26 November 2025 · Shift 1

Average of a group Total output of the first 10 workers $=10\times 43.75=437.5$ units. Total output of all 13 workers $=13\times 46.5=604.5$ units. So the three new workers together make 604.5-437.5=167 units a day. Two of them make 50.2+46.4=96.6 units a day. Third new worker =167-96.6=70.4 units a day. Hence option (c) is correct.

Complementary angles Use $\sin \theta =\cos \left({90}^{\circ }-\theta \right)$: if a sine equals a cosine, the two angles add up to ${90}^{\circ }$. ${\left(3x+60\right)}^{\circ }+{\left(3y-60\right)}^{\circ }={90}^{\circ }$. $3x+3y={90}^{\circ }$, so $x+y={30}^{\circ }$, which indeed lies between $0^{\circ }$ and ${90}^{\circ }$. $\cot \left(x+y\right)=\cot {30}^{\circ }=\sqrt{3}$. Hence option (a) is correct.

Mixture of alloys Write each alloy in parts, taking 3 measures of alloy A and 2 measures of alloy B. Alloy A has l:m=5:3, so 3 measures give $l=5\times 3=15$ and $m=3\times 3=9$. Alloy B has l:m=4:5, so 2 measures give $l=4\times 2=8$ and $m=5\times 2=10$. In alloy C, l=15+8=23, m=9+10=19 and the total =23+19=42. Percentage of $l=\frac{23}{42}\times 100=54.76\%$. Hence option (d) is correct.

Length of an arc A full circle measures ${360}^{\circ }$ at the centre and its whole boundary is $2\pi r$ long. An arc that makes an angle $\theta$ at the centre is exactly that fraction of the boundary: $\mathrm{arc}=\frac{\theta }{{360}^{\circ }}\times 2\pi r$. Here $\theta ={45}^{\circ }$, so arc $AB=\frac{45}{360}\times 2\pi r=\frac{1}{8}\times 2\pi r$. Arc $AB=\frac{\pi r}{4}$. Hence option (b) is correct.

Train crossing problems Let the length of the train be L m and its speed v m/s. Passing the man: the man moves the same way, so the relative speed is (v-3) and L=40(v-3)=40v-120. Crossing the platform: the platform is 1.5L long, so L+1.5L=50v, i.e. 2.5L=50v and L=20v. Equating the two values of L: 20v=40v-120, so 20v=120 and v=6 m/s. Then $L=20\times 6=120$ m. The tractor comes from the opposite side, so the relative speed is 6+2=8 m/s and the length to be covered is 120+40=160 m. Time $=\frac{160}{8}=20$ seconds. Hence option (a) is correct.

Successive transactions Let the trader's cost price be C. The trader sells to A at $40\%$ profit, so A pays $C\times \frac{140}{100}=1.4C$. A sells to B at $20\%$ loss, so B pays $1.4C\times \frac{80}{100}=1.12C$. B sells to C at $10\%$ profit, so C pays $1.12C\times \frac{110}{100}=1.232C$. 1.232C=3388, so $C=\frac{3388}{1.232}=2750$. Hence option (c) is correct.

Simple interest rate The sum and the rate are the same in both cases, so the two periods simply add: 4+2=6 years. Formula: $SI=\frac{P\times R\times T}{100}$. $672=\frac{1400\times x\times 6}{100}$. 672=84x. $x=\frac{672}{84}=8$. Hence option (b) is correct.

Partnership share Let R's capital be x. Q's capital is five times R's, so Q=5x. One-third of P's capital equals Q's capital, so $\frac{P}{3}=5x$, which gives P=15x. P:Q:R=15x:5x:x=15:5:1, that is 21 equal parts in all. The capitals stay in the business for the same time, so the profit is divided in this ratio. P's share $=\frac{15}{21}\times 42000=30000$. Hence option (c) is correct.

Division in a ratio Let A's share be x. A gets one-third of B's share, so B=3x. B gets one-third of C's share, so $C=3\times 3x=9x$. x+3x+9x=1300, so 13x=1300 and x=100. B's share $=3x=3\times 100=300$. Hence option (b) is correct.

Mark-up and discount The sale price is fixed only by the mark-up and the discount, so take the cost price as 100. Marked price $=100+35\%$ of 100=135. A flat discount of $30\%$ leaves $70\%$ of the marked price: $135\times \frac{70}{100}=94.5$. The selling price 94.5 is below the cost price 100, so there is a loss of 100-94.5=5.5. Loss $\%=\frac{5.5}{100}\times 100=5.5\%$. The earlier sale at ₹X is only extra information; it does not enter this calculation. Hence option (a) is correct.

Ratio after removal Let the red and the blue marbles be 7x and 5x. No red marble is removed, so in the new ratio 7:4 the 7 parts still stand for 7x; hence the blue marbles left are 4x. 5x-12=4x x=12 Red marbles $=7x=7\times 12=84$. So option (a) is correct.

Compound interest amount Formula: $A=P{\left(1+\frac{R}{100}\right)}^n$ $A=12000{\left(1+\frac{12}{100}\right)}^3=12000\times {\left(1.12\right)}^3$ ${\left(1.12\right)}^3=1.404928$ $A=12000\times 1.404928=16859.136$ The amount is ₹16,859.13, so option (b) is correct.

