Geometry 91 MCQ part b

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geometry 91 mutliple choice questions with answers

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MULTIPLE CHOICE QUESTIONS 1. In below figure area of  ABC  100 and area of  PQR  225 . What is the value of r ' if r  8cm ? a. 8 cm b. 12 cm c. 18 cm d. 4 cm Sol: Correct option is (b) Here ratio of sides = 225  1.5 100  r '  1.5  r  1.5  8  12cm 2. The measure of angles of a quadrilateral taken in order are proportional to 1:2:4:5, then the quadrilateral is a. parallelogram b. square c. trapezium d. none of the above Sol: Correct option is (c) Let the angles be x, 2 x, 4 x, 5 x  x  2 x  4 x  5 x  360  12 x  360  x  30  Angles are 30, 60, 120, and 150 As SR II PQ  PQRS is a trapezium 3. Find the measure of the smallest angle of a quadrilateral if the measures of its interior angles are in the ratio 2:3:3:4? a. 60 o b. 120o c. 90o d. 180o Sol: Correct option is (a) Let the angles be 2 x,3 x, 3 x, 4 x  2 x  3 x  3 x  4 x  360  12 x  360  x  30  Smallest angles = 2 x  60 o 4. In below figure PQRS is a parallelogram. Also PQ= 8 cm and PS = 6 cm. What is the area of this parallelogram? a. 12cm 2 b. 48cm 2 c. 24cm 2 d. 8cm 2 Sol: Correct option is (c) Area = PQ  ST Here PQ = 8 cm and ST 1  sin 30o  ST  6   3cm PS 2 Area = 8  3  24cm 2 5. In below figure PQRS is a rhombus with each side 5 cm. What is the length of diagonal PR if SR = 8 cm a. 6 cm b. 5 cm c. 8 cm d. 2 cm Sol: Correct option is (a) Here PQ= 5 cm 1 Also OQ= SQ  4cm 2  PO  PQ 2  OQ 2  52  42  3  PQ  2  3  6cm 6. In below figure PQRS is a parallelogram PO and OQ are bisectors of P and Q . What is the measure of POQ ? a. 60 o b. 30o c. 90o d. 45 o Sol: Here Correct option is (c) 1 1 P  Q  POQ  180o 2 2  1 (P  Q )  POQ  180o 2 1 (180o )  POQ  180o 2   POQ  90o 7. What is the value of x in below figure if PQRS is a parallelogram a. 60 o b. 30o c. 70 o d. 75o Sol: Correct option is (d) SRQ  SPQ  60o  45o  60o  x  180o  x  180o  45o  60o  x  75o 8. Two parallelogram stand on same base and between the same parallel lines. What is the ratio of their areas? a. 1:2 b. 1:3 c. 2:5 d. 1:1 Sol: Correct option is (d) As both have same base and also same height, therefore they have equal area and hence ration = 1:1 9. In below figure PQRS is a parallelogram and STUR is a rectangle. Which of the following statement is true? a. perimeter of PQRS < perimeter of STUR b. perimeter of PQRS > perimeter of STUR c. perimeter of PQRS = perimeter of STUR d. none of the above Sol: Correct option is (b) Perimeter of PQRS = 2 (SR+PS) Perimeter of STUR = 2 (SR+ST) We know that in  PTS PS >ST as PS is hypotenuse  Perimeter of PQRS > Perimeter of STUR 10. For a regular polygon having n sides. What is the sum of all interior angles? a. (n  4)  180 b. (n  2)  180 c. ( n  2)  90 d. (n  3)  90 Sol: Correct option is (b) Sum of all interior angles = (n  2)  180 For triangles n  3  sum= (3  2) 180  180 o For quadrilateral n  4  sum= (4  2)  180  360o 11. What is the number of sides in a regular polygon if its exterior angle is x ? a. 360  x b. x 360 360 x c. d. 180 x Sol: Correct option is (c) 3 6 0 3 6 0  E x te r io r x a n g le Number of sides = 12. What is the number of diagonals in the regular polygon having n sides? a. n( n  3) 2 n( n  3) 4 n(n  2) 3 b. c. d. n( n  1) Sol: Correct option is (a) Number of diagonals = nC2  n  n! n(n  1) n 2  3n n(n  3) n  n   (n  2)!2! 