29 November 2025 · Shift 3

Algebraic identities Expand the two squares inside the bracket first. ${\left(a+b\right)}^2=a^2+2ab+b^2$ and ${\left(a-b\right)}^2=a^2-2ab+b^2$. Adding them, the 2ab terms cancel: ${\left(a+b\right)}^2+{\left(a-b\right)}^2=2a^2+2b^2=2\left(a^2+b^2\right)$. Now square that: ${\left(2\left(a^2+b^2\right)\right)}^2=4{\left(a^2+b^2\right)}^2$. Expanding, $4\left(a^4+2a^2{b}^2+b^4\right)=4a^4+8a^2{b}^2+4b^4$. Hence the correct option is (c).

Profit and selling price Formula: selling price = cost price $\times \frac{100+\mathrm{profit}\%}{100}$. Here the cost price is ₹320 and the profit is 25%. So $SP=320\times \frac{125}{100}=320\times \frac{5}{4}$. $SP=80\times 5=400$. The selling price is ₹400, so the correct option is (a).

Speed, distance and time Speed is distance divided by time. The distance covered is 20 km and the time taken is 4 hours. Speed $=\frac{20}{4}=5$ km/h. Hence the correct option is (b).

Time and work Take the whole task as the LCM of 20 and 15, that is 60 units. Ram does $\frac{60}{20}=3$ units a day and Ajay does $\frac{60}{15}=4$ units a day. In the first 6 days Ram alone does $6\times 3=18$ units. Work still left =60-18=42 units. Working together they do 3+4=7 units a day, so they need $\frac{42}{7}=6$ more days. Total time =6+6=12 days. Hence the correct option is (c).

Volume of a cuboid Volume of a cuboid = length $\times$ breadth $\times$ height. Volume of the room $=6\times 3\times 2.5=45$ cubic metres. Each carton needs 3 cubic metres of space. Number of cartons $=\frac{45}{3}=15$. Hence the correct option is (c).

Lateral surface of a prism For any right prism, lateral surface area = perimeter of the base $\times$ height. The base is a regular pentagon of side 5 cm, so its perimeter $=5\times 5=25$ cm. Lateral surface area $=25\times 8=200$ cm². Hence the correct option is (b).

Similar triangles In two similar triangles every pair of corresponding sides is in the same ratio, say k. So each side of the second triangle is k times the matching side of the first. The perimeter is the sum of the three sides, so the second perimeter $=k\times$ (first perimeter). Therefore the ratio of the perimeters is the same as the ratio of the corresponding sides. The square $k^2$ belongs to the ratio of the areas, not of the perimeters, so option (c) is not the answer. Hence the correct option is (b).

Cylinder - volume and surface area Let the radius be 2x and the height 5x, so that they are in the ratio 2 : 5. Volume $=\pi r^{2}h=3\times {\left(2x\right)}^2\times 5x=60x^3$. Total surface area - curved surface area $=2\pi r^2=2\times 3\times {\left(2x\right)}^2=24x^2$. A: $60x^3=7500$ gives $x^3=125$, so x=5; then $24x^2=24\times 25=600$, exactly the second number. B: $60x^3=1875$ gives $x^3=31.25$, so $x\approx 3.15$ and $24x^2\approx 238$, not 375. C: $60x^3=3000$ gives $x^3=50$, so $x\approx 3.68$ and $24x^2\approx 326$, not 600. D: $60x^3=3750$ gives $x^3=62.5$, so $x\approx 3.97$ and $24x^2\approx 378$, not 750. Only set A makes the statement true, so the correct option is (a).

Ratio and division of money B gets half of what C gets, so C=2B; taking B=3 units makes C=6 units, which keeps everything whole. A gets one-third of what B gets, so $A=\frac{3}{3}=1$ unit. Therefore A:B:C=1:3:6. Total =1+3+6=10 units, and this stands for Rs. 800. So 1 unit $=\frac{800}{10}=$ Rs. 80. B's share $=3\times 80=$ Rs. 240. Hence the correct option is (b).

