22 November 2025 · Shift 3

Average Sum of the 11 numbers $=11\times 50=550$. When the largest number is removed the average of the remaining 10 numbers falls by 4, so it becomes 50-4=46. Sum of those 10 numbers $=10\times 46=460$. Largest number =550-460=90. The largest exceeds the smallest by 60, so the smallest number =90-60=30. Hence, option (c) is the correct answer.

Perfect square To make a number a perfect square by adding to it, look for the first perfect square that is bigger than it. ${41}^2=1681$, which is smaller than 1720. ${42}^2=1764$, which is the first perfect square above 1720. Number to be added =1764-1720=44. Check: $1720+44=1764={42}^2$. Hence, option (c) is the correct answer.

Square roots Take each square root first, because the bracket has to be simplified before it is squared. $\sqrt{169}=13$ and $\sqrt{64}=8$. So ${\left(\sqrt{169}+\sqrt{64}\right)}^2={\left(13+8\right)}^2$. $={21}^2=441$. Hence, option (c) is the correct answer.

Square root of a decimal Write the decimal as a fraction: $0.0016=\frac{16}{10000}$. $\sqrt{0.0016}=\sqrt{\frac{16}{10000}}=\frac{\sqrt{16}}{\sqrt{10000}}=\frac{4}{100}=0.04$. Short cut: a decimal with an even number of decimal places has a root with half that many, so 4 places give 2 places. Check: $0.04\times 0.04=0.0016$. Hence, option (c) is the correct answer.

Ratio from an equation Given: 11P=9Q. In such an equation the ratio is obtained by cross-writing the coefficients, so divide both sides by 11Q. $\frac{P}{Q}=\frac{9}{11}$ So P:Q=9:11. Hence, option (b) is the correct answer.

Compound ratio b is the term common to both ratios: it is 6 in a:b=4:6 and 7 in b:c=7:5. Make the two values of b equal by multiplying the first ratio by 7 and the second by 6. $a:b=4\times 7:6\times 7=28:42$ and $b:c=7\times 6:5\times 6=42:30$. Now b is 42 in both, so a:b:c=28:42:30. Dividing each term by 2 gives a:b:c=14:21:15. Hence, option (b) is the correct answer.

Ratio and change Let the number of boys be 5x and the number of girls be 3x. Only girls change, so after 24 girls join, boys =5x and girls =3x+24. $\frac{5x}{3x+24}=\frac{5}{4}$ 20x=15x+120 5x=120, so x=24. Number of boys $=5x=5\times 24=120$. Hence, option (b) is the correct answer.

Ratio Given a:b=4:7, so take a=4k and b=7k for some common multiple k. b=28 gives 7k=28, so k=4. $a=4k=4\times 4=16$. Hence, option (c) is the correct answer.

Average of algebraic terms Average of three terms is their sum divided by 3, so $\frac{3x+\left(2x+2\right)+\left(3x+2\right)}{3}=20$. Add the like terms in the numerator: 3x+2x+3x=8x and 2+2=4. $\frac{8x+4}{3}=20$ 8x+4=60 8x=56, so x=7. Check: the terms become 21, 16 and 23, whose sum is 60 and average is 20. Hence, option (c) is the correct answer.

Average Total marks in the first 4 subjects $=4\times 65=260$. For an average of 90 in 5 subjects, the total needed $=5\times 90=450$. Marks required in the fifth subject =450-260=190. Hence, option (a) is the correct answer.

Discount on marked price Discount is always calculated on the marked price, which here is ₹3,000. Formula: Discount = Marked price $\times \frac{r}{100}$, where r is the rate of discount. Discount $=3000\times \frac{20}{100}=30\times 20=600$ So the discount amounts to ₹600 and the customer actually pays ₹2,400. Hence the correct option is (c).

Successive discounts, profit % Let the cost price be 100. Marked 200% above cost price, so Marked price $=100+\frac{200}{100}\times 100=300$ First discount of $40\%$: $300\times \frac{60}{100}=180$ Second discount of $10\%$ is taken on this reduced price: $180\times \frac{90}{100}=162$ Selling price = 162, so Profit = 162-100 = 62 Profit $\%=\frac{62}{100}\times 100=62\%$ Hence the correct option is (c).

Time and work Take the total work as the LCM of 20 and 25, that is 100 units. One day's work of (A + B) $=\frac{100}{20}=5$ units. One day's work of A $=\frac{100}{25}=4$ units. So one day's work of B = 5-4 = 1 unit. Time taken by B alone $=\frac{100}{1}=100$ days. Hence the correct option is (c).

Angle subtended by an arc The angle an arc makes at the centre is twice the angle the same arc makes at any point on the remaining part of the circle. Here arc AB subtends ${200}^{\circ }$ at the centre O, and the friend's viewpoint C lies on the other side of AB. Angle at C $=\frac{1}{2}\times {200}^{\circ }={100}^{\circ }$ Note that the rule uses the whole ${200}^{\circ }$ arc, not the remaining ${360}^{\circ }-{200}^{\circ }={160}^{\circ }$. Hence the correct option is (c).

