20 November 2025 · Shift 3

Square of a number For any number that ends in 5 there is a one-line rule: drop the 5, multiply what is left by the next whole number, and write 25 after the product. In 175 the part left after dropping the 5 is 17, and the next whole number is 18. $17\times 18=306$, so ${175}^2=30625$. Check by expanding: ${175}^2={\left(170+5\right)}^2={170}^2+2\times 170\times 5+5^2$. =28900+1700+25=30625. Hence the correct option is (a).

Trigonometric ratios Put in the standard values: $\cos {45}^{\circ }=\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$, $\cot {60}^{\circ }=\frac{1}{\sqrt{3}}$, $\csc {30}^{\circ }=2$, $\sec {30}^{\circ }=\frac{2}{\sqrt{3}}$ and $\tan {45}^{\circ }=1$. First term: $\cos {45}^{\circ }\sin {45}^{\circ }=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{1}{2}$. Second term: $\cot {60}^{\circ }\csc {30}^{\circ }=\frac{1}{\sqrt{3}}\times 2=\frac{2}{\sqrt{3}}$. Third term: $\sec {30}^{\circ }\tan {45}^{\circ }=\frac{2}{\sqrt{3}}\times 1=\frac{2}{\sqrt{3}}$. Adding, the value $=\frac{1}{2}+\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{1}{2}+\frac{4}{\sqrt{3}}$. Hence the correct option is (b).

Heights and distances Let the pole stand at a horizontal distance d metres from the hill and let its height be h metres. From the top of the hill the line of sight to the foot of the pole makes ${60}^{\circ }$ with the horizontal, and the whole 300 m of the hill lies below it. $\tan {60}^{\circ }=\frac{300}{d}\Rightarrow \sqrt{3}=\frac{300}{d}\Rightarrow d=\frac{300}{\sqrt{3}}=100\sqrt{3}$. The line of sight to the top of the pole makes ${30}^{\circ }$, and the drop to that point is only 300-h. $\tan {30}^{\circ }=\frac{300-h}{d}\Rightarrow \frac{1}{\sqrt{3}}=\frac{300-h}{100\sqrt{3}}\Rightarrow 300-h=100$. So h=200 m and $d=100\sqrt{3}$ m. Required difference $=h-d=200-100\sqrt{3}=100\left(2-\sqrt{3}\right)$ metres. Hence the correct option is (d).

Volume of a sphere The container is already full, so the water that spills out is exactly equal to the volume of the solid that goes in. The sphere has diameter $2\times 6=12$ cm, the same as the side of the cube, so the whole sphere is submerged. Volume of the sphere $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(6\right)}^3$. $=\frac{4}{3}\pi \times 216=288\pi$ cubic centimetres. The cube itself holds ${12}^3=1728$ cubic centimetres, which is more than $288\pi \approx 904.8$, so the sphere really does fit. Hence the correct option is (a).

Median to the hypotenuse In $△ABC$ the longest side is CA=145 cm, so CA is the hypotenuse and the right angle is the one at B. M is the midpoint of this hypotenuse, and BM is the median drawn from the right angle to the hypotenuse. The midpoint of the hypotenuse of a right-angled triangle is the centre of its circumcircle, so it is the same distance from all three vertices: MA=MC=MB. Hence $BM=\frac{1}{2}\times CA=\frac{1}{2}\times 145=72.5$ cm. Hence the correct option is (b).

Radian to degree $\pi$ radians $={180}^{\circ }$, so to change radians into degrees multiply by $\frac{{180}^{\circ }}{\pi }$. $\frac{-4\pi }{3}$ radians $=\frac{-4\pi }{3}\times \frac{{180}^{\circ }}{\pi }$. The $\pi$ cancels and $\frac{180}{3}=60$, so the value $=-4\times {60}^{\circ }$. $=-{240}^{\circ }$, and since the answer is a whole number of degrees it is written $-{240}^{\circ }0\prime 0″$. Hence the correct option is (a).

Triangle inequality Three lengths can close up into a triangle only if the sum of the two shorter ones is greater than the longest one. (a) 2+4=6, which is greater than 5 — a triangle is possible. (b) 6+6=12, which is greater than 6 — this is an equilateral triangle. (d) 6+8=14, which is greater than 11 — a triangle is possible. (c) 3+2=5, which is less than 7, so the two shorter sticks can never meet over the longest one. Hence the correct option is (c).

Diagonals of a polygon From each vertex of an n-sided polygon you can draw a line to n-3 other vertices — all except the vertex itself and its two neighbours. That gives n(n-3) lines, but every diagonal has been counted from both of its ends, so the number of diagonals is $\frac{n\left(n-3\right)}{2}$. A heptagon has n=7. Number of diagonals $=\frac{7\left(7-3\right)}{2}=\frac{7\times 4}{2}=14$. Hence the correct option is (b).

