28 November 2025 · Shift 3

Loss percentage At a loss of $20\%$, the selling price is $80\%$ of the cost price. $SP=CP\times \frac{80}{100}$ $240=CP\times \frac{80}{100}$ $CP=\frac{240\times 100}{80}=300$ So the cost price is Rs. 300, option (b).

Average price of a mixture Total cost of the rice $=3\times 60+2\times 70=180+140=320$ Total quantity =3+2=5 kg Price per kg of the mixture $=\frac{320}{5}=64$ So the mixture costs ₹64 per kg, option (b).

Time, speed and distance Time = distance $÷$ speed. $=\frac{240}{60}$ =4 So the car takes 4 hours, option (c).

Time and work P is twice as efficient as Q, so take their one-day efficiencies as 2 and 1. Together they do 2+1=3 units of work in a day. Total work $=3\times 18=54$ units. P alone needs $\frac{54}{2}=27$ days. So P finishes the work in 27 days, option (b).

Discount Discount $=15\%$ of $1200=\frac{15}{100}\times 1200=180$ Selling price = marked price - discount. =1200-180=1020 So the selling price is Rs. 1020, option (b).

Volume of a cuboid Melting keeps the total volume unchanged, so the number of cubes is the volume of the cuboid divided by the volume of one cube. Volume of the cuboid $=50\times 40\times 10=20000$ cubic cm. Volume of one cube $=2^3=8$ cubic cm. Number of cubes $=\frac{20000}{8}=2500$ So 2500 cubes can be formed, option (b).

Congruence of triangles The SSS criterion says that if the three sides of one triangle equal the three corresponding sides of another triangle, the triangles are congruent. Here every side of the first plot equals the corresponding side of the second plot, so all three pairs of sides match. SAS would need two sides and the included angle, and no angle has been measured here. So the two plots are congruent by SSS, option (a).

Volume of a prism For a right prism, volume = base area $\times$ height. 2400= base area $\times 30$ Base area $=\frac{2400}{30}=80$ So the base area is 80 sq. cm, option (c).

Ratio in a mixture The ratio 3:1 has 3+1=4 parts, so one part $=\frac{80}{4}=20$ litres. Milk $=3\times 20=60$ litres and water $=1\times 20=20$ litres. Adding water changes only the water: water =20+10=30 litres, milk stays 60 litres. New ratio =60:30=2:1 So the new ratio of milk to water is 2 : 1, option (d).

Tangent to a circle The radius drawn to the point of contact is perpendicular to the tangent, so $\angle OBA={90}^{\circ }$. In right-angled triangle OBA, $O{A}^2=O{B}^2+A{B}^2$. ${17}^2=8^2+A{B}^2$ $A{B}^2=289-64=225$ $AB=\sqrt{225}=15$ So the tangent AB is 15 cm long, option (c).

Chords of a circle The perpendicular from the centre bisects a chord, so a chord of length 2l lies at a distance $d=\sqrt{r^2-l^2}$ from the centre. For the 24 cm chord, l=12: $d_1=\sqrt{{13}^2-{12}^2}=\sqrt{169-144}=\sqrt{25}=5$ cm. For the 10 cm chord, l=5: $d_2=\sqrt{{13}^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ cm. Both chords lie on the same side of the centre, so the two distances are subtracted. Distance between the chords =12-5=7 cm. Hence the correct option is (b).

Volume of sphere and cube Let the radius of the sphere be R, so the edge of the cube is $a=\frac{R}{2}$. Volume of the sphere $=\frac{4}{3}\pi R^3$. Volume of the cube $=a^3={\left(\frac{R}{2}\right)}^3=\frac{R^3}{8}$. Required percentage $=\frac{R^3/8}{\frac{4}{3}\pi R^3}\times 100=\frac{3}{32\pi }\times 100$. $=\frac{300}{32\times 3.14}=\frac{300}{100.48}\approx 2.99$ So the block occupies about $3.0\%$ of the sphere's volume, option (c).

Algebraic identity Use the identity ${\left(a+b\right)}^2+{\left(a-b\right)}^2=2\left(a^2+b^2\right)$ — the cross terms cancel. Here $a=\sqrt{5}$ and $b=\sqrt{7}$, so $a^2=5$ and $b^2=7$. ${\left(\sqrt{5}+\sqrt{7}\right)}^2+{\left(\sqrt{5}-\sqrt{7}\right)}^2=2\left(5+7\right)$ $=2\times 12=24$ Hence the correct option is (c).

Partnership profit share Both capitals stay invested for the same time, so the profit is divided in the ratio of the capitals. A:B=50000:70000=5:7 Total number of parts =5+7=12. B's share $=48000\times \frac{7}{12}=4000\times 7=28000$ So B gets Rs. 28,000 — option (c).

