20 November 2025 · Shift 1

Partnership — profit sharing In a partnership the profit is shared in the ratio of capital × time. $A:B:C=6000\times 10:4000\times 8:5000\times 8=60000:32000:40000$. Dividing throughout by 4000, A:B:C = 15:8:10, so there are 15+8+10=33 equal parts. B's share $=\frac{8}{33}\times 330000=8\times 10000=80000$. So B gets ₹80,000.

Algebraic identities First find ab from the identity ${\left(a+b\right)}^2=a^2+b^2+2ab$. ${15}^2=117+2ab$, so 2ab = 225-117 = 108 and ab = 54. Now use $a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right)$. $a^3+b^3={15}^3-3\times 54\times 15=3375-2430$. $a^3+b^3=945$.

Quadratic equation word problem Each of the R students gets (R+6) candies and nothing is left over. Therefore R(R+6) = 432. $R^2+6R-432=0$, which factorises as (R+24)(R-18)=0. R is a number of students, so it cannot be negative; hence R = 18. Check: 18 students get 18+6=24 candies each and $18\times 24=432$.

Volume of a cone Volume of a cone $=\frac{1}{3}\pi r^{2}h$. $\frac{1}{3}\times \frac{22}{7}\times r^2\times 9=4158$. $\frac{66}{7}{r}^2=4158$. $r^2=\frac{4158\times 7}{66}=441$. $r=\sqrt{441}=21$, so the radius of the base is 21 cm.

Interior angle of a regular polygon In a regular polygon the interior angle and the exterior angle add up to 180°. Smaller polygon: exterior angle =180°-150°=30°. Number of its sides $=\frac{360°}{30°}=12$. The larger polygon has three times as many sides, i.e. $3\times 12=36$ sides. Its exterior angle $=\frac{360°}{36}=10°$. Interior angle of the larger polygon =180°-10°=170°. The perimeter and the side ratio are not needed here, because the angles of a regular polygon depend only on the number of sides.

Similar and congruent triangles AD = AE, so $△ADE$ is isosceles and $\angle ADE=\angle AED$. $\angle ADB$ and $\angle AEC$ are the supplements of these two equal angles, so $\angle ADB=\angle AEC$. Together with the given $\angle BAD=\angle CAE$ this gives $△ABD\sim △ACE$ by AA similarity. The corresponding sides AD and AE are equal, so the ratio of similarity is 1 and the two triangles are in fact congruent. Hence AB = AC and BD = CE, i.e. 2p+4 = 3q-8 and 3p = q. Putting q = 3p: $2p+4=9p-8\Rightarrow 7p=12\Rightarrow p=\frac{12}{7}$. Then $q=3p=\frac{36}{7}$, so $p+q=\frac{12}{7}+\frac{36}{7}=\frac{48}{7}$.

Volume of a cuboid A cuboid has 4 lengths, 4 breadths and 4 heights, so the total length of all edges =4(l+b+h). $4\left(l+b+h\right)=280\Rightarrow l+b+h=70$. Let $l=3x,b=5x,h=6x$; then 3x+5x+6x = 14x = 70, so x = 5. l = 15 cm, b = 25 cm and h = 30 cm. Volume $=15\times 25\times 30=11250$ cubic cm.

Melting and recasting solids Melting does not change the volume, so volume of the sphere $=n\times$ volume of one hemisphere. Volume of the sphere $=\frac{4}{3}\pi {\left(21\right)}^3=\frac{4}{3}\pi \times 9261=12348\pi$. Volume of one hemisphere $=\frac{2}{3}\pi {\left(7\right)}^3=\frac{2}{3}\pi \times 343=\frac{686\pi }{3}$. $n=12348\pi ÷\frac{686\pi }{3}=\frac{12348\times 3}{686}=54$. Curved surface area of one hemisphere $=2\pi r^2=2\pi {\left(7\right)}^2=98\pi$. Total curved surface area $=54\times 98\pi =5292\pi$ cm².