2 2 2 13. What is the area of polygon having n sides and length of each side as x ? a. nx  180  tan   2  n   360  b. nx 2 tan    n  c. nx 2  360  sin   4  n  nx 2  180  cot   4  n  Correct option is (d) d. Sol: Area of regular polygon having n sides and length of each side as x = nx 2  180  cot   4  n  For example Area of equilateral triangle having length of each side x is = 3x 2 3x 2 1 3x 2  cot 60o    4 4 4 3 14. Each interior angle of a regular polygon is 120 o . What is the number of sides of the polygon? a. 2 b. 7 c. 8 d. 6 Sol: Correct option is (d) Interior angle = 120o  Exterior angle = 180 o  120 o  60o  Number of sides = 360o 6 60o 15. Each angle of a regular octagon is a. 150 o b. 135o c. 120 o d. 90o Sol: Correct option is (b) In octagon number of sides = 8  Exterior angle = 360  45o 8  Interior angle = 135 o 16. The sum of all interior angles of a polygon is 900o . The number of sides of the polygon is a. 5 b. 6 c. 7 d. 8 Sol: Correct option is (c) Here sum of all interior angles = 900o  (n  2) 180  900 n 900 2 180 n7 17. If each interior angle of a regular polygon is 9  8  times of one of the interior angles of a regular hexagon, then the number of sides of the polygon is a. 4 b. 6 c. 5 d. 8 Sol: Correct option is (d) Interior angle of regular hexagon = 120o  Interior angle of regular polygon =  Exterior angle of polygon = 45 o 360o  Number of sides = 8 45o 9 120  135o 8 18. If each interior angle of a regular polygon is double its exterior angle, then the number of sides of the polygon is a. 6 b. 4 c. 8 d. 12 Sol: Correct option is (a) Here Interior angle = 2 Exterior angle We know that interior angle = 180  360 360  2 n n 360 n 360 360 and exterior angle = n n  180   180  3  n 3  360 6 180 19. The interior angle of a regular polygon is 90o more than its exterior angle. What are the numbers of sides of a regular polygon? a. 4 b. 6 c. 8 d. 10 Sol: Correct option is (c) 360 360 and exterior angle = n n Interior angle = 180   180  360 360   90o n n  2 360  90o n 2  360 8 90 n 20. What are the numbers of diagonals of a regular polygon having 7 sides? a. 10 b. 12 c. 11 d. 14 Sol: Correct option is (d) n( n  3) 2 7(7  3) 7  4   14 2 2 Number of diagonals of polygon of n sides =  Number of diagonals of regular polygon of 7 sides = 21. A regular polygon has 27 diagonals. What are the number of sides of this regular polygon has? a. 9 b. 6 c. 7 d. 10 Sol: Correct option is (a) n( n  3) 2 Number of diagonals of regular polygon of n sides = n(n  3)  27 2   n(n  3)  54  n 2  3n  54  0  (n  9)(n  6)  0 n9 22. The sum of the number of sides of two regular polygons is 14. The difference between the interior angles of two polygon is 15o . What are the number of sides the two polygons have? a. 4, 6 b. 6, 9 c. 6, 8 d. 8, 11 Sol: Correct option is (c) Let n and n ' be the number of sides of 2 polygons  n  n '  14 360   360   Also  180     180  '   15 n   n    360 360   15 n' n Also n '  