Ratio of ages Let the present ages of X and Y be 3x and 4x. After 5 years, $\frac{3x+5}{4x+5}=\frac{4}{5}$. Cross-multiplying, 5(3x+5)=4(4x+5). 15x+25=16x+20, which gives x=5. So the present ages are $X=3\times 5=15$ years and $Y=4\times 5=20$ years. Three years ago they were 15-3=12 years and 20-3=17 years. Required ratio =12:17. Hence the correct option is (c).

Percentage change in area Let the original length be l and the original breadth be b, so the area is lb. The breadth becomes 1.2b, and the new length l' must keep the area the same. $l\prime \times 1.2b=l\times b$ $l\prime =\frac{l}{1.2}=\frac{5l}{6}$ Reduction in length $=l-\frac{5l}{6}=\frac{l}{6}$. Percentage reduction $=\frac{l/6}{l}\times 100=\frac{100}{6}=16.67\%$. Hence the correct option is (d).

Discount and marked price A discount of 20% means the selling price is 80% of the marked price. Formula: $SP=MP\times \left(1-\frac{d}{100}\right)$. $800=MP\times \frac{80}{100}=MP\times \frac{4}{5}$ $MP=800\times \frac{5}{4}=1000$ So the original marked price of the pant was ₹1,000. Hence the correct option is (a).

Alligation - mixture ratio Here the cheaper rice costs ₹30 per kg, the dearer rice ₹42 per kg and the required mean price is ₹36 per kg. Rule of alligation: $\frac{\mathrm{quantity}\mathrm{of}\mathrm{cheaper}}{\mathrm{quantity}\mathrm{of}\mathrm{dearer}}=\frac{\mathrm{dearer}\mathrm{cost}-\mathrm{mean}\mathrm{price}}{\mathrm{mean}\mathrm{price}-\mathrm{cheaper}\mathrm{cost}}$. $=\frac{42-36}{36-30}=\frac{6}{6}=\frac{1}{1}$ So the two varieties must be mixed in equal weights, i.e. in the ratio 1:1. Hence the correct option is (a).

Mixture replaced by water Take the original mixture as 100 units, so milk $=\frac{3}{5}\times 100=60$ units and water $=\frac{2}{5}\times 100=40$ units. Let x units of the mixture be taken out; it carries away 0.6x units of milk and 0.4x units of water, and the same x units are returned as pure water. Milk left =60-0.6x and water =40-0.4x+x=40+0.6x. Given $\frac{60-0.6x}{40+0.6x}=\frac{1}{2}$. 2(60-0.6x)=40+0.6x $120-1.2x=40+0.6x\Rightarrow 80=1.8x$ $x=\frac{800}{18}=44.44$ Since the whole mixture was 100 units, 44.44% of it was replaced. Hence the correct option is (b).

Tangential quadrilateral A quadrilateral in which a circle touches all four sides is called a tangential quadrilateral. The two tangents drawn from an external point to a circle are equal, so the tangent lengths from A, B, C and D pair off along the sides. Adding these equal pairs gives Pitot's theorem: the two pairs of opposite sides have equal sums. AB+CD=BC+AD 18+CD=14+10 18+CD=24 CD=6 cm Hence the correct option is (b).

Parallel tangents and diameter A tangent to a circle is perpendicular to the radius drawn to its point of contact. The tangents at P and at Q are parallel, so the radii OP and OQ are perpendicular to the same direction. Two such radii from the same centre must lie along one straight line, so P, O and Q are collinear. Hence the chord PQ passes through the centre and is a diameter. $2r=18\Rightarrow r=\frac{18}{2}=9$ cm Hence the correct option is (a).

