27 November 2025 · Shift 2

Percentage Let the required percentage be x. Then $\frac{x}{100}\times 280=70$. $x=\frac{70\times 100}{280}$ x=25 Hence, option (d) 25% is the most appropriate answer.

Profit percentage Profit is always taken as a percentage of the cost price, so first find the cost price. Let the selling price be 100; then the profit is 20% of it, i.e. 20. Cost price = selling price − profit =100-20=80. Profit % on cost price $=\frac{20}{80}\times 100=25\%$. Hence, option (a) is the most appropriate answer.

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$, where P is the principal, R the rate and T the time. Here P=1500, R=4 and T=3. $SI=\frac{1500\times 4\times 3}{100}=\frac{18000}{100}$ SI=180 So the simple interest is Rs. 180.00, i.e. option (c).

Perimeter of octagon The border of the frame runs all along the boundary, so its length is the perimeter of the octagon. A regular octagon has 8 equal sides, so perimeter $=8\times \mathrm{side}$. Perimeter $=8\times 15=120$ cm. Hence, option (b) 120 cm is the most appropriate answer.

Cone surface area For a right circular cone of radius r and slant height l: curved surface area $=\pi rl$ and area of the base $=\pi r^2$. Given: $\pi rl=2\times \pi r^2$ Cancel $\pi r$ from both sides (as $r\ne 0$): l=2r. Hence, option (c) is the most appropriate answer.

Percentage change in area The surface area of a hemisphere is $3\pi r^2$, so the area is proportional to $r^2$ whatever formula for a hemisphere is used. Let the old radius be r and the new radius be R; a rise of 69% makes the new area 169 for an old area of 100. $\frac{R^2}{r^2}=\frac{169}{100}$ $\frac{R}{r}=\frac{13}{10}$ Increase in radius $=\frac{13-10}{10}\times 100=30\%$. Hence, option (b) is the most appropriate answer.

Surface area of pyramid Total surface area of a pyramid = area of the base + lateral surface area. The base is an isosceles triangle with sides 26 cm, 26 cm and 20 cm; its altitude falls on the mid-point of the 20 cm side. Altitude $=\sqrt{{26}^2-{10}^2}=\sqrt{676-100}=\sqrt{576}=24$ cm. Area of the base $=\frac{1}{2}\times 20\times 24=240$ sq. cm. Perimeter of the base =26+26+20=72 cm. Lateral surface area $=\frac{1}{2}\times$ perimeter of base $\times$ slant height $=\frac{1}{2}\times 72\times 35=1260$ sq. cm. Total surface area =1260+240=1500 sq. cm. Hence, option (a) is the most appropriate answer.

Slant height of pyramid Stacking the second pyramid changes nothing inside a single pyramid, so work with one pyramid only. In a right pyramid the vertical height, half of a base side and the slant height form a right-angled triangle. Half of the base side $=\frac{10}{2}=5$ cm and the height is 12 cm. Slant height $=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}$ =13 cm. Hence, option (d) is the most appropriate answer.

Trigonometric ratios $\cot \theta =\frac{\mathrm{base}}{\mathrm{perpendicular}}=\frac{7}{24}$, so take base =7k and perpendicular =24k. By Pythagoras theorem, hypotenuse $=\sqrt{{\left(7k\right)}^2+{\left(24k\right)}^2}=\sqrt{49+576}k=\sqrt{625}k=25k$. $\cos \theta =\frac{\mathrm{base}}{\mathrm{hypotenuse}}=\frac{7k}{25k}=\frac{7}{25}$ Hence, option (b) is the most appropriate answer.

Profit sharing ratio Both partners invest for the same period of one year, so the profit is divided in the ratio of their investments, X : Y =3:5. Total number of parts =3+5=8, and these 8 parts stand for Rs. 1,60,000. 1 part $=\frac{160000}{8}=20000$ Y's share $=5\times 20000=100000$, i.e. Rs. 1,00,000. Hence, option (a) is the most appropriate answer.