14  n  360 360   15  n  8 14  n n n '  14  8  6 Hence the number of sides are 6, 8 23. The sum of all interior angles of a regular polygon exceeds the sum of all exterior angles by 180 o . What are the number or sides this polygon has? a. 4 b. 5 c. 6 d. 8 Sol: Correct option is (b) Sum of exterior angles = 360o Sum of interior angles = 360 o  180 o  540 o  (n  2) 180  540  n2 3 n5 24. The sum of all interior angles of a regular polygon is 3 times the sum of all exterior angle. What are the number of side this polygon has? a. 8 b. 4 c. 6 d. 7 Sol: Correct option is (a) Sum of exterior angles = 360o Sum of interior angles = 3  360 o  1080o  (n  2)  180  1080  n26 n 8 25. What is the ratio of the measure of interior angle to the measure of exterior angle of a regular hexagon? a. 1:4 b. 3:1 c. 2:3 d. 2:1 Sol: Correct option is (d) Here n  6  Exterior angle = 360  60o 6  Interior angle = 180 o  60o  120o  Ratio = 120  2 :1 60 26. A regular polygon has 27 diagonals. What is the measure of each interior angle of this regular polygon? a. 120 o b. 108o c. 140o d. 90o Sol: Correct option is (c) Here number of diagonals = 27  n(n  3)  27 2  n 2  3n  54  0  (n  9)(n  6)  0 n9  Interior angle = 180  360  140 o 9 9 27. The interior angles of a polygon P is   times, the interior angle of 4 ' polygon P . What is the ratio of the number of sides of these polygons? Given sum of sides = 11 a. 3:8 b. 7:3 c. 3:4 d. 5:8 Sol: Correct option is (a) Let polygon P has n sides and polygon P' has n ' sides  180  360 9  360    180  '  4 n n  360 810  405  ' n n  180  Also n  n '  11  180  360 810  405  n 11  n  n  8  n'  3 n' 3    3:8 n 8 28. Which of the following statement is true for the two congruent circles? a. they have equal radius b. they have equal perimeter c. they have equal area d. all of the above Sol: Correct option is (d) Congruent circles have equal radius  they have equal area and perimeter 29. In below figure PQRS is a quadrilateral inscribed in a circle with centre O. What is the value of x in the below figure? a. 100 o b. 110 o c. 10o d. 90o Sol: Correct option is (b) In a cyclic quadrilateral the sum of opposite angles equal to 180 o .  x  70o  180o  x  110o 30. In below figure PQ=RS, then which of the statement is true? a. x  2 y b. x  y 2 c. x  y d. xy  1 Sol: x y Correct option is (c) Equal chords of a circle subtend equal angles at the centre 31. What is the value of x in the below figure? a. 60 o b. 120o c. 90o d. 70 o Sol: Correct option is (b) Here RSQ  180  QST  RSQ  180  120  60o Also x  RSQ  180o  x  180  RSQ  180  60  x  120o 32. In below figure PS and RQ are two chords intersecting at T. Which of the following is true? a. PT  QT  RT  TS b. PT  TS  QT  TR c. PS  RQ d. PT  TS  RQ  TQ Sol: Correct option is (b) By the property of intersecting chords PT  TS  QT  TR 33. In below figure PR is secant and PT is tangent to a circle. Which of the following statement is true? a. PT 2  PR  PQ b. PQ 2  PR 2  PT 2 c. PQ 2  QR 2  PT 2 d. PT 2  PR 2 Sol: Correct option is (a) PT 2  PR  PQ 34. Two circles having radii r1 and r2 touch externally. What is the distance between their centres? a. r12  r2 2 b. r2  r1 c. r1  r2 d. r12  r2 2 Sol: