29 November 2025 · Shift 1

Profit and loss When there is a loss, $\mathrm{S.P.}=\mathrm{C.P.}\times \frac{100-\mathrm{loss}\%}{100}$. Here the selling price is ₹105 and the loss is $30\%$, so $105=\mathrm{C.P.}\times \frac{100-30}{100}=\mathrm{C.P.}\times \frac{70}{100}$. $\mathrm{C.P.}=\frac{105\times 100}{70}=150$. So the cost price of the article is ₹150. Hence the correct option is (b).

Time, speed and distance Formula: $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$. $\mathrm{Time}=\frac{800}{40}$ =20 hours. Hence the correct option is (b).

Time and work Take the total work as the LCM of 10 and 20, that is 20 units. A does $\frac{20}{10}=2$ units a day and B does $\frac{20}{20}=1$ unit a day. Working together for 5 days they finish $\left(2+1\right)\times 5=15$ units. Work left =20-15=5 units, and B alone needs $\frac{5}{1}=5$ days for it. Hence the correct option is (c).

Area and tiling Area of the wall $=8\times 4=32$ square metres. Area of one rectangular tile $=2\times 1=2$ square metres. Number of tiles $=\frac{\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{wall}}{\mathrm{area}\mathrm{of}\mathrm{one}\mathrm{tile}}=\frac{32}{2}=16$. Hence the correct option is (c).

Volume of a prism The base of the prism is a trapezium, so first find its area. $\mathrm{Area}=\frac{1}{2}\times \left(\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{height}$ $=\frac{1}{2}\times \left(5+9\right)\times 6=\frac{1}{2}\times 14\times 6=42$ square centimetres. Volume of a right prism = area of the base $\times$ height of the prism $=42\times 10=420$ cubic centimetres. Hence the correct option is (c).

Similarity of triangles Two triangles are similar when their corresponding angles are equal and their corresponding sides are proportional; three standard tests settle this. AA: if two angles of one triangle equal two angles of the other, the third pair is automatically equal, so the triangles are similar. SSS: if all three pairs of corresponding sides are in the same ratio, the triangles are similar. SAS: if one pair of angles is equal and the two sides containing that angle are proportional, the triangles are similar. Each of the three tests is by itself sufficient. Hence the correct option is (d).

Hollow cylinder Outer radius $R=\frac{4.2}{2}=2.1$ cm, and the thickness 7 mm =0.7 cm. Inner radius $r=R-\mathrm{thickness}=2.1-0.7=1.4$ cm. Volume of the hollow rod $=\pi \left(R^2-r^2\right)h$ $=\frac{22}{7}\times \left({2.1}^2-{1.4}^2\right)\times 15$ $=\frac{22}{7}\times \left(4.41-1.96\right)\times 15=\frac{22}{7}\times 2.45\times 15=115.5$ cubic centimetres. Weight = volume $\times$ weight of 1 cubic centimetre $=115.5\times 0.4=46.2$ kg. Hence the correct option is (a).

Algebraic identity Use $a^2-b^2=\left(a+b\right)\left(a-b\right)$ with $a=\sqrt{3}+2$ and $b=\sqrt{3}-2$. $a+b=\left(\sqrt{3}+2\right)+\left(\sqrt{3}-2\right)=2\sqrt{3}$ $a-b=\left(\sqrt{3}+2\right)-\left(\sqrt{3}-2\right)=4$ So the value $=2\sqrt{3}\times 4=8\sqrt{3}$. Hence the correct option is (b).

Partnership All three partners kept their money in the business for the same period of 1 year, so the profit is divided in the ratio of the investments. P:Q:R=60000:50000:40000=6:5:4. Total number of parts =6+5+4=15. Q's share $=\frac{5}{15}\times 150000=50000$, that is ₹50,000. Hence the correct option is (a).

Ratio and proportion First : second =3:4 and second : third =4:5; the second number is already 4 in both ratios, so they can be joined directly. First : second : third =3:4:5. Total number of parts =3+4+5=12, and these 12 parts stand for 180. One part $=\frac{180}{12}=15$. First number $=3\times 15=45$. Hence the correct option is (c).

Percentage change in volume Let the original radius be r and the height be h, so the volume is $V=\pi r^{2}h$. New radius $=r+150\%$ of r=2.5r; new height $=h-20\%$ of h=0.8h. New volume $=\pi {\left(2.5r\right)}^2\left(0.8h\right)=\pi \times 6.25r^2\times 0.8h=5\pi r^{2}h$. The volume becomes 5 times the original, so the rise is 5V-V=4V. Percentage increase $=\frac{4V}{V}\times 100=400\%$. इसलिए सही विकल्प (a) है।

Successive discounts Let the marked price be ₹100. After a single discount of $20\%$ the price is $100\times \frac{80}{100}=80$. After discounts of $10\%$ and $x\%$ the price is $100\times \frac{90}{100}\times \left(1-\frac{x}{100}\right)$. Both must be equal: $90\left(1-\frac{x}{100}\right)=80\Rightarrow 1-\frac{x}{100}=\frac{8}{9}$. $\frac{x}{100}=\frac{1}{9}\Rightarrow x=\frac{100}{9}=11.11$. इसलिए सही विकल्प (b) है।

