19 November 2025 · Shift 3

Incentre angle property For the incentre O of a triangle ABC the standard result is $\angle AOC={90}^{\circ }+\frac{\angle B}{2}$. Substituting the given value: ${160}^{\circ }={90}^{\circ }+\frac{\angle B}{2}$. $\frac{\angle B}{2}={160}^{\circ }-{90}^{\circ }={70}^{\circ }$. $\angle B={140}^{\circ }$. Check: $\angle A+\angle C={180}^{\circ }-{140}^{\circ }={40}^{\circ }$, so $\angle AOC={180}^{\circ }-\frac{{40}^{\circ }}{2}={160}^{\circ }$.

Mensuration: cylinder CSA Let the length and the breadth be 22x and 7x. Perimeter: $2\left(22x+7x\right)=174\Rightarrow 58x=174\Rightarrow x=3$. So length =66 cm, breadth =21 cm and area $=66\times 21=1386{\mathrm{cm}}^2$. The top of a solid cylinder is a circle, so $\pi r^2=1386$. $r^2=1386\times \frac{7}{22}=441\Rightarrow r=21$ cm. The radius is 84% of the height: $21=\frac{84}{100}h\Rightarrow h=25$ cm. Curved surface area $=2\pi rh=2\times \frac{22}{7}\times 21\times 25=3300{\mathrm{cm}}^2$.

Rhombus diagonals and area Let the diagonals be 5k and 12k. In a rhombus the diagonals bisect each other at right angles. So the half-diagonals $\frac{5k}{2}$ and 6k form a right triangle with the side: ${\left(\frac{5k}{2}\right)}^2+{\left(6k\right)}^2={13}^2$. $\frac{25k^2}{4}+36k^2=169\Rightarrow \frac{169k^2}{4}=169$. $k^2=4\Rightarrow k=2$. Diagonals $=5\times 2=10$ cm and $12\times 2=24$ cm. Area $=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 10\times 24=120{\mathrm{cm}}^2$.

Partnership profit share Both partners keep their money in for the same 8 months, so the time cancels and the profit is shared in the ratio of the amounts invested. A:B=80000:120000=2:3. Total parts =2+3=5, and 5 parts = Rs. 60,000, so 1 part = Rs. 12,000. B’s share $=3\times 12000=$ Rs. 36,000.

Average of a new group Total score of the first 40 students $=40\times 42=1680$. After the new students join there are 40+16=56 students whose average is 50, so their total $=56\times 50=2800$. Total score of the 16 new students =2800-1680=1120. Their average $=\frac{1120}{16}=70$.

Average salary Let the original number of employees be n. The new average is 60000-6000=54000. Equating the total salary before and after: $\frac{60000n+6\times 45000}{n+6}=54000$. 60000n+270000=54000n+324000. $6000n=54000\Rightarrow n=9$. So the company originally had 9 employees.

Speed from part of a journey Distance covered : distance still left =5:7, so the whole journey is 5+7=12 equal parts. 12 parts =54 km, hence 1 part $=\frac{54}{12}=4.5$ km. Distance covered $=5\times 4.5=22.5$ km. Time taken =3 hours 30 minutes =3.5 hours. Speed $=\frac{22.5}{3.5}=\frac{45}{7}\approx 6.43$ km/hr, which is the value listed in option (b).

Speed conversion Distance covered : distance still left =5:9, so the journey is 5+9=14 equal parts. 14 parts =140 km, hence 1 part =10 km and the distance covered $=5\times 10=50$ km. Time taken =3 hours 20 minutes $=\frac{10}{3}$ hours. Speed $=\frac{50}{10/3}=15$ km/hr. Converting to metres per second: $15\times \frac{5}{18}=\frac{25}{6}\approx 4.16$ m/s.