Correct option is (c) Distance between O and O ' = r1  r2 35. Two circles having radii r1 and r2 touch internally. What is the distance between their centres if r2  r1 ? a. r1  r2 b. r2 2  r12 c. r12  r2 2 d. r2  r1 Sol: Correct option is (d) Distance between O and O ' = r2  r1 36. In below figure SR  PQ . What is the length of PQ? a. 18 cm b. 7 cm c. 12 cm d. 5 cm Sol: Correct option is (a) Here  PSR QSR  PR  RQ  PQ  2 PR Also PR  152  122  81  9cm  PQ  2 PR  2  9  18cm 37. What is the length of the perpendicular from the centre of a circle having radius 13 cm to the chord of a circle 10 cm long? a. 5 cm b. 13 cm c. 12 cm d. 10 cm Sol: Correct option is (c) Here PO= 13 cm and PQ=10 cm Also OR  PQ  PR  1 PQ  5cm 2  OR  PO2  PR 2  132  52  12cm 38. The radius of a circle makes an angle of 50o with a chord of circle. What is the angle subtended by this chord at centre? a. 70 o b. 50o c. 45 o d. 80o Sol: Correct option is (d) Here OPQ  OQP  50o  POQ  180  50  80o 39. In below figure RPQ  30o and PQR  100o . What is the value of x in below figure? a. 40 o b. 100 o c. 50o d. 30o Sol: Correct option is (c) Here PRQ  180  100  30  50o Also x  PRQ  50o  x  50o 40. In below figure, what is the value of x if POR  100o ? a. 120 o b. 130o c. 100 o d. 110 o Sol: Correct option is (b) 1 Here PSR  POR 2 1  PSR   100  50o 2 Also x  PSR  180o  x  180 o  50 o  130 o 41. In below figure O is the centre. What is the value of x in below figure? a. 40 o b. 90o c. 150 o d. 50o Sol: Correct option is (d) Here PQR  90o  x  180  90  40  50 o 42. A tangent of 3 cm is drawn to a circle from a point which is 5 cm from the centre of a circle. What is the radius of a circle? a. 3 cm b. 5 cm c. 4 cm d. 8 cm Sol: Correct option is (c) Here OP= 5 cm and PQ = 3 cm  OQ  52  32  16  4cm Radius  4cm 43. In below figure PS = 12 cm and PQ = 4 cm. What is the value of QR? a. 4 cm b. 12 cm c. 48 cm d. 32 cm Sol: Correct option is (d) Here PS 2  PQ  PQR  PR  PS 2 144   36cm PQ 4  RQ  36  4  32cm 44. In below figure SPR  45o . What is the value of x in below figure? a. 60 o b. 45 o c. 90o d. 30o Sol: Correct option is (b) Here PQR  SPR  45o Also RPQ  90o  x  180  90  45  45o 45. In below figure PT = 4 cm, TR = 7 cm and TQ = 2 cm. What is the value of ST? a. 4 cm b. 8 cm c. 7 cm d. 14 cm Sol: Correct option is (d) Here ST  TQ  PT  TR  ST  PT  TR 4  7   14cm TQ 2 46. In below figure PQ = 3 cm, QT = 12 cm, PR = 5 cm. What is the value of RS? a. 4 cm b. 5 cm c. 3 cm d. 1 cm Sol: Correct option is (a) Here PQ  PT  PR  PS  3 15  5  PS  PS  9cm  RS  9  5  4cm 47. In below figure PQ = 4 cm, and PR = 2 cm. What is the value of RS? a. 3 cm b. 4 cm c. 6 cm d. 8 cm Sol: Correct option is (c) Here PQ 2  PR  PS  PS  4 4  8cm 2  RS  8  2  6cm 48. Two circles of radii 8 cm and 11 cm touch each other externally. What is the distance between the centres of the two circles? a. 8 cm b. 11 cm c. 3 cm d. 19 cm Sol: Correct option is (d) Distance between O and O ' = 11+8= 19 cm 49. Two circles of radii 8 cm and 11 cm touch each other internally. What is the distance between the centres of the two circles? a. 8 cm b. 3 cm c. 11 cm d. 19 cm Sol: Correct option is (b) In above figure Distance between