Lateral surface area of a pyramid For a pyramid, lateral surface area $=\frac{1}{2}\times$ (perimeter of the base) $\times$ (slant height). Perimeter of the rectangular base =2(14+9) = 46 m. Lateral surface area $=\frac{1}{2}\times 46\times 20$. $=23\times 20=460$, so the lateral surface area is 460 m².

Profit and loss — break even There is no overall gain or loss only when the rupee profit on the first article equals the rupee loss on the second. Let the cost prices be x and y; then $20\%$ of x must equal $30\%$ of y. $\frac{20x}{100}=\frac{30y}{100}\Rightarrow 20x=30y$. $\frac{x}{y}=\frac{30}{20}=\frac{3}{2}$. So the cost prices must be in the ratio CP = 3:2.

Partnership and profit ratio When all partners keep their money in the business for the same period, the profit is divided in the ratio of the amounts invested. So investment of A : investment of B = 7:8. Putting the investment of A as 70000: $\frac{70000}{B}=\frac{7}{8}$. $B=\frac{70000\times 8}{7}=80000$. So B invested ₹80,000; hence option (b) is correct.

Chain of ratios Q is common to both ratios, so make its value the same in each. P:Q=3:5=6:10 and Q:R=2:3=10:15. So P:Q:R=6:10:15, that is 6+10+15=31 equal parts in all. 31 parts $=6200\Rightarrow$ 1 part $=\frac{6200}{31}=200$. $Q=10\times 200=2000$. Hence option (b) is correct.

Time and work, men and boys Both groups finish the same work, so their total work is equal. 12(6M+5B)=4(15M+20B) $72M+60B=60M+80B\Rightarrow 12M=20B\Rightarrow M:B=5:3$. Take one man's one-day work as 5 units and one boy's as 3 units. Total work $=12\left(6\times 5+5\times 3\right)=12\times 45=540$ units. One day's work of 9 men and 15 boys $=9\times 5+15\times 3=45+45=90$ units. Time $=\frac{540}{90}=6$ days. Hence option (d) is correct.

Successive discount on marked price Let the marked price be M. A discount of $18\%$ leaves $82\%$ of it. $\frac{82}{100}\times M=8364$ $M=\frac{8364\times 100}{82}=10200$ With a discount of $30\%$ the selling price is $70\%$ of the marked price. Selling price $=\frac{70}{100}\times 10200=7140$. So the product would sell for ₹7,140; hence option (b) is correct.

Simple interest, multiple of sum Under simple interest the interest earned in every year is the same. Becoming 7 times means the interest earned is 7P-P=6P, and this takes 18 years. So the interest of one year $=\frac{6P}{18}=\frac{P}{3}$. To become 10 times, the interest required is 10P-P=9P. Time $=9P÷\frac{P}{3}=27$ years. Shortcut: $\frac{T_2}{T_1}=\frac{10-1}{7-1}\Rightarrow T_2=\frac{9}{6}\times 18=27$ years. Hence option (b) is correct.

Rate of simple interest If the sum triples, the amount is 3P, so the simple interest earned is 3P-P=2P. Formula: $SI=\frac{P\times R\times T}{100}$. $2P=\frac{P\times R\times 10}{100}$ $R=\frac{2\times 100}{10}=20$. So the rate is $20\%$ per annum; hence option (c) is correct.

Alligation on cost price The ratio is decided by the cost price of the mixture, so find it first. A profit of $25\%$ on the cost gives $CP=\frac{225\times 100}{125}=180$. So 1 kg of the mixture must cost ₹180. Now apply alligation between ₹170 and ₹210 about the mean ₹180. Cheaper : dearer =(210-180):(180-170)=30:10=3:1. So the two kinds must be mixed in the ratio 3:1; hence option (d) is correct.