Marked price and discount A discount of $10\%$ means the selling price is $90\%$ of the marked price. Formula: $\mathrm{S.P.}=\mathrm{M.P.}\times \frac{100-\mathrm{discount}}{100}$. $1350=\mathrm{M.P.}\times \frac{90}{100}$ $\mathrm{M.P.}=1350\times \frac{100}{90}=1500$ So the marked price is ₹1500; the $25\%$ profit only fixes the cost price at $1350\times \frac{100}{125}=₹1080$ and is not needed here. इसलिए सही विकल्प (a) है।

Mixture replacement Formula: if x litres are drawn from V litres and replaced by water n times, milk left $=V{\left(1-\frac{x}{V}\right)}^n$. Here V=50, x=5 and n=2. Milk $=50{\left(1-\frac{5}{50}\right)}^2=50\times \frac{9}{10}\times \frac{9}{10}=40.5$ litres. Water =50-40.5=9.5 litres. Milk : Water =40.5:9.5=81:19. इसलिए सही विकल्प (a) है।

Parallel chords of a circle For a chord of length 2l at distance d from the centre of a circle of radius r: $r^2=d^2+l^2$. For AB=6 cm the half-chord is 3 cm, so $d_1=\sqrt{5^2-3^2}=\sqrt{25-9}=4$ cm. The chords are on opposite sides of the centre, so $d_1+d_2=7\Rightarrow d_2=7-4=3$ cm. For CD: half-chord $=\sqrt{5^2-3^2}=\sqrt{25-9}=4$ cm. $CD=2\times 4=8$ cm, so x=8. इसलिए सही विकल्प (b) है।

Tangent to a circle A tangent is perpendicular to the radius drawn to the point of contact, so $\angle OBA={90}^{\circ }$. In right-angled triangle OBA, by Pythagoras: $O{A}^2=O{B}^2+A{B}^2$. $O{A}^2=8^2+{15}^2=64+225=289$ $OA=\sqrt{289}=17$ cm. इसलिए सही विकल्प (c) है।

Areas of similar figures In two similar figures the areas are in the ratio of the squares of the corresponding sides. Side ratio (smaller : larger) =1:2. Area ratio $=1^2:2^2=1:4$. Area of the smaller park $=\frac{1}{4}\times 144=36m^2$. इसलिए सही विकल्प (c) है।

Algebraic identities Given x+y=13 and x-y=7. Adding the two: $2x=20\Rightarrow x=10$; subtracting: $2y=6\Rightarrow y=3$. $x^2-y^2=\left(x+y\right)\left(x-y\right)=13\times 7=91$ $x^2+y^2={10}^2+3^2=100+9=109$ $x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=91\times 109=9919$ इसलिए सही विकल्प (b) है।

Simplification by identities First note that $18.1956=2\times 2.57\times 3.54$. So the numerator $={\left(2.57\right)}^2+2\left(2.57\right)\left(3.54\right)+{\left(3.54\right)}^2={\left(2.57+3.54\right)}^2={\left(6.11\right)}^2$. The denominator $={\left(3.54\right)}^2-{\left(2.57\right)}^2=\left(3.54+2.57\right)\left(3.54-2.57\right)=6.11\times 0.97$. Expression $=\frac{{\left(6.11\right)}^2}{6.11\times 0.97}=\frac{6.11}{0.97}$ $\frac{6.11}{0.97}=6.29$ (to two decimal places, the nearest option). इसलिए सही विकल्प (a) है।

Surds Simplify the surd: $\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$. So $x=2\sqrt{5}+\sqrt{5}=3\sqrt{5}$. $x^2={\left(3\sqrt{5}\right)}^2=9\times 5=45$ इसलिए सही विकल्प (c) है।

Compound interest Under compound interest, a year's interest is $20\%$ of that year's opening amount. Amount after 1 year $=5000\times \frac{120}{100}=6000$, so interest of the 2nd year $=6000\times \frac{20}{100}=1200$. Amount after 2 years $=6000\times \frac{120}{100}=7200$, so interest of the 3rd year $=7200\times \frac{20}{100}=1440$. Required difference =1440-1200=240. इसलिए सही विकल्प (d) है।

Simple interest Simple interest = Amount - Principal =24000-16000=8000. Formula: $\mathrm{S.I.}=\frac{P\times R\times T}{100}$. $8000=\frac{16000\times 5\times T}{100}=800T$ $T=\frac{8000}{800}=10$ years. इसलिए सही विकल्प (c) है।

Average of consecutive numbers For an odd count of equally spaced numbers the average equals the middle term, so the third number is 57. Consecutive odd numbers differ by 2, so the five numbers are 53, 55, 57, 59, 61. The five consecutive odd numbers just before these are 43, 45, 47, 49, 51. Their average is again the middle term =47. इसलिए सही विकल्प (c) है।

Average age Total age of the 20 players $=20\times x=20x$. Let the total age of the 10 new players be T; the team now has 20+10=30 players. The average is unchanged, so $\frac{20x+T}{30}=x$. $20x+T=30x\Rightarrow T=10x$ इसलिए सही विकल्प (c) है।

Pipes and cisterns Take the capacity of the tank = LCM of 24, 30 and 40 =120 units. $A+B=\frac{120}{24}=5$, $B+C=\frac{120}{30}=4$, $C+A=\frac{120}{40}=3$ units/min. Adding all three: $2\left(A+B+C\right)=5+4+3=12\Rightarrow A+B+C=6$ units/min. So C=(A+B+C)-(A+B)=6-5=1 unit/min. In the last 5 minutes all three pipes run together: $6\times 5=30$ units. Hence pipe C alone had already filled 120-30=90 units. $L=\frac{90}{1}=90$ minutes. इसलिए सही विकल्प (c) है।