18 November 2025 · Shift 3

Irrational numbers An irrational number cannot be written as a ratio of two integers, and the set of irrational numbers is not closed under addition. Take $a=\sqrt{2}$ and $b=\sqrt{3}$: then $a+b=\sqrt{2}+\sqrt{3}$, which is irrational. Now take b=-a: the two irrational parts cancel and $a+b=\sqrt{2}+\left(-\sqrt{2}\right)=0$, which is rational. So the sum is neither always irrational nor always rational, and it does become rational when a=-b. Hence the correct option is (c).

Trigonometric ratios 8,15,17 is a Pythagorean triplet, since $8^2+{15}^2=64+225=289={17}^2$. So the angle opposite the longest side is a right angle: $\angle Q={90}^{\circ }$ and PR is the hypotenuse. For $\angle R$ the opposite side is PQ=8 cm, the adjacent side is QR=15 cm and the hypotenuse is PR=17 cm. So $\tan R=\frac{8}{15}$ and $\sin R=\frac{8}{17}$. For $\angle P$ the opposite side is QR=15 cm and the adjacent side is PQ=8 cm, hence $\cot P=\frac{8}{15}$. $\tan R+\sin R+2\cot P=\frac{8}{15}+\frac{8}{17}+\frac{16}{15}=\frac{24}{15}+\frac{8}{17}$ =1.6+0.470=2.070 Hence the correct option is (a).

Time and work efficiency Sonu is 30% less efficient than Monu, so if Monu's efficiency is 100, Sonu's is 70. Efficiency ratio Sonu : Monu =70:100=7:10. Working together they do 7+10=17 units a day, and they finish in 40 days, so the total work $=17\times 40=680$ units. 70% of the work $=\frac{70}{100}\times 680=476$ units. Sonu alone does 7 units a day, so he needs $\frac{476}{7}=68$ days. Hence the correct option is (d).

Chord and radius Join the two ends of the chord to the centre; the two radii and the chord form a triangle. Here OA=OB=r and $\angle AOB={90}^{\circ }$, so $△AOB$ is a right-angled isosceles triangle with AB=12 cm as its hypotenuse. By Pythagoras' theorem: $r^2+r^2={12}^2$ $2r^2=144\Rightarrow r^2=72$ $r=\sqrt{72}=6\sqrt{2}$ cm. Hence the correct option is (b).

Length of a chord The perpendicular drawn from the centre of a circle to a chord bisects that chord. So in right-angled $△OMA$, OA=17 cm is the radius and OM=8 cm is the given distance. $AM=\sqrt{O{A}^2-O{M}^2}=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ cm. Length of the chord $AB=2\times AM=2\times 15=30$ cm. Hence the correct option is (c).

Tangent-chord angle By the alternate segment theorem, the angle between a tangent and a chord drawn from the point of contact equals the angle that the chord subtends in the alternate segment. The chord AB subtends $\angle AOB={100}^{\circ }$ at the centre, and the angle at the circumference is half the angle at the centre. So the angle in the alternate segment $=\frac{{100}^{\circ }}{2}={50}^{\circ }$. Therefore the angle between the chord AB and the tangent at A is ${50}^{\circ }$. Hence the correct option is (b).

Cylinder surface and volume For a cylinder of radius r and height h: curved surface area $=2\pi rh$ and total surface area $=2\pi r\left(h+r\right)$. $\frac{2\pi rh}{2\pi r\left(h+r\right)}=\frac{3}{4}\Rightarrow \frac{h}{h+r}=\frac{3}{4}$ $4h=3h+3r\Rightarrow h=3r$ Volume $=\pi r^{2}h=\pi r^2\left(3r\right)=3\pi r^3=3234$ $r^3=\frac{3234\times 7}{3\times 22}=343\Rightarrow r=7$ cm, so $h=3\times 7=21$ cm. Curved surface area $=2\pi rh=2\times \frac{22}{7}\times 7\times 21=924$ ${cm}^2$. Hence the correct option is (a).