Mixture, replacement by water Let the milk be 5k litres and the water 3k litres, so the vessel holds 8k litres. The 8 litres drawn off are part of the same mixture, so they contain $\frac{5}{8}\times 8=5$ litres of milk and 3 litres of water. After 8 litres of pure water is poured back: milk =5k-5 and water =3k-3+8=3k+5. $\frac{5k-5}{3k+5}=\frac{5}{4}$ $4\left(5k-5\right)=5\left(3k+5\right)\Rightarrow 20k-20=15k+25\Rightarrow 5k=45$, so k=9. Milk at the start =5k=45 litres (check: 45:27 becomes 40:32=5:4). Hence option (c) is correct.

Distance between parallel chords The perpendicular from the centre bisects a chord, so radius, half the chord and that perpendicular form a right triangle. For the 10 cm chord: distance $=\sqrt{{13}^2-5^2}=\sqrt{169-25}=12$ cm. For the 24 cm chord: distance $=\sqrt{{13}^2-{12}^2}=\sqrt{169-144}=5$ cm. If the two chords lie on opposite sides of the centre, the gap is 12+5=17 cm. If they lie on the same side, the gap is 12-5=7 cm, and this is the smaller value. So the minimum distance is 7 cm; hence option (c) is correct.

Trapezium circumscribing a circle A quadrilateral drawn around a circle is a tangential quadrilateral, and the two tangents from any vertex are equal. Adding those equal tangents gives Pitot's result: the sums of the two pairs of opposite sides are equal. So (sum of the parallel sides) = (sum of the non-parallel sides) =9+11=20 cm. Perimeter = (sum of parallel sides) + (sum of non-parallel sides) =20+20=40 cm. Hence option (c) is correct.

Slope of a straight line Bring the equation to the slope form y=mx+c; the coefficient m is the slope. $5x-7y=8\Rightarrow 7y=5x-8$ $y=\frac{5}{7}x-\frac{8}{7}$ So the slope is $\frac{5}{7}$. The same follows from ax+by+c=0, where slope $=-\frac{a}{b}=-\frac{5}{-7}=\frac{5}{7}$. Hence option (c) is correct.

Areas of similar triangles In similar triangles the perimeters are in the same ratio as the corresponding sides. $\frac{\mathrm{side}\mathrm{of}M}{\mathrm{side}\mathrm{of}N}=\frac{48}{60}=\frac{4}{5}$ The areas are in the ratio of the squares of the sides: $\frac{16}{25}$. $\frac{80}{\mathrm{area}\mathrm{of}N}=\frac{16}{25}$ Area of $N=\frac{80\times 25}{16}=125$. So the area of Triangle N is $125{\mathrm{cm}}^2$; hence option (c) is correct.

Similar triangles, ratio of perimeters For similar triangles, $\frac{A_1}{A_2}={\left(\frac{s_1}{s_2}\right)}^2$, where s is any pair of corresponding sides. Here $\frac{A_{ABC}}{A_{DEF}}=\frac{9}{1}$, so $\frac{s_1}{s_2}=\sqrt{\frac{9}{1}}=\frac{3}{1}$. Perimeters of similar triangles are in the same ratio as the corresponding sides. So the ratio of the perimeters is 3:1; hence option (c) is correct.

Common tangents to two circles When two circles touch each other externally, two direct (external) common tangents can be drawn, one on each side. One more tangent can be drawn at the point of contact itself, which is common to both circles. So the total number of common tangents is 2+1=3. For comparison: 4 tangents when the circles lie apart, 2 when they intersect, 1 when they touch internally and none when one lies inside the other. Hence option (b) is correct.

Profit percentage and selling price The cost price stays the same, so find it from the first sale. $CP=\frac{SP\times 100}{100+\mathrm{profit}\%}=\frac{345000\times 100}{115}=300000$ So the car cost ₹3,00,000. For a profit of $30\%$: $SP=300000\times \frac{130}{100}=390000$. So it must be sold for ₹3,90,000; hence option (d) is correct.