O and O ' = r2  r1 =11-8 = 3 cm 60. Three circles touch each other externally. The distance between their centres is 4 cm, 6 cm and 8 cm. What are the radii of the circles? a. 1 cm, 3 cm, 5 cm b. 1 cm, 3 cm, 6 cm c. 1 cm, 2 cm, 4 cm d. 4 cm, 2 cm, 6 cm Sol: Correct option is (a) PQ = 4, QR = 6 and PR = 8 Here PQ+QR+PR= 2( x  y  z )  2( x  y  z )  4  6  8  x yz 9 Also x  z  4  y5 Also z  y  6  z 1  x3 Hence radii are (1, 3, 5) 61. A regular polygon having n sides has 2n diagonals. What are the number of sides of this polygon has? a. 4 b. 7 c. 9 d. 11 Sol: Correct option is (b) n( n  3) 2 Number of diagonals =  n(n  3)  2n 2  n 2  3n  4n  n2  7n n7 62. A rectangular field has its length and breadth in the ration 3:4. What is the area of rectangular field if its perimeter is 28 cm? a. 24cm 2 b. 48cm 2 c. 40cm 2 d. 12cm 2 Sol: Correct option is (b) 4x 3 Let length be x  breadth = 4x  7x   2  x    28  2   28 3  3  x6  Breadth = 8 cm and Length = 6 cm Also Area = 6  8  48cm 2 63. A rectangular playground costs Rs. 18 at Re. 1 per square metre for leveling. What is the length of the playground if the ratio of length and breadth is 1:2? a. 2 m b. 1 m c. 3.8 m d. 3 m Sol: Correct option is (d) Breadth = 2x Let length = x Also Area = Rs.18  18m 2 2 1Re/ m  2 x  x  18m 2  2 x 2  18m 2  x  3m 64. The ratio of the length and breadth of a rectangular playground is 2:3 and its area is 2400 m 2 . How long will it take to walk around the playground at the rate of 2 km/hr? a. 20 min b. 10 min c. 12 min d. 6 min Sol: Correct option is (d) Let length be 2x and breadth be 3x  2 x  3 x  2400  x 2  400  x  20m  Perimeter = 2(2  20  3  20)  200m Time = 200 m 1 1  hr   60  6 min 2000m / hr 10 10 65. If the length of the rectangle is doubled and the breadth remains same then what is the percentage change in the area? a. 100% b. 200% c. 50% d. 150% Sol: Correct option is (a) If length is doubled then area also gets doubled  A'  2 A Percentage change = A'  A 2A  A 100   100  100% A A 66. How many tiles of length 6 cm and breadth 2 cm are required to cover the rectangular wall of length 18 cm and breadth 8 cm? a. 10 b. 12 c. 6 d. 18 Sol: Correct option is (b) Area of tile = 6  2  12cm 2 Area of wall = 18  8  144cm 2  Number of tiles = 144cm2  12 12cm 2 67. If the area of the square is 144m 2 , then what is the length of the diagonal of the square? a. 6 2m b. 6m c. 12m d. 12 2m Sol: Correct option is (d) Area = 144m 2  Side = 144  12m  Diagonal = Side 2  12 2m 68. If the length of a diagonal of a square is 8.484 m, then what is the area of the square? a. 36m 2 b. 24m 2 c. 16m 2 d. 25m 2 Sol: Correct option is (a) Diagonal = 8.484 m Also diagonal = side 2  side 2  8.484  side  6m  Area  6  6  36 m 2 69. If the side of square is increased by 100% then by how much percentage does its area get increased? a. 100% b. 200% c. 300% d. 50% Sol: Correct option is (c) Let side be l  Area  l  l  l 2 After increment new side = 2l  AreaA'  (2l )  (2l )  4l 2  Percentage increase = 4l 2  l 2  100  300% l2 70. What is the length of the base of an isosceles triangle whose area is 6m 2 and length of one of equal side is 5 m? a. 6 m b. 8 m c. 4 m d. 12 m Sol: Correct option