Division in a ratio Given ratio $=\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$. Multiply every term by the LCM of the denominators, 12, to clear the fractions. $\frac{1}{2}\times 12:\frac{1}{3}\times 12:\frac{1}{4}\times 12=6:4:3$ Sum of the parts =6+4+3=13. First portion $=2600\times \frac{6}{13}=200\times 6=1200$ So the first portion is Rs. 1200 — option (b).

Election percentage Let the total number of voters be 100x and work down stage by stage. Votes polled $=80\%$ of 100x=80x. Valid votes $=90\%$ of 80x=72x. Winner $=55\%$ of 72x=39.6x and the rival $=45\%$ of 72x=32.4x. Winning margin =39.6x-32.4x=7.2x $7.2x=1800\Rightarrow x=250$ Total voters $=100\times 250=25000$, i.e. 25,000 — option (c).

Simple interest Principal P=5000 and the amount after 4 years is 7000. Simple interest =7000-5000=2000 $R=\frac{SI\times 100}{P\times T}=\frac{2000\times 100}{5000\times 4}=10\%$ New rate $=10+3=13\%$. New simple interest $=\frac{5000\times 13\times 4}{100}=50\times 52=2600$ So the interest would be ₹2,600 — option (b).

Similar triangles In similar triangles the ratio of the areas equals the square of the ratio of corresponding sides. $\frac{A_1}{A_2}=\frac{81}{49}\Rightarrow \frac{s_1}{s_2}=\sqrt{\frac{81}{49}}=\frac{9}{7}$ Perimeter is a length, so it is in the same ratio as the sides: $\frac{P_1}{P_2}=\frac{9}{7}$. $\frac{63}{P_2}=\frac{9}{7}\Rightarrow P_2=\frac{63\times 7}{9}=49$ The smaller triangle has a perimeter of 49 units — option (c).

Reciprocal identity $3x+\frac{3}{x}=5$ Take 3 common: $3\left(x+\frac{1}{x}\right)=5$, so $x+\frac{1}{x}=\frac{5}{3}$. Squaring gives ${\left(x+\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}+2$. $x^2+\frac{1}{x^2}={\left(\frac{5}{3}\right)}^2-2=\frac{25}{9}-2=\frac{25-18}{9}$ $=\frac{7}{9}$ Hence the correct option is (b).

Mixed fraction addition Split each mixed number into its whole part and its fractional part and add the two groups separately. Whole parts: 4+44+444+4444+44444=49380. The fractions are $\frac{1}{2},\frac{1}{6},\frac{1}{12},\frac{1}{20},\frac{1}{30}$, i.e. $\frac{1}{n\left(n+1\right)}$ for n=1,2,3,4,5. Since $\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$, the sum telescopes. Sum of fractions $=1-\frac{1}{6}=\frac{5}{6}$. Total $=49380+\frac{5}{6}=49380\frac{5}{6}$ — option (a).

Surds simplification First write $\sqrt{50}$ in terms of $\sqrt{2}$: $\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$. $x=5\sqrt{2}+\sqrt{2}=6\sqrt{2}$ $x^2={\left(6\sqrt{2}\right)}^2=36\times 2=72$ Note that 52 comes from wrongly adding 50+2 without combining the like surds. Hence the correct option is (b).

Compound interest With annual compounding, a year's interest is $15\%$ of the amount standing at the beginning of that year. First year: $8000\times \frac{15}{100}=1200$, so the amount becomes 9200. Second year: $9200\times \frac{15}{100}=1380$, so the amount becomes 10580. Third year: $10580\times \frac{15}{100}=1587$. Difference =1587-1380=207 So the difference is Rs. 207 — option (d).

Average of even numbers Consecutive even numbers form an AP, and for an odd count the average is the middle term. So the middle number is 34 and the five numbers are 30,32,34,36,38. The five consecutive even numbers just before these are 20,22,24,26,28. Each new number is 10 less than the corresponding old one, so the average also falls by 10: 34-10=24. Hence the correct option is (b).

Combined average Total age = average $\times$ number of students, and the totals simply add up. Group A: $25\times 18=450$ years. Group B: $50\times x=50x$ years. Both groups together: $\left(25+50\right)\times 22=75\times 22=1650$ years. 450+50x=1650 $50x=1200\Rightarrow x=24$ Hence the correct option is (b).

Time and work Take the total work as the LCM of 20, 30 and 60, i.e. 60 units. So A does 3 units/day, B does 2 units/day and C does 1 unit/day. Let the work finish in T days. A works only the first 5 days, C works on all T days, and B leaves 3 days before the end, so B works (T-3) days. $3\times 5+2\left(T-3\right)+1\times T=60$ 15+2T-6+T=60 $3T+9=60\Rightarrow 3T=51$ T=17 The whole work was completed in 17 days — option (c).