15 November 2025 · Shift 1

Classification of numbers A rational number can be written as $\frac{p}{q}$ with p,q integers and $q\ne 0$; an irrational number cannot be written in that form. So the two families are completely separate — no number can be rational and irrational at the same time. Hence 'Every rational number is an irrational number' is false, which makes option (d) the required answer. Every whole number is an integer (the integers merely add the negative numbers), so (a) is true. Every integer n can be written as $\frac{n}{1}$, so (b) is true, and the real numbers are exactly the rationals together with the irrationals, so (c) is true.

Decimal expansion of numbers A number is rational exactly when its decimal expansion either terminates or repeats a fixed block of digits for ever. An irrational number cannot be written as $\frac{p}{q}$, so its expansion can do neither of these two things. Hence its decimal expansion is non-terminating and non-repeating, as in $\sqrt{2}=1.41421356\dots$ or $\pi =3.14159265\dots$ Terminating expansions such as 0.25 and repeating expansions such as $0.333\dots$ belong to rational numbers, so option (c) is correct.

Profit and loss on two articles Let the cost price of the second item be x; the first item costs $50\%$ more, so its cost price is 1.5x. Profit on the first item $=\frac{20}{100}\times 1.5x=0.3x$. Loss on the second item $=\frac{10}{100}\times x=0.1x$. Overall profit =0.3x-0.1x=0.2x. Given that this equals ₹200, $0.2x=200\Rightarrow x=\frac{200}{0.2}=1000$. So the second item costs ₹1,000 and the first costs $1.5\times 1000=1500$, i.e. ₹1,500 — option (d).

Overall profit percentage Total cost price $=30\times 500=15000$, i.e. ₹15,000. First lot: 20 kettles cost $20\times 500=10000$ and are sold at $20\%$ profit, so their selling price $=10000\times \frac{120}{100}=12000$. Second lot: 10 kettles cost $10\times 500=5000$ and are sold at $10\%$ loss, so their selling price $=5000\times \frac{90}{100}=4500$. Total selling price =12000+4500=16500, so the overall profit =16500-15000=1500. Profit percentage $=\frac{1500}{15000}\times 100=10\%$, so option (c) is correct.

Discount percentage The listed price of one plate is Rs. 320, so the marked price of 15 plates $=15\times 320=4800$. They are actually sold for Rs. 4000, so the discount =4800-4000=800. Discount percentage is always taken on the marked price: $\mathrm{discount}\%=\frac{\mathrm{discount}}{\mathrm{marked}\mathrm{price}}\times 100$. $=\frac{800}{4800}\times 100=\frac{100}{6}=16.67\%$, so option (d) is correct.

Mark-up, discount and profit Cost price =12000. Marked price $=12000\times \frac{130}{100}=15600$. A discount of $20\%$ is allowed on this, so selling price $=15600\times \frac{80}{100}=12480$. Profit =12480-12000=480. Profit percentage $=\frac{480}{12000}\times 100=4\%$, so option (c) is correct. Shortcut: the net effect of a $30\%$ mark-up and a $20\%$ discount is $\frac{130\times 80}{100}=104$, i.e. a profit of $4\%$.

Average with a corrected entry Wrong total of marks $=20\times 65=1300$. One student's marks were entered as 45 instead of 85, so the total fell short by 85-45=40. Correct total =1300+40=1340. Correct average $=\frac{1340}{20}=67$, so option (b) is correct. Quick check: the average rises by $\frac{40}{20}=2$, and 65+2=67.

Average and percentage Let the marks in English be x. History marks are $25\%$ more than this: $x+\frac{25}{100}x=\frac{5x}{4}$. The average of the two subjects is 81, so their total $=2\times 81=162$. Therefore $x+\frac{5x}{4}=162$, i.e. $\frac{9x}{4}=162$. $x=\frac{162\times 4}{9}=72$. So the marks in English are 72 (and in History 90), which makes option (a) correct.

Percentage: union of two sets Students who passed in both subjects have been counted once in English and once in Maths, so they are counted twice. Use $n\left(E\cup M\right)=n\left(E\right)+n\left(M\right)-n\left(E\cap M\right)$. Passed in at least one subject $=55\%+65\%-40\%$. $=120\%-40\%=80\%$, so option (a) is correct.