is (b) Area = 12m 2 Let QR = 2a  QS  a Also PS  52  a 2 1 1  Area   QS  PS   a  52  a 2 2 2 1   a  52  a 2  6m 2 2  a  52  a 2  12  a 2 (52  a 2 )  122  25a 2  a 4  144 a  4m  Base  QR  2a  8m 71. If every side of a triangle is tripled, then the ratio of the new area to the old area is a. 3:1 b. 1:3 c. 9:1 d. 2:3 Sol: Correct option is (c) Let a, b, c be sides of triangle Area = s( s  a)(s  b)(s  c) Also a  3a ' , b  3b' , c  3c '  s '  3s  Area A'  3s (3s  3a)(3s  3b)(3s  3c)  9 s ( s  a)(s  b)(s  c )  A'  9 :1 A 72. What is the area of an equilateral triangle whose altitude is 4 3cm long? a. 8 3cm 2 b. 4 3cm 2 c. 16 3cm2 d. 8cm 2 Sol: Correct option is (c) Here PS = a2  a  2 2  3a 2  3a  4 3  a  8cm 2 1  8  4 3  16 3cm 2 2  Area  73. The adjacent sides of a parallelogram are 4 cm and 9 cm and the angle between them is 60 o . What is the area of the parallelogram? a. 18 3cm2 b. 36cm 2 c. 18cm 2 d. 36 3cm 2 Sol: Correct option is (c) Area = 9  4  sin 60 o  9 4 3 2  18 3cm 2 74. Area of a rhombus is 20cm 2 . One of the diagonal of a rhombus exceeds the other diagonal by 3 cm. What is the length of the smaller diagonal? a. 4 cm b. 6 cm c. 8 cm d. 5 cm Sol: Correct option is (d) Let smaller diagonal = xcm Larger diagonal = ( x  3)cm 1  Area   x( x  3) 2 1   x( x  3)  20 2  x 2  3 x  40  0  ( x  5)( x  8)  0  x  5cm 75. The length of two parallel sides of a trapezium are 8 cm and 14 cm. What is the area of this trapezium if distance between the parallel lines is 7 cm? a. 11cm 2 b. 77cm 2 c. 144cm 2 d. 154cm 2 Sol: Correct option is (b) 1 (8  14)  7 2 Area of trapezium = 1 (22)  7  77cm 2 2  76. If the circumference of the circle is 14 cm , then what is the area of this circle? a. 49 cm 2 b. 7 cm 2 c. 49cm 2 d. 7cm2 Sol: Correct option is (a) Circumference = 2 r  14  r  7cm  Area   r 2   (7) 2  49 cm 2 77. The sum of the radius and circumference of a circle, is 153cm. What is 22 the area of the circle? (   ) 7 a. 21 cm 2 b. 441cm 2 c. 441 cm 2 d. 21cm 2 Sol: Correct option is (c) Here r  2 r  153cm  44   r 1    153 7    r  21cm  Area   r 2   (21)2  441 cm 2 78. The radius of a circle is increased by 100%. What is the percentage increase in the area of the circle? a. 100% b. 50% c. 200% d. 300% Sol: Correct option is (d) Let initial radius = r Then radius after increment = 2r Let initial area = A =  r 2 Incremented Area =  (2r ) 2  4 r 2  A' Percentage increase = A'  A 4 r 2   r 2 100  100  300% A  r2 79. If the circumference of the circle is increased by 200% what is the percentage increase in the area of the circle? a. 400% b. 200% c. 800% d. 100% Sol: Correct option is (c) Let initial circumference = C = 2 r Incremented circumference = C '  2 r  6 r  r '  3r  A'   (3r ) 2  9 r 2  Percentage increase = A'  A 9 r 2   r 2  100   100  800% A  r2 80. How long will it take to walk around a circle of radius 14 cm at the rate 22 of 4 cm/min? (   ) 7 a. 22 min b. 12 min c. 11 min d. 88 min Sol: Correct option is (a) Circumference = 2 r  2 22 14  88cm 7 88cm  22 min 4cm / min Time = 81. A rope is bent in the form a square of side 22 cm. It is cut again and bent 22 in the form a semicircle. The radius of this semicircle is (   ) 7 a. 14 