29 November 2025 · Shift 2

Perfect square area If the side of the square plot is a whole number a yards, then the area is $a^2$ square yards. So the area must itself be a perfect square. Squares near the given values are ${33}^2=1089$, ${34}^2=1156$ and ${35}^2=1225$. Only 1156 appears among the options, and it corresponds to a side of 34 yards. The numbers 1150, 1160 and 1165 all lie strictly between ${34}^2$ and ${35}^2$, so none of them is a perfect square. Hence option (a) is correct.

Cost price from loss A loss means the article was sold for less than it cost, so the cost price is greater than the selling price. Formula: Cost price = Selling price + Loss. Cost price = ₹159 + ₹30 = ₹189. Check: selling ₹189 worth of goods for ₹159 does give a loss of ₹30. Hence option (c) is correct.

Compound interest, one year For a period of one year, interest is added only once, so compound interest equals simple interest. Formula: $CI=P\left({\left(1+\frac{R}{100}\right)}^n-1\right)$ $CI=8000\left({\left(1+\frac{9}{100}\right)}^1-1\right)$ $=8000\times \frac{9}{100}=80\times 9=720$ So the compound interest is ₹720. Hence option (b) is correct.

Interior angle of a polygon The window is a regular octagon, so all eight interior angles are equal. Formula: each interior angle of a regular polygon of n sides $=\frac{\left(n-2\right)\times {180}^{\circ }}{n}$ For an octagon, n=8, so the angle $=\frac{\left(8-2\right)\times {180}^{\circ }}{8}$ $=\frac{6\times {180}^{\circ }}{8}=\frac{{1080}^{\circ }}{8}={135}^{\circ }$ Hence option (c) is correct.

Total surface area of a cylinder The ends are flat, so the total surface is the curved surface plus the two circular ends. Formula: total surface area $=2\pi rh+2\pi r^2=2\pi r\left(r+h\right)$ Here r=0.3 cm and h=3 cm, so r+h=3.3 cm. $=2\times \frac{22}{7}\times 0.3\times 3.3$ $=\frac{44\times 0.99}{7}=\frac{43.56}{7}=6.22$ So the total surface area is about 6.22 cm${}^2$. Hence option (d) is correct.

Volume of a triangular prism First identify the triangular base: $6^2+8^2=36+64=100={10}^2$, so it is right-angled with legs 6 cm and 8 cm. Area of the base $=\frac{1}{2}\times 6\times 8=24$ cm${}^2$ Volume of one prism = area of base $\times$ length $=24\times 15=360$ cm${}^3$ Volume of 25 such blocks $=25\times 360=9000$ So the total volume of wood required is 9000 cm${}^3$. Hence option (a) is correct.

Trigonometric identity Take out the common factor from the numerator and from the denominator. Numerator: $2{\cos }^{3}A-\cos A=\cos A\left(2{\cos }^{2}A-1\right)=\cos A\cdot \cos 2A$ Denominator: $\sin A-2{\sin }^{3}A=\sin A\left(1-2{\sin }^{2}A\right)=\sin A\cdot \cos 2A$ So the expression $=\frac{\cos A\cdot \cos 2A}{\sin A\cdot \cos 2A}=\cot A$ Therefore $\cot A=1$, that is $\tan A=1$. $\tan {45}^{\circ }=1$, so $A={45}^{\circ }$. Hence option (c) is correct.

Trigonometric identity Square each expression and then add, so that the cross terms cancel. $x^2={\left(7\cos A+9\sin A\right)}^2=49{\cos }^{2}A+126\sin A\cos A+81{\sin }^{2}A$ $y^2={\left(7\sin A-9\cos A\right)}^2=49{\sin }^{2}A-126\sin A\cos A+81{\cos }^{2}A$ Adding, the terms $\pm 126\sin A\cos A$ cancel. $x^2+y^2=49\left({\cos }^{2}A+{\sin }^{2}A\right)+81\left({\sin }^{2}A+{\cos }^{2}A\right)$ Using ${\sin }^{2}A+{\cos }^{2}A=1$, this becomes 49+81=130. Hence option (c) is correct.