Common tangents to circles The number of common tangents depends on how the distance between the centres compares with the radii. Sum of the radii = 6+3 = 9 units, distance between the centres d = 10 units. Since $d>r_1+r_2$, neither circle touches or cuts the other — they lie completely outside each other. Two circles lying apart have 2 direct (external) tangents and 2 transverse (internal) tangents. Total common tangents = 2+2 = 4 Hence the correct option is (a).

Surface area of hemisphere The curved surface of a hemisphere is half the surface of the whole sphere, so CSA $=\frac{1}{2}\times 4\pi r^2=2\pi r^2$ Put r = 14 and $\pi =\frac{22}{7}$: CSA $=2\times \frac{22}{7}\times 14\times 14$ $=2\times 22\times 2\times 14=1232$ So the curved surface area is $1232{cm}^2$. Hence the correct option is (d).

Regular polygon angles At every vertex, interior angle + exterior angle $={180}^{\circ }$. Exterior angle $={180}^{\circ }-{144}^{\circ }={36}^{\circ }$ The exterior angles of any polygon add up to ${360}^{\circ }$, and in a regular polygon they are all equal. $n=\frac{{360}^{\circ }}{{36}^{\circ }}=10$ So each sensor pad is a regular decagon with 10 sides. Hence the correct option is (b).

Similar triangles, shadows At the same moment the sun's rays fall at the same angle everywhere, so the two right triangles formed are similar. In similar triangles the corresponding sides are proportional: $\frac{h_1}{s_1}=\frac{h_2}{s_2}$ $\frac{4}{3}=\frac{h}{18}$ $h=\frac{4\times 18}{3}=\frac{72}{3}=24$ The skyscraper is 24 m high. Hence the correct option is (b).

Rate of discount Discount = Marked price - Selling price = 18000-12000 = 6000 The rate of discount is always taken as a percentage of the marked price. Discount $\%=\frac{6000}{18000}\times 100$ $=\frac{1}{3}\times 100=33\frac{1}{3}\%=33.33\%$ The exact value is $33.33\%$, so the rounded figure $33\%$ in option (a) is not the accurate answer. Hence the correct option is (b).

Marked price from profit First find the selling price from the cost price and the profit. Selling price $=1200\times \left(1+\frac{20}{100}\right)=1200\times \frac{6}{5}=1440$ A discount of $22\%$ is allowed on the marked price, so the selling price is $78\%$ of the marked price. $MP\times \frac{78}{100}=1440$ $MP=\frac{1440\times 100}{78}=\frac{144000}{78}=1846.15$ Hence the correct option is (a).

Wheel revolutions and radius In one revolution a wheel covers a distance equal to its circumference, $2\pi r$. Total distance = 2.75 km $=2.75\times 100000=275000$ cm. So $875\times 2\pi r=275000$ $2\pi r=\frac{275000}{875}=314.2857$ $r=\frac{314.2857}{2\times 3.1416}=\frac{314.2857}{6.2832}$ r = 50.02 The radius of the wheel is 50.02 cm. Hence the correct option is (a).

Pythagoras theorem Let the two legs be a and b with a-b = 7 and, by Pythagoras, $a^2+b^2={13}^2=169$. ${\left(a-b\right)}^2=a^2+b^2-2ab$, so 49 = 169-2ab, giving 2ab = 120 ${\left(a+b\right)}^2=a^2+b^2+2ab=169+120=289$ $a+b=\sqrt{289}=17$ The sides are 12 cm and 5 cm — the familiar 5, 12, 13 triplet, whose difference is 7 cm and sum is 17 cm. Hence the correct option is (a).

Trigonometric identity Use the algebraic identity ${\left(A-B\right)}^2=A^2-2AB+B^2$ with $A=\sin x$ and $B=\cos x$. ${\left(\sin x-\cos x\right)}^2={\sin }^{2}x-2\sin x\cos x+{\cos }^{2}x$ Group the two squares and apply ${\sin }^{2}x+{\cos }^{2}x=1$: ${\left(\sin x-\cos x\right)}^2=1-2\sin x\cos x$ Option (a) would be right only if the middle term vanished, which it does not. Hence the correct option is (d).

Central and inscribed angle The angle a chord subtends at the centre is twice the angle it subtends at any point on the remaining arc. The camera stands at a point C on the circle and sees the two points A and B at ${60}^{\circ }$. Angle at the centre $=2\times {60}^{\circ }={120}^{\circ }$ Hence the correct option is (d).

Time and work Take the total work as the LCM of 20 and 30, that is 60 units. (A + B) together do $\frac{60}{20}=3$ units a day, and A alone does $\frac{60}{30}=2$ units a day. So B alone does 3-2 = 1 unit a day. Time taken by B alone $=\frac{60}{1}=60$ days. Hence the correct option is (d).