Board Previous Year Questions (Board PYQs)

Board Previous Year Questions

PYQ 1 (1 mark): The area of a sector of angle $p$ (degrees) of a circle of radius $R$ is: (A) $\tfrac{p}{180}\,2\pi R$ (B) $\tfrac{p}{180}\,\pi R^2$ (C) $\tfrac{p}{360}\,2\pi R$ (D) $\tfrac{p}{720}\,2\pi R^2$. [CBSE] Solution: $\tfrac{p}{360}\pi R^2=\tfrac{p}{720}\,2\pi R^2$. Answer: (D).

PYQ 2 (1 mark): If the circumference of a circle equals the perimeter of a square, the ratio of their areas is: (A) $22:7$ (B) $14:11$ (C) $7:22$ (D) $11:14$. [CBSE] Solution: $2\pi r=4a\Rightarrow a=\tfrac{\pi r}{2}$. Areas: $\pi r^2 : a^2=\pi r^2:\tfrac{\pi^2 r^2}{4}=4:\pi=4:\tfrac{22}{7}=28:22=14:11$. Answer: (B).

PYQ 3 (2 marks): Find the area of a sector of a circle of radius 6 cm whose angle is $60^\circ$. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: $\tfrac16\times\tfrac{22}{7}\times36=\tfrac{132}{7}\approx18.86$ cm$^2$. Answer: ~18.86 cm$^2$.

PYQ 4 (2 marks): Find the area of a quadrant of a circle whose circumference is 22 cm. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: $r=\tfrac{22\times7}{2\times22}=3.5$; quadrant $=\tfrac14\times\tfrac{22}{7}\times12.25=9.625$ cm$^2$. Answer: 9.625 cm$^2$.

PYQ 5 (2 marks): The minute hand of a clock is 14 cm long. Find the area swept in 5 minutes. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: $5$ min $=30^\circ$; $\tfrac{30}{360}\times\tfrac{22}{7}\times196=51.33$ cm$^2$. Answer: ~51.33 cm$^2$.

PYQ 6 (3 marks): In a circle of radius 21 cm, an arc subtends $60^\circ$. Find (i) the arc length, (ii) the sector area, (iii) the segment area. $(\pi=\tfrac{22}{7},\sqrt3=1.73)$ [CBSE] Solution: (i) arc $=\tfrac16\times2\times\tfrac{22}{7}\times21=22$ cm. (ii) sector $=\tfrac16\times\tfrac{22}{7}\times441=231$ cm$^2$. (iii) triangle (equilateral) $=\tfrac{\sqrt3}{4}\times441\approx190.7$; segment $=231-190.7=40.3$ cm$^2$. Answers: 22 cm, 231 cm$^2$, ~40.3 cm$^2$.

PYQ 7 (3 marks): A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of (i) the minor segment, (ii) the major sector. $(\pi=3.14)$ [CBSE] Solution: (i) minor segment $=\tfrac14\times3.14\times100-\tfrac12\times100=78.5-50=28.5$ cm$^2$. (ii) major sector $=\tfrac{270}{360}\times3.14\times100=235.5$ cm$^2$. Answers: 28.5 cm$^2$, 235.5 cm$^2$.

PYQ 8 (3 marks): A car has two wipers (non-overlapping), each blade 25 cm sweeping $115^\circ$. Find the total area cleaned per sweep. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: one $=\tfrac{115}{360}\times\tfrac{22}{7}\times625\approx627.48$; two $\approx1254.96$ cm$^2$. Answer: ~1254.96 cm$^2$.

PYQ 9 (3 marks): A horse is tied at a corner of a 15 m square field with a 5 m rope. Find (i) the grazing area, (ii) the increase if the rope becomes 10 m. $(\pi=3.14)$ [CBSE] Solution: (i) $\tfrac14\times3.14\times25=19.625$ m$^2$; (ii) increase $=\tfrac14\times3.14\times(100-25)=58.875$ m$^2$. Answers: 19.625 m$^2$, 58.875 m$^2$.

PYQ 10 (3 marks): A brooch is a circle of diameter 35 mm, with 5 diameters dividing it into 10 equal sectors. Find (i) the total length of wire, (ii) the area of each sector. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: (i) circumference $=\tfrac{22}{7}\times35=110$ mm; 5 diameters $=5\times35=175$ mm; total $=285$ mm. (ii) each sector angle $=36^\circ$; area $=\tfrac{36}{360}\times\tfrac{22}{7}\times17.5^2=\tfrac{1}{10}\times\tfrac{22}{7}\times306.25\approx96.25$ mm$^2$. Answers: 285 mm, ~96.25 mm$^2$.

PYQ 11 (3 marks): An umbrella has 8 equally spaced ribs; radius 45 cm. Find the area between two consecutive ribs. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: angle $=45^\circ$; $\tfrac18\times\tfrac{22}{7}\times2025\approx795.54$ cm$^2$. Answer: ~795.54 cm$^2$.

PYQ 12 (3 marks): A lighthouse spreads light over a sector of $80^\circ$ to 16.5 km. Find the sea area warned. $(\pi=3.14)$ [CBSE] Solution: $\tfrac{80}{360}\times3.14\times272.25\approx189.97$ km$^2$. Answer: ~189.97 km$^2$.

PYQ 13 (1 mark): The area of the largest circle inscribed in a square of side 14 cm is: (A) 154 (B) 196 (C) 44 (D) 616 cm$^2$. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: $r=7$; $\tfrac{22}{7}\times49=154$ cm$^2$. Answer: (A).

PYQ 14 (2 marks): Find the area of the shaded region if a circle of radius 7 cm is inscribed in a square of side 14 cm. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: $196-154=42$ cm$^2$. Answer: 42 cm$^2$.

PYQ 15 (3 marks): A round table cover of radius 28 cm has six equal designs (each a $60^\circ$ segment). Find the cost of making them at Rs 0.35 per cm$^2$. $(\sqrt3=1.7,\pi=\tfrac{22}{7})$ [CBSE] Solution: one design $=\tfrac16\times\tfrac{22}{7}\times784-\tfrac{1.7}{4}\times784=410.67-333.2=77.47$; six $\approx464.8$ cm$^2$; cost $=464.8\times0.35\approx$ Rs 162.68. Answer: ~Rs 162.68.

PYQ 16 (2 marks): Two circles have radii 19 cm and 9 cm. Find the radius of the circle whose circumference equals the sum of their circumferences. [CBSE] Solution: $R=19+9=28$ cm. Answer: 28 cm.

PYQ 17 (2 marks): Find the radius of the circle whose area equals the sum of the areas of two circles of radii 8 cm and 6 cm. [CBSE] Solution: $R^2=64+36=100\Rightarrow R=10$ cm. Answer: 10 cm.

PYQ 18 (3 marks): A chord of a circle of radius 15 cm subtends $60^\circ$. Find the areas of the minor and major segments. $(\pi=3.14,\sqrt3=1.73)$ [CBSE] Solution: minor $=117.75-97.31=20.44$ cm$^2$; major $=706.5-20.44=686.06$ cm$^2$. Answers: 20.44 cm$^2$, 686.06 cm$^2$.

PYQ 19 (1 mark): If the perimeter of a semicircular protractor is 36 cm, its diameter is: (A) 14 (B) 12 (C) 7 (D) 16 cm. $(\pi=\tfrac{22}{7})$ [CBSE] Solution: $\pi r+2r=36\Rightarrow r(\tfrac{22}{7}+2)=36\Rightarrow r\times\tfrac{36}{7}=36\Rightarrow r=7$, diameter $=14$. Answer: (A).