Rationalising surds Simplify the three parts one at a time. For the first fraction note that ${\left(\sqrt{5}+\sqrt{2}\right)}^2=5+2+2\sqrt{10}=7+2\sqrt{10}$, so multiply above and below by $\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)$. $\frac{2\sqrt{10}\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)}{{\left(\sqrt{5}+\sqrt{2}\right)}^2-{\left(\sqrt{7}\right)}^2}=\frac{2\sqrt{10}\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)}{7+2\sqrt{10}-7}=\sqrt{5}+\sqrt{2}+\sqrt{7}$ In the second part multiply inside the root by $\frac{\sqrt{5}-2}{\sqrt{5}-2}$: $\sqrt{\frac{{\left(\sqrt{5}-2\right)}^2}{5-4}}=\sqrt{5}-2$. In the third part rationalise: $\frac{3}{\sqrt{7}-2}=\frac{3\left(\sqrt{7}+2\right)}{7-4}=\sqrt{7}+2$. Now put the three values back: $\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)-\left(\sqrt{5}-2\right)-\left(\sqrt{7}+2\right)$ $=\sqrt{2}+2-2=\sqrt{2}$ Hence the correct option is (c).

Areas of similar triangles In two similar triangles the ratio of the areas equals the square of the ratio of any pair of corresponding sides. Ratio of the corresponding sides $=\frac{9}{27}=\frac{1}{3}$. Ratio of the areas $={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$. $\frac{40}{\mathrm{area}\mathrm{of}M}=\frac{1}{9}$ Area of Triangle M $=40\times 9=360{cm}^2$. Hence the correct option is (b).

Decimal simplification Numerator: 0.071-0.060=0.011. Denominator: 0.61+0.02+0.03=0.66. $\frac{0.011}{0.66}=\frac{11}{660}=\frac{1}{60}$ $\frac{1}{60}=0.01666\dots \approx 0.0167$ Hence the correct option is (d).

Simplifying surds First write $\sqrt{20}$ in its simplest form: $\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$. So $y=2\sqrt{5}+\sqrt{5}=3\sqrt{5}$. $y^2={\left(3\sqrt{5}\right)}^2=9\times 5=45$ Hence the correct option is (c).

Compound interest - sum For compound interest the amount after n years is $A=P{\left(1+\frac{r}{100}\right)}^n$. $P{\left(1+\frac{r}{100}\right)}^2=12100$ …(i) $P{\left(1+\frac{r}{100}\right)}^4=14641$ …(ii) Dividing (ii) by (i): ${\left(1+\frac{r}{100}\right)}^2=\frac{14641}{12100}=1.21$ So $1+\frac{r}{100}=\sqrt{1.21}=1.1$, i.e. $r=10\%$. From (i): $P\times 1.21=12100\Rightarrow P=\frac{12100}{1.21}=10000$ Hence the correct option is (a).

Simple interest rate Let the sum be P. Becoming triple means the amount is 3P, so the simple interest earned is 3P-P=2P. 6 years 3 months $=6+\frac{3}{12}=6.25$ years. Formula: $SI=\frac{P\times r\times t}{100}$. $2P=\frac{P\times r\times 6.25}{100}$ $r=\frac{200}{6.25}=32$ So the annual rate of interest is about 32%. Hence the correct option is (b).

Average of an AP The numbers 4, 6, 8, …, N are consecutive even numbers, so they form an AP with common difference 2. For an AP the average of all the terms equals the average of the first and the last term. $\frac{4+N}{2}=52$ 4+N=104 N=100 Hence the correct option is (a).

Average of five AP terms Let the five terms be a, a+d, a+2d, a+3d, a+4d. Their sum is 5a+10d=5(a+2d), so the average is a+2d, which is the middle term. The first and the last term add up to a+(a+4d)=2(a+2d). $2\left(a+2d\right)=80\Rightarrow a+2d=40$ So the average of the five numbers is 40. Hence the correct option is (d).

Pipes and cisterns Take the capacity of the tank as the LCM of 40, 60 and 48, i.e. $LCM\left(40,60,48\right)=240$ units. $P+Q+R=\frac{240}{40}=6$ units per minute. $Q+R=\frac{240}{60}=4$ units per minute. $R+P=\frac{240}{48}=5$ units per minute. Subtract the third rate from the first: Q=(P+Q+R)-(R+P)=6-5=1 unit per minute. Time taken by Q alone $=\frac{240}{1}=240$ minutes. Hence the correct option is (b).