18 November 2025 · Shift 1

Sets of numbers Natural numbers start from 1: $1,2,3,4,\dots$ Whole numbers are the natural numbers together with 0: $0,1,2,3,4,\dots$ So every natural number is also a whole number — the set of natural numbers lies completely inside the set of whole numbers. The intersection of a set with a larger set that contains it is the smaller set itself, so the common part is exactly the set of natural numbers. Hence, option (a) is the most appropriate answer.

Real and imaginary numbers A square root gives a real number only when the quantity under the root sign is zero or positive. $\sqrt{8}\approx 2.828$, $-\sqrt{8}\approx -2.828$ and $\frac{1}{\sqrt{8}}\approx 0.354$ — all three are irrational, but every one of them is a real number. In $\sqrt{-8}$ the number under the root is negative, and no real number multiplied by itself can give -8. It equals $2i\sqrt{2}$, which is imaginary, not real. Hence, option (b) is the most appropriate answer.

Density of irrational numbers Take any two rational numbers a and b with a<b. The number $a+\frac{b-a}{\sqrt{2}}$ lies between a and b, because $0<\frac{1}{\sqrt{2}}<1$. It is irrational, since a non-zero rational divided by $\sqrt{2}$ is irrational and rational + irrational is always irrational. The same construction can now be repeated inside the smaller gap a to that new number, and so on without end. For example, between 1 and 2 we already have $\sqrt{2}$, $\sqrt{3}$, $\sqrt{2}+0.1$, $\pi -1$, and endlessly many more. So the number of irrational numbers between two rational numbers is infinite. Hence, option (d) is the most appropriate answer.

Medians and centroid The medians of a triangle meet at the centroid, and the centroid divides every median in the ratio 2:1 measured from the vertex. So $AO=\frac{2}{3}\times AE=\frac{2}{3}\times 15=10$ cm. And $BO=\frac{2}{3}\times BF=\frac{2}{3}\times 24=16$ cm. The medians cut at right angles, so triangle AOB is right-angled at O and Pythagoras applies. $A{B}^2=A{O}^2+B{O}^2={10}^2+{16}^2=100+256=356$ $AB=\sqrt{356}=\sqrt{4\times 89}=2\sqrt{89}$ cm Hence, option (a) is the most appropriate answer.

Men and days (inverse proportion) For a fixed amount of work, men and days are inversely proportional, so $M_1{D}_1=M_2{D}_2$. Total work $=18\times 25=450$ man-days. With 15 men, days required $=\frac{450}{15}=30$ days. The check is that fewer men must take more days, and 30 is indeed larger than 25. Hence, option (d) is the most appropriate answer.

Time and work Let the whole work be 100 units. A finishes 70 units in 14 days, so A's one day's work $=\frac{70}{14}=5$ units. Work still left =100-70=30 units, and A and B together clear it in 3 days. So (A + B)'s one day's work $=\frac{30}{3}=10$ units. B's one day's work =10-5=5 units. Time taken by B alone $=\frac{100}{5}=20$ days. Hence, option (b) is the most appropriate answer.

Position of two circles Let the radii be R and r with R>r, and let d be the distance between the two centres. The standard comparisons are: d=R+r means the circles touch externally; d=R-r means they touch internally; R-r<d<R+r means they cut at two points; and d=0 means they are concentric. Here we are given d=R-r. The smaller circle then lies inside the bigger one and the two boundaries meet at exactly one point, which is internal contact. Hence, option (c) is the most appropriate answer.

Angle between two tangents Let O be the centre and let the tangents drawn at A and at B meet at the external point C. A radius is always perpendicular to the tangent at the point of contact, so $\angle OAC=\angle OBC={90}^{\circ }$. OACB is a quadrilateral, and the four angles of a quadrilateral add up to ${360}^{\circ }$. $\angle AOB+\angle OAC+\angle ACB+\angle OBC={360}^{\circ }$ ${80}^{\circ }+{90}^{\circ }+\angle ACB+{90}^{\circ }={360}^{\circ }$ $\angle ACB={360}^{\circ }-{260}^{\circ }={100}^{\circ }$ Useful shortcut: the angle between the two tangents $={180}^{\circ }-$ (angle subtended at the centre). Hence, option (b) is the most appropriate answer.

Circumradius of a right triangle First test the sides with the converse of Pythagoras' theorem: $8^2+{15}^2=64+225=289$ and ${17}^2=289$. Since the two sums agree, the triangle is right-angled and 17 cm is its hypotenuse. The circle through all three vertices is the circumcircle. An angle in a semicircle is ${90}^{\circ }$, so in a right-angled triangle the hypotenuse itself is the diameter of the circumcircle. Circumradius $=\frac{17}{2}=8.5$ cm. Hence, option (d) is the most appropriate answer.

