12 November 2025 · Shift 3

Partnership ratio In a partnership the profit is shared in the ratio of investment × time. So investment $=\frac{\mathrm{profit}\mathrm{share}}{\mathrm{time}}$. Ratio of investments $=\frac{5}{15}:\frac{7}{21}:\frac{3}{6}=\frac{1}{3}:\frac{1}{3}:\frac{1}{2}$. Multiplying every part by 6 gives 2:2:3. Hence the required ratio of their investments is 2:2:3.

Trigonometric identity Use the identity $1+{\cot }^{2}x={cosec}^{2}x$, which gives ${cosec}^{2}x-{\cot }^{2}x=1$. So the expression becomes $1-{\cos }^{2}x$. By the basic identity ${\sin }^{2}x+{\cos }^{2}x=1$, we get $1-{\cos }^{2}x={\sin }^{2}x$. Hence the simplified value is ${\sin }^{2}x$. Option (d) would be right only if the term ${\cos }^{2}x$ were absent, so it is the tempting wrong choice.

Profit and cost price Let the cost price be 7 units, so that the profit is 4 units in the same units. Selling price = cost price + profit =7+4=11 units. Therefore SP : CP =11:7. Option (c) is the answer; option (b) is the ratio of SP to profit, not to cost price.

Simple ratio Mangoes : grapes =5:8, so let the numbers be 5k and 8k. Given 5k=60, hence $k=\frac{60}{5}=12$. Number of grapes $=8k=8\times 12=96$. Option (d) comes from taking k=6 by mistake, that is from halving the number of mangoes.

CI and SI difference For 2 years the difference between compound and simple interest is one year's interest on the first year's interest. Formula: difference $=P{\left(\frac{r}{100}\right)}^2$. Substituting: $720=P{\left(\frac{12}{100}\right)}^2=P\times \frac{144}{10000}$. So $P=\frac{720\times 10000}{144}=50000$. The principal is ₹50,000.

Compound interest with withdrawal At $20\%$ per annum a sum becomes $\frac{120}{100}$ times itself in one year. Amount at the end of the first year $=300000\times \frac{120}{100}=360000$. After the withdrawal the balance is 360000-30000=330000. Only this balance earns interest in the second year: $330000\times \frac{120}{100}=396000$. So ₹3,96,000 remains at the end of the second year. Option (a) comes from deducting the ₹30,000 after the second year's interest instead of before it.

Corrected average Sum actually used $=40\times 28=1120$. The two wrong entries add up to 23+17=40, while the correct ones add up to 13+7=20. So the sum was 40-20=20 too large. Correct sum =1120-20=1100. Correct average $=\frac{1100}{40}=27.5$.

HCF and common remainder If a divisor leaves the same remainder in each case, it divides the difference of any two of the numbers exactly. 16115-15827=288 16703-16115=588 16703-15827=876 The greatest such divisor is the HCF of these differences: $288=2^5\times 3^2$, $588=2^2\times 3\times 7^2$, so their HCF is 12, and $876=12\times 73$ as well. Hence l=12; dividing, $15827=12\times 1318+11$, so n=11. Therefore l+n=12+11=23.

Successive percentage rise Let the original price be x. A rise of $10\%$ multiplies it by $\frac{11}{10}$ and a rise of $12.5\%$ multiplies it by $\frac{9}{8}$. So $x\times \frac{11}{10}\times \frac{9}{8}=69300$, that is $x\times \frac{99}{80}=69300$. $x=69300\times \frac{80}{99}=700\times 80=56000$. The original price of the car was ₹56,000.

Average speed and time Let the one-way distance be d km; the same d is covered each way. Time $=\frac{\mathrm{distance}}{\mathrm{speed}}$, so the total time is $\frac{d}{60}+\frac{d}{40}=15$. Taking the LCM 120: $\frac{2d+3d}{120}=15$. $\frac{5d}{120}=15$, so $d=\frac{15\times 120}{5}=360$. The one-way distance to the town is 360 km. Option (d) is the total distance of the round trip, not the one-way distance.

Time and work Take the total work as the LCM of the two given times, i.e. $LCM\left(20,30\right)=60$ units. One day's work of P and Q together $=\frac{60}{20}=3$ units. One day's work of Q alone $=\frac{60}{30}=2$ units. So one day's work of P alone =3-2=1 unit. Time taken by P alone $=\frac{60}{1}=60$ days, so option (c) is correct.

Similar triangles in a trapezium PQ and RS are parallel, so in $△POQ$ and $△ROS$ the alternate angles are equal: $\angle OPQ=\angle ORS$ and $\angle OQP=\angle OSR$. Hence $△POQ\sim △ROS$ by AA similarity, with PQ and RS as corresponding sides. For similar triangles, $\frac{\mathrm{area}\mathrm{of}△POQ}{\mathrm{area}\mathrm{of}△ROS}={\left(\frac{PQ}{RS}\right)}^2$. Given PQ=2RS, the ratio of the areas $={\left(\frac{2}{1}\right)}^2=4$. Area of $△POQ=4\times 16=64$, i.e. 64 cm${}^2$, so option (a) is correct.

Discount and profit ₹765 is the amount paid after a $15\%$ discount, so it is $85\%$ of the marked price. Marked price $=765\times \frac{100}{85}=900$, i.e. ₹900. Even at ₹765 the shopkeeper gains $25\%$, so ₹765 is $125\%$ of the cost price. Cost price $=765\times \frac{100}{125}=612$, i.e. ₹612. Without any discount the notebooks would be sold at the marked price, so the selling price would be ₹900. Profit =900-612=288, i.e. ₹288, so option (c) is correct.

