24 November 2025 · Shift 1

Partnership profit share Both partners keep their money in the business for the same time, so the profit is divided in the ratio of the capitals. Ratio of capitals =12000:20000=3:5, so the profit falls into 3+5=8 equal parts. Rahul's share $=\frac{3}{8}\times 16000=6000$. Hence Rahul gets ₹6,000.

Average after removal Sum of the 8 numbers $=8\times 25=200$. The two numbers taken away add up to 35+45=80, so the six numbers left add up to 200-80=120. New average $=\frac{120}{6}=20$. Hence the new average is 20.

Percentage comparison Shyam earns ₹80,000, so Ram earns $80000\times \frac{125}{100}=100000$, that is ₹1,00,000. Shyam earns 100000-80000=20000 less than Ram, and this shortfall must be compared with Ram's income, not with Shyam's. Required percentage $=\frac{20000}{100000}\times 100=20\%$. Short cut: if A is $x\%$ more than B, then B is $\frac{x}{100+x}\times 100$ percent less than A, here $\frac{25}{125}\times 100=20\%$. Hence Shyam's income is 20% less than Ram's.

Discount and selling price A discount is always reckoned on the marked price, which is Rs. 1,500 here. Selling price $=1500\times \left(1-\frac{20}{100}\right)=1500\times \frac{80}{100}=1200$. Hence the selling price is Rs. 1,200.

Rhombus diagonals The diagonals of a rhombus bisect each other at right angles, so they cut it into four equal right triangles. In one such triangle the legs are the halves of the diagonals and the hypotenuse is the side, 13 cm. Half of the known diagonal $=\frac{10}{2}=5$ cm, so half of the other diagonal $=\sqrt{{13}^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$ cm. Other diagonal $=2\times 12=24$ cm. Hence the length of the other diagonal is 24 cm.

Volume of a pyramid Volume of a pyramid $=\frac{1}{3}\times$ base area $\times$ height. The base is a square of side 5 cm, so its area $=5\times 5=25$ sq. cm. Volume $=\frac{1}{3}\times 25\times 6=50$. Hence the volume of the crystal is $50{\mathrm{cm}}^3$.

Cone and cylinder volumes Volume of a cone $=\frac{1}{3}\pi r^{2}h$ and volume of a cylinder $=\pi r^{2}h$. Here both solids have the same radius r=9 cm and the same height h=15 cm, so the quantity $\pi r^{2}h$ is the same for both. Ratio $=\frac{1}{3}\pi r^{2}h:\pi r^{2}h=\frac{1}{3}:1=1:3$. Hence the required ratio is 1:3, and it does not depend on the actual radius or height at all.

Rational and irrational numbers Test option (c) with an example: $\sqrt{2}$ and $\sqrt{8}$ are both irrational, yet $\sqrt{2}\times \sqrt{8}=\sqrt{16}=4$, which is rational. So the product of two irrational numbers need not be irrational, and this statement is the false one. The other three are true: the sum of two rationals is rational; a non-zero rational times an irrational is always irrational; and every integer n can be written as $\frac{n}{1}$, so it is rational. Hence the false statement is option (c).

Profit and loss in lots Every novel costs the same on an average, ₹160, so the cost of any lot is 160 times the number of novels in it. Novels left for the combo deal =150-50-70=30. Cost of these 30 novels $=30\times 160=4800$, i.e. ₹4,800. The 20% profit is reckoned on this cost alone, so the combo deal fetches $4800\times \frac{120}{100}=5760$. Selling price per novel $=\frac{5760}{30}=192$. Hence each novel in the combo deal is sold for ₹192.00; the profit on the first lot and the loss on the second do not enter this calculation.

Angles of a triangle The three angles of a triangle always add up to ${180}^{\circ }$. As the angles are in the ratio 3:5:7, let them be 3x, 5x and 7x. Then $3x+5x+7x={180}^{\circ }$, so $15x={180}^{\circ }$ and $x={12}^{\circ }$. The largest angle answers to the largest part: $7x=7\times {12}^{\circ }={84}^{\circ }$. Hence the largest angle of the panel is ${84}^{\circ }$.

Midpoint theorem, area ratio The midpoints X, Y and Z of the three sides of $△PQR$ are joined. By the midpoint theorem every side of $△XYZ$ is parallel to a side of $△PQR$ and equal to half of it, so $△XYZ\sim △PQR$ with the sides in the ratio 1:2. In similar triangles the areas are in the ratio of the squares of the corresponding sides. $\frac{ar\left(△XYZ\right)}{ar\left(△PQR\right)}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$ The medial triangle in fact cuts $△PQR$ into four congruent triangles, so the required ratio is 1:4, option (c).

Centroid of a triangle The centroid is the point where the three medians meet, and its coordinates are simply the averages of the coordinates of the three vertices. Formula: $G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$ Here $\left(x_1,y_1\right)=\left(1,4\right)$, $\left(x_2,y_2\right)=\left(7,4\right)$ and $\left(x_3,y_3\right)=\left(4,10\right)$. x-coordinate of $G=\frac{1+7+4}{3}=\frac{12}{3}=4$ y-coordinate of $G=\frac{4+4+10}{3}=\frac{18}{3}=6$ Hence the flower is planted at (4,6), option (a).

Alligation and mixture Take 5 parts of the first mixture and 4 parts of the second, so the final mixture has 9 parts. The alcohol taken from the two mixtures must add up to the alcohol in the final mixture: $5\times 30+4\times x=9\times 22$ 150+4x=198 4x=198-150=48 $x=\frac{48}{4}=12$ Alligation gives the same thing: (22-x):(30-22)=5:4, so 4(22-x)=40 and 22-x=10. Hence $x=12\%$, option (c).