Simple interest Under simple interest the amount grows by the same sum every year, so the growth between two years tells us the yearly interest. Interest of the 4th to the 6th year =21800-20500=1300, i.e. ₹1,300 for 2 years. Interest for one year $=\frac{1300}{2}=650$, i.e. ₹650, so the interest of the first 4 years $=4\times 650=2600$, i.e. ₹2,600. Principal =20500-2600=17900, i.e. ₹17,900. Now use $SI=\frac{P\times R\times T}{100}$ with P=17900, R=10 and T=5. $SI=\frac{17900\times 10\times 5}{100}=\frac{895000}{100}=8950$, i.e. ₹8,950. Hence the correct option is (c).

Simple interest Let the principal be P. The nine years split as 2 years, 4 years and the rest. Remaining time =9-2-4=3 years, which carries the rate of $12\%$. Interest of the first 2 years at $7\%=\frac{P\times 7\times 2}{100}=\frac{14P}{100}$. Interest of the next 4 years at $10\%=\frac{P\times 10\times 4}{100}=\frac{40P}{100}$. Interest of the last 3 years at $12\%=\frac{P\times 12\times 3}{100}=\frac{36P}{100}$. Total interest $=\frac{14P+40P+36P}{100}=\frac{90P}{100}$. $\frac{90P}{100}=8100\Rightarrow P=\frac{8100\times 100}{90}=9000$, i.e. ₹9,000. Hence the correct option is (c).

Simple interest The loan is repaid with simple interest, so first work out the interest for the whole period. Simple interest $=\frac{P\times R\times T}{100}=\frac{45000\times 10\times 3}{100}$ $=45000\times \frac{30}{100}=13500$, i.e. ₹13,500. Amount to be paid = principal + interest =45000+13500=58500. So ₹58,500 has to be repaid after 3 years. Hence the correct option is (a).

Tangent and chord Let O be the common centre of the two circles. PQ touches the smaller circle at M, and the radius drawn to the point of contact is perpendicular to the tangent, so $OM\perp PQ$. But OM is now the perpendicular dropped from the centre of the bigger circle on to its chord PQ, and such a perpendicular bisects the chord. $MQ=\frac{PQ}{2}=\frac{20}{2}=10$ cm Hence the correct option is (c).

Parallel chords of a circle The perpendicular from the centre bisects a chord, so the half-chords are $\frac{30}{2}=15$ cm and $\frac{16}{2}=8$ cm. Of two chords the shorter one is farther from the centre, so the 16 cm chord is the farther of the two. Let the 30 cm chord be at a distance d cm from the centre; then the 16 cm chord is at (d+7) cm. Both distances belong to the same radius r: $r^2=d^2+{15}^2={\left(d+7\right)}^2+8^2$ $d^2+225=d^2+14d+49+64$ $14d=112\Rightarrow d=8$ $r^2=8^2+{15}^2=64+225=289$ Area $=\pi r^2=289\pi$ square cm. Hence the correct option is (a).

Surface area and volume of a hemisphere A solid hemisphere has a curved surface $2\pi r^2$ and a flat circular face $\pi r^2$, so its total surface area is $3\pi r^2$. $3\pi \left({\left(r+14\right)}^2-r^2\right)=7392$ $3\times \frac{22}{7}\times \left(28r+196\right)=7392$ $28r+196=\frac{7392\times 7}{66}=784$ $28r=588\Rightarrow r=21$ cm Volume of the original hemisphere $=\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times {21}^3$ $=\frac{2}{3}\times \frac{22}{7}\times 9261=19404$ So the volume is 19404 cubic cm. Hence the correct option is (c).

Fractions of a whole Let the original length of the rope be L metres. The first piece is $\frac{5}{7}L$, so the second piece is $L-\frac{5}{7}L=\frac{2}{7}L$. The second piece measures 10 metres, so $\frac{2}{7}L=10$. $L=10\times \frac{7}{2}=35$ So the rope was 35 metres long. Hence the correct option is (d).

Similar triangles In two similar triangles the perimeters are in the same ratio as the corresponding sides. Here the perimeter is doubled, so the corresponding sides are in the ratio 2:1. The areas of similar triangles are in the ratio of the squares of the corresponding sides, i.e. $2^2:1^2=4:1$. Area of $A\prime B\prime C\prime =4\times 13=52$ So the larger triangle has an area of 52 square cm. Hence the correct option is (c).