Simple interest rate Formula: $SI=\frac{P\times R\times T}{100}$. Substituting P=2500, SI=500 and T=3: $500=\frac{2500\times R\times 3}{100}$. 500=75R $R=\frac{500}{75}=\frac{20}{3}$ So the rate is $\frac{20}{3}\%$ per annum (that is $6\frac{2}{3}\%$), and option (b) is correct.

Volume of a cylinder When the sheet is rolled with the 20 cm side as the height, the 88 cm side wraps round and becomes the circumference of the base. $2\pi r=88$ $r=\frac{88\times 7}{2\times 22}=14$ cm, and the height h=20 cm. Volume $=\pi r^{2}h=\frac{22}{7}\times 14\times 14\times 20$ Volume $=22\times 2\times 14\times 20=12320$ cubic cm, so option (d) is correct.

Sum of two squares First find the value asked for: ${12}^2+{65}^2=144+4225=4369$. Option (a): ${15}^2+{42}^2=225+1764=1989$. Option (b): ${20}^2+{63}^2=400+3969=4369$, which is exactly the required value. Option (c): ${18}^2+{45}^2=324+2025=2349$ and option (d): ${82}^2+{72}^2=6724+5184=11908$. Only option (b) matches, so option (b) is correct.

Mixture and alligation Tin in the given alloy $=20\%$ of 75=15 kg, so the lead =75-15=60 kg. Only pure tin is added, so the quantity of lead stays 60 kg throughout. In the new mixture tin is $40\%$, so lead is $100-40=60\%$ of it. New total weight $=\frac{60\times 100}{60}=100$ kg. Tin to be added =100-75=25 kg, so option (b) is correct.

Volume of a hemispherical shell The material is the outer hemisphere minus the hollow inner hemisphere, so volume $=\frac{2}{3}\pi R^3-\frac{2}{3}\pi r^3=\frac{2}{3}\pi \left(R^3-r^3\right)$. Substituting R=11 cm and r=9 cm: $=\frac{2}{3}\pi \left({11}^3-9^3\right)$. $=\frac{2}{3}\pi \left(1331-729\right)=\frac{2}{3}\pi \times 602$ $=\frac{1204\pi }{3}$ cm${}^3$, so option (b) is correct.

Fraction of capacity Adding 30 litres raises the water from $\frac{2}{5}$ of the tank to $\frac{3}{5}$ of the tank. So the 30 litres is exactly $\frac{3}{5}-\frac{2}{5}=\frac{1}{5}$ of the capacity. $\frac{1}{5}$ of the capacity =30 litres. Capacity $=30\times 5=150$ litres, so option (c) is correct.

Centroid of a triangle Formula: the centroid $G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$ — simply the average of the three vertices. x-coordinate $=\frac{4+6+8}{3}=\frac{18}{3}=6$ y-coordinate $=\frac{7+8+6}{3}=\frac{21}{3}=7$ So the centroid is (6,7), so option (c) is correct.

Areas of similar triangles In two similar triangles every pair of corresponding lengths — sides, altitudes, medians, angle bisectors, perimeters — is in the same ratio. So altitudes in the ratio p:q means the corresponding sides are also in the ratio p:q. The ratio of the areas of two similar triangles is the square of the ratio of any pair of corresponding lengths. Ratio of the areas $=p^2:q^2$, so option (b) is correct.

Rationalising a surd denominator To rationalise, multiply the numerator and the denominator by the conjugate of the denominator, i.e. by $\sqrt{5}-\sqrt{3}$. $\frac{8}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{8\left(\sqrt{5}-\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}$ The denominator uses $\left(a+b\right)\left(a-b\right)=a^2-b^2$ and becomes 5-3=2, which is free of surds. $=\frac{8\left(\sqrt{5}-\sqrt{3}\right)}{2}=4\left(\sqrt{5}-\sqrt{3}\right)$ So the rationalised form is $4\left(\sqrt{5}-\sqrt{3}\right)$, and option (b) is correct.

Algebraic identity Identity: $a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$. Substituting the given values: $405=15\times \left(a^2+b^2+c^2-ab-bc-ca\right)$. $a^2+b^2+c^2-ab-bc-ca=\frac{405}{15}=27$ Second identity: ${\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=2\left(a^2+b^2+c^2-ab-bc-ca\right)$. Required value $=2\times 27=54$, so option (b) is correct.

Trigonometric simplification Expand the numerator: ${\left(1+\cos \theta \right)}^2+{\sin }^2\theta =1+2\cos \theta +{\cos }^2\theta +{\sin }^2\theta$. Using ${\sin }^2\theta +{\cos }^2\theta =1$, the numerator $=1+2\cos \theta +1=2\left(1+\cos \theta \right)$. In the denominator use ${\csc }^2\theta -1={\cot }^2\theta$, so it becomes ${\cot }^2\theta \cdot {\sin }^2\theta$. ${\cot }^2\theta \cdot {\sin }^2\theta =\frac{{\cos }^2\theta }{{\sin }^2\theta }\times {\sin }^2\theta ={\cos }^2\theta$. So the expression $=\frac{2\left(1+\cos \theta \right)}{{\cos }^2\theta }=\frac{2}{{\cos }^2\theta }+\frac{2\cos \theta }{{\cos }^2\theta }$. $=2{\sec }^2\theta +2\sec \theta =2\sec \theta \left(1+\sec \theta \right)$, so option (d) is correct.

Equation of a straight line Slope $m=\frac{y_2-y_1}{x_2-x_1}=\frac{7-5}{2-3}=\frac{2}{-1}=-2$. Point-slope form through (3,5): y-5=-2(x-3). y-5=-2x+6 y=-2x+11, which rearranges to 2x+y=11. Check both points: 2(3)+5=11 and 2(2)+7=11, so option (d) is correct.