21 November 2025 · Shift 1

Terminating decimals A fraction in its lowest terms gives a terminating decimal only when its denominator can be written as $2^m\times 5^n$. If the denominator carries any prime other than 2 or 5, the decimal never ends and its digits repeat in a fixed block. Here $20=2^2\times 5$, $16=2^4$ and $40=2^3\times 5$, so all three of those fractions terminate. But 19 is a prime other than 2 and 5, so $\frac{1}{19}=0.052631578\dots$ goes on for ever with a repeating block of 18 digits. Hence, option (b) is correct.

Profit and loss One dozen means 12 calculators, so the cost price of one calculator is $\frac{1800}{12}=150$, that is ₹150. Each one is sold for ₹180, so the profit on one calculator is 180-150=30, that is ₹30. Number of calculators $=\frac{\mathrm{Total}\mathrm{profit}}{\mathrm{Profit}\mathrm{on}\mathrm{one}}=\frac{420}{30}=14$. Check: 14 calculators cost $14\times 150=2100$ and fetch $14\times 180=2520$, a profit of 2520-2100=420. Hence, option (d) is correct.

Volume displacement The sphere sinks completely, so the water it pushes up has exactly the same volume as the sphere. Radius of the cylinder $R=\frac{40}{2}=20$ cm and the rise in the water level h=3 cm. Volume of the water pushed up $=\pi R^{2}h=\pi \times {20}^2\times 3=1200\pi$ cubic cm. Volume of the sphere $=\frac{4}{3}\pi r^3$, so $\frac{4}{3}\pi r^3=1200\pi$. Cancelling $\pi$: $r^3=\frac{1200\times 3}{4}=900$. $r=\sqrt[3]{900}\approx 9.65$ cm. Hence, option (b) is correct.

Discount and profit The marked price is 25% above the cost price, so $\mathrm{MP}=\mathrm{CP}\times \frac{125}{100}$. Therefore $\mathrm{CP}=4500\times \frac{100}{125}=3600$, that is ₹3,600. A discount of 20% is allowed on the marked price, so $\mathrm{SP}=4500\times \frac{80}{100}=3600$, that is ₹3,600. The selling price equals the cost price, so the profit is 3600-3600=0. Profit percentage $=\frac{0}{3600}\times 100=0\%$ — the deal is neither a gain nor a loss. Hence, option (b) is correct.

Average of a changed group Work with total ages, because averages cannot be added or subtracted directly. Total age of the original 40 students $=40\times 20=800$ years. Ten leave and ten join, so the hostel again has 40 students; their total age $=40\times 18.2=728$ years. Total age of the 10 students who left $=10\times 22.4=224$ years. Total age of the 30 students who stayed =800-224=576 years. Total age of the 10 students who joined =728-576=152 years. Their average age $=\frac{152}{10}=15.2$ years. Hence, option (b) is correct.

Percentage, defect rate Each of the three machines produces 500 units, so every defect rate is a percentage of 500. Defective units of A $=2.5\%$ of $500=\frac{2.5\times 500}{100}=12.5$. Defective units of B $=3.2\%$ of $500=\frac{3.2\times 500}{100}=16$. The three machines together give 42 defective units, so C contributes 42-(12.5+16)=13.5. Defect rate of C $=\frac{13.5}{500}\times 100=2.7\%$. Hence, option (b) is correct.

Averages and equations Let the marks of the first, second and third tests be A, B and C. $\frac{A+B}{2}=84\Rightarrow A+B=168$ … (i) $\frac{B+C}{2}=82\Rightarrow B+C=164$ … (ii) $\frac{A+C}{2}=80\Rightarrow A+C=160$ … (iii) Adding (i), (ii) and (iii): 2(A+B+C)=492, so A+B+C=246. Subtracting (iii) from this total: B=246-160=86. The extra clue agrees: (i) - (ii) gives A-C=4, exactly the 4 marks by which the third test fell short of the first. Hence, option (b) is correct.

Partnership profit sharing All three keep their money in the business for the same period, so the profit is divided in the ratio of the capitals alone. 40000:60000:70000=4:6:7, which makes 4+6+7=17 equal parts. Value of one part $=\frac{51000}{17}=3000$, that is ₹3,000. A holds 4 parts, so A's share $=4\times 3000=12000$, that is ₹12,000. Hence, option (c) is correct.