Partnership profit share In a partnership the profit is shared in the ratio of (capital $\times$ time) of each partner. K stays for the whole year: $5000\times 12=60000$ L joins after 3 months, so his money works for 9 months: $7000\times 9=63000$ M joins after 6 months, so his money works for 6 months: $9000\times 6=54000$ K:L:M=60000:63000:54000=20:21:18 Total parts =20+21+18=59, and these 59 parts stand for ₹2,360. One part $=\frac{2360}{59}=40$ Share of M $=18\times 40=720$, i.e. ₹720. Hence the correct option is (c).

Radian to degree A straight angle is $\pi$ radians $={180}^{\circ }$, so 1 radian $=\frac{{180}^{\circ }}{\pi }$. Degree measure $=\frac{5\pi }{2}\times \frac{{180}^{\circ }}{\pi }$ The $\pi$ cancels: $=\frac{5\times {180}^{\circ }}{2}=5\times {90}^{\circ }$ $={450}^{\circ }$ Hence the correct option is (c).

Average with correction Total marks of the original 20 students $=20\times 48=960$ The 5 new students bring $5\times 60=300$ marks, so 25 students now have 960+300=1260 marks. One mark was entered as 28 in place of 68, so the total is short by 68-28=40. Corrected total =1260+40=1300 Final average $=\frac{1300}{25}=52$ Hence the correct option is (a).

Alternate-day work Take the total work as the LCM of 20 and 30, i.e. 60 units. K's one day work $=\frac{60}{20}=3$ units and L's one day work $=\frac{60}{30}=2$ units. They work on alternate days, so every pair of days (K then L) finishes 3+2=5 units. Number of such pairs needed $=\frac{60}{5}=12$, and the work ends exactly at the end of a pair. Total time $=12\times 2=24$ days Hence the correct option is (d).

Cone total surface area For a right circular cone, total surface area $=\pi r\left(r+l\right)$, where r is the radius and l the slant height. $\frac{22}{7}\times 7\times \left(7+l\right)=396$ $22\left(7+l\right)=396$ $7+l=\frac{396}{22}=18$ l=18-7=11 So the slant height is 11 cm. Hence the correct option is (b).

Algebraic identity Use the identity ${\left(x-y\right)}^2=x^2+y^2-2xy$. Rearranging it gives $x^2+y^2={\left(x-y\right)}^2+2xy$. $={\left(5\right)}^2+2\times 36$ =25+72=97 Hence the correct option is (d).

Mixture of milk and water Let the original mixture hold m litres of milk and w litres of water. Adding 10 litres of water leaves the milk unchanged, and milk is then $40\%$: $\frac{m}{m+w+10}=\frac{40}{100}\Rightarrow 5m=2\left(m+w+10\right)\Rightarrow 3m=2w+20$ …(i) Next 10 litres of milk are added and milk becomes $50\%$, i.e. milk equals water: $\frac{m+10}{m+w+20}=\frac{1}{2}\Rightarrow 2m+20=m+w+20\Rightarrow m=w$ …(ii) Putting w=m in (i): $3m=2m+20\Rightarrow m=20$, so w=20. Required ratio of water to milk =20:20=1:1 Hence the correct option is (c).

Area of a circular ring Inner radius r=35 m, and the outer embankment is 7 m wider, so R=35+7=42 m. Area of the ring $=\pi R^2-\pi r^2=\pi \left(R^2-r^2\right)$ $=\frac{22}{7}\times \left({42}^2-{35}^2\right)=\frac{22}{7}\times \left(1764-1225\right)$ $=\frac{22}{7}\times 539=22\times 77$ =1694 So the region between the two embankments measures 1694 sq m. Hence the correct option is (b).

Algebraic identity Factorise: $a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)=\left(a+b\right)\left(a-b\right)\left(a^2+b^2\right)$ Adding the two given equations: $2a=5+3=8\Rightarrow a=4$ Subtracting them: $2b=5-3=2\Rightarrow b=1$ $a^2+b^2=4^2+1^2=16+1=17$ $a^4-b^4=5\times 3\times 17=255$ Hence the correct option is (a).

