21 November 2025 · Shift 3

Square root 53824 ends in 4, so its square root must end in 2 or 8. ${230}^2=52900$ and ${240}^2=57600$, so the root lies between 230 and 240. The only options in that range ending in 2 or 8 are 232 and 238, so test 232. ${232}^2={\left(230+2\right)}^2={230}^2+2\times 230\times 2+2^2$ =52900+920+4=53824 So $\sqrt{53824}=232$. Hence the correct option is (b).

Square roots First simplify the right-hand side: ${58}^2=3364$, so $\sqrt{3364}=58$. $\sqrt{?}+19=58$ $\sqrt{?}=58-19=39$ Squaring both sides, $?={39}^2=1521$. Hence the correct option is (c).

Simplification Take out each square root first: $\sqrt{441}=21$, $\sqrt{225}=15$ and $\sqrt{64}=8$. $\left(\frac{168}{21}\right)\times \left(\frac{15}{5}\right)\times \left(\frac{8}{4}\right)$ $=8\times 3\times 2$ =48 Hence the correct option is (b).

Average of a series When an odd number of consecutive even numbers is taken, the average equals the middle term. So the 4th of the seven numbers is 52. Three numbers lie above it, each greater than the previous one by 2. Largest number $=52+3\times 2=52+6=58$. Hence the correct option is (a).

Average and totals Total runs in 4 innings $=4\times 60=240$. The two known scores add up to 55+65=120. So highest + lowest =240-120=120 ... (i) It is given that highest - lowest =40 ... (ii) Adding (i) and (ii): $2\times$ highest =160. Highest =80 (and the lowest is 120-80=40). Hence the correct option is (c).

Percentage increase Increase in price =650-500=150, i.e. ₹150. Percentage increase $=\frac{\mathrm{increase}}{\mathrm{original}\mathrm{price}}\times 100$ $=\frac{150}{500}\times 100$ $=30\%$ Hence the correct option is (b).

Successive percentage change Take the number as 100. After a 50% increase it becomes 100+50=150. The 40% decrease is now taken on 150: $150-\frac{40}{100}\times 150=150-60=90$. Net change =90-100=-10 on 100, i.e. a decrease of 10%. Shortcut: net change $=a+b+\frac{ab}{100}=50-40-\frac{50\times 40}{100}=-10\%$. Hence the correct option is (b).

Discount Selling price = marked price $\times \left(1-\frac{\mathrm{discount}\%}{100}\right)$ $=20000\times \left(1-\frac{15}{100}\right)$ $=20000\times \frac{85}{100}$ =17000 So the selling price is Rs. 17,000. Hence the correct option is (b).

Discount A discount of 20% leaves $100-20=80\%$ of the marked price to be paid. Selling price $=5000\times \frac{80}{100}$ =4000 So the selling price is Rs. 4,000. Hence the correct option is (c).

Discount A discount of 15% leaves $100-15=85\%$ of the marked price to be paid. Selling price $=8000\times \frac{85}{100}$ =6800 So the selling price is Rs. 6,800. Hence the correct option is (a).

Time and work — together Take the total work as the LCM of 6 and 4, that is 12 units. X's one day work $=\frac{12}{6}=2$ units and Y's one day work $=\frac{12}{4}=3$ units. Working together they finish 2+3=5 units in one day. Time taken together $=\frac{12}{5}=2.4$ days Hence option (b) is correct.

Time and work — one alone Take the total work as the LCM of 8 and 12, that is 24 units. (Ram + Shyam) do $\frac{24}{8}=3$ units a day, while Ram alone does $\frac{24}{12}=2$ units a day. Shyam's one day work =3-2=1 unit Time taken by Shyam alone $=\frac{24}{1}=24$ days Hence option (c) is correct.

Regular polygon — sides At every vertex of a polygon the interior angle and the exterior angle form a straight line. Exterior angle $={180}^{\circ }-{144}^{\circ }={36}^{\circ }$ The exterior angles of any polygon add up to ${360}^{\circ }$, so the number of sides is $n=\frac{{360}^{\circ }}{{36}^{\circ }}=10$. Check with the direct formula: $\frac{\left(n-2\right)\times {180}^{\circ }}{n}=\frac{8\times {180}^{\circ }}{10}={144}^{\circ }$ Hence option (a) is correct.

Similar triangles — shadows At the same moment the sun's rays fall at the same angle, so the pole with its shadow and the tower with its shadow form similar right triangles. In similar triangles corresponding sides are proportional, so the ratio of height to shadow is the same for both. $\frac{6}{4}=\frac{h}{28}$ $h=\frac{6\times 28}{4}=42$ The tower is 42 m high, so option (b) is correct.

