50=250$ m. Check: the policeman runs $15\times 50=750$ m, which is exactly 500+250 m. Hence option (d) is correct.

Area and polygon angles Statement I: area of a triangle $=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}=\frac{1}{2}\times 10\times 8=40{\mathrm{cm}}^2$, so I is true. Statement II: the interior angles of an n-sided polygon add up to $\left(n-2\right)\times {180}^{\circ }$. A hexagon has n=6, so the sum $=\left(6-2\right)\times {180}^{\circ }=4\times {180}^{\circ }={720}^{\circ }$, so II is true as well. Both statements are true, so option (b) is correct.

Slant height of a pyramid The slant height of a right pyramid is measured on a lateral face — from the apex straight down to the midpoint of a base edge; the crawling insect is only a story and does not affect it. That slant height is the hypotenuse of the right triangle formed by the vertical height of the pyramid and half of a base side. Here the height h=20 cm and half the base side $=\frac{a}{2}=\frac{10}{2}=5$ cm. Slant height $l=\sqrt{h^2+{\left(\frac{a}{2}\right)}^2}=\sqrt{{20}^2+5^2}=\sqrt{400+25}=\sqrt{425}$. $\sqrt{425}=5\sqrt{17}\approx 5\times 4.1231=20.61$ cm. Hence option (a) is correct.

Successive discount Successive discounts are not added; each one acts on the price that is left after the previous one, so take the marked price as ₹100 and go step by step. After $15\%$: $100\times \frac{85}{100}=85$. After $20\%$: $85\times \frac{80}{100}=68$. After $25\%$: $68\times \frac{75}{100}=51$. The customer finally pays ₹51, so the net discount =100-51=49, i.e. $49\%$. Hence option (b) is correct; simply adding 15+20+25=60 is the trap to avoid.

Parallel chords of a circle The perpendicular from the centre bisects a chord, so half of the smaller chord $=\frac{10}{2}=5$ cm. That half-chord, its distance from the centre and the radius form a right triangle: $r=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$ cm. For the longer chord, half of it $=\frac{24}{2}=12$ cm, so its distance from the centre $=\sqrt{{13}^2-{12}^2}=\sqrt{169-144}=\sqrt{25}=5$ cm. The chords are on opposite sides of the centre, so the distance between them is the sum =12+5=17 cm. Hence option (c) is correct.

Radius of a circle The two circles are concentric and the comparison given is between their radii, not their diameters or circumferences. Radius of the drainage circle =35-15=20 metres. Hence option (a) is correct.

Comparing surds The three surds are of different orders, so bring them all to one common order — the LCM of $2,3$ and 4 is 12. $\sqrt{7}=7^{1/2}=7^{6/12}=\sqrt[12]{7^6}=\sqrt[12]{117649}$. $\sqrt[3]{12}={12}^{1/3}={12}^{4/12}=\sqrt[12]{{12}^4}=\sqrt[12]{20736}$. $\sqrt[4]{125}={125}^{1/4}={125}^{3/12}=\sqrt[12]{{125}^3}=\sqrt[12]{1953125}$. All three are now twelfth roots, so the smallest number is the one with the smallest number inside the root, namely 20736. Therefore $\sqrt[3]{12}\approx 2.29$ is the smallest, against $\sqrt{7}\approx 2.65$ and $\sqrt[4]{125}\approx 3.34$. Hence option (c) is correct.

Intercept of a line The x-intercept is the x-coordinate of the point where the line meets the x-axis, and every point of the x-axis has y=0. Put y=0 in 5x+2y=10: 5x+2(0)=10, i.e. 5x=10. $x=\frac{10}{5}=2$. So the line cuts the x-axis at $\left(2,0\right)$ and the required intercept is 2. Hence option (b) is correct.

Length of a chord The perpendicular drawn from the centre to a chord bisects the chord, so it forms a right triangle whose hypotenuse is the radius. Half of the chord $=\sqrt{r^2-d^2}=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$ cm. Length of the chord $=2\times 4=8$ cm. Hence option (b) is correct; (3,4,5) is the triplet at work here.

Side of a rhombus The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle whose legs are the two half-diagonals. Half-diagonals $=\frac{12}{2}=6$ cm and $\frac{10}{2}=5$ cm. Side $=\sqrt{6^2+5^2}=\sqrt{36+25}=\sqrt{61}$. $\sqrt{61}\approx 7.81$ cm, since ${7.81}^2=60.99$. Hence option (a) is correct.

Common tangents to circles When two circles touch each other externally, the distance between their centres equals the sum of the radii. Such a pair has two direct (external) common tangents, one on each side of the line of centres. There is also one transverse tangent — the line drawn at the single point of contact, which touches both circles there. Total number of common tangents =2+1=3. Hence option (d) is correct; the count is 4 when the circles lie completely apart and 2 when they intersect at two points.