Arithmetic progression The daily heights rise by the same amount every day, so they form an arithmetic progression with first term a=1.6 and common difference d=0.4. Formula: $a_n=a+\left(n-1\right)d$. For the 10th day put n=10: $a_{10}=1.6+\left(10-1\right)\times 0.4$. $a_{10}=1.6+9\times 0.4=1.6+3.6=5.2$ cm. Note the common trap — there are nine increases between day 1 and day 10, not ten. Hence, option (c) is correct.

Circumradius of a triangle For a triangle with sides a, b, c and area $\Delta$, the radius of the circumscribed circle is $R=\frac{abc}{4\Delta }$. So the area is needed first; use Heron's formula with $s=\frac{13+14+15}{2}=21$. $\Delta =\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\sqrt{21\times 8\times 7\times 6}$. $\Delta =\sqrt{7056}=84$ square cm. Now $R=\frac{13\times 14\times 15}{4\times 84}=\frac{2730}{336}$. $R=8.125\approx 8.1$ cm. Hence, option (d) is correct.

Simple interest The amount is $184\%$ of the sum, so the interest alone is $184\%-100\%=84\%$ of the sum in 6 years. Interest for one year $=\frac{84\%}{6}=14\%$ of the sum, hence r=14. Formula: $SI=\frac{P\times R\times T}{100}$ $SI=\frac{3500\times 14\times 3}{100}=\frac{147000}{100}=1470$ So the interest earned is Rs. 1470, and option (d) is correct.

Simple interest and ratio Let the distributed money be 4x for Kavya and 5x for Aryan, so the distributed part is 9x. Formula: $SI=\frac{P\times R\times T}{100}$ Kavya's interest $=\frac{4x\times 8\times 5}{100}=1.6x$ and Aryan's interest $=\frac{5x\times 7\times 5}{100}=1.75x$. Difference =1.75x-1.6x=0.15x, and this equals 2691. $0.15x=2691\Rightarrow x=\frac{2691}{0.15}=17940$ Distributed money $=9x=9\times 17940=161460$, and this is $90\%$ of Aman's total money. Total $=\frac{161460\times 100}{90}=179400$ So Aman had Rs. 1,79,400, and option (b) is correct.

Heights and distances The wall, the ground and the ladder form a right triangle in which $\tan \theta =\frac{\mathrm{height}\mathrm{on}\mathrm{the}\mathrm{wall}}{\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{foot}}=\frac{8}{15}$. So take the height =8k and the base =15k. Hypotenuse (the ladder) $=\sqrt{{\left(8k\right)}^2+{\left(15k\right)}^2}=\sqrt{289k^2}=17k$ The ladder is 17 m long, so $17k=17\Rightarrow k=1$. Height at which the ladder touches the wall =8k=8 metres. So option (c) is correct.

Compound interest Under compound interest the money is multiplied by the same factor in every equal block of time. In 6 years the growth factor is $\frac{9000}{4000}=\frac{9}{4}$. Six years is two blocks of three years, so if k is the 3-year factor then $k^2=\frac{9}{4}$. $k=\sqrt{\frac{9}{4}}=\frac{3}{2}$ Amount on Rs. 3,200 after 3 years $=3200\times \frac{3}{2}=4800$ So the amount received is Rs. 4,800, and option (b) is correct.

Problems on ages Let the present ages of the son and the mother be 3x and 7x years. After 6 years they become 3x+6 and 7x+6, and this new ratio is 3:5. $\frac{3x+6}{7x+6}=\frac{3}{5}$ $5\left(3x+6\right)=3\left(7x+6\right)\Rightarrow 15x+30=21x+18$ $6x=12\Rightarrow x=2$ Son's age $=3\times 2=6$ years and mother's age $=7\times 2=14$ years. So option (a) is correct.

Tangent–chord angle Alternate segment theorem: the angle between a tangent and a chord drawn from the point of contact equals the inscribed angle that the same chord makes in the alternate segment. Here the tangent at A and the chord AC make ${50}^{\circ }$, so the angle in the alternate segment standing on chord AC is also ${50}^{\circ }$. The angle at the centre is twice the inscribed angle standing on the same arc. Angle subtended by minor arc AC at the centre $=2\times {50}^{\circ }={100}^{\circ }$ The ${85}^{\circ }$ at C belongs to chord AB and is not needed for this part. So option (c) is correct.