Trigonometric ratio The vertical height 8 is the side opposite $\theta$ and 17 is the hypotenuse, so the third side is the base. Base $=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ Formula: $\tan \theta =\frac{P}{B}$, where P is the side opposite the angle and B is the base. $\tan \theta =\frac{8}{15}$ $\theta$ lies in the first quadrant, where every trigonometric ratio is positive, so no sign change is needed. Hence option (b) is correct.

Average of odd numbers The first 25 odd numbers are $1,3,5,\dots ,49$, which form an AP with first term 1 and common difference 2. The sum of the first n odd numbers is $n^2$, so the sum here is ${25}^2=625$. Average $=\frac{625}{25}=25$ So the average of the first n odd numbers is always n itself; the same value also comes from $\frac{\mathrm{first}+\mathrm{last}}{2}=\frac{1+49}{2}=25$. Hence option (b) is correct.

Simple interest — principal Under simple interest the same amount of interest is added every year, so the growth from year to year is uniform. Interest for (4-2)=2 years =2100-1800=300 Interest for 1 year $=\frac{300}{2}=150$ Principal = amount after 2 years - interest of 2 years $=1800-2\times 150=1500$ The principal is ₹1,500, so option (d) is correct.

Indices and surds Write every base as a power of 2. First factor of the numerator: ${64}^{\frac{5}{6}}={\left(2^6\right)}^{\frac{5}{6}}=2^5=32$ Second factor: ${\left(\frac{1}{32}\right)}^{-\frac{4}{5}}={32}^{\frac{4}{5}}={\left(2^5\right)}^{\frac{4}{5}}=2^4=16$ Numerator $=32\times 16=512$ For the denominator use the identity ${\left(a+b\right)}^2-{\left(a-b\right)}^2=4ab$ with $a=\sqrt{18}$ and $b=\sqrt{2}$. Denominator $=4\sqrt{18}\times \sqrt{2}=4\sqrt{36}=4\times 6=24$ Value $=\frac{512}{24}=\frac{64}{3}$ Hence option (c) is correct.

Tangent and secant The radius drawn to the point of contact is perpendicular to the tangent, so $△OTP$ is right-angled at T. $OP=\sqrt{O{T}^2+P{T}^2}=\sqrt{5^2+{12}^2}=\sqrt{25+144}=\sqrt{169}=13$ The secant passes through the centre, so B, the nearer point of the circle, lies on OP at a distance of one radius from O. PB=OP-OB=13-5=8 PB is 8 cm, so option (d) is correct.

Centroid of a triangle The centroid divides each median in the ratio 2:1, and its coordinates are simply the averages of the coordinates of the three vertices. Formula: $G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$ Only the y-coordinate is asked, so use $y_1=5$, $y_2=9$ and $y_3=7$. $y_G=\frac{5+9+7}{3}=\frac{21}{3}=7$ Hence option (c) is correct.

Circumcentre property The circumcentre is the point where the perpendicular bisectors of the three sides meet. Every point on the perpendicular bisector of a segment is equally far from the two end points of that segment. So O is equally far from P, Q and R, and OP=OQ=OR is the circumradius of the circle drawn through the three vertices. The point equidistant from the three sides is the incentre, and the circumcentre lies outside the triangle when the triangle is obtuse and on the hypotenuse when it is right-angled, so the other options are not always true. Hence option (b) is correct.

Angle bisector theorem Angle bisector theorem: the bisector of an angle of a triangle divides the opposite side in the ratio of the two sides containing that angle. $\frac{BD}{DC}=\frac{AB}{AC}=\frac{10}{15}=\frac{2}{3}$ So BC is split into 2+3=5 equal parts, of which BD takes 2 parts. $BD=\frac{2}{5}\times BC=\frac{2}{5}\times 20=8$ BD is 8 cm, so option (c) is correct.

Common tangents — touching The number of common tangents is decided by comparing the distance between the centres with the sum and the difference of the radii. $r_1+r_2=7+4=11$ and d=11, so $d=r_1+r_2$. The circles therefore touch each other externally at exactly one point. Externally touching circles have 2 direct common tangents plus 1 transverse tangent drawn at the point of contact. Total =2+1=3, so option (c) is correct.

Common tangents — apart Compare the distance between the centres with the sum of the radii. $r_1+r_2=5+3=8$ while d=10, so $d>r_1+r_2$. The circles lie completely outside each other and do not cut or touch. Such circles have 2 direct (external) common tangents and 2 transverse (internal) common tangents. Total =2+2=4, so option (a) is correct.

Direct common tangent Formula: length of the direct common tangent $L=\sqrt{d^2-{\left(r_1-r_2\right)}^2}$, where d is the distance between the centres. $12=\sqrt{d^2-{\left(8-3\right)}^2}$ Squaring both sides: $144=d^2-25$ $d^2=144+25=169\Rightarrow d=13$ The centres are 13 cm apart, so option (b) is correct.