Melting sphere into cylinder Melting and recasting changes the shape but not the amount of material, so the volume stays the same. Volume of the sphere $=\frac{4}{3}\pi R^3=\frac{4}{3}\pi {\left(12\right)}^3=\frac{4}{3}\pi \times 1728=2304\pi$ Volume of the rod $=\pi r^{2}h=\pi r^2\times 16$ Equating: $16\pi r^2=2304\pi$ $r^2=\frac{2304}{16}=144$, so $r=\sqrt{144}=12$ So the radius of the rod is 12 cm. Hence option (a) is correct.

Height and area of a trapezium The non-parallel sides are equal, so the trapezium is isosceles; dropping perpendiculars from the ends of the shorter side cuts off two congruent right triangles. The two triangle bases together make up the extra length of the longer side. Base of each triangle $=\frac{26-14}{2}=\frac{12}{2}=6$ cm In one right triangle the hypotenuse is 13 cm, so the height $=\sqrt{{13}^2-6^2}=\sqrt{169-36}=\sqrt{133}$ cm Area $=\frac{1}{2}\times$ (sum of the parallel sides) $\times$ height $=\frac{1}{2}\times \left(26+14\right)\times \sqrt{133}=\frac{1}{2}\times 40\times \sqrt{133}=20\sqrt{133}$ cm${}^2$ Hence option (c) is correct.

Volume of solids For a cylinder of base radius r and height h, curved surface area $=2\pi rh$ and total surface area $=2\pi r\left(h+r\right)$. So $\frac{2\pi rh}{2\pi r\left(h+r\right)}=\frac{2}{5}$, which gives 5h=2h+2r, i.e. $h=\frac{2r}{3}$. Melting does not change the volume, so volume of hemisphere = volume of cylinder. $\frac{2}{3}\pi {\left(21\right)}^3=\pi r^{2}h=\pi r^2\times \frac{2r}{3}=\frac{2}{3}\pi r^3$. Hence $r^3={21}^3$, so r=21 cm. Hence option (b) is correct.

Partnership profit share In a partnership the profit is divided in the ratio of (capital $\times$ time). Let the starting capitals be 4x, 3x and 2x; for the first 4 months these hold. For the remaining 8 months P has $4x\times \frac{150}{100}=6x$, Q has $2\times 3x=6x$ and R still has 2x. P's equivalent capital $=4x\times 4+6x\times 8=16x+48x=64x$. Q's equivalent capital $=3x\times 4+6x\times 8=12x+48x=60x$. R's equivalent capital $=2x\times 12=24x$. Total =64x+60x+24x=148x, so Q's share $=72000\times \frac{60}{148}=₹29,189$ (to the nearest rupee). Hence option (b) is correct.

Ratio and proportion Let X have x and Y have y, so x+y=8000. Given $\frac{4}{9}x=\frac{3}{8}y$, so $\frac{x}{y}=\frac{3}{8}\times \frac{9}{4}=\frac{27}{32}$. So x:y=27:32 and the whole amount is 27+32=59 parts. 59 parts =8000, hence Y's amount $=8000\times \frac{32}{59}=\frac{256000}{59}=₹4339$ (to the nearest rupee). Hence option (d) is correct.

Complementary angles Use the complementary-angle identity $\cos \theta =\sin \left({90}^{\circ }-\theta \right)$ to bring both sides to the same ratio. $\cos \left(x+{30}^{\circ }\right)=\sin \left({90}^{\circ }-\left(x+{30}^{\circ }\right)\right)=\sin \left({60}^{\circ }-x\right)$. So $\sin 3x=\sin \left({60}^{\circ }-x\right)$, which gives $3x={60}^{\circ }-x$. $4x={60}^{\circ }$, so $x={15}^{\circ }$. Check: $\sin {45}^{\circ }=\cos {45}^{\circ }$, so the value fits. Hence option (a) is correct.

LCM property The LCM of two numbers equals the larger one only when the larger is a multiple of the smaller. LCM(p,q)=q therefore means q is a multiple of p, i.e. p divides q. LCM(q,r)=r likewise means r is a multiple of q, i.e. q divides r. Divisibility is transitive, so p divides r as well; thus r is a common multiple of p, q and r. No smaller number can be a common multiple, so LCM(p,q,r)=r. Hence option (c) is correct.