Time and work Take the whole work as the LCM of 30 and 50, i.e. 150 units. Then A and B together do $\frac{150}{30}=5$ units a day, and A alone does $\frac{150}{50}=3$ units a day. So B alone does 5-3=2 units a day. Time taken by B $=\frac{150}{2}=75$ days.

Successive discounts After a discount of 25% the price becomes $\frac{75}{100}$ of the marked price, and the second discount of 20% leaves $\frac{80}{100}$ of that. So $\mathrm{MP}\times \frac{75}{100}\times \frac{80}{100}=1200$. $\mathrm{MP}\times \frac{3}{5}=1200$. $\mathrm{MP}=1200\times \frac{5}{3}=2000$, i.e. ₹2,000. Note that the two discounts are not simply 45%: the single equivalent discount here is 40%.

Discount and marked price A discount of $22\%$ is allowed on the marked price, so the selling price is only $\left(100-22\right)\%=78\%$ of the marked price. Formula: $SP=MP\times \frac{100-d}{100}$ $1560=MP\times \frac{78}{100}$ $MP=\frac{1560\times 100}{78}=2000$ Hence the marked price of the product is ₹2,000, so option (b) is correct.

Perimeter and area of a rectangle Let the width be x cm; the length then is (x+3) cm. Perimeter: $2\left(l+b\right)=66\Rightarrow 2\left(\left(x+3\right)+x\right)=66$ $2\left(2x+3\right)=66\Rightarrow 2x+3=33\Rightarrow x=15$ So width = 15 cm and length = 15+3 = 18 cm. Area $=l\times b=18\times 15=270{\mathrm{cm}}^2$ Hence the area of the rectangle is $270{\mathrm{cm}}^2$, option (d).

Mixture and ratio Milk : water = 3:5, so the mixture has 3+5 = 8 equal parts in 80 litres. One part $=\frac{80}{8}=10$ litres, so milk = 30 litres and water = 50 litres. Only water is poured in, so the milk stays 30 litres. Let x litres of water be added. $\frac{30}{50+x}=\frac{4}{7}$ $210=200+4x\Rightarrow 4x=10\Rightarrow x=2.5$ Hence 2.5 litres of water must be added, option (d).

Chord, radius and diameter The perpendicular drawn from the centre to a chord bisects the chord, so half the chord $=\frac{30}{2}=15$ cm. In right triangle OMA: $O{A}^2=O{M}^2+M{A}^2$ $O{A}^2=8^2+{15}^2=64+225=289$ $OA=\sqrt{289}=17$ cm, and this OA is the radius. Diameter $=2\times 17=34$ cm, so option (c) is correct.

Intersecting chords theorem When two chords of a circle cut each other inside it, the products of the two parts of each chord are equal. Formula: $PA\times AQ=SA\times AN$ $8\times 12=6\times AN$ $AN=\frac{96}{6}=16$ cm SN = SA + AN = 6 + 16 = 22 cm, so option (b) is correct.

Parallel chords of a circle Drop $OM\perp PQ$ and $ON\perp RS$; each perpendicular bisects its chord, so PM = 12 cm and RN = 9 cm. Let the radius be r. Then $O{M}^2=r^2-{12}^2$ and $O{N}^2=r^2-9^2$. $O{N}^2-O{M}^2=\left(r^2-81\right)-\left(r^2-144\right)=63$ The chords lie on opposite sides of O, so ON + OM = 21; hence $ON-OM=\frac{63}{21}=3$. Solving the pair: ON = 12 cm and OM = 9 cm. $r^2=O{M}^2+P{M}^2=9^2+{12}^2=81+144=225$ $r=\sqrt{225}=15$ cm, so option (a) is correct.

Chord distance and diameter The perpendicular from the centre bisects the chord, so $PM=\frac{64}{2}=32$ cm and OM = 24 cm. In right triangle OMP: $O{P}^2=O{M}^2+P{M}^2$ $O{P}^2={24}^2+{32}^2=576+1024=1600$ $OP=\sqrt{1600}=40$ cm, which is the radius. Diameter $=2\times 40=80$ cm, so option (d) is correct.

