26 November 2025 · Shift 3

Algebraic identity Use the identity ${\left(a-b\right)}^2=a^2-2ab+b^2$ with a=4x and b=5. $a^2={\left(4x\right)}^2=16x^2$. $2ab=2\times 4x\times 5=40x$. $b^2=5^2=25$. So ${\left(4x-5\right)}^2=16x^2-40x+25$, and the middle term carries a minus sign. Hence option (b) is correct.

Partnership profit share In a partnership the profit is shared in the ratio of (capital $\times$ time) of each partner. For P this product is $3500\times 10=35000$. For Q it is $2500\times 14=35000$. The ratio is 35000:35000=1:1, so the two shares are exactly the same. Neither partner gets more, so option (c) is correct.

Combining ratios To join two ratios, the common term B must be given the same value in both. Multiply A:B=5:7 by 8 and B:C=8:9 by 7, so that B becomes 56 in each. This gives A:B=40:56 and B:C=56:63. Hence A:B:C=40:56:63. So the required ratio is A:C=40:63, and option (a) is correct.

Average of numbers Sum = average $\times$ number of terms. Sum of the five numbers $=58\times 5=290$. Sum of the remaining four numbers $=55\times 4=220$. The number removed is the difference of the two sums: 290-220=70. Hence option (a) is correct.

Percentage of a number Let the number be x. $80\%$ of x is 72, so $\frac{80}{100}\times x=72$. $x=\frac{72\times 100}{80}$. x=90. Hence option (c) is correct.

Profit and loss Let the cost price be x. Selling at a $15\%$ loss gives the first selling price =0.85x. Selling at a $10\%$ gain gives the second selling price =1.10x. The two selling prices differ by Rs. 125, so 1.10x-0.85x=125. That is 0.25x=125, hence $x=\frac{125}{0.25}=500$. The cost price is Rs. 500, so option (a) is correct.

Discount and marked price The cost price is Rs. 800 and the gain is $12\%$, so the selling price $=800\times \frac{112}{100}=896$. A discount of $20\%$ means the selling price is $80\%$ of the marked price M. So $M\times \frac{80}{100}=896$. $M=\frac{896\times 100}{80}=1120$. The marked price is Rs. 1120, so option (b) is correct.

Time and work Take the whole work as 1 unit and add the three given one-day rates. $\left(A+B\right)+\left(B+C\right)+\left(C+A\right)=\frac{1}{24}+\frac{1}{40}+\frac{1}{30}=\frac{5+3+4}{120}=\frac{12}{120}=\frac{1}{10}$. This sum counts each person twice, so $A+B+C=\frac{1}{20}$. $A=\left(A+B+C\right)-\left(B+C\right)=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$, so A alone needs 40 days. $B=\left(A+B+C\right)-\left(C+A\right)=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so B alone needs 60 days. $C=\left(A+B+C\right)-\left(A+B\right)=\frac{1}{20}-\frac{1}{24}=\frac{1}{120}$, so C alone needs 120 days. Hence option (c) is correct.

Simple interest rate Let the sum be P; when it triples, the amount becomes 3P. So the simple interest earned in 20 years =3P-P=2P. Formula: $SI=\frac{P\times R\times T}{100}$, hence $2P=\frac{P\times R\times 20}{100}$. Cancelling P gives $R=\frac{2\times 100}{20}=10$. The rate is $10\%$ per annum, so option (c) is correct.

Relative speed of trains While one train passes another completely, the distance covered is the sum of the two lengths. Total distance =500+400=900 m. Relative speed $=\frac{900}{54}=\frac{50}{3}$ m/s. In km/h this is $\frac{50}{3}\times \frac{18}{5}=60$ km/h. Train A supplies 36 km/h of this, so Train B's speed =60-36=24 km/h. Hence option (a) is correct.

Common tangents of circles Here both radii are $r_1=r_2=4$ units and the distance between the centres is d=5 units. Sum of the radii =4+4=8 units and difference of the radii =4-4=0. Since 0<5<8, that is $\left(r_1-r_2\right)<d<r_1+r_2$, the two circles cut each other at two distinct points. A pair of intersecting circles has only the two direct (external) common tangents; a transverse tangent is impossible because the circles overlap. Hence option (a) is correct.

Diagonal of a rectangle The diagonal, the length and the breadth of a rectangle form a right-angled triangle, so $d^2=l^2+b^2$. $d^2={15}^2+{20}^2$ $d^2=225+400=625$ $d=\sqrt{625}=25$ cm. Hence option (a) is correct.

Volume of a cuboid A rectangular parallelepiped is a cuboid, and its volume $=l\times b\times h$. $V=100\times 70\times 60$ $V=7000\times 60$ V=420000 cubic centimetres. Hence option (b) is correct.

Volume of a pyramid For a right pyramid, volume $=\frac{1}{3}\times \left(\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{base}\right)\times \left(\mathrm{vertical}\mathrm{height}\right)$. The square base of the tent has an area of $16m^2$ and the vertical height is 6 m. $V=\frac{1}{3}\times 16\times 6$ $V=32m^3$ Hence option (b) is correct.