Ratio of savings Let the savings of Arjun and Karan be 5x and 4x. Each saves ₹6,000 more, so $\frac{5x+6000}{4x+6000}=\frac{7}{6}$ 6(5x+6000)=7(4x+6000) 30x+36000=28x+42000 $2x=6000\Rightarrow x=3000$ Karan's saving $=4x=4\times 3000=12000$, i.e. ₹12,000, so option (b) is correct.

LCM of numbers A number divisible by all three must be a multiple of their LCM. $15=3\times 5$, $20=2^2\times 5$, $25=5^2$ LCM $=2^2\times 3\times 5^2=300$ So the numbers wanted are the multiples of 300: 300, 600, 900, 1200, 1500, … Of the options, 6400 and 12500 are not divisible by 3, and 3000 — though divisible — is bigger than 1500. The least such option is $1500=300\times 5$, so option (a) is correct.

Average speed both ways The two distances are equal, so the average speed is the harmonic mean of the two speeds. Formula: average speed $=\frac{2uv}{u+v}$ $=\frac{2\times 20\times 16}{20+16}=\frac{640}{36}=\frac{160}{9}$ $=17.77\dots$, i.e. about 17.7 km/h So option (c) is correct.

LCM with equal shortfall In each case (divisor - remainder) is the same: 8-3=5, 12-7=5, 20-15=5. So the number falls short of a common multiple of 8, 12 and 20 by 5. LCM of 8, 12 and 20 =120 Required number =120-5=115 Check: $115=8\times 14+3=12\times 9+7=20\times 5+15$. So option (b) is correct.

Change in average Sum of the 15 numbers $=15\times 30=450$ $40\%$ of 15=6 numbers, each raised by 5, so the sum rises by $6\times 5=30$. $60\%$ of 15=9 numbers, each lowered by 3, so the sum falls by $9\times 3=27$. New sum =450+30-27=453 New average $=\frac{453}{15}=30.2$ So option (c) is correct.

Two bodies moving towards each other When two bodies move towards each other their speeds are added. Relative (closing) speed =65+45=110 km/h Time to meet $=\frac{\mathrm{distance}}{\mathrm{closing}\mathrm{speed}}=\frac{550}{110}=5$ hours Distance covered by the car from City M $=65\times 5=325$ km They meet 325 km from City M, so option (c) is correct.

Time and work, pairs Take the whole work as the LCM of 28, 20 and 10, i.e. 140 units. $\left(A+B\right)=\frac{140}{28}=5$ units/day, $\left(B+C\right)=\frac{140}{20}=7$ units/day, $\left(A+B+C\right)=\frac{140}{10}=14$ units/day. C=(A+B+C)-(A+B)=14-5=9 units/day A=(A+B+C)-(B+C)=14-7=7 units/day A+C=7+9=16 units/day Time taken by A and C $=\frac{140}{16}=\frac{35}{4}=8\frac{3}{4}$ days. So option (b) is correct.

Work left after one leaves Take the whole work as 60 units, the LCM of 20 and 30. A does $\frac{60}{20}=3$ units/day and B does $\frac{60}{30}=2$ units/day. Work done together in 3 days $=3\times \left(3+2\right)=15$ units Work left =60-15=45 units B alone needs $\frac{45}{2}=22.5$ more days. So option (d) is correct.

Percentage discount Discount = marked price - selling price =9000-7200=1800, i.e. ₹1,800. Formula: discount $\%=\frac{MP-SP}{MP}\times 100$ $=\frac{1800}{9000}\times 100$ $=20\%$ So option (d) is correct.

Simple interest, find principal Formula: $SI=\frac{P\times R\times T}{100}$ $4800=\frac{P\times 8\times 3}{100}$ $4800=\frac{24P}{100}$ $P=\frac{4800\times 100}{24}=20000$ The sum is ₹20,000, so option (d) is correct.

Rate from two amounts Under simple interest the interest earned in every year is the same. Interest of (5-3)=2 years =3500-3000=₹500 Interest of 1 year $=\frac{500}{2}=₹250$ Principal = amount after 3 years - interest of 3 years $=3000-3\times 250=2250$, i.e. ₹2,250. Rate $=\frac{\mathrm{yearly}\mathrm{interest}}{P}\times 100=\frac{250}{2250}\times 100=\frac{100}{9}$ $\frac{100}{9}=11\frac{1}{9}\%$, so option (b) is correct.

Mixture: replacement Milk $=\frac{3}{5}\times 50=30$ L and water $=\frac{2}{5}\times 50=20$ L. The 10 L taken out is mixture, so it too is in the ratio 3:2, i.e. 6 L milk and 4 L water. Milk left =30-6=24 L, water left =20-4=16 L. Now 10 L of pure water is added: water =16+10=26 L, milk stays 24 L. New ratio =24:26=12:13, so option (b) is correct.

Tangent-secant property Tangent-secant property: the square of the tangent equals the product of the whole secant and its external part. $A{B}^2=AC\times AD$ ${12}^2=8\times AD$ $144=8AD$ AD=18 cm CD=AD-AC=18-8=10 cm So y=10 cm and option (a) is correct.