Simple interest rate Under simple interest, Amount = Principal + SI. Let the principal be 4 units, so the amount $=\frac{9}{4}\times 4=9$ units. SI for 4 years =9-4=5 units. Rate $=\frac{SI\times 100}{P\times T}=\frac{5\times 100}{4\times 4}=\frac{500}{16}=31.25\%$. Hence option (c) is correct.

Mixture and alligation Choose capacities that are divisible by the sums of the milk-water ratios (3, 5 and 7). Take the capacities as 30, 40 and 70 units, which keep the ratio 3:4:7; the total is 140 units. Milk in A $=30\times \frac{1}{3}=10$ units. Milk in B $=40\times \frac{2}{5}=16$ units. Milk in C $=70\times \frac{3}{7}=30$ units. Total milk =10+16+30=56 units, so water =140-56=84 units. Milk : water =56:84=2:3. Hence option (d) is correct.

Successive percentage Start from the salary that everything is compared with, so let R =100. Q is $10\%$ more than R, so Q $=100\times \frac{110}{100}=110$. P is $30\%$ more than Q, so P $=110\times \frac{130}{100}=143$. Required percentage $=\frac{143-100}{100}\times 100=43\%$. Note that the two rises do not simply add to $40\%$, because the second rise acts on the already increased salary. Hence option (b) is correct.

Comparing fractions Give all four fractions the same denominator; the LCM of 12, 18, 9 and 36 is 36. $\frac{5}{12}=\frac{15}{36}$, $\frac{7}{18}=\frac{14}{36}$, $\frac{4}{9}=\frac{16}{36}$ and $\frac{11}{36}$ stays as it is. With equal denominators only the numerators decide: 16>15>14>11. So the descending order is $\frac{4}{9}>\frac{5}{12}>\frac{7}{18}>\frac{11}{36}$. Hence option (a) is correct.

Recurring decimals A purely recurring decimal with a two-digit period equals that two-digit number over 99: $0.ab=\frac{ab}{99}$. So $0.76=\frac{76}{99}$, and the options are $\frac{79}{99}$, $\frac{59}{99}$, $\frac{61}{99}$ and $\frac{57}{99}$. All five share the denominator 99, so only the numerators need comparing. Only 79>76; the numerators 59, 61 and 57 are all smaller than 76. Hence option (a) is correct.

Surds and roots Simplify each root separately. $\sqrt[3]{125}=5$, since $5^3=125$. $\sqrt[4]{1296}=6$, since $6^4=1296$. $\sqrt{64}=8$. So the expression $=\frac{5+6}{8}=\frac{11}{8}$. Hence option (d) is correct.

Simultaneous surd equations Treat $\sqrt{a}$ and $\sqrt{b}$ as the two unknowns and solve the pair. Adding: $\left(\sqrt{a}+\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)=17+3$, so $2\sqrt{a}=20$ and $\sqrt{a}=10$. Subtracting: $\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)=17-3$, so $2\sqrt{b}=14$ and $\sqrt{b}=7$. Squaring, a=100 and b=49. Therefore a+b=100+49=149. Hence option (a) is correct.

Simple interest time Simple interest = Amount - Principal =19500-15000=₹4500. Use $SI=\frac{P\times R\times T}{100}$. $4500=\frac{15000\times 5\times T}{100}=750T$. $T=\frac{4500}{750}=6$ years. Hence option (b) is correct.

Combined average Total age = average $\times$ number of students. Class A: $10\times 30=300$ years; class B: $20\times x=20x$ years. The two classes together have 10+20=30 students with average 44, so $\frac{300+20x}{30}=44$. 300+20x=1320, so 20x=1020. $x=\frac{1020}{20}=51$ years. Hence option (b) is correct.

Discount and marked price Discount is the amount cut from the marked price, so Marked price = Selling price + Discount. Let the discount be ₹100 (this makes percentages of the discount easy to read). Selling price $=160\%$ of $100=\frac{160}{100}\times 100=₹160$. Marked price =160+100=₹260. Marked price as a percentage of the discount $=\frac{260}{100}\times 100=260\%$. Hence option (a) is correct.