Partnership: profit sharing In a partnership the profit is divided in the ratio of (capital $\times$ time) of each partner. P invests for the whole year: $20000\times 12=240000$. Q also invests for the whole year: $30000\times 12=360000$. R joins after 5 months, so he stays for 12-5=7 months: $40000\times 7=280000$. Ratio =240000:360000:280000=24:36:28. Dividing throughout by 4 gives 6:9:7, so option (d) is correct.

Partnership profit sharing When all the partners keep their money in the business for the same time, the profit is divided in the ratio of the capitals. K:L:M=30000:40000:70000=3:4:7 Total number of parts =3+4+7=14, and these 14 parts together stand for the whole profit of ₹84,000. Value of one part $=\frac{84000}{14}=6000$ L's share $=4\times 6000=24000$ Hence L receives ₹24,000, so option (c) is correct.

Ratio of cylinder volumes Volume of a cylinder $=\pi r^{2}h$, so for two cylinders $V_1:V_2=r12h_1:r22h_2$ (the $\pi$ cancels). Take $r_1=2k,r_2=3k$ and $h_1=5m,h_2=3m$. $V_1:V_2={\left(2\right)}^2\times 5:{\left(3\right)}^2\times 3$ $=4\times 5:9\times 3=20:27$ Hence the required ratio is 20:27, so option (b) is correct.

Circle inscribed in a square A circle inscribed in a square touches all four sides, so its diameter is exactly the side of the square. Side =20 cm, hence radius $r=\frac{20}{2}=10$ cm. Area of the square $=20\times 20=400$ sq cm Area of the circle $=\pi r^2=3.14\times {10}^2=314$ sq cm Uncovered area =400-314=86 sq cm Hence the required area is $86{\mathrm{cm}}^2$, so option (b) is correct.

Algebraic identity Use the identity ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)$. The question already gives both pieces on the right-hand side, so nothing needs to be solved for a,b,c separately. ${\left(a+b+c\right)}^2=300+2\times 92$ =300+184=484 Hence the required value is 484, so option (c) is correct.

Curved surface areas Curved surface area of a cone $=\pi Rl$, where R is its base radius and l its slant height. Curved surface area of a hemisphere $=2\pi r^2$. Here the cone's radius is R=2r, so its curved surface area $=\pi \left(2r\right)l=2\pi rl$. The two areas are equal: $2\pi rl=2\pi r^2$ Cancelling $2\pi r$ from both sides gives l=r. Hence the slant height of the cone is l=r, so option (b) is correct.

Trigonometric simplification Add the two fractions over the common denominator $\left(1+\sin x\right)\cos x$. $\frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}=\frac{{\cos }^{2}x+{\left(1+\sin x\right)}^2}{\left(1+\sin x\right)\cos x}$ Numerator $={\cos }^{2}x+1+2\sin x+{\sin }^{2}x$, and ${\sin }^{2}x+{\cos }^{2}x=1$, so it becomes $2+2\sin x$. $=\frac{2\left(1+\sin x\right)}{\left(1+\sin x\right)\cos x}=\frac{2}{\cos x}$ $=2\sec x$ Hence the expression equals $2\sec x$, so option (d) is correct.

Trigonometric equation The coefficients are $\sqrt{3}$ and 1, whose resultant is $\sqrt{3+1}=2$, so divide the whole equation by 2. $\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta =\frac{1}{2}$ $\sin {60}^{\circ }\cos \theta +\cos {60}^{\circ }\sin \theta =\frac{1}{2}$, i.e. $\sin \left({60}^{\circ }+\theta \right)=\frac{1}{2}$ Since $0^{\circ }\le \theta \le {90}^{\circ }$, the angle ${60}^{\circ }+\theta$ lies between ${60}^{\circ }$ and ${150}^{\circ }$, so it must be ${150}^{\circ }$. $\theta ={150}^{\circ }-{60}^{\circ }={90}^{\circ }$ Check: $\sqrt{3}\cos {90}^{\circ }+\sin {90}^{\circ }=0+1=1$, which satisfies the equation. Hence $\theta ={90}^{\circ }$, so option (a) is correct.

Wages in ratio of efficiency Wages are always shared in the ratio of the work done, that is, in the ratio of the efficiencies. Woman : girl =3:1, so the money of each day is cut into 3+1=4 equal parts. The pair earned Rs. 800 for 4 days, so their daily earning together $=\frac{800}{4}=200$ Girl's daily wage $=\frac{1}{4}\times 200=50$ Hence the girl gets Rs. 50 per day, so option (a) is correct.