Triangle with incircle The inscribed circle only fixes the points of contact S, U and T; the sides themselves come from the two given relations and the perimeter. Let PQ=x cm. $PQ=QR+7\Rightarrow QR=x-7$ $PQ=PR+4\Rightarrow PR=x-4$ Perimeter: x+(x-7)+(x-4)=34 $3x-11=34\Rightarrow 3x=45\Rightarrow x=15$ cm. Therefore PR=15-4=11 cm. Hence the correct option is (c).

Simple interest Simple interest $=\frac{P\times R\times T}{100}$, and here the rate is 8% for both, so the interest depends only on the product $P\times T$. Equating the two interests: $P_A\times 5=P_B\times 7$. B invested ₹12,000 less than A, so $P_B=P_A-12000$. $5P_A=7\left(P_A-12000\right)$ $5P_A=7P_A-84000$ $2P_A=84000\Rightarrow P_A=42000$ So A's investment is ₹42,000, and the correct option is (c).

Partnership profit sharing When all the partners invest for the same period, the profit is divided in the ratio of their investments. 40000:30000:80000=4:3:8 Total number of parts =4+3+8=15. B's share $=\frac{3}{15}\times 30000=6000$ So B receives ₹6,000, and the correct option is (b).

Ratio and proportion Let the two numbers be 3x and 7x. Given: $\frac{3x+14}{7x+14}=\frac{13}{17}$ Cross-multiplying: 17(3x+14)=13(7x+14) 51x+238=91x+182 $91x-51x=238-182\Rightarrow 40x=56$ $x=\frac{56}{40}=\frac{7}{5}$ Sum of the original numbers $=3x+7x=10x=10\times \frac{7}{5}=14$, so option (a) is correct.

LCM and divisibility A number divisible by 7, 11 and 15 together must be a multiple of their LCM. 7, 11 and 15 have no common factor, so $LCM=7\times 11\times 15=1155$. Divide 5757 by 1155: $1155\times 4=4620$ and $1155\times 5=5775$. So 5757 lies between these two, and the next multiple of 1155 is 5775. Least number to be added =5775-5757=18, so option (b) is correct.

HCF and LCM of two numbers The HCF is 13, so let the numbers be 13a and 13b where a and b are co-prime. Then $LCM=13ab$. Given $\frac{LCM}{HCF}=\frac{7}{1}$, so $\frac{13ab}{13}=7\Rightarrow ab=7$. As 7 is prime and a,b are co-prime, a=1 and b=7. The numbers are $13\times 1=13$ and $13\times 7=91$. Their sum 13+91=104 matches the condition, so option (b) is correct.

Average: uniform change Rule: if the same quantity is subtracted from every observation, the average falls by exactly that quantity. Sum of the 20 numbers $=20\times 55=1100$. After subtracting 8 from each, the sum falls by $20\times 8=160$, giving 1100-160=940. New average $=\frac{940}{20}=47$, which is simply 55-8. Hence the new average is 47, option (b).

Average: new member Let the original number of employees be n. Total salary of these employees =30000n. After the new employee joins, there are n+1 employees and the average is 30000+300=30300. $\frac{30000n+60000}{n+1}=30300$ 30000n+60000=30300n+30300 $60000-30300=30300n-30000n\Rightarrow 29700=300n$ $n=\frac{29700}{300}=99$ So the company originally had 99 employees, option (a).

Percentage comparison Take a convenient value Q=100. P is 60% more than Q, so P=100+60=160. P+Q=160+100=260 R is 80% less than P+Q, so R is 20% of 260. $R=\frac{20}{100}\times 260=52$ Required percentage $=\frac{R}{P}\times 100=\frac{52}{160}\times 100=32.5\%$ Hence option (a) is correct.

Time, speed and distance Formula: $\mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}}$ First stretch: $\frac{200}{40}=5$ hours. Reduced speed $=40-50\%$ of 40=40-20=20 km/h. Second stretch: $\frac{100}{20}=5$ hours. Total time =5+5=10 hours, so option (d) is correct.

