19 November 2025 · Shift 1

Types of decimals A decimal is called mixed recurring when a few digits just after the decimal point do not repeat and the repetition begins only after them. 0.75 comes to an end after two digits, so it is a terminating decimal, not a recurring one at all. In $0.3333\dots$ the repetition starts at the very first decimal place, and in $0.272727\dots$ the repeating block 27 also starts at the very first decimal place, so both are pure recurring. In $2.1454545\dots$ the digit 1 stands alone and never repeats, and only the block 45 repeats after it, which is exactly what a mixed recurring decimal looks like. Hence, option (d) is correct.

Discount and marked price A discount of $20\%$ is taken off the marked price, so the buyer pays $100\%-20\%=80\%$ of it. Let the marked price be M. $M\times \frac{80}{100}=6500$ $M=6500\times \frac{100}{80}$ M=8125 Hence, option (a) is correct.

Incentre and orthocentre angles Two standard results are needed here, both measured on the same side BC. At the incentre: $\angle BIC={90}^{\circ }+\frac{\angle A}{2}$. At the orthocentre: $\angle BOC={180}^{\circ }-\angle A$. Substitute them in the given condition $\angle BIC-\angle BOC={60}^{\circ }$: $\left({90}^{\circ }+\frac{\angle A}{2}\right)-\left({180}^{\circ }-\angle A\right)={60}^{\circ }$ $\frac{3\angle A}{2}-{90}^{\circ }={60}^{\circ }$ $\frac{3\angle A}{2}={150}^{\circ }$, so $\angle A={100}^{\circ }$. Hence, option (a) is correct.

Trigonometric ratios $\cot \theta =\frac{5}{12}$ means $\tan \theta =\frac{12}{5}$, so take the side opposite $\theta$ as 12 and the side adjacent as 5. The hypotenuse is $\sqrt{{12}^2+5^2}=\sqrt{169}=13$, and since $\theta$ lies in the first quadrant every ratio is positive. So $\sin \theta =\frac{12}{13}$ and $\cos \theta =\frac{5}{13}$. First bracket: $\frac{1+\sin \theta }{\cos \theta }=\frac{1+\frac{12}{13}}{\frac{5}{13}}=\frac{25}{13}\times \frac{13}{5}=5$. Second bracket: $\frac{1-\sin \theta }{\cos \theta }=\frac{1-\frac{12}{13}}{\frac{5}{13}}=\frac{1}{13}\times \frac{13}{5}=\frac{1}{5}$. The required value $=5-\frac{1}{5}=\frac{24}{5}$. The same thing in one line: $\frac{1+\sin \theta }{\cos \theta }-\frac{1-\sin \theta }{\cos \theta }=\frac{2\sin \theta }{\cos \theta }=2\tan \theta =2\times \frac{12}{5}=\frac{24}{5}$. Hence, option (b) is correct.

Tangents from an external point The two tangents drawn from an outside point are equal and make equal angles with the line joining that point to the centre. So OA bisects the ${90}^{\circ }$ angle between them, leaving ${45}^{\circ }$ on each side. A tangent is perpendicular to the radius at its point of contact, so triangle OPA is right-angled at P. $\sin {45}^{\circ }=\frac{OP}{OA}=\frac{7}{R}$ $R=\frac{7}{\sin {45}^{\circ }}=7\sqrt{2}$ $R=7\times 1.414=9.89$ cm (to two decimal places). Hence, option (c) is correct.

Chord and perpendicular from centre The perpendicular drawn from the centre of a circle to a chord always bisects that chord. So $AB=\frac{AC}{2}=\frac{30}{2}=15$ cm. OA is a radius, and triangle OBA is right-angled at B, so by Pythagoras' theorem $O{A}^2=O{B}^2+A{B}^2$. ${17}^2=O{B}^2+{15}^2$ $O{B}^2=289-225=64$ OB=8 cm Hence, option (b) is correct.