Speed, time and distance Speed of the train $=\frac{480}{6}=80$ km/h. Train : car =8:3, so the car's speed $=80\times \frac{3}{8}=30$ km/h. Formula: $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$ Time taken by the car $=\frac{480}{30}=16$ hours. So option (b) is correct.

Ratio and percentage Let the number of boys be 5x and the number of girls be 7x. $30\%$ of the boys fail, so $70\%$ of them pass: boys passed $=\frac{70}{100}\times 5x=3.5x$. $50\%$ of the girls fail, so $50\%$ of them pass: girls passed $=\frac{50}{100}\times 7x=3.5x$. Required ratio =3.5x:3.5x=1:1 So option (c) is correct.

Complementary angles Complementary-angle rule: $\sec \left({90}^{\circ }-A\right)=\mathrm{cosec}A$, so a secant equals a cosecant when the two angles add up to ${90}^{\circ }$. $\mathrm{cosec}{52}^{\circ }=\sec \left({90}^{\circ }-{52}^{\circ }\right)=\sec {38}^{\circ }$ Given $\sec \theta =\mathrm{cosec}{52}^{\circ }$, so $\sec \theta =\sec {38}^{\circ }$. Comparing the two sides, $\theta ={38}^{\circ }$. So option (b) is correct.

Laws of indices Write every factor as a power of 5. ${\left(\sqrt{5}\right)}^6={\left(5^{\frac{1}{2}}\right)}^6=5^3$ and ${25}^2={\left(5^2\right)}^2=5^4$ LHS $=5^3\times 5^4=5^{3+4}=5^7$ RHS $=5^n\times 5\sqrt{5}=5^n\times 5^1\times 5^{\frac{1}{2}}=5^{n+\frac{3}{2}}$ The bases are equal, so compare the powers: $n+\frac{3}{2}=7$. $n=7-\frac{3}{2}=\frac{14-3}{2}=\frac{11}{2}$ So option (c) is correct.

Surds and rationalisation Given $y=5-\sqrt{24}$, so $\frac{1}{y}=\frac{1}{5-\sqrt{24}}$. Rationalise by multiplying the numerator and the denominator by $5+\sqrt{24}$. $\frac{1}{y}=\frac{5+\sqrt{24}}{{\left(5\right)}^2-{\left(\sqrt{24}\right)}^2}=\frac{5+\sqrt{24}}{25-24}=5+\sqrt{24}$ $y+\frac{1}{y}=\left(5-\sqrt{24}\right)+\left(5+\sqrt{24}\right)$ The surd terms cancel, leaving $y+\frac{1}{y}=10$. So option (c) is correct.

Similar triangles: areas In similar triangles the ratio of the areas is the square of the ratio of corresponding sides. Let the smaller side be s=3 cm and the corresponding larger side be S. ${\left(\frac{s}{S}\right)}^2=\frac{6}{48}=\frac{1}{8}$ $\frac{s}{S}=\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}$ $S=2\sqrt{2}\times s=2\sqrt{2}\times 3=6\sqrt{2}$ cm So option (b) is correct.

Similar triangles: area ratio The triangle cut off at the top is similar to the whole triangle, and its height is $\frac{1}{4}$ of the whole height. For similar figures, ratio of areas = (ratio of heights)${}^2$. $\frac{\mathrm{ar}\left(\mathrm{small}\right)}{\mathrm{ar}\left(\mathrm{large}\right)}={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$ Take the area of the larger triangle as 16 units, so the small triangle is 1 unit. Area of the trapezoid =16-1=15 units. Trapezoid : larger triangle =15:16 So option (b) is correct.

Algebraic identities Identity: ${\left(a+b\right)}^2=a^2+b^2+2ab$, so $a^2+b^2={\left(a+b\right)}^2-2ab$. $a^2+b^2={\left(12\right)}^2-2\times 14$ =144-28 =116 So option (d) is correct.

Algebraic identities Identity: $x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)$, so $\frac{x^3+y^3}{x+y}=x^2-xy+y^2$. Adding the two given equations: $\left(x+y\right)+\left(x-y\right)=16+8\Rightarrow 2x=24\Rightarrow x=12$. Then y=16-12=4. $x^2-xy+y^2={\left(12\right)}^2-12\times 4+{\left(4\right)}^2$ =144-48+16 =112 So option (c) is correct.