26 September 2025 · Shift 3

Square of a decimal Use (a + b)² with a = 12 and b = 0.12. = 144 + 2(12)(0.12) + 0.0144 = 144 + 2.88 + 0.0144 = 146.8944. 146.8944 − 396. = −249.1056 — option (a).

Rate of growth Growth compounds, so 52,000 = 32,000(1 + $\frac{r}{100}$)². $\frac{52000}{32000}$ = 1.625, so (1 + $\frac{r}{100}$)² = 1.625. 1 + $\frac{r}{100}$ = $\sqrt{1.625}$ ≈ 1.2747. r ≈ 27.47% — option (d).

Simple interest amount SI = $\frac{P\times R\times T}{100}$ = $\frac{3000\times 3\times 3}{100}$. = ₹270. Amount = principal + interest. = 3,000 + 270 = ₹3,270 — option (d).

Distance in seconds 90 km/h = 90 × $\frac{5}{18}$ = 25 m/s. Distance = speed × time. = 25 × 22. = 550 m — option (a).

Dimensions in AP In an AP the three terms are (a − d), a and (a + d), so their sum is 3a = 42 and a = 14. Volume: (14 − d)(14)(14 + d) = 1344 ⇒ 196 − d² = 96. d² = 100, so d = 10. Dimensions = 4, 14, 24 cm — option (d).

Arc length $\frac{\mathrm{arc}\mathrm{length}}{\mathrm{circumference}}$ = $\frac{\mathrm{central}\mathrm{angle}}{360°}$. $\frac{11}{2\mathrm{\pi r}}$ = $\frac{36}{360}$ = $\frac{1}{10}$, so 2πr = 110. r = $\frac{110\times 7}{22\times 2}$. = 17.5 cm — option (b).

Savings percentage Total spending = 20 + 30 + 40 = 90% of the salary. So the savings are 100 − 90 = 10%. 10% of the salary = ₹90,000. Salary = ₹9,00,000 — option (c).

Discount then VAT After the 50% discount: 3,200 × $\frac{50}{100}$ = ₹1,600. VAT of 10% is charged on this discounted price. 1,600 × $\frac{110}{100}$. = ₹1,760 — option (a).

Average speed For three equal distances, average speed = $\frac{3}{\frac{1}{60}+\frac{1}{90}+\frac{1}{180}}$. $\frac{1}{60}$ + $\frac{1}{90}$ + $\frac{1}{180}$ = $\frac{3+2+1}{180}$ = $\frac{6}{180}$ = $\frac{1}{30}$. Average = 3 × 30. = 90 km/h — option (b).

Replacement mixture At the start: acid = $\frac{4}{5}$ × 80 = 64 L and water = 16 L. Removing 10 L takes away $\frac{10}{80}$ of each: acid 8 L and water 2 L, leaving 56 L and 14 L. The 10 L added in the ratio 1 : 4 brings 2 L acid and 8 L water. Final = 58 : 22 = 29 : 11 — option (a).

Work with a third person Take the work as 40 units: Rahul does 2 a day and Rajesh 1 a day. In 8 days together they finish 8 × 3 = 24 units, leaving 16 units. Rahul and Suresh clear 16 units in 4 days, i.e. 4 units a day; Rahul’s 2 leaves 2 units a day for Suresh. Suresh alone: $\frac{40}{2}$ = 20 days — option (d).

Cylinder ratios Volume = πr²h, so $\frac{V\mathrm{₁}}{V\mathrm{₂}}$ = $\frac{r\mathrm{₁²}h\mathrm{₁}}{r\mathrm{₂²}h\mathrm{₂}}$. $\frac{7}{9}$ = $\frac{\mathrm{1²}\times h\mathrm{₁}}{\mathrm{3²}\times h\mathrm{₂}}$ = $\frac{h\mathrm{₁}}{9h\mathrm{₂}}$. So $\frac{h\mathrm{₁}}{h\mathrm{₂}}$ = $\frac{7\times 9}{9}$ = 7. Ratio = 7 : 1 — option (a).

Prism volume Volume of a prism = base area × height. With the height fixed, the volume varies directly with the base area. Doubling the base area therefore doubles the volume. Hence option (b).

Recast cone Melting keeps the volume the same, so r²h stays constant. Halving the radius divides r² by 4. So the height must become 4 times as large. 24 × 4 = 96 cm — option (b).

Complementary angles Since C is the right angle, A + B = 90°, so A and B are complementary. cosB = cos(90° − A) = sinA. sinA is given as $\frac{3}{5}$. So cosB = $\frac{3}{5}$ — option (c).

Degrees to radians Radians = degrees × $\frac{\pi }{180}$. = 6 × $\frac{\pi }{180}$. = $\frac{6\pi }{180}$. = $\frac{\pi }{30}$ — option (d).

Complementary angles A and B are complementary, so B = 90° − A. cosB = cos(90° − A) = sinA. sinA is given as $\frac{35}{37}$. So cosB = $\frac{35}{37}$ — option (c).

Vertically opposite angles When two lines cross, the angles opposite each other are equal. This is the vertically opposite angles theorem. So the other angle has the same measure. = 55° — option (b).

Inradius The incircle touches all three sides of the triangle. So the perpendicular distance from the incentre to any side IS the inradius. That distance is given as 3 units. r = 3 units — option (c).

Algebraic identity (y + 1)² + (y − 1)² = 2y² + 2, since the cross terms cancel. y = $\sqrt{5}$, so y² = 5. 2(5) + 2. = 12 — option (a).

Basic proportionality By the basic proportionality theorem, $\frac{\mathrm{PD}}{\mathrm{DQ}}$ = $\frac{\mathrm{PE}}{\mathrm{ER}}$. $\frac{1}{2}$ = $\frac{6}{\mathrm{ER}}$. ER = 6 × 2. = 12 cm — option (b).

Equal chords Equal chords of a circle subtend equal angles at the centre. AB and CD are given as equal. So the two angles must be the same. x = 35° — option (d).

Area of a segment Area of the segment = area of the sector − area of the triangle. Sector = $\frac{90}{360}$ × π × 5² = $\frac{25\pi }{4}$. Triangle = $\frac{1}{2}$ × 5 × 5 × sin90° = 12.5. = $\frac{25\pi }{4}$ − 12.5 cm² — option (a).

sinA from tanA Perpendicular = 7 and base = 24. Hypotenuse = $\sqrt{\mathrm{7²}+\mathrm{24²}}$ = $\sqrt{49+576}$ = $\sqrt{625}$ = 25. sinA = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$. = $\frac{7}{25}$ — option (a).

Work with a third person Take the work as 120 units: Arun does 3 a day and Veena 4 a day. In 5 days together they finish 5 × 7 = 35 units, leaving 85 units. All three clear 85 units in 5 days, i.e. 17 units a day; Arun and Veena give 7, so Charu does 10 a day. Charu alone: $\frac{120}{10}$ = 12 days — option (a).