Rectangle — perimeter and area Let the length be l and the breadth be b. Perimeter: $2\left(l+b\right)=14\sqrt{2}$, so $l+b=7\sqrt{2}$. Area: lb=18. The difference follows from the identity ${\left(l-b\right)}^2={\left(l+b\right)}^2-4lb$, so neither side has to be found separately. ${\left(l-b\right)}^2={\left(7\sqrt{2}\right)}^2-4\times 18$ ${\left(l-b\right)}^2=98-72=26$ $l-b=\sqrt{26}$ cm Hence, option (c) is correct.

Area of rhombus — cost The area of a rhombus is half the product of its diagonals: $\frac{1}{2}\times d_1\times d_2$. Area of one tile $=\frac{1}{2}\times 15\times 18=135$ sq cm. Area of the whole floor $=500\times 135=67500$ sq cm. Change the unit: 1 sq m =10000 sq cm, so the floor is $\frac{67500}{10000}=6.75$ sq m. Cost $=6.75\times 20=135$, that is Rs. 135. Hence, option (a) is correct.

Rhombus — diagonals and area Let the diagonals be 7k and 24k. The diagonals of a rhombus bisect each other at right angles, so each side, together with the two half-diagonals, forms a right-angled triangle. ${\left(\frac{7k}{2}\right)}^2+{\left(\frac{24k}{2}\right)}^2={25}^2$ $\frac{49k^2+576k^2}{4}=625$, that is $\frac{625k^2}{4}=625$. $k^2=4$, so k=2. The diagonals are $7\times 2=14$ cm and $24\times 2=48$ cm. Area $=\frac{1}{2}\times 14\times 48=336$ sq cm. Hence, option (b) is correct.

Partnership — profit sharing Both partners keep their money in for the same time, so the profit is divided in the ratio of the investments, P:Q=5:7. Total number of parts =5+7=12. One part $=\frac{120000}{12}=10000$. P's share $=5\times 10000=50000$, that is Rs. 50,000. Hence, option (b) is correct.

LCM and divisibility A number divisible by each of 8, 12, 15 and 20 must be a multiple of their LCM. $8=2^3,12=2^2\times 3,15=3\times 5,20=2^2\times 5$ Taking the highest power of each prime, $\mathrm{LCM}=2^3\times 3\times 5=120$. The greatest four-digit number is 9999, and $9999÷120$ leaves the remainder 39. So the required number =9999-39=9960, i.e. $120\times 83$. Hence, option (c) is correct.

Average of a group Sum of the 9 numbers $=9\times 57=513$. Sum of the first 4 numbers $=4\times 54=216$. Sum of the last 4 numbers $=4\times 60=240$. The first four and the last four together cover eight numbers; only the fifth number is left out of both groups. Fifth number =513-(216+240)=513-456=57. Hence, option (c) is correct.

Change in average Runs scored in 30 innings $=50\times 30=1500$. The new average must be 50+2=52, over 31 innings. Total runs then needed $=52\times 31=1612$. Runs required in the 31st innings =1612-1500=112. Shortcut: required runs = new average + (old innings) $\times$ increase $=52+30\times 2=112$. Hence, option (d) is correct.

Speed, time and distance Formula: $\mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}}$ First part: $t_1=\frac{80}{20}=4$ hours. New speed $=20\times \frac{120}{100}=24$ km/h. Second part: $t_2=\frac{60}{24}=2.5$ hours, i.e. 2 hours 30 minutes. Total time =4+2.5=6.5 hours = 6 hours 30 minutes. Hence, option (a) is correct.

Races: beat distance Bablu's speed $=\frac{300}{32}=9.375$ m/s. Rajkumar finishes the race in 28 seconds, so the winning moment is at t=28 s. Distance covered by Bablu in 28 s $=9.375\times 28=262.5$ m. Beat distance =300-262.5=37.5 m. Shortcut: beat distance $=\mathrm{race}\mathrm{length}\times \frac{\mathrm{time}\mathrm{difference}}{\mathrm{loser’s}\mathrm{time}}=300\times \frac{32-28}{32}=37.5$ m. Hence, option (a) is correct.

Time and work Take the total work as $\mathrm{LCM}\left(7,14\right)=14$ units. A and B together do $\frac{14}{7}=2$ units a day, and A alone does $\frac{14}{14}=1$ unit a day. So B alone does 2-1=1 unit a day. Time taken by B alone $=\frac{14}{1}=14$ days. Hence, option (d) is correct.