Trigonometric identity Simplify the surd first by multiplying inside by $\left(1+\sin \theta \right)$: $\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}=\sqrt{\frac{{\left(1+\sin \theta \right)}^2}{1-{\sin }^2\theta }}=\frac{1+\sin \theta }{\cos \theta }$ For the fraction, multiply numerator and denominator by $\left(\sin \theta +\cos \theta +1\right)$. Numerator $={\left(\sin \theta +\cos \theta \right)}^2-1=1+2\sin \theta \cos \theta -1=2\sin \theta \cos \theta$ Denominator $=\left(\sin \theta -\cos \theta +1\right)\left(\sin \theta +\cos \theta +1\right)=2\sin \theta \left(1+\sin \theta \right)$ So $\frac{\sin \theta +\cos \theta -1}{\sin \theta -\cos \theta +1}=\frac{2\sin \theta \cos \theta }{2\sin \theta \left(1+\sin \theta \right)}=\frac{\cos \theta }{1+\sin \theta }$ Now multiply the two simplified parts: $\frac{\cos \theta }{1+\sin \theta }\times \frac{1+\sin \theta }{\cos \theta }=1$ Hence the correct option is (a).

Time, speed and distance Let the usual speed be S km/h. The journey of 300 km takes $2\frac{1}{2}=\frac{5}{2}$ hours less when the speed becomes (S+20) km/h. $\frac{300}{S}-\frac{300}{S+20}=\frac{5}{2}$ $300\times \frac{\left(S+20\right)-S}{S\left(S+20\right)}=\frac{5}{2}\Rightarrow \frac{300\times 20}{S\left(S+20\right)}=\frac{5}{2}$ $S\left(S+20\right)=\frac{6000\times 2}{5}=2400$ $S^2+20S-2400=0\Rightarrow \left(S+60\right)\left(S-40\right)=0$ Speed cannot be negative, so the usual speed is S=40 km/h. Time to cover 192 km at the usual speed $=\frac{192}{40}=4.8$ hours. Hence the correct option is (a).

Simplification by BODMAS Change the mixed numbers to improper fractions: $6\frac{1}{4}=\frac{25}{4}$, $1\frac{2}{7}=\frac{9}{7}$, $11\frac{1}{4}=\frac{45}{4}$. Inside the bracket 'of' is done first: $\frac{2}{5}$ of $\frac{25}{4}=\frac{2\times 25}{5\times 4}=\frac{5}{2}$ Then the division: $\frac{5}{2}÷\frac{3}{7}=\frac{5}{2}\times \frac{7}{3}=\frac{35}{6}$ Now take 'of' $1\frac{2}{7}$: $\frac{35}{6}\times \frac{9}{7}=\frac{315}{42}=\frac{15}{2}$ Divide by $11\frac{1}{4}$: $\frac{15}{2}÷\frac{45}{4}=\frac{15}{2}\times \frac{4}{45}=\frac{2}{3}$ x is the square of this number, so $x={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$. $81x=81\times \frac{4}{9}=9\times 4=36$ Hence the correct option is (a).

Simplification of surds Write each surd in its simplest form: $\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$ and $\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}$. $\frac{5\sqrt{2}-3\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}}$ =2 Hence the correct option is (c).

Compound interest principal For compound interest, amount $=P{\left(1+\frac{r}{100}\right)}^n$. Dividing the 5-year amount by the 3-year amount kills P and leaves two years of growth: ${\left(1+\frac{r}{100}\right)}^2=\frac{16105.09}{13310}=1.21$ $1+\frac{r}{100}=\sqrt{1.21}=1.1$, so $r=10\%$. Now use the 3-year amount: $13310=P\times {\left(1.1\right)}^3=P\times 1.331$ $P=\frac{13310}{1.331}=10000$ So the sum is ₹10,000. Hence the correct option is (d).

Marked price and discount Everything is compared with the discount, so let the discount be ₹100. Marked price $=120\%$ of the discount $=\frac{120}{100}\times 100=120$ Selling price = marked price - discount =120-100=20 Required percentage $=\frac{\mathrm{MP}}{\mathrm{SP}}\times 100=\frac{120}{20}\times 100$ $=6\times 100=600\%$ Hence the correct option is (a).