cm b. 7 cm c. 21 cm d. 28 cm Sol: Correct option is (d) Perimeter of square = 4  22  88cm Circumference of semicircle =  r   r  88cm  22 r  88 7  r  28cm 82. What is the lateral surface area of cuboid having length 2 cm, breadth 4 cm and height 3 cm? a. 18cm 2 b. 36cm 2 c. 9cm 2 d. 4cm 2 Sol: Correct option is (b) Lateral surface area = 2(l  b) h  2(2  4)3  36cm2 83. What is the total surface area of cuboid having length 1 cm, breadth 2 cm and height 4 cm? a. 14cm 2 b. 7cm 2 c. 28cm 2 d. 35cm 2 Sol: Correct option is (c) Total surface area = 2(lb  bh  hl )  2(1 2  2  4  4 1)  2(2  8  4)  28cm 2 84. The length of the cuboid is 3cm, breadth 2 cm and height 36 cm. What is the length of each cube which will have the same capacity as that of given cuboid? a. 4 cm b. 2 cm c. 8 cm d. 6 cm Sol: Correct option is (d) Volume of cuboid = 2  36  3  216 Volume of cube= a 3  a 3  216  a  6cm 85. The length, breadth and height of a box are 3 cm, 4 cm, and 6 cm respectively. What is the cost of covering the box fully if the cost of covering is Rs. 4/ cm 2 . a. Rs. 432 b. Rs. 108 c. Rs. 216 d. Rs. 324 Sol: Correct option is (a) Surface area of box = 2(3  4  4  6  6  3)  2(12  24  18)  108cm 2 Cost = 4 108  Rs.432 86. When each side of a cube is increased by 3 cm, then the volume is increased by 279 cm3 . What is the length of side or original cube? a. 3 cm b. 4 cm c. 7 cm d. 9 cm Sol: Correct option is (b) Let the length of side of original cube = x Volume of original cube = x 3 Incremented volume = ( x  3)3  ( x  3)3  x3  279  x 3  27  9 x ( x  3)  x 3  279  9 x 2  27 x  252  0  x 2  3 x  28  0  ( x  4)( x  7)  0  x  4cm 87. A cube having each edge 5 cm is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 5 cm x 5 cm, what is the rise in the water level? a. 25 cm b. 125 cm c. 5 cm d. 2.5 cm Sol: Correct option is (c) Volume of cube = (5)3  125cm3 Rise = 125  5cm 5 5 88. A rectangular water reservoir is 10 m by 12 m at the base. Water flows into it through a pipe whose cross section is 4 m by 3 m at the rate of 2 m/sec. what is the height to which the water will rise in the reservoir in 10 seconds? a. 1 m b. 3 m c. 2 m d. 1.5 m Sol: Correct option is (c) Volume of water flown = 4  3  2  10  240m3  Height  240  2m 10 12 89. The lateral surface area of a cylinder is 56 cm 2 . What is the volume of cylinder if its height is 7 cm? a. 56 cm3 b. 28 cm 3 c. 16 cm3 d. 112 cm3 Sol: Correct option is (d) Lateral surface area = 2 rh  2 rh  56  rh  28 As h  7cm  r  4m  Volume    (4) 2  7  112 cm3 90. The curved surface area of a cylinder is 60 cm 2 and its base radius is 5 cm, then its total surface area is? a. 55cm 2 b. 22cm 2 c. 110cm 2 d. 30cm 2 Sol: Correct option is (c) Curved surface area = 2 rh  2 rh  60 cm 2  rh  30 As r = 5 cm  h  6cm  Total surface area = 2 r ( h  r )  2 (5)(6  5)  110 cm2 91. The ratio between the radius and height of cylinder is 2:3. What is the curved surface area of the cylinder if the volume is 96 cm3 ? Sol: Correct option is (a) Let radius = 2x and height = 3x  Volume   (2 x )2 (3x )  12 x 3  12 x 3  96  x3  8 x2 Radius = 4 cm and Height = 6 cm Surface area = 48 cm 2

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