Discount and marked price A discount of $20\%$ means the selling price is $80\%$ of the marked price. $\frac{80}{100}\times \mathrm{M.P.}=960$ $\mathrm{M.P.}=\frac{960\times 100}{80}=1200$ Check: $20\%$ of ₹1,200 is ₹240, and 1200-240=960. Hence the marked price is ₹1,200 and option (b) is correct.

Mixture ratio In the first container the parts add to 6+5=11 and in the second to 4+7=11, and the containers are equal, so take each as 11 units. Milk in the final mixture =6+4=10 units. Water in the final mixture =5+7=12 units. The question asks for water : milk, so the ratio is 12:10=6:5. Hence, option (d) is correct.

Parallel chords of a circle The perpendicular from the centre bisects a chord, so each chord gives a right triangle whose hypotenuse is the radius. For the 16 cm chord: $d_1=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ cm. For the 30 cm chord: $d_2=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8$ cm. The chords lie on opposite sides of the centre, so the required distance is $d_1+d_2$. =15+8=23 cm Hence, option (a) is correct.

Chords and radius The perpendicular from the centre bisects a chord, so the half-chords are $\frac{48}{2}=24$ cm and $\frac{40}{2}=20$ cm. Let the centre be x cm from PQ; as the chords are on opposite sides, it is (22-x) cm from RS. From the right triangle on PQ: $r^2=x^2+{24}^2$ From the right triangle on RS: $r^2={\left(22-x\right)}^2+{20}^2$ $x^2+576={\left(22-x\right)}^2+400$ $x^2+576=484-44x+x^2+400$ 44x=884-576=308, so x=7. $r^2=7^2+{24}^2=49+576=625$, hence r=25 cm. Hence, option (a) is correct.

Tangents from a point OP and OQ are radii and the chord PQ also equals the radius, so OP=OQ=PQ. Therefore $△OPQ$ is equilateral and $\angle POQ={60}^{\circ }$. A tangent is perpendicular to the radius at the point of contact, so $\angle OPA=\angle OQA={90}^{\circ }$. In the quadrilateral OPAQ the four angles add up to ${360}^{\circ }$: $\angle PAQ={360}^{\circ }-{90}^{\circ }-{90}^{\circ }-{60}^{\circ }={120}^{\circ }$ Hence, option (d) is correct.

Length of a tangent The radius drawn to the point of contact is perpendicular to the tangent, so the triangle formed by the centre O, the point of contact T and the outside point P is right-angled at T. Applying Pythagoras' theorem to the two given lengths 8 cm and 15 cm: $8^2+{15}^2=64+225=289$ $\sqrt{289}=17$, so the required length is 17 cm. (8, 15, 17) is a standard Pythagorean triplet, worth memorising along with (3, 4, 5), (5, 12, 13) and (7, 24, 25). Hence, option (a) is correct.

Equation of a line Slope $m=\frac{y_2-y_1}{x_2-x_1}=\frac{8-4}{4-2}=\frac{4}{2}=2$ Using the point-slope form $y-y_1=m\left(x-x_1\right)$ with the point (2,4): y-4=2(x-2) y-4=2x-4, so y=2x. Check: (4,8) satisfies it, since $2\times 4=8$. Hence, option (a) is correct.

Chord and central angle Let BC be the chord and O the centre, so that $\angle BOC={60}^{\circ }$. OB=OC=9 cm because both are radii, so $△BOC$ is isosceles and its base angles are equal. Each base angle $=\frac{{180}^{\circ }-{60}^{\circ }}{2}={60}^{\circ }$, so all three angles are ${60}^{\circ }$. The triangle is therefore equilateral, giving BC=OB=9 cm. Hence, option (c) is correct.

Profit and selling price A $5\%$ profit means the selling price is $105\%$ of the cost price. $\mathrm{C.P.}=\frac{5250\times 100}{105}=5000$ For a $20\%$ profit, $\mathrm{S.P.}=5000\times \frac{120}{100}$ =6000 Shortcut: $\mathrm{S.P.}=5250\times \frac{120}{105}=6000$. Hence the selling price should be ₹6,000 and option (c) is correct.