Time and work Total work $=\mathrm{men}\times \mathrm{days}=16\times 28=448$ man-days. Work done in the first 8 days $=16\times 8=128$ man-days. Remaining work =448-128=320 man-days. Men still on the job =16-8=8. Further days needed $=\frac{320}{8}=40$ days, so option (d) is correct.

Discount and saving The amount saved on a purchase is exactly the discount allowed on the listed value. Formula: $\mathrm{saving}=\mathrm{listed}\mathrm{value}\times \frac{\mathrm{rate}\mathrm{of}\mathrm{discount}}{100}$ $=3050\times \frac{36}{100}$ $=30.5\times 36=1098$ So his total saving is ₹1,098, option (d).

Marked price and discount Let the marked price be $\mathrm{MP}$. A 35% discount means the buyer pays 65% of the marked price. $\frac{65}{100}\times \mathrm{MP}=7000\Rightarrow \mathrm{MP}=\frac{7000\times 100}{65}=\frac{140000}{13}$ With a 40% discount the selling price is 60% of the marked price. New SP $=\frac{60}{100}\times \frac{140000}{13}=\frac{84000}{13}$ Equivalently, New SP $=7000\times \frac{60}{65}=\frac{84000}{13}=6461.54$ So the new selling price is ₹6,461.54, option (b).

Mixture: milk and water Container A holds 40 litres in the ratio 3:5, so the parts total 3+5=8. Milk $=\frac{3}{8}\times 40=15$ L and water $=\frac{5}{8}\times 40=25$ L. Container B holds 60 litres in the ratio 5:1, so the parts total 5+1=6. Milk $=\frac{5}{6}\times 60=50$ L and water $=\frac{1}{6}\times 60=10$ L. After mixing, milk =15+50=65 L and water =25+10=35 L. Required ratio =65:35=13:7, so option (c) is correct.

Comparison of surds Surds of different orders can be compared only after they are written with a common index. The orders are 3, 2 and 6; their LCM is 6. $\sqrt[3]{3}=3^{\frac{1}{3}}=3^{\frac{2}{6}}=\sqrt[6]{9}$ $\sqrt{5}=5^{\frac{1}{2}}=5^{\frac{3}{6}}=\sqrt[6]{125}$ $\sqrt[6]{7}=7^{\frac{1}{6}}=\sqrt[6]{7}$ All three now have index 6, so compare the radicands: 125>9>7. Hence the largest value is $\sqrt{5}$, and option (b) is correct.

Equation of a straight line Formula: point-slope form $y-y_1=m\left(x-x_1\right)$ Here m=-4 and $\left(x_1,y_1\right)=\left(4,-2\right)$. y-(-2)=-4(x-4) y+2=-4x+16 y=-4x+14 Only this line gives y=-2 when x=4, so option (b) is correct.

Similar triangles: medians In two similar triangles every pair of corresponding lengths is in the same ratio — sides, perimeters, altitudes, medians and angle bisectors alike. Only the areas behave differently: they are in the square of that ratio. Given $\frac{\mathrm{perimeter}\mathrm{of}△ABC}{\mathrm{perimeter}\mathrm{of}△MNQ}=\frac{7}{8}$, the scale factor is $\frac{7}{8}$. Let AD be the median from A and MP the median from M; these are corresponding medians. So $\frac{AD}{MP}=\frac{7}{8}$. Hence the required ratio is 7:8, option (a). The ratio 49:64 would be that of the areas.

Basic proportionality theorem Since $DE\parallel BC$, the basic proportionality (Thales) theorem applies to $△ABC$. Formula: $\frac{AD}{DB}=\frac{AE}{EC}$ $\frac{8}{9}=\frac{16}{EC}$ $8\times EC=9\times 16=144$ $EC=\frac{144}{8}=18$ cm, so option (b) is correct.