Perimeter of a semicircle Perimeter of the rectangle =2(l+b)=2(16+14)=60 cm The square has the same perimeter, so its side $=\frac{60}{4}=15$ cm. This side is the diameter of the semicircle, hence $r=\frac{15}{2}=7.5$ cm. The circumference of a semicircle includes the bounding diameter: $\pi r+2r$. $=\frac{22}{7}\times 7.5+15=23.57+15$ =38.57 cm Hence the required circumference is 38.57 cm, so option (a) is correct.

Total surface area of cuboid Let the base edges be l and b, with height h=18 cm. Base diagonal: $\sqrt{l^2+b^2}=21$, so $l^2+b^2=441$; base area: lb=92. ${\left(l+b\right)}^2=l^2+b^2+2lb=441+2\times 92=625$ l+b=25 Total surface area =2(lb+bh+hl)=2lb+2h(l+b) $=2\times 92+2\times 18\times 25$ =184+900=1084 Hence the total surface area is $1084{\mathrm{cm}}^2$, so option (c) is correct.

Value of a surd expression Here $y=2+\sqrt{2}$; build $y^2$ first and then $y^3$. $y^2={\left(2+\sqrt{2}\right)}^2=4+4\sqrt{2}+2=6+4\sqrt{2}$ $y^3=y\cdot y^2=\left(2+\sqrt{2}\right)\left(6+4\sqrt{2}\right)=12+8\sqrt{2}+6\sqrt{2}+8$ $=20+14\sqrt{2}$ $3y=3\left(2+\sqrt{2}\right)=6+3\sqrt{2}$ $y^3+3y=\left(20+14\sqrt{2}\right)+\left(6+3\sqrt{2}\right)=26+17\sqrt{2}$ Hence the required value is $26+17\sqrt{2}$, so option (a) is correct.

Square pyramid surface area In a right pyramid on a square base, the slant height l, the vertical height h and half of a base edge form a right triangle. $l=\sqrt{h^2+{\left(\frac{a}{2}\right)}^2}=\sqrt{{21}^2+{10}^2}$ $=\sqrt{441+100}=\sqrt{541}\approx 23.25$ cm Lateral surface area $=\frac{1}{2}\times \left(\mathrm{perimeter}\mathrm{of}\mathrm{base}\right)\times l=\frac{1}{2}\times 80\times 23.25$ $\approx 930$ sq cm Area of the square base $=20\times 20=400$ sq cm Total surface area $\approx 930+400=1330$ sq cm Hence the slant height is about 23.25 cm and the total surface area about $1330{\mathrm{cm}}^2$, so option (a) is correct.

Angles in a circle Join OB. Since OB is a radius and BC equals the radius, OB=BC, so $△OBC$ is isosceles. Therefore $\angle BOC=\angle BCO={32}^{\circ }$. $\angle OBA$ is the exterior angle of $△OBC$ at B, so $\angle OBA={32}^{\circ }+{32}^{\circ }={64}^{\circ }$. In $△OAB$, OA=OB (radii), hence $\angle OAB=\angle OBA={64}^{\circ }$. In $△OAC$: $\angle AOC={180}^{\circ }-{64}^{\circ }-{32}^{\circ }={84}^{\circ }$. D lies on CO produced, so $\angle AOD$ and $\angle AOC$ make a straight line at O. $\angle AOD={180}^{\circ }-{84}^{\circ }={96}^{\circ }$ Hence $\angle AOD={96}^{\circ }$, so option (d) is correct.

Mixture of two vessels The two vessels are equally full, so both totals must be written as the same number before adding. The ratios have 5+1=6 and 6+5=11 parts; the LCM of 6 and 11 is 66, so take 66 units in each vessel. First vessel: milk $=\frac{5}{6}\times 66=55$, water $=\frac{1}{6}\times 66=11$ Second vessel: milk $=\frac{6}{11}\times 66=36$, water $=\frac{5}{11}\times 66=30$ After mixing, milk =55+36=91 and water =11+30=41. Hence milk : water =91:41, so option (c) is correct.

Compound interest In compound interest the second year earns more than the first only because the first year's interest itself now earns interest. Extra interest =3600-3000=600, and this is one year's interest on ₹3,000. Rate $=\frac{600}{3000}\times 100=20\%$ For the first year compound interest equals simple interest, so $20\%$ of the sum =3000. Sum $=\frac{3000\times 100}{20}=15000$, i.e. ₹15,000. $40\%$ of the sum $=\frac{40}{100}\times 15000=6000$ Hence the required amount is ₹6,000, so option (c) is correct.