Angle between two tangents A tangent is perpendicular to the radius at its point of contact, so $\angle OPT=\angle OQT={90}^{\circ }$. OPTQ is a quadrilateral, and the angles of a quadrilateral add up to ${360}^{\circ }$. $\angle POQ+{90}^{\circ }+{90}^{\circ }+{50}^{\circ }={360}^{\circ }$ $\angle POQ={360}^{\circ }-{230}^{\circ }={130}^{\circ }$ Note the short cut: $\angle POQ={180}^{\circ }-\angle PTQ$. Hence option (b) is correct.

Equal tangents from a point The two tangents drawn to a circle from the same external point are always equal in length. Proof in one line: OP = OQ (radii), OT is common and $\angle OPT=\angle OQT={90}^{\circ }$, so $△OPT\cong △OQT$ (RHS). Therefore TQ = TP = 10 units, so option (d) is correct.

Tangent length and shortest distance Let P be the point of contact. The tangent is perpendicular to the radius there, so $\angle OPT={90}^{\circ }$. In right triangle OPT: $O{T}^2=O{P}^2+P{T}^2$ $O{T}^2=8^2+{15}^2=64+225=289\Rightarrow OT=17$ cm The point of the circle nearest to T is S, where the line OT meets the circle. Shortest distance = ST = OT - OS = 17 - 8 = 9 cm, so option (d) is correct.

Centroid of a triangle The centroid is the average of the three vertices: $G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$. x-coordinate: $\frac{4+\left(-5\right)+x}{3}=0\Rightarrow -1+x=0\Rightarrow x=1$ y-coordinate: $\frac{6+0+y}{3}=0\Rightarrow 6+y=0\Rightarrow y=-6$ Hence the third vertex is (1,-6), so option (b) is correct.

Slope of a straight line Bring the equation to the slope-intercept form y = mx + c, in which m is the slope. $x+6y=12\Rightarrow 6y=-x+12$ $y=-\frac{1}{6}x+2$ Comparing with y = mx + c gives $m=-\frac{1}{6}$, so the slope has the value $\frac{1}{6}$, the sign only telling us that the line falls from left to right. The same result comes straight from ax+by=c, where slope $=-\frac{a}{b}=-\frac{1}{6}$. Hence option (b) is correct.

Angle in a semicircle PQ is a diameter and R lies on the circle, so $\angle PRQ={90}^{\circ }$ — the angle in a semicircle is a right angle. In $△PRQ$ the three angles add up to ${180}^{\circ }$: $\angle RPQ+\angle PRQ+\angle RQP={180}^{\circ }$ ${50}^{\circ }+{90}^{\circ }+\angle RQP={180}^{\circ }$ $\angle RQP={40}^{\circ }$, so option (a) is correct.

Cyclic quadrilateral and central angle Opposite angles of a cyclic quadrilateral are supplementary, so $\angle PSR+\angle PQR={180}^{\circ }$. $\angle PQR={180}^{\circ }-{115}^{\circ }={65}^{\circ }$ $\angle PQR$ and $\angle POR$ stand on the same arc PR, and the angle at the centre is twice the angle at the circumference. $\angle POR=2\times \angle PQR=2\times {65}^{\circ }={130}^{\circ }$ Hence $\angle POR={130}^{\circ }$, so option (c) is correct.

Profit and loss on cost price Both the loss and the profit are reckoned on the cost price, so find the cost price first. At a $30\%$ loss the selling price is $\left(100-30\right)\%=70\%$ of the cost price. $14000=CP\times \frac{70}{100}\Rightarrow CP=\frac{14000\times 100}{70}=20000$ For a $25\%$ profit: $SP=CP\times \frac{100+25}{100}=20000\times \frac{125}{100}$ SP = 25000 Hence the item should be sold for ₹25,000, so option (a) is correct.