Compound interest, principal Formula: $A=P{\left(1+\frac{R}{100}\right)}^n$ Here the amount A=26620, R=10 and n=3. $26620=P{\left(1+\frac{10}{100}\right)}^3=P{\left(\frac{11}{10}\right)}^3$ $26620=P\times \frac{1331}{1000}$ $P=\frac{26620\times 1000}{1331}=20000$ Hence the principal is ₹20,000, option (a).

LCM, HCF and ratio The ratio 2:3 is already in its lowest terms, so let the numbers be 2k and 3k; then their HCF is k. Given HCF =30, therefore k=30. The numbers are $2\times 30=60$ and $3\times 30=90$. Check with LCM $\times$ HCF = product of the numbers: $180\times 30=5400$ and $60\times 90=5400$. Hence the larger number is 90, option (b).

Relative speed, chase Both run in the same direction, so the dog closes the gap at the relative speed 30-24=6 m/s. Time to catch = head start ÷ relative speed $=\frac{150}{6}=25$ s. Distance run by the rabbit = its own speed × time $=24\times 25=600$ m. (The dog covers $30\times 25=750$ m, which is exactly 150+600 — a useful check.) Hence the rabbit runs 600 m before it is caught, option (b).

Square root by identity Split the number so that the pattern shows: $100020001={10}^8+2\times {10}^4+1$. This is of the form $a^2+2a+1={\left(a+1\right)}^2$ with $a={10}^4=10000$. So $100020001={\left(10000+1\right)}^2={10001}^2$. A quick check: ${10001}^2=10001\times 10001=100010000+10001=100020001$. Hence $\sqrt{100020001}=10001$, option (b).

Slope-intercept form The slope-intercept form of a straight line is y=mx+c, where m is the slope and c is the intercept on the y-axis. Comparing y=-3x+5 with y=mx+c, the coefficient of x gives m=-3 (and c=5). A negative slope simply means the line falls as x increases. Hence the slope of the line is -3, option (b).

Direct common tangent Drop a perpendicular from the smaller centre onto the radius of the bigger circle; it is parallel to the tangent and forms a right triangle with the line of centres. Formula: length of the direct common tangent $=\sqrt{d^2-{\left(r_1-r_2\right)}^2}$, where d is the distance between the centres. Substituting d=12, $r_1=7$ and $r_2=4$: length $=\sqrt{{12}^2-{\left(7-4\right)}^2}$ $=\sqrt{144-9}=\sqrt{135}$ $\sqrt{135}=\sqrt{9\times 15}=3\sqrt{15}$ Hence the tangent is $3\sqrt{15}$ cm long, option (b).

Percentage increase relations '$50\%$ more than S' means $R=S+\frac{50}{100}S=\frac{3}{2}S$. With S=150: $R=\frac{3}{2}\times 150=225$, so R+S=225+150=375. '$200\%$ more than' means the original plus twice itself, i.e. three times it: $P+Q=3\times 375=1125$. Similarly 'P is $300\%$ more than Q' means P=Q+3Q=4Q. So $4Q+Q=1125\Rightarrow 5Q=1125\Rightarrow Q=225$. $P=4\times 225=900$ Hence P=900, option (a).

Compound interest ratio Let the sum be P. At $25\%$ per annum the amount is multiplied by $1+\frac{25}{100}=\frac{5}{4}$ every year. Amount at the end of the 2nd year $=P{\left(\frac{5}{4}\right)}^2$ Amount at the end of the 3rd year $=P{\left(\frac{5}{4}\right)}^3$ Required ratio $=\frac{P{\left(\frac{5}{4}\right)}^2}{P{\left(\frac{5}{4}\right)}^3}=\frac{1}{\frac{5}{4}}=\frac{4}{5}$ Hence the ratio is 4:5, option (c).

Simple interest, time In simple interest the interest earned each year is the same, so 8 years give $32\%$ of the principal. Interest for 1 year $=\frac{32}{8}=4\%$ of the principal, i.e. the rate is $4\%$ per annum. The interest becomes equal to the principal when it reaches $100\%$ of the principal. Required time $=\frac{100}{4}=25$ years. Hence 25 years, option (b).

Discount and selling price The discount is $\frac{3}{20}$ of the marked price, so take the marked price as 20 units. Discount $=\frac{3}{20}\times 20=3$ units. Selling price = marked price − discount =20-3=17 units. Discount : Selling price =3:17. Hence option (a) is correct.

Wages in ratio of work Take the total work as LCM(24,36)=72 units. Aman does $\frac{72}{24}=3$ units a day and Rakesh does $\frac{72}{36}=2$ units a day. As they work together throughout, the work each finally does — and hence the money each earns — is in the ratio of their daily rates, 3:2. 5 equal parts of the wage are ₹6,000, so 1 part is ₹1,200. Aman gets $3\times 1200=3600$ and Rakesh gets $2\times 1200=2400$, that is ₹3,600 and ₹2,400. Difference =3600-2400=1200, so the answer is ₹1,200, option (d).

Work on alternate days Take the total work as LCM(6,12)=12 units. R does $\frac{12}{6}=2$ units a day and S does $\frac{12}{12}=1$ unit a day. Working alternately starting with R, one pair of days (R then S) finishes 2+1=3 units. $\frac{12}{3}=4$ such pairs are needed, that is $4\times 2=8$ days, and the work ends exactly on S's day. Hence the whole work is completed in 8 days, option (b).