Areas of similar triangles In two similar triangles the ratio of the areas equals the square of the ratio of the corresponding sides. $\frac{\mathrm{ar}\left(ABC\right)}{\mathrm{ar}\left(DEF\right)}={\left(\frac{5}{7}\right)}^2=\frac{25}{49}$ $\frac{150}{\mathrm{ar}\left(DEF\right)}=\frac{25}{49}$ $\mathrm{ar}\left(DEF\right)=\frac{150\times 49}{25}=6\times 49$ $\mathrm{ar}\left(DEF\right)=294{\mathrm{cm}}^2$ Hence option (d) is correct.

Direct common tangent Drop a perpendicular from the smaller centre to the bigger radius: the tangent, that perpendicular and the line of centres form a right triangle whose legs are the tangent and $r_1-r_2$. So, length of the direct common tangent $=\sqrt{d^2-{\left(r_1-r_2\right)}^2}$. $=\sqrt{{15}^2-{\left(6-2\right)}^2}$ $=\sqrt{225-16}=\sqrt{209}$ $\sqrt{209}\approx 14.45$ units. Hence option (c) is correct.

Trigonometric identities Note first that $\cot A\cdot \tan A=1$, whatever the acute angle A may be. Use the identity ${\left(x+y\right)}^3=x^3+y^3+3xy\left(x+y\right)$ with $x=\cot A$ and $y=\tan A$. ${\left(\cot A+\tan A\right)}^3={\cot }^{3}A+{\tan }^{3}A+3\left(1\right)\left(\cot A+\tan A\right)$ $8^3={\cot }^{3}A+{\tan }^{3}A+3\times 8$ $512={\cot }^{3}A+{\tan }^{3}A+24$ ${\cot }^{3}A+{\tan }^{3}A=512-24=488$ Hence option (c) is correct.

SI and CI difference For 2 years the compound interest exceeds the simple interest only by the interest earned on the first year's interest, so the difference $=P{\left(\frac{r}{100}\right)}^2$. Difference $=35000\times {\left(\frac{15}{100}\right)}^2$ $=35000\times \frac{225}{10000}$ $=\frac{35\times 225}{10}=787.5$ So the two schemes differ by Rs. 787.5. Hence option (a) is correct.

Medians and centroid The centroid divides every median in the ratio 2:1, measured from the vertex. $RO=\frac{2}{3}\times RF=\frac{2}{3}\times 24=16$ cm $OE=\frac{1}{3}\times PE=\frac{1}{3}\times 30=10$ cm In $△ROE$ the angle between these two segments is ${60}^{\circ }$, so its area $=\frac{1}{2}\times RO\times OE\times \sin {60}^{\circ }$. $=\frac{1}{2}\times 16\times 10\times \frac{\sqrt{3}}{2}=40\sqrt{3}{\mathrm{cm}}^2$ The three medians divide a triangle into 6 triangles of equal area, and $△ROE$ is one of them. Area of $△PQR=6\times 40\sqrt{3}=240\sqrt{3}{\mathrm{cm}}^2$ Hence option (b) is correct.

Relative speed chase The gap closes at the relative speed =5.75-4.5=1.25 km/h. The head start is 2,400 metres =2.4 km, so the time taken to catch the thief $=\frac{2.4}{1.25}=1.92$ hours. Distance run by the thief in this time $=4.5\times 1.92=8.64$ km $=8.64\times 1000=8,640$ metres. Hence option (b) is correct.

Cylinder: volume and CSA Volume of a cylinder $=\pi r^{2}h$, so first find the height. $1540=\frac{22}{7}\times 7^2\times h$ 1540=154h, so h=10 cm. Curved surface area $=2\pi rh=2\times \frac{22}{7}\times 7\times 10$ $=440{\mathrm{cm}}^2$ Hence option (a) is correct.

Squaring a surd Square both sides using ${\left(x+y\right)}^2=x^2+2xy+y^2$. $a^2={\left(\sqrt{5}+\sqrt{2}\right)}^2={\left(\sqrt{5}\right)}^2+2\times \sqrt{5}\times \sqrt{2}+{\left(\sqrt{2}\right)}^2$ $=5+2\sqrt{10}+2$ $=7+2\sqrt{10}$ Hence option (b) is correct.

Trigonometric simplification From the identity ${\sin }^{2}x+{\cos }^{2}x=1$, we get $1-{\sin }^{2}x={\cos }^{2}x$. $\frac{1-{\sin }^{2}x}{\cos x}=\frac{{\cos }^{2}x}{\cos x}$ $=\cos x$ Hence option (a) is correct.

Compound ratio The ratio A:B is not needed here, because $\frac{B}{D}=\frac{B}{C}\times \frac{C}{D}$. $\frac{B}{D}=\frac{6}{7}\times \frac{8}{9}=\frac{48}{63}$ $\frac{48}{63}=\frac{16}{21}$, that is B:D=16:21. Hence option (d) is correct.

Trigonometric ratios $\cos \theta =\frac{3}{5}$ means base =3 and hypotenuse =5, so the perpendicular $=\sqrt{5^2-3^2}=4$ and $\sin \theta =\frac{4}{5}$. ${\sec }^2\theta =\frac{1}{{\cos }^2\theta }=\frac{25}{9}$ ${\mathrm{cosec}}^2\theta =\frac{1}{{\sin }^2\theta }=\frac{25}{16}$ ${\sec }^2\theta -{\mathrm{cosec}}^2\theta =\frac{25}{9}-\frac{25}{16}=\frac{400-225}{144}$ $=\frac{175}{144}$ Hence option (c) is correct.