Chord and central angle Let AB be the chord and O the centre, so that $\angle AOB={60}^{\circ }$. OA and OB are radii of the same circle, so OA=OB=15 cm and $△OAB$ is isosceles. The two base angles are therefore equal: $\angle OAB=\angle OBA=\frac{{180}^{\circ }-{60}^{\circ }}{2}={60}^{\circ }$. All three angles are ${60}^{\circ }$, so $△OAB$ is equilateral and the chord equals the radius. AB=15 cm Hence the correct option is (b).

Angle at centre and circumference The arc PQ subtends $\angle POQ$ at the centre and $\angle PRQ$ at the point R on the remaining part of the circle. By the angle-at-the-centre theorem, the angle at the centre is twice the angle at any point of the remaining arc. $\angle PRQ=\frac{1}{2}\angle POQ=\frac{1}{2}\times {100}^{\circ }$ $\angle PRQ={50}^{\circ }$ Hence the correct option is (a).

Common tangents to circles An internal (transverse) common tangent is one that passes between the two circles and cuts the line joining their centres. Such a tangent exists only when neither circle lies inside the other, that is when the distance between the centres is at least the sum of the radii. When two circles touch internally, the smaller one lies completely inside the larger and they meet at a single point. No straight line can then pass between them, so the number of internal common tangents is 0. Such a pair has only one common tangent altogether, the one drawn at the point of contact, and that is a direct (external) tangent. Hence the correct option is (d).

Number of common tangents Let d be the distance between the centres and $r_1,r_2$ the radii. $d=r_1+r_2$ is exactly the condition for the two circles to touch each other externally. Two direct (external) common tangents can be drawn, one on either side of the pair. At the single point of contact one more tangent is common to both circles, and it is the only transverse one. Total number of common tangents =2+1=3. Hence the correct option is (d).

Length of internal tangent For two circles that lie outside each other, the length of a common internal (transverse) tangent is $\sqrt{d^2-{\left(r_1+r_2\right)}^2}$, where d is the distance between the centres. Here d=17 cm, $r_1=3$ cm and $r_2=5$ cm, so $r_1+r_2=8$ cm. Length $=\sqrt{{17}^2-8^2}=\sqrt{289-64}$ $=\sqrt{225}=15$ cm Hence the correct option is (c).

Length of internal tangent Length of a common internal tangent $=\sqrt{d^2-{\left(r_1+r_2\right)}^2}$. Here $r_1+r_2=10+8=18$ cm and d=20 cm, and since 20>18 the circles lie apart, so the tangent exists. Length $=\sqrt{{20}^2-{18}^2}=\sqrt{400-324}$ $=\sqrt{76}=\sqrt{4\times 19}=2\sqrt{19}$ cm Hence the correct option is (a).

External and internal tangents Common external tangent $=\sqrt{d^2-{\left(r_1-r_2\right)}^2}$ and common internal tangent $=\sqrt{d^2-{\left(r_1+r_2\right)}^2}$, with d=13 cm. From the external tangent: ${12}^2={13}^2-{\left(r_1-r_2\right)}^2$ ${\left(r_1-r_2\right)}^2=169-144=25\Rightarrow r_1-r_2=5$ From the internal tangent: $5^2={13}^2-{\left(r_1+r_2\right)}^2$ ${\left(r_1+r_2\right)}^2=169-25=144\Rightarrow r_1+r_2=12$ Adding the two results: $2r_1=17\Rightarrow r_1=8.5$ cm $r_2=12-8.5=3.5$ cm So the radii are 8.5 cm and 3.5 cm. Hence the correct option is (d).

Speed, time and distance For a fixed distance the time taken is inversely proportional to the speed. The speeds are in the ratio 7:10, so the times are in the ratio 10:7. The difference of 10-7=3 parts stands for the 36 minutes saved, so 1 part =12 minutes. Time at 7 km/h $=10\times 12=120$ minutes =2 hours. Distance = speed $\times$ time $=7\times 2=14$ km. Hence the correct option is (c).

Mixture and alligation Let the vessel hold 5x litres of juice and 3x litres of soda, that is 8x litres in all. The 8 litres taken out is mixture, so it carries juice and soda in the ratio 5:3, i.e. 5 litres of juice and 3 litres of soda. Juice left =5x-5; soda left =3x-3, and adding 8 litres of soda makes it 3x-3+8=3x+5. $\frac{5x-5}{3x+5}=\frac{5}{7}$ 7(5x-5)=5(3x+5) 35x-35=15x+25 $20x=60\Rightarrow x=3$ Juice at the beginning $=5x=5\times 3=15$ litres. Hence the correct option is (b).