Volume of a cone The cone has the same radius as the circle, so first pull the radius out of the circumference. Circumference $=2\pi r$, which gives $r=\frac{49}{2}=24.5$ cm, and the height is h=14 cm. Volume of a cone $=\frac{1}{3}\pi r^{2}h$ $=\frac{1}{3}\times \frac{22}{7}\times \frac{49}{2}\times \frac{49}{2}\times 14$ $=\frac{22\times 49\times 49\times 2}{3\times 2\times 2}=\frac{26411}{3}=8803.67$ So the volume is approximately 8803 cm³. Hence, option (a) is the most appropriate answer.

Incircle tangent lengths Tangents drawn to a circle from an external point are equal, so let PA=PC=x, QA=QB=y and RB=RC=z. Then PQ=x+y, QR=y+z and PR=x+z. Perimeter =(x+y)+(y+z)+(x+z)=2(x+y+z)=39, so x+y+z=19.5. PQ-QR=(x+y)-(y+z)=x-z=4 and PQ-PR=(x+y)-(x+z)=y-z=2. Substituting x=z+4 and y=z+2: $\left(z+4\right)+\left(z+2\right)+z=19.5\Rightarrow 3z=13.5\Rightarrow z=4.5$. Hence x=4.5+4=8.5 and y=4.5+2=6.5. AQ+PC=y+x=6.5+8.5=15, so the required value is 15 cm and option (d) is correct.

Simple interest in two parts Let the whole sum lent be P; $30\%$ of it earns $7.5\%$ p.a. and $70\%$ of it earns $11.25\%$ p.a. Replace the two rates by one equivalent rate on the whole sum: $\frac{30\times 7.5+70\times 11.25}{100}=\frac{225+787.5}{100}=10.125\%$ per annum. For 2.5 years the interest is $\frac{10.125\times 2.5}{100}=\frac{25.3125}{100}$ of P, i.e. about 0.2531P. So $0.2531P=7825\Rightarrow P=\frac{7825}{0.2531}$. $P\approx 30916.63$, so the sum lent was about Rs. 30,916.63 and option (a) is correct.

Income, expenditure and savings Ankush spends $85\%$, so he saves $100\%-85\%=15\%$ of his income, and that saving is Rs. 9,000. Income of Ankush $=9000\times \frac{100}{15}=60000$, i.e. Rs. 60,000. Ankit spends $70\%$, so he saves $30\%$ of his income, again Rs. 9,000. Income of Ankit $=9000\times \frac{100}{30}=30000$, i.e. Rs. 30,000. Income of Mohit $=\frac{1}{3}\left(60000+30000\right)=\frac{90000}{3}=30000$, i.e. Rs. 30,000. Mohit spends $\frac{7}{9}$ of his income, so he saves $1-\frac{7}{9}=\frac{2}{9}$ of it. Savings of Mohit $=\frac{2}{9}\times 30000=\frac{60000}{9}=6666.67$, so option (a) is correct.

Partnership capital ratio The capitals of P and Q are in the ratio 8:6, which reduces to 4:3. Let the capital of Q be x, so $\frac{80000}{x}=\frac{4}{3}$. $x=80000\times \frac{3}{4}=60000$. Hence Q invested Rs 60,000 and option (a) is correct.

Ratio from percentages Let the total score of the test be T. First student's score $=70\%$ of $T=\frac{70T}{100}$; second student's score $=80\%$ of $T=\frac{80T}{100}$. Required ratio $=\frac{70T}{100}:\frac{80T}{100}=70:80$. Dividing both terms by 10 gives 7:8, so option (b) is correct. Note the order asked for — first to second — which rules out the reversed option 8:7.

Average age with replacement In 3 years the same 12 people grow older by $12\times 3=36$ years in total. The average today equals the average of 3 years ago, so the total age today equals the total age of 3 years ago. Therefore the replacement must wipe out that gain of 36 years. The 4 players who left have a present total age $=4\times 30=120$ years. Total age of the 4 new players =120-36=84 years. Average age of the new players $=\frac{84}{4}=21$ years, so option (b) is correct.

Change in average Average of the 17 numbers =32, so their sum $=17\times 32=544$. Doubling every number doubles the sum: new sum $=2\times 544=1088$. Adding 8 to each of the 17 numbers adds $17\times 8=136$, so the final sum =1088+136=1224. New average $=\frac{1224}{17}=72$. Shortcut: whatever is done to every number is done to the average as well, so the new average $=2\times 32+8=72$ and option (b) is correct.

Speed, time and percentage Original speed $=\frac{1900}{4}=475$ km/h. Speed needed to cover the same distance in 5 hours $=\frac{1900}{5}=380$ km/h. Decrease in speed =475-380=95 km/h. Percentage decrease $=\frac{95}{475}\times 100=20\%$. Shortcut: distance being fixed, speed varies inversely with time, so time 4:5 gives speed 5:4 and the drop is $\frac{1}{5}=20\%$, so option (b) is correct.

Successive discounts Marked price = ₹3,500 and the first discount is $30\%$, so $70\%$ of the marked price is paid. Price after the first discount $=3500\times \frac{70}{100}=2450$, i.e. ₹2,450. The second discount of $15\%$ is given on this already reduced price, so $85\%$ of it remains. Final selling price $=2450\times \frac{85}{100}=2082.5$. Hence the article is finally sold for ₹2,082.5 and option (a) is correct.

Successive discounts Successive discounts multiply, they do not add — the second one acts on the reduced price. Price after the $20\%$ discount $=50000\times \frac{80}{100}=40000$, i.e. ₹40,000. Price after the further $30\%$ discount $=40000\times \frac{70}{100}=28000$. Check with the single equivalent discount: $100-\frac{80\times 70}{100}=100-56=44\%$, and $50000\times \frac{56}{100}=28000$. So the final selling price is ₹28,000 and option (c) is correct.

Removal and replacement in a mixture In 80 litres with wine : water =5:3, wine $=\frac{5}{8}\times 80=50$ litres and water $=\frac{3}{8}\times 80=30$ litres. The 16 litres taken out is $\frac{16}{80}=\frac{1}{5}$ of the mixture, so it carries away one-fifth of each liquid and $\frac{4}{5}$ of each is left. Wine left $=50\times \frac{4}{5}=40$ litres; water left $=30\times \frac{4}{5}=24$ litres. Now 16 litres of pure water is poured in, so water =24+16=40 litres while the wine stays 40 litres. New ratio of wine to water =40:40=1:1, so option (b) is correct.

Comparing surds Roots of different orders cannot be compared directly; raise them all to the LCM of the orders. $LCM\left(4,3,4\right)=12$, so compare the 12th power of each number. ${\left(\sqrt[4]{4}\right)}^{12}=4^3=64$, ${\left(\sqrt[3]{6}\right)}^{12}=6^4=1296$, ${\left(\sqrt[4]{7}\right)}^{12}=7^3=343$. Since 64<343<1296, the same order holds for the numbers themselves: $\sqrt[4]{4}<\sqrt[4]{7}<\sqrt[3]{6}$. So the ascending order is $\sqrt[4]{4},\sqrt[4]{7},\sqrt[3]{6}$ and option (a) is correct.

Equation of a line Use the point-slope form of a straight line: $y-y_1=m\left(x-x_1\right)$. Here m=-3 and $\left(x_1,y_1\right)=\left(4,6\right)$. y-6=-3(x-4) y-6=-3x+12 y=-3x+18, so option (a) is correct. A quick check: putting x=4 gives y=-12+18=6, so the line does pass through (4,6).

Similar triangles: area and perimeter In similar triangles the ratio of the areas equals the square of the ratio of any pair of corresponding lengths. $\frac{\mathrm{smaller}\mathrm{area}}{\mathrm{larger}\mathrm{area}}=\frac{36}{64}=\frac{9}{16}$. Ratio of corresponding sides $=\sqrt{\frac{9}{16}}=\frac{3}{4}$. Perimeters are corresponding lengths, so they are also in the ratio 3:4. $\frac{30}{p}=\frac{3}{4}\Rightarrow p=30\times \frac{4}{3}=40$. Hence the perimeter of the larger triangle is 40 m and option (a) is correct.

Basic proportionality and areas AB=AD+DB=5+15=20 cm and AC=AE+EC=4+12=16 cm. $\frac{AD}{AB}=\frac{5}{20}=\frac{1}{4}$ and $\frac{AE}{AC}=\frac{4}{16}=\frac{1}{4}$, which agrees with $DE\parallel BC$. Because $DE\parallel BC$, $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$, so $△ADE\sim △ABC$ by AA similarity. For similar triangles the areas are in the square of the ratio of corresponding sides: $\frac{\left(△ADE\right)}{\left(△ABC\right)}={\left(\frac{AD}{AB}\right)}^2={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$. Hence the